Answers to Selected Exercises

Worked solutions to the daggered (†) and odd-numbered exercises from each chapter. Try every problem before reading its solution.

Chapter 1

Chapter 1 — Solutions to † and odd-numbered exercises

Constants used: $\mu = 3.986\times10^{14}\ \text{m}^3/\text{s}^2$, $R_E = 6{,}371\ \text{km}$, $g_0 = 9.81\ \text{m/s}^2$, sound speed $343\ \text{m/s}$.

1.1 † At $300\ \text{km}$, $r = 6{,}671\ \text{km} = 6.671\times10^6\ \text{m}$. $$v = \sqrt{\mu/r} = \sqrt{3.986\times10^{14}/6.671\times10^{6}} = \sqrt{5.975\times10^{7}} = 7{,}730\ \text{m/s} \approx 7.73\ \text{km/s}.$$ This is essentially the ~$7.8\ \text{km/s}$ quoted for LEO — and slightly higher than the value at a taller orbit, because lower orbits are faster (a theme of Ch. 6). 7.73 km/s.

1.3 † Escape velocity is $\sqrt{2}$ times the circular speed at the same radius: $$v_{esc} = \sqrt{2}\times 7.7 = 1.414\times 7.7 = 10.9\ \text{km/s}.$$ (The same factor turns the ~$7.9\ \text{km/s}$ surface circular speed into the ~$11.2\ \text{km/s}$ surface escape speed.) ≈ 10.9 km/s.

1.5 † Orbits per day $= 1440\ \text{min} / 92\ \text{min} = 15.65$. The ISS circles Earth about 15–16 times per day (≈ 15.5), so its crew sees roughly that many sunrises and sunsets daily.

1.7 † At GEO, $r = 6{,}371 + 35{,}786 = 42{,}157\ \text{km} = 4.2157\times10^7\ \text{m}$. $$g = \mu/r^2 = 3.986\times10^{14}/(4.2157\times10^{7})^2 = 3.986\times10^{14}/1.777\times10^{15} = 0.224\ \text{m/s}^2.$$ As a fraction of surface gravity: $0.224/9.81 = 0.023 = \mathbf{2.3\%}$. Even at GEO, $35{,}786\ \text{km}$ up, gravity is small but decidedly not zero — a geostationary satellite is in free fall, held in its orbit by that residual pull. "Zero gravity" is never literally true anywhere near Earth.

1.9 † Dose $= 0.7\ \text{mSv/day} \times 182\ \text{days} = 127\ \text{mSv}$. In years of ground-level background: $127 / 3 = 42$. A six-month ISS expedition delivers roughly 127 mSv, about 42 years' worth of sea-level background radiation — which is why cumulative dose limits, not just mission length, constrain astronaut careers.

1.11 † circular_velocity(mu, r) = math.sqrt(mu/r). For (3.986e14, 6.871e6) (Earth at 500 km): $$\sqrt{3.986\times10^{14}/6.871\times10^{6}} = \sqrt{5.801\times10^{7}} = 7{,}617\ \text{m/s}.$$ # Expected output: 7617. (Higher altitude than 400 km → slightly slower than the 7,673 m/s of the in-chapter example, as expected.)

1.13 † mass_ratio(dv, ve) = math.exp(dv/ve). For (9400, 3400): $$e^{9400/3400} = e^{2.765} = 15.87.$$ # Expected output: 15.87. Meaning: to reach orbit on a kerosene engine the fully fueled rocket must be about sixteen times heavier than the empty one — i.e. ~94% propellant. This is the tyranny of the rocket equation as a single number.

1.14 † (Find the error.) The claim "the ISS is in zero gravity, and that is why astronauts float" is wrong on the physics. At $400\ \text{km}$, $g = \mu/r^2 \approx 8.7\ \text{m/s}^2$ — about 89% of surface gravity, nowhere near zero. Astronauts float because they and the station are in continuous free fall (orbiting), so nothing pushes them against a floor; it is a state of falling, not an absence of gravity. The student confused the effect (weightlessness) with a false cause (no gravity).

1.15 (Find the error.) Two mistakes. (1) Altitude is not orbit. Crossing the $100\ \text{km}$ Kármán line makes you suborbital: without ~$7.8\ \text{km/s}$ of horizontal speed you simply fall back down (like a thrown ball). A cannon fired straight up gives altitude and zero sideways speed, so its shot returns to the ground. (2) A balloon cannot do it at all. A balloon rises only until the surrounding air density matches its own; the air near $100\ \text{km}$ is far too thin to float in, and in any case a balloon supplies no horizontal velocity. Orbit requires sideways speed, which neither method provides.

1.16 † (Mission project — model answer + rubric.) Example (Track C): "A small science orbiter that enters a stable orbit around Mars to map its surface mineralogy and monitor its weather, extending the record built by past Mars orbiters." Major delta-v legs to list: (a) launch to LEO (~9.4 km/s, the launcher's job); (b) trans-Mars injection from LEO (~3.6 km/s); (c) mid-course corrections (small); (d) Mars orbit insertion (~1–2 km/s, or less with aerobraking). Rubric: full credit for a clear one-sentence purpose, a plausible destination/orbit, and a leg-by-leg delta-v list with units; partial credit if any leg (especially arrival/insertion) is omitted. Any of the four tracks is acceptable — the point is to commit and to see the mission as a sum of delta-v costs.

1.17 (Ranking, with justification — model answer.) The ranking is mission-specific; two contrasting examples: Track C (Mars orbiter): (1) no-repair — it is months from Earth with 4–24 min light delay, so it must be autonomous and reliable; (2) radiation — a long cruise outside Earth's magnetosphere doses electronics and (if crewed) people; (3) thermal — solar flux is weaker at Mars, driving heater power. Track A (GEO comsat): (1) radiation — GEO sits above the belts in harsh electron/solar flux for 15+ years; (2) thermal — long eclipse-free sun exposure plus deep eclipses; (3) no-repair — unreachable at $35{,}786\ \text{km}$. Rubric: credit for a defensible ranking whose top three each carry a one-line, mission-specific reason; the contrast between tracks (not a single "right" order) is the learning goal.

1.18 † For $1\ \text{kg}$ at $v = 7{,}700\ \text{m/s}$: $$\text{KE} = \tfrac12 m v^2 = \tfrac12(1)(7{,}700)^2 = 2.96\times10^7\ \text{J} \approx 29.6\ \text{MJ}.$$ Lifting $1\ \text{kg}$ to $400\ \text{km}$ (exact potential change): $$\text{PE} = \mu\left(\tfrac{1}{R_E} - \tfrac{1}{r}\right) = 3.986\times10^{14}\left(\tfrac{1}{6.371\times10^{6}} - \tfrac{1}{6.771\times10^{6}}\right) = 3.70\times10^{6}\ \text{J} \approx 3.7\ \text{MJ}.$$ (The rough $\overline{g}h \approx 9\times400{,}000 = 3.6\ \text{MJ}$ agrees.) Ratio $= 29.6/3.7 \approx \mathbf{8}$. Conclusion: about eight times as much energy goes into sideways speed as into lifting the payload — a quantitative proof that "orbit is sideways, not up."

1.19 (Why can't you just fly a jet or a balloon to orbit?) An orbit requires two things at once: (1) altitude above the sensible atmosphere, and (2) ~$7.8\ \text{km/s}$ of horizontal speed. A jet fails on the speed: air-breathing engines and wings need air, so they quit in the thin upper atmosphere, and no jet approaches even Mach 3, far below the Mach ~23 of orbit — and none can operate in vacuum at all. A balloon provides only buoyant altitude (and cannot even reach $100\ \text{km}$, where there is too little air to float in) and zero horizontal speed. Neither can supply the sideways velocity, which is the actual requirement — only a rocket, carrying its own reaction mass, can.

1.20 † (Why can't you just build a bigger rocket to go faster?) Because delta-v depends on the mass ratio, not the size. From $\Delta v = v_e \ln(m_0/m_f)$: scale every dimension of a rocket up by a factor of ten while keeping the proportions the same, and $m_0$ and $m_f$ both grow by the same factor ($\sim 1000\times$), so the ratio $m_0/m_f$ — and therefore $\Delta v$ — is unchanged. You get a rocket a thousand times heavier that goes exactly as fast. The only routes to more delta-v are a higher exhaust velocity $v_e$ (better engine/propellant) or a higher mass ratio (lighter structure, more propellant fraction), and the latter is capped by how light real tanks can be (the ceiling of Case Study 2).

1.21 † (Themes in tension.) The rocket equation demands that everything be as light as possible — every kilogram of structure is a kilogram of payload lost and fuel spent lifting it (theme 4, mass is the enemy). But the unforgiving environment and the no-repair rule demand adding mass — radiation shielding, thermal protection, and redundant backup components so nothing fails the one time it must work. Concrete example: shielding a spacecraft's electronics against radiation makes it more likely to survive, but the shielding is heavy, so the designer must trade radiation dose against launch mass — protect too little and it dies in orbit; protect too much and it never reaches orbit. Living in that trade-off is the discipline.

1.23 † (Is space still hard?) Two quantified facts that have not changed: (1) reaching orbit still costs ~$9.4\ \text{km/s}$ of delta-v and forces a rocket to be ~90% propellant — a limit set by gravity and chemistry, not by engineering fashion; (2) the environment is still lethal and most spacecraft still cannot be repaired, so each is a one-shot that must work perfectly — which is why launch failures still happen and still make the news. What has changed is the economics: reusability (theme 5) — landing and reflying boosters — has cut launch cost on the order of tenfold, so we launch far more often and more cheaply. But cheaper access has not repealed the physics; the $9.4\ \text{km/s}$ wall is exactly as tall as it was in 1957. Space is more affordable, not less hard.

1.25 † Payload fraction $= 15\ \text{t} / 549\ \text{t} = 0.0273 = \mathbf{2.7\%}$. About 97% of the launcher's lift-off mass is rocket and propellant; only ~2.7% is the payload it exists to deliver. This is mass is the enemy in one number: because the rocket equation charges an exponential price to reach orbit, nearly all of what you launch is there only to launch the tiny remainder — so every kilogram shaved from structure is a kilogram that can become payload.

Chapter 2 — Newton's Laws in Space (solutions to † and odd-numbered exercises)

2.1 $p = m v = 12{,}000\ \text{kg} \times 7{,}800\ \text{m/s} = 9.36\times10^{7}\ \text{kg}\cdot\text{m/s}$.

2.3 $T = \dot m_p\, v_e = 300\ \text{kg/s} \times 3{,}100\ \text{m/s} = 930{,}000\ \text{N} = 930\ \text{kN}$. (Units: $\text{kg/s}\times\text{m/s} = \text{kg}\cdot\text{m/s}^2 = \text{N}$.)

2.5 $v_{\text{esc}} = \sqrt{2\mu/R} = \sqrt{2(4.283\times10^{4})/3{,}390}\ \text{km/s} = \sqrt{25.27} = 5.03\ \text{km/s}$ (keeping $\mu$ in $\text{km}^3/\text{s}^2$ and $R$ in km gives $\text{km/s}$). About half of Earth's $11.2\ \text{km/s}$.

2.7 $a = \mu/r^2 = (3.986\times10^{14})/(4.2164\times10^{7})^2 = (3.986\times10^{14})/(1.778\times10^{15}) = 0.224\ \text{m/s}^2$. As a fraction of surface gravity, $0.224/9.82 \approx 0.023$, i.e. about $2.3\%$. (Check: $(R/r)^2 g = (6371/42164)^2 (9.82) = (0.151)^2(9.82) = 0.224$. ✓) Gravity at GEO is real but weak.

2.9 Set $\mu/r^2 = \tfrac12\,\mu/R_\oplus^2 \Rightarrow r^2 = 2R_\oplus^2 \Rightarrow r = R_\oplus\sqrt{2} = 6{,}371 \times 1.4142 = 9{,}010\ \text{km}$. Altitude $= 9{,}010 - 6{,}371 = 2{,}639\ \text{km}$. Gravity is still half-strength nearly $2{,}600\ \text{km}$ up — the inverse square falls off gently.

2.10 $v_{\text{esc}} = \sqrt{2(4.903\times10^{3})/1{,}737} = \sqrt{5.645} = 2.38\ \text{km/s}$. As a fraction of Earth's: $2.38/11.2 \approx 0.21$, about one-fifth. Because escape velocity — and hence the delta-v to leave — is so low, the Apollo ascent stage needed only a single small engine and little propellant to lift off the Moon, whereas leaving Earth needs a giant multistage launcher.

2.11 $\dfrac{v_{\text{esc}}}{v_{\text{circ}}} = \dfrac{\sqrt{2\mu/r}}{\sqrt{\mu/r}} = \sqrt{2}$ at any radius, since the $\mu/r$ under the roots is identical. Numerically escape exceeds orbit by $\sqrt{2}-1 = 0.414$, i.e. about $41\%$.

2.12 Gravity-well depth for Mars: $\mu/R = (4.283\times10^{13})/(3.390\times10^{6}) = 1.263\times10^{7}\ \text{J/kg} \approx 12.6\ \text{MJ/kg}$. Compare $\tfrac12 v_{\text{esc}}^2 = \tfrac12(5{,}027)^2 = \tfrac12(2.527\times10^{7}) = 1.264\times10^{7}\ \text{J/kg}$. They agree — as they must, because $v_{\text{esc}} = \sqrt{2\mu/R}$ is defined precisely by $\tfrac12 v_{\text{esc}}^2 = \mu/R$.

2.13 escape_velocity(3.986e14, 6.371e6) $= \sqrt{2(3.986\times10^{14})/(6.371\times10^{6})} = \sqrt{1.2513\times10^{8}} = 11{,}186\ \text{m/s}$. # Expected output: 11186.

2.15 separation_velocity(1000, 10, 3000) $= (10 \times 3000)/1000 = 30.0\ \text{m/s}$. # Expected output: 30.0. (Momentum from rest: $0 = (m_{\text{total}}-m_{\text{gas}})V + m_{\text{gas}}(V-v_e) \Rightarrow V = m_{\text{gas}} v_e / m_{\text{total}}$.)

2.16 Two errors. (1) The velocity in the thrust term must be the exhaust speed relative to the rocket, $v_e$, not the rocket's ground speed $v$; thrust is frame-independent, but $v\,\dot m$ changes with the observer, so it cannot be the thrust. (2) For a body that is shedding mass, $d(mv)/dt$ is not simply the external force on it — you must also track the momentum carried off by the departing exhaust (apply Newton's laws to the closed rocket + exhaust system). Done correctly, the thrust is $T = v_e\,|\dot m|$, which is nonzero on the pad ($v = 0$), as observation confirms.

2.17 At $420\ \text{km}$, $r = 6{,}791\ \text{km} = 6.791\times10^{6}\ \text{m}$, so $g = \mu/r^2 = (3.986\times10^{14})/(6.791\times10^{6})^2 = (3.986\times10^{14})/(4.612\times10^{13}) = 8.64\ \text{m/s}^2$ — about $88\%$ of surface gravity, nowhere near "escaped." The astronauts float because they and the station are in free fall together (only gravity acts, with nothing pushing back), not because gravity is weak. The article confuses weightlessness with the absence of gravity.

2.18 (Mission project — answer depends on track.) Example (Track A, GEO comsat): primary body Earth, $\mu = 3.986\times10^{5}\ \text{km}^3/\text{s}^2$, $R = 6{,}371\ \text{km}$, surface $g = 9.82\ \text{m/s}^2$, $v_{\text{esc}} = 11.2\ \text{km/s}$. Design consequence: Earth's high escape velocity means the launch vehicle must supply a large delta-v (~9.4 km/s just to LEO), which dominates the whole mission's mass budget. (Track B: use the Moon's $2.38\ \text{km/s}$ — a small ascent stage suffices to leave. Track C: Mars, $5.03\ \text{km/s}$. Track D: heliocentric, $\mu_\odot$, with a target body whose escape velocity is a slow walk.)

2.19 Escape from the same altitude is $\sqrt{2}$ times the local orbital speed: $\sqrt{2}\times 7.7 \approx 10.9\ \text{km/s}$. But you are already moving at $7.7\ \text{km/s}$, so the extra speed you must add is only about $10.9 - 7.7 \approx 3.2\ \text{km/s}$. From the surface you start from rest and need the full $11.2\ \text{km/s}$. Escaping from orbit is far cheaper because the orbital motion already carries most of the required energy — the delta-v-map lesson "in orbit, halfway to anywhere."

2.21 $\dot m_p = T/v_e = (5.0\times10^{5}\ \text{N})/(3{,}000\ \text{m/s}) = 166.7\ \text{kg/s}$. Propellant in a $120\ \text{s}$ burn: $m_p = \dot m_p\,t = 166.7 \times 120 = 20{,}000\ \text{kg}$. (Sanity: $20\ \text{t}$ of propellant for a two-minute, half-meganewton burn is a plausible upper-stage scale.)

2.23 Gravity follows $g = \mu/r^2$, which approaches zero only as $r\to\infty$ — it is never actually zero at any finite distance. For $g$ to fall to $1\%$ of its surface value, $(R_\oplus/r)^2 = 0.01 \Rightarrow r = 10\,R_\oplus \approx 63{,}700\ \text{km}$ from Earth's center (about $57{,}000\ \text{km}$ altitude). You cannot "climb above" gravity; you can only get far enough that it weakens — and even the Moon, $60\,R_\oplus$ away, still feels Earth's pull strongly enough to orbit it.

2.25 Gravity at the ISS is about $8.6\ \text{m/s}^2$ — roughly $88\%$ of its surface value — so it is plainly not "zero gravity." Astronauts float because they, the station, and everything in it fall together along the same orbit; with nothing pushing back, they feel weightless. "Microgravity" is the honest term: what remains is only a tiny residual acceleration (from drag, tides, and vibration) superimposed on a very real free fall driven by very real gravity.

2.26 A rocket throws propellant not all at once but continuously, expelling an infinitesimal mass each instant at exhaust velocity $v_e$. Conservation of momentum for one such instant gives $m\,dv = -v_e\,dm$ (the rocket gains the momentum the exhaust carries away), and — crucially — each increment pushes a rocket that is lighter than the last, because earlier propellant is already gone. Integrating that relation from the full mass $m_0$ to the empty mass $m_f$ yields $\Delta v = v_e \ln(m_0/m_f)$, the Tsiolkovsky rocket equation. The "little at a time, continuously" is the continuous mass loss with a shrinking vehicle mass; the logarithm is what that shrinking produces. This is exactly the derivation carried out in Chapter 3.

Chapter 3 — The Rocket Equation

3.1 $v_e = I_{sp}\,g_0 = 311 \times 9.81 \approx 3{,}050\ \text{m/s}$ (about 3.05 km/s).

3.3 Mass ratio $= e^{\Delta v/v_e} = e^{2500/3400} = e^{0.735} \approx 2.09$.

3.5 $I_{sp} = v_e/g_0 = 4{,}400/9.81 \approx 449\ \text{s}$. That is in the LOX/liquid-hydrogen range — the highest-$I_{sp}$ chemical combination, because hydrogen's light exhaust molecules leave fastest.

3.7 At $v_e = 3.0$ km/s: for $\Delta v = 3$ km/s, ratio $= e^{1} = 2.72$; for $\Delta v = 6$ km/s, ratio $= e^{2} = 7.39$. The mass ratio grew by a factor of $7.39/2.72 = 2.72$ (that is, by $e$), not by 2. Doubling the delta-v does not double the mass ratio because the mass ratio is an exponential of delta-v — the extra delta-v multiplies the ratio by $e^{(\Delta v)/v_e}$ again.

3.9 Ratio $= e^{10000/4500} = e^{2.22} \approx 9.2$; propellant fraction $= 1 - 1/9.2 \approx 89\%$. A single chemical stage would need to be ~89% propellant just for this maneuver — which is why deep-space missions favor electric propulsion, whose far higher $v_e$ shrinks the required mass ratio dramatically (at the cost of very low thrust; see Chapter 20).

3.10 delta_v(3400, 500000, 100000) $= 3400 \ln(5) = 3400 \times 1.6094 = 5472$. # Expected output: 5472.

3.12 Stage 1: $2900\ln(500000/150000) = 2900\ln(3.333) = 3492$. Stage 2: $3400\ln(120000/25000) = 3400\ln(4.8) = 5333$. Total $\approx 8825$ m/s. # Expected output: 8825.

3.13 The flaw is physical, not mathematical. Yes, $\ln$ is unbounded — but the mass ratio is $m_0/m_f$, and $m_f$ can never be zero: it includes the tanks, engines, and structure, which have mass and cannot be made arbitrarily light (the structural coefficient $\varepsilon$). As you add propellant, you must add tank to hold it, so $m_f$ grows too, and the mass ratio saturates near $1/\varepsilon$. Making the rocket "bigger" keeps the proportions the same, so the ratio — and the delta-v — do not change at all. The real escape is staging, not size.

3.15 Set $m_f = 1000 + 0.08\,m_p$ and $m_0 = 1000 + 1.08\,m_p$. The required mass ratio is $R = e^{5000/3400} = 4.35$. Solving $\dfrac{1000 + 1.08\,m_p}{1000 + 0.08\,m_p} = 4.35$ gives $m_p \approx 4{,}580\ \text{kg}$, with $m_s = 0.08\,m_p \approx 366\ \text{kg}$ and lift-off mass $m_0 \approx 5{,}950\ \text{kg}$. (Sanity check: $R = 5946/1366 = 4.35$. ✓)

3.17 Each stage needs $R = e^{4500/3300} = e^{1.364} = 3.91$. With dry mass 10% of propellant, $\varepsilon = 0.1/1.1 = 0.091$, so the single-stage ceiling is $1/\varepsilon = 11.0$. Since $3.91 < 11.0$, each stage is comfortably buildable. (An even split is near-optimal here because both stages share the same $v_e$.)

3.18 The overall ideal mass ratio is $e^{9400/3200} \approx 19$. Carrying a 1,000 kg payload with a realistic two-stage structure (overall payload fraction of order 2–3%, as Case Study 2 found) puts the lift-off mass in the tens of tonnes — roughly $30$–$50\ \text{t}$. The exact number depends on the structural coefficient and the split; the order of magnitude ($10^4$–$10^5$ kg) is the point.

3.20 Because delta-v depends on the mass ratio $m_0/m_f$, not on absolute size. Scale every part of the rocket up by ten and the ratio is unchanged, so the delta-v is identical — you have a rocket ten times heavier that goes exactly as fast. More delta-v requires a higher $v_e$ (better engine/propellant) or a higher mass ratio (lighter structure), never mere bigness.

3.21 LEO → escape is only about $3.2\ \text{km/s}$, far less than the $11.2\ \text{km/s}$ escape velocity from the surface, because in LEO you are already moving at ~$7.8\ \text{km/s}$. Escape is about total energy: you have already "paid" most of the kinetic and potential energy climbing to orbit, so only the remaining increment to reach escape energy is needed. (This is the delta-v map's lesson: once in orbit, you are halfway to anywhere.)

3.23 Thrust $= \dot m\, v_e = 250\ \text{kg/s} \times 3{,}000\ \text{m/s} = 750{,}000\ \text{N} = 750\ \text{kN}$ (this is essentially the momentum form of thrust, from conservation of momentum). Thrust does not appear in the rocket equation: it sets how fast the stage spends its delta-v (and thus its burn time and gravity losses), but the amount of delta-v depends only on $v_e$ and the mass ratio. Two engines with the same $v_e$ and mass ratio give the same delta-v even if one has ten times the thrust.

Chapter 4 — Getting to Orbit

4.1 $v = \sqrt{\mu/r} = \sqrt{3.986\times10^5 / 6{,}671} = \sqrt{59.75} \approx 7.73\ \text{km/s}$. It is smaller than the $7.9\ \text{km/s}$ at the surface — higher orbits are slower.

4.3 The launch delta-v exceeds the orbital speed because the rocket must also pay two losses on the way up. Gravity loss ($\sim 1.5\ \text{km/s}$): delta-v spent holding the vehicle up against gravity during the climb ($\int g\sin\gamma\,dt$). Drag loss ($\sim 0.1\ \text{km/s}$): delta-v spent against air resistance. So $9.4 \approx 7.8 + 1.5 + 0.1$.

4.5 $\text{drop} = d^2/(2R) = (7{,}900)^2 / (2\times6{,}371{,}000) = 6.241\times10^7 / 1.274\times10^7 \approx 4.9\ \text{m}$. A free body falls $\tfrac12 g t^2 = \tfrac12(9.81)(1)^2 = 4.9\ \text{m}$ in one second. The two are equal: the orbiting body falls exactly as far as the round Earth curves away beneath it, so its altitude never changes. That equality is the orbit, written in meters.

4.7 Straight up the whole burn: $\Delta v_{\text{grav}} = g\,t = 9.81 \times 160 = 1{,}570\ \text{m/s} \approx 1.57\ \text{km/s}$. The real loss is smaller because a gravity turn tilts the velocity toward horizontal, so $\sin\gamma < 1$ for most of the burn and the integral $\int g\sin\gamma\,dt$ is reduced — horizontal flight ($\sin\gamma = 0$) costs no gravity loss at all.

4.9 $\Delta v = g\,t = 9.81 \times 5 = 49\ \text{m/s}$, spent to gain zero speed and zero altitude — a $100\%$ gravity loss (the "hover trap"). Lift-off thrust-to-weight must exceed 1 so the vehicle actually accelerates upward and stops the gravity-loss clock; at $T/W = 1$ all the thrust merely holds the rocket in place, and at $T/W < 1$ it cannot leave the pad.

4.11 Drag is $D = \tfrac12\rho v^2 C_d A$. The density $\rho$ is large only low down, where the rocket is still slow ($v^2$ tiny); by the time $v$ is large (Mach 20+), the rocket is high up where $\rho$ has thinned to almost nothing (scale height $\sim 8\ \text{km}$; negligible above $\sim 100\ \text{km}$). So $\rho$ and $v$ are never both large — drag gets only a brief, low-speed window. Gravity, by contrast, acts at full strength ($g \approx 9.81\ \text{m/s}^2$) for the entire burn. Hence drag loss ($\sim 0.1\ \text{km/s}$) is far below gravity loss ($\sim 1.5\ \text{km/s}$).

4.12 circular_velocity(3.986e5, 6771) $= \sqrt{398600/6771} = \sqrt{58.87} = 7.673\ \text{km/s}$, rounded to $7.67$. # Expected output: 7.67.

4.13 launch_delta_v(7800, 1500, 100, 0) $= 7800 + 1500 + 100 - 0 = 9400$. launch_delta_v(7800, 1500, 100, 465) $= 9400 - 465 = 8935$. # Expected output: 9400 then 8935.

4.14 gravity_loss(9.81, 0.6, 162) $= 9.81 \times 0.6 \times 162 = 953.5 \to 954$. gravity_loss(9.81, 1.0, 162) $= 9.81 \times 162 = 1589.2 \to 1589$. # Expected output: 954 then 1589.

4.15 $\sqrt{2gh} = \sqrt{2 \times 9.81 \times 400{,}000} = \sqrt{7.848\times10^6} = 2{,}801\ \text{m/s} \approx 2.8\ \text{km/s}$ gets the vehicle to $400\ \text{km}$ altitude with zero horizontal velocity — the top of a ballistic arc, from which it falls straight back down. That is not an orbit. To orbit at $400\ \text{km}$ ($r = 6{,}771\ \text{km}$) the vehicle needs $\sqrt{\mu/r} = 7.67\ \text{km/s}$ of sideways speed. The student had it backwards: the altitude ($\sim 2.8\ \text{km/s}$ worth) is the cheap part, and the "wasted" sideways delta-v is almost the entire reason a launch is hard.

4.17 Equatorial, due-east LEO: $\Delta v = 7.8 + 1.5 + 0.1 - 0.46 = 8.94\ \text{km/s}$, versus the no-credit $9.4\ \text{km/s}$ — a saving of $0.46\ \text{km/s}$. Low latitudes are favored because the eastward surface speed is $0.465\cos\phi\ \text{km/s}$, largest at the equator ($\phi = 0$); since delta-v is exponentially costly (Chapter 3), even a few hundred m/s of free velocity is worth building a spaceport near the equator to collect.

4.19 Vehicle X ($T/W = 1.2$) climbs more sluggishly, lengthening the burn and keeping $\sin\gamma$ sizable for longer, so it pays more gravity loss. Vehicle Y ($T/W = 1.5$) accelerates harder and reaches high speed while still low in the dense atmosphere, so it suffers more drag and a higher max-Q (and heavier acceleration/g-loads). The designer picks a value between the extremes (~$1.2$–$1.5$): high enough to curb gravity loss, low enough to keep aerodynamic and structural loads manageable.

4.20 Energy ratio $= (v_{\text{orbit}}/v_{\text{hop}})^2 = (7{,}800/1{,}400)^2 = (5.57)^2 \approx 31$. Reaching orbit takes about thirty times the energy of reaching $100\ \text{km}$ of altitude. The implication: "reaching space" is a modest hop, while "reaching orbit" is a vastly larger investment, almost all of it in kinetic energy (sideways speed) rather than height. This is the quantitative core of "space is hard."

4.21 A balloon or jet buys altitude and, for a jet, a few hundred m/s of speed — but orbit needs $\sim 7.8\ \text{km/s}$ of sideways speed. At the edge of space there is no air, so wings generate no lift and air-breathing engines no thrust; you would still need essentially the full $7.8\ \text{km/s}$ from a rocket. Altitude, however you get it, is not an orbit: without the sideways speed the vehicle simply falls back down. The balloon/jet saves almost none of the hard part.

4.22 Gravity loss is $\int g\sin\gamma\,dt$. Rising slowly and vertically means $\sin\gamma \approx 1$ for a long time $t$, so the integral is large — you spend delta-v at rate $g \approx 9.81\ \text{m/s}$ per second while gaining little speed. In the hover limit ($T/W = 1$) you gain zero speed for nonzero delta-v: the hover trap. Efficient ascent does the opposite — accelerate briskly and pitch over promptly so $\sin\gamma$ falls quickly and the integral stays small.

4.23 Mass ratio $= e^{\Delta v/v_e} = e^{9400/3000} = e^{3.13} \approx 23$. The single-stage ceiling is $1/\varepsilon \approx 12.5$ (Chapter 3, with $\varepsilon = 0.08$). Because $23 > 12.5$, no realistic single stage can supply the launch delta-v — so the vehicle must stage, splitting the delta-v across stages whose contributions add.

4.25 Chapter 1's $\sim 9.4\ \text{km/s}$ is the launch delta-v the rocket must supply; Chapter 3's $\sim 7.8\ \text{km/s}$ is the orbital speed the payload ends up at. They differ by the ascent losses of this chapter: $9.4 = 7.8\ (\text{orbital speed}) + \sim 1.5\ (\text{gravity loss}) + \sim 0.1\ (\text{drag loss})$. The $\sim 1.6\ \text{km/s}$ gap is exactly gravity + drag (before any Earth-rotation credit, which can pull the $9.4$ back toward $\sim 8.9\ \text{km/s}$ for an eastward equatorial launch).

Chapter 5 — Aerodynamics of Ascent (solutions to † and odd-numbered exercises)

Use $\rho_0 = 1.225\ \text{kg/m}^3$, $H = 8\ \text{km}$, $g_0 = 9.81\ \text{m/s}^2$.

5.1 † $q = \tfrac12\rho v^2 = \tfrac12(1.225)(100)^2 = 0.5\times1.225\times10{,}000 = 6{,}125\ \text{Pa} \approx 6.1\ \text{kPa}$. (A brisk sea-level dash is only ~6% of atmospheric pressure.)

5.3 † $\rho(11\,\text{km}) = 1.225\,e^{-11/8} = 1.225\,e^{-1.375} = 1.225\times0.2528 = 0.3097\ \text{kg/m}^3$. Then $q = \tfrac12(0.3097)(450)^2 = 0.5\times0.3097\times202{,}500 \approx 31{,}360\ \text{Pa} \approx 31.4\ \text{kPa}$ — about $31\%$ of sea-level pressure ($101\ \text{kPa}$), i.e. near max-Q levels.

5.5 † $D = q\,C_d\,A = 30{,}000\times0.4\times10.75 = 129{,}000\ \text{N} = 129\ \text{kN}$. As a weight, $129{,}000/9.81 \approx 1.32\times10^4\ \text{kg} \approx 13\ \text{t}$ — the air pushes back on the nose with the weight of a loaded truck.

5.7 † $v_{\text{maxQ}} = \sqrt{2aH} = \sqrt{2\times12\times8000} = \sqrt{192{,}000} \approx 438\ \text{m/s}$. Near the tropopause (speed of sound $\approx 295\ \text{m/s}$) that is Mach $438/295 \approx 1.5$ — right in the expected max-Q band.

5.9 † Under constant net acceleration from rest, vertical, kinematics give $v^2 = 2ah$. The max-Q condition is $v_{\text{maxQ}}^2 = 2aH$. Setting the two equal: $2ah = 2aH \Rightarrow h = H$ — max-Q at exactly one scale height. Real max-Q is usually a bit higher than $H$ because acceleration is not constant: a real vehicle lifts off at low thrust-to-weight ($a$ small, ~$4\ \text{m/s}^2$) and only speeds up as it burns off mass, so it is slower low down than the constant-$a$ idealization and reaches the max-Q speed at greater altitude; the climb also isn't perfectly vertical ($\sin\gamma < 1$).

5.11 † density(12000) $= 1.225\,e^{-1.5} = 0.27333\ \text{kg/m}^3$; dynamic_pressure(12000, 485) $= 0.5\times0.27333\times485^2 = 0.5\times0.27333\times235{,}225 \approx 32{,}148\ \text{Pa}$. # Expected output: 32148.

5.13 † ballistic_coefficient(350000, 0.5, 10.75) $= 350{,}000/(0.5\times10.75) = 350{,}000/5.375 = 65{,}116\ \text{kg/m}^2$. `drag_decel(33000, 65116)` $= 33{,}000/65{,}116 = 0.507\ \text{m/s}^2$. # Expected output: 65116 0.507.

5.14 † The error: they set the deceleration equal to the drag force in newtons, forgetting to divide by mass. Correct: $a_{\text{drag}} = D/m = 1{,}170\ \text{N}/500\ \text{kg} = 2.34\ \text{m/s}^2$. The wrong number ($1{,}170\ \text{m/s}^2 \approx 119\,g$) is physically absurd — a launch vehicle survives only a few $g$. Note the physics too: this small rocket decelerates $\sim 5\times$ harder than the big launcher at the same $q$, because its ballistic coefficient is far smaller ($\beta = 500/(0.5\times0.071) \approx 1.4\times 10^4$ vs $6.5\times10^4\ \text{kg/m}^2$). Always divide force by mass; always sanity-check against $g$.

5.15 (odd) $q = \tfrac12\rho v^2$. Near orbital velocity the vehicle is above the sensible atmosphere, where $\rho$ has collapsed exponentially to a few millionths of sea level (§5.1). So although $v^2$ is enormous, the product $\rho v^2$ — and thus $q$ — is negligible. Max-Q is not where $v$ is largest but where $\rho v^2$ is largest: low in the atmosphere, around Mach $1.5$–$2$. The blog confuses "fastest" with "highest dynamic pressure."

5.16 † At a fixed altitude, $\rho$ is fixed and $q \propto v^2$. To bring $q$ from $45$ to $35\ \text{kPa}$ requires $v^2$ to scale by $35/45 = 0.778$, so $v$ must scale by $\sqrt{0.778} = 0.882$ — a $\sim 12\%$ reduction in speed at the max-Q altitude. Throttling down achieves this: with less thrust the vehicle accelerates more gently through the dense air, so its speed (and hence $q$) at the critical altitude is lower.

5.17 (odd, mission project) No single numeric answer. A complete response records, in the MDR's Ascent Loads & Environment note: (i) a peak dynamic pressure to survive ($\sim 30$–$35\ \text{kPa}$); (ii) the $q\alpha$/lateral-load case (hold $\alpha\approx0$; a gust budget); (iii) the acoustic/buffeting vibration spectrum for payload qualification; (iv) the fairing jettison condition (heating $< \sim 1{,}135\ \text{W/m}^2$, $\sim 110$–$140\ \text{km}$). These become launch-vehicle interface requirements (Ch. 30) and structural load cases (Ch. 23).

5.18 † Relative heating $\rho v^3$ (SI), with $\rho = 1.225\,e^{-h/8000}$: - $100\ \text{km}$: $\rho = 1.225\,e^{-12.5} = 4.57\times10^{-6}$; $\rho v^3 = 4.57\times10^{-6}\times2500^3 = 4.57\times10^{-6}\times1.5625\times10^{10} \approx 7.1\times10^{4}$. - $120\ \text{km}$: $\rho = 1.225\,e^{-15} = 3.75\times10^{-7}$; $\rho v^3 = 3.75\times10^{-7}\times3000^3 = 3.75\times10^{-7}\times2.7\times10^{10} \approx 1.0\times10^{4}$. - $140\ \text{km}$: $\rho = 1.225\,e^{-17.5} = 3.08\times10^{-8}$; $\rho v^3 = 3.08\times10^{-8}\times3500^3 = 3.08\times10^{-8}\times4.29\times10^{10} \approx 1.3\times10^{3}$.

Heating is lowest (safest) at $140\ \text{km}$ — jettison there. Note that even though speed is rising ($v^3$ up by $2.7\times$ from 100 to 140 km), the density falls by $\sim 150\times$ over the same span, so the product plummets by $\sim 50\times$. The exponential utterly dominates — the whole point of §5.5. (Actual flux $\tfrac12\rho v^3$ is half these; the $\sim 1{,}135\ \text{W/m}^2$ threshold is crossed near $140\ \text{km}$ in this simple model, which over-estimates high-altitude density, so real jettison is a bit lower, ~110–130 km.)

5.19 † Why can't you just fly slowly to keep $q$ low? Because flying slowly runs up an enormous gravity loss (Chapter 4). Every second the engines burn while the vehicle climbs slowly, gravity drains delta-v at up to $g \approx 9.81\ \text{m/s}$ per second; in the hover limit (thrust-to-weight $=1$) you spend delta-v at rate $g$ for zero speed gain — the hover trap. Avoiding max-Q by flying the whole ascent slowly would trade a brief, manageable structural load for a ruinous, permanent delta-v penalty. The right move is the opposite: accelerate briskly, and throttle down only briefly, only through the worst of max-Q.

5.21 † $C_N = C_{N\alpha}\,\alpha = 2\times(5^\circ\text{ in rad}) = 2\times0.0873 = 0.1745$. Side force $N = q\,C_N\,A = 33{,}000\times0.1745\times10.75 \approx 6.19\times10^4\ \text{N} \approx 62\ \text{kN}$. This is dangerous because a launch vehicle is a long, slender tube — strong along its axis but weak in bending. A $62\ \text{kN}$ side load applied far from the center of mass creates a large bending moment that can buckle or snap the airframe. Because $N \propto q\alpha$, the load grows in direct proportion to the angle, so a gust that briefly doubles $\alpha$ doubles the load — which is exactly why $\alpha$ is held near zero through max-Q, where $q$ is greatest.

5.22 † Drag-loss fraction $= 0.1/9.4 \approx 0.011 = 1.1\%$ of the launch delta-v — nearly negligible. Yet max-Q is a top-tier design driver because it is a structural concern, not a delta-v one: the dynamic-pressure force (and especially the $q\alpha$ bending load) can physically break the vehicle, even though the delta-v it costs is a rounding error. Force and delta-v are different accountants (§5.3).

5.23 (odd) Extra mass-ratio factor $= e^{\Delta v_{\text{extra}}/v_e} = e^{50/3000} = e^{0.0167} \approx 1.017$ — the required mass ratio grows about $1.7\%$. That is a cheap insurance premium: a ~1.7% propellant penalty to avoid either a structural failure or a heavier airframe (whose added dry mass the rocket equation would tax far more, all the way to orbit). Throttling through max-Q is one of the best trades in launch.

5.24 † Across a thin layer of thickness $dh$, the pressure drop equals the weight of the layer: $dp = -\rho g\,dh$. For an ideal gas at (roughly) constant temperature, density is proportional to pressure, $\rho = pM/(R_uT) = p/(R_{\text{air}}T)$, so $dp = -\dfrac{g}{R_{\text{air}}T}\,p\,dh$, i.e. $\dfrac{dp}{p} = -\dfrac{dh}{H}$ with $H = R_{\text{air}}T/g$ constant. A constant fractional pressure drop per unit height integrates to an exponential, $p \propto e^{-h/H}$, and since $\rho \propto p$, so does density. The key idea: the restoring weight of each layer is proportional to the local density (hence pressure), so each step of height removes a fixed fraction — the signature of exponential decay.

5.25 (odd) Two ways the atmosphere is unforgiving on ascent: (1) the crushing/bending aerodynamic loads of max-Q (dynamic pressure, drag, and the $q\alpha$ side load that can snap a slender airframe); (2) aerodynamic heating plus buffeting and the ferocious acoustic environment that shake and heat the vehicle and payload. One way the vacuum above is easier, aerodynamically: there are no aerodynamic forces at all — no drag, no dynamic pressure, no buffeting — so the vehicle's shape and orientation no longer matter. That is precisely why the fairing can be jettisoned and why upper stages and spacecraft need no aerodynamic protection or streamlining once above the sensible atmosphere.

Chapter 6 — Energy in Space (solutions to † and odd-numbered exercises)

Uses $\mu_\oplus = 3.986\times10^{5}\ \text{km}^3/\text{s}^2$, $R_\oplus = 6{,}371\ \text{km}$.

6.1 At $800\ \text{km}$ altitude, $r = 6{,}371 + 800 = 7{,}171\ \text{km}$. $v_{\text{circ}} = \sqrt{\mu/r} = \sqrt{398{,}600/7{,}171} = \sqrt{55.59} = 7.46\ \text{km/s}$.

6.3 $v^2 = \mu(2/r - 1/a) = 398{,}600\,(2/10{,}000 - 1/15{,}000) = 398{,}600\,(2.000\times10^{-4} - 6.667\times10^{-5}) = 398{,}600 \times 1.3333\times10^{-4} = 53.15\ \text{km}^2/\text{s}^2$, so $v = 7.29\ \text{km/s}$.

6.5 $\varepsilon = -\mu/(2a) \Rightarrow a = -\mu/(2\varepsilon) = -398{,}600/(2\times(-15)) = 398{,}600/30 = 13{,}287\ \text{km}$. (Positive $a$, as it must be for a bound orbit with $\varepsilon<0$.)

6.7 Perigee $r_p = 6{,}371+500 = 6{,}871\ \text{km}$; apogee $r_a = 6{,}371+39{,}900 = 46{,}271\ \text{km}$. (a) $a = (r_p + r_a)/2 = (6{,}871 + 46{,}271)/2 = 26{,}571\ \text{km}$. (b) $v_p = \sqrt{\mu(2/6{,}871 - 1/26{,}571)} = \sqrt{398{,}600 \times 2.5344\times10^{-4}} = \sqrt{101.02} = 10.05\ \text{km/s}$. (c) $v_a = \sqrt{\mu(2/46{,}271 - 1/26{,}571)} = \sqrt{398{,}600 \times 5.581\times10^{-6}} = \sqrt{2.225} = 1.49\ \text{km/s}$. Check angular momentum: $r_p v_p = 6{,}871 \times 10.05 \approx 69{,}060\ \text{km}^2/\text{s}$; $r_a v_a = 46{,}271 \times 1.49 \approx 69{,}000\ \text{km}^2/\text{s}$. Equal to within rounding. ✓ (This is a 12-hour Molniya orbit; note the huge speed spread — fast, brief perigee; slow, lingering apogee, which is exactly why Molniya orbits "hang" over high latitudes.)

6.9 $\varepsilon = +3.0\ \text{km}^2/\text{s}^2$ (hyperbolic). (a) $v_\infty = \sqrt{2\varepsilon} = \sqrt{6} = 2.45\ \text{km/s}$. (b) $a = -\mu/(2\varepsilon) = -398{,}600/6 = -66{,}433\ \text{km}$ (negative → hyperbola). (c) $v^2 = \mu(2/r - 1/a) = 398{,}600\,(2/6{,}771 - 1/(-66{,}433)) = 398{,}600\,(2.9538\times10^{-4} + 1.5053\times10^{-5}) = 398{,}600 \times 3.1043\times10^{-4} = 123.74$, so $v = 11.12\ \text{km/s}$. Check: $v_{\text{esc}}^2(6{,}771) + v_\infty^2 = 117.74 + 6.0 = 123.74$. ✓

6.11 circular_velocity(3.986e14, 6.771e6) $= \sqrt{3.986\times10^{14}/6.771\times10^{6}} = \sqrt{5.887\times10^{7}} = 7{,}672.6 \to$ `round` $= 7673$. # Expected output: 7673.

6.13 vis_viva(3.986e14, 6.771e6, 2.4468e7) $= \sqrt{3.986\times10^{14}(2/6.771\times10^{6} - 1/2.4468\times10^{7})} = \sqrt{1.0145\times10^{8}} = 10{,}072\ \text{m/s}$. # Expected output: 10072.

6.15 The backwards clause is "losing energy makes it go slower." Losing energy makes the orbit smaller (more negative $\varepsilon$ → smaller $a$), and a lower orbit is faster ($v_{\text{circ}} = \sqrt{\mu/r}$ increases as $r$ falls). So as drag saps its energy, the ISS actually speeds up as it spirals down — the famous "drag paradox." The claim that it falls is correct; the claim that it slows is exactly wrong. (Drag does slow it instantaneously at each moment, but that lowers the orbit, and the lower orbit's steady speed is higher — energy down, speed up.)

6.16 Hohmann-style raise from LEO ($r_1 = 6{,}771\ \text{km}$) to MEO ($r_2 = 26{,}571\ \text{km}$). - $v_{\text{circ},1} = \sqrt{398{,}600/6{,}771} = 7.673\ \text{km/s}$. - Transfer semi-major axis $a = (6{,}771 + 26{,}571)/2 = 16{,}671\ \text{km}$. - $v_p = \sqrt{\mu(2/6{,}771 - 1/16{,}671)} = \sqrt{93.85} = 9.688\ \text{km/s}$. - $v_a = \sqrt{\mu(2/26{,}571 - 1/16{,}671)} = \sqrt{6.093} = 2.468\ \text{km/s}$. - $v_{\text{circ},2} = \sqrt{398{,}600/26{,}571} = 3.873\ \text{km/s}$. - Burn 1 (perigee raise): $9.688 - 7.673 = 2.015\ \text{km/s}$. - Burn 2 (apogee circularize): $3.873 - 2.468 = 1.405\ \text{km/s}$. - Total $\approx 3.42\ \text{km/s}$. (Both prograde; sanity: the total is comparable to the LEO→GEO Hohmann because MEO is a large but not extreme raise.)

6.17 (Mission project — pattern shown for Track A, GEO comsat.) With $\mu_\oplus$ and $a = r = 42{,}164\ \text{km}$: $\varepsilon = -\mu/(2a) = -4.73\ \text{MJ/kg}$; $v_{\text{circ}} = \sqrt{\mu/r} = 3.07\ \text{km/s}$. A GEO comsat does not escape Earth, so no escape line is needed (record instead the GTO transfer speeds, $10.1$ and $1.6\ \text{km/s}$, and the $\approx1.5\ \text{km/s}$ apogee kick). Track C (Mars orbiter) and Track D (asteroid) additionally record Earth escape from their parking orbit; Track B (lunar) records the Moon's $\mu = 4.903\times10^{3}\ \text{km}^3/\text{s}^2$ and its gentle escape. Grade on: correct $\mu$, correct $a$, units, and whether an escape line is (or isn't) appropriate.

6.18 From a $300\ \text{km}$ parking orbit, $r = 6{,}671\ \text{km}$. $v_{\text{circ}} = \sqrt{398{,}600/6{,}671} = 7.730\ \text{km/s}$; $v_{\text{esc}} = \sqrt{2\times398{,}600/6{,}671} = \sqrt{119.50} = 10.932\ \text{km/s}$. Injection speed $v = \sqrt{v_\infty^2 + v_{\text{esc}}^2} = \sqrt{3.5^2 + 10.932^2} = \sqrt{12.25 + 119.50} = \sqrt{131.75} = 11.478\ \text{km/s}$. Injection burn $= 11.478 - 7.730 = 3.75\ \text{km/s}$.

6.19 Firing prograde (toward the target) adds energy, which raises your orbit; a higher orbit is slower and has a longer period, so — counterintuitively — you drift farther behind the station, not closer. To actually catch up you must burn retrograde: this lowers your orbit, where you move faster and complete each lap in less time, so you gain on the target; once alongside (or below and ahead) you raise back up to match orbits. On the ground "go faster" and "catch up" are the same command; in orbit they are opposites (§6.6). This is the exact problem Gemini pilots had to master before the first docking.

6.21 Transfer ellipse from LEO to lunar distance: $r_p = 6{,}771\ \text{km}$, $r_a = 384{,}400\ \text{km}$, so $a = (6{,}771 + 384{,}400)/2 = 195{,}586\ \text{km}$. $v_p = \sqrt{\mu(2/6{,}771 - 1/195{,}586)} = \sqrt{398{,}600 \times 2.9026\times10^{-4}} = \sqrt{115.7} = 10.76\ \text{km/s}$. Departure burn $= 10.76 - 7.673 = 3.08\ \text{km/s} \approx 3.1\ \text{km/s}$, matching the delta-v map's "LEO → trans-lunar injection $\approx 3.1\ \text{km/s}$." ✓ (Note this ignores the Moon's own gravity, which slightly lowers the true figure — a patched-conics refinement for Ch. 11.)

6.23 Rocket equation (Ch. 3): mass ratio $= e^{\Delta v/v_e} = e^{1450/3050} = e^{0.4754} = 1.61$. Propellant fraction $= 1 - 1/1.61 = 1 - 0.622 = 0.378 \approx 38\%$. So a kick stage supplying the GTO→GEO burn must be about $38\%$ propellant — modest, because $1.45\ \text{km/s}$ is small next to $v_e = 3.05\ \text{km/s}$.

6.25 Both start in the same $400\ \text{km}$ circular orbit. - A (prograde $+50\ \text{m/s}$): energy $\uparrow$, semi-major axis $\uparrow$, period $\uparrow$ (longer). It rises into a higher, slower orbit. After one revolution it takes longer to come around, so relative to its start point (and to B) it ends up behind. - B (retrograde $-50\ \text{m/s}$): energy $\downarrow$, semi-major axis $\downarrow$, period $\downarrow$ (shorter). It drops into a lower, faster orbit and comes around sooner, ending up ahead. - The satellite that slowed down (B) pulls ahead; the one that sped up (A) falls behind — the threshold concept of §6.6 made numerical. B, with the shorter period, next passes over a given ground point first. (This is precisely why a chaser burns retrograde to catch a target ahead of it.)

Chapter 7 — Atmospheric Re-Entry

Solutions to the daggered (†) and odd-numbered exercises. Use $g_0 = 9.81\ \text{m/s}^2$, $\mu = 3.986\times10^5\ \text{km}^3/\text{s}^2$, $c_p = 1{,}005\ \text{J/(kg·K)}$, $H = 7{,}200\ \text{m}$, $e = 2.718$ (except where a problem states $H = 7{,}000$).

7.1 $\tfrac12 v^2 = \tfrac12 (7{,}600)^2 = \tfrac12 (5.776\times10^7) = 2.89\times10^7\ \text{J/kg} \approx 28.9\ \text{MJ/kg}$.

7.3 $\Delta T \approx v^2/(2c_p) = (6{,}000)^2/(2\cdot1{,}005) = 3.6\times10^7/2{,}010 \approx 17{,}910\ \text{K} \approx 18{,}000\ \text{K}$ (ideal). Real air never reaches this: at a few thousand kelvin the $\mathrm{O_2}$ and $\mathrm{N_2}$ dissociate, and higher still the atoms ionize, both absorbing energy endothermically without a proportional temperature rise, so the real gas caps near several thousand K.

7.5 $f_p = 8.98\sqrt{10^{18}} = 8.98\times10^9\ \text{Hz} = 8.98\ \text{GHz}$. The S-band link at $2.2\ \text{GHz}$ is below $8.98\ \text{GHz}$, so it is reflected by the sheath — yes, blacked out. (Only a link above ~9 GHz would ride over this plasma frequency.)

7.7 Convective flux $\dot q \propto v^3$, so the ratio is $(11/7.8)^3 = (1.410)^3 \approx 2.8$. The peak flux is about 2.8× higher for the lunar-return entry — and once radiative heating (which grows even faster with speed) is included, the real penalty is larger still. This is why fast returns need robust ablators.

7.9 The estimate $\Delta T = v^2/(2c_p)$ assumes calorically perfect air with a constant $c_p$, so it converts all the flow's kinetic energy into a temperature rise and predicts ~$30{,}000\ \text{K}$. Real air is not calorically perfect at those energies: the energy first goes into dissociating $\mathrm{O_2}$ and $\mathrm{N_2}$ into atoms (a few thousand K) and then ionizing the atoms (stripping electrons), both of which soak up large amounts of energy without a matching temperature rise — effectively the gas's heat capacity balloons. The real shock layer settles near $6{,}000$–$11{,}000\ \text{K}$. The ionization is also the direct cause of the plasma sheath and the radio blackout of §7.6.

7.10 $a_{\max} = v_E^2 \sin\gamma_E/(2eH) = (7{,}800)^2 (\sin 2^\circ)/(2\cdot2.718\cdot7{,}200) = (6.084\times10^7)(0.0349)/(39{,}143) = 2.123\times10^6/39{,}143 = 54.2\ \text{m/s}^2 \approx 5.5\ \text{g}$.

7.11 With $\gamma_E = 5^\circ$: $a_{\max} = (6.084\times10^7)(\sin 5^\circ)/(39{,}143) = (6.084\times10^7)(0.0872)/39{,}143 = 5.303\times10^6/39{,}143 = 135.5\ \text{m/s}^2 \approx 13.8\ \text{g}$. The peak g grew by $13.8/5.5 = 2.5\times$, exactly matching $\sin 5^\circ/\sin 2^\circ = 0.0872/0.0349 = 2.5$. Confirms $a_{\max}\propto \sin\gamma_E$.

7.12 They reach the same peak g — the Allen–Eggers $a_{\max} = v_E^2\sin\gamma_E/(2eH)$ contains no $\beta$, so mass and drag area cancel out of the peak. What differs is altitude: from $D/m = \rho v^2/(2\beta)$, a high-$\beta$ (dense) body needs a larger $\rho$ to produce the same deceleration, so it penetrates deeper (lower altitude, denser air) before peaking; the low-$\beta$ body peaks higher up. The vehicle sets where the peak occurs, not how large it is.

7.13 peak_g(7800, 6, 7000): $\sin 6^\circ = 0.1045$; $a_{\max} = (7{,}800)^2(0.1045)/(2\cdot2.718\cdot 7{,}000) = 6.360\times10^6/38{,}056 = 167.1\ \text{m/s}^2$; $/9.81 = 17.0\ \text{g}$. # Expected output: 17.0.

7.15 plasma_frequency(1e16) $= 8.98\sqrt{10^{16}} = 8.98\times10^8 = 8.98\times10^8\ \text{Hz}$ (898 MHz). # Expected output: 898000000.0. A $400\ \text{MHz}$ UHF link is below 898 MHz, so it is blacked out.

7.16 Original circular orbit: $r = 6{,}371 + 500 = 6{,}871\ \text{km}$, $v_{\text{circ}} = \sqrt{\mu/r} = \sqrt{3.986\times10^5/6{,}871} = \sqrt{58.01} = 7.617\ \text{km/s}$. New (transfer) ellipse: apogee $r_a = 6{,}871\ \text{km}$ (the burn point), perigee $r_p = 6{,}371 + 60 = 6{,}431\ \text{km}$, so $a = (6{,}871 + 6{,}431)/2 = 6{,}651\ \text{km}$. Speed at apogee (by vis-viva): $v_a = \sqrt{\mu(2/r_a - 1/a)} = \sqrt{3.986\times10^5(2/6{,}871 - 1/6{,}651)} = \sqrt{3.986\times10^5(1.407\times10^{-4})} = \sqrt{56.09} = 7.490\ \text{km/s}$. Deorbit $\Delta v = 7.617 - 7.490 = 0.127\ \text{km/s} \approx 127\ \text{m/s}$. (Sanity check: a ~$130\ \text{m/s}$ deorbit burn is exactly the order of real LEO deorbits — the atmosphere does the other ~7.5 km/s of braking for free.)

7.17 (Mission project — rubric; example answers.) Track A (GEO comsat): "No atmospheric entry; end-of-life disposal to a graveyard orbit above GEO." Track B (lunar lander): "No atmosphere at the Moon → fully propulsive descent, ~2 km/s of landing delta-v, no TPS; the free brake of §7.1 is unavailable." Track C (Mars orbiter): "Thin Mars atmosphere → optional aerocapture/aerobraking on arrival to save propellant; note arrival speed and whether the atmosphere is used." Track D (asteroid sample return): "No atmosphere at the target, but the return capsule performs the fastest, hottest Earth entry (≥12 km/s) → robust ablative shield, low-$\beta$ capsule, lifting or skip entry." Full credit: names whether entry occurs, the entry speed, the TPS strategy, and the ballistic-coefficient class, and records it in the MDR.

7.18 A ballistic entry at $12\ \text{km/s}$ is untenable: even a moderate $\gamma_E = 5^\circ$ gives $a_{\max} = (12{,}000)^2(\sin 5^\circ)/(2eH) = (1.44\times10^8)(0.0872)/39{,}143 = 320.6\ \text{m/s}^2 \approx 32.7\ \text{g}$ — far beyond crew tolerance. So fly a **lifting** entry ($L/D$ from an offset CG) to spread the deceleration and cap peak g near a survivable few g, and manage the narrow high-speed corridor — possibly with a controlled skip — since the skip-out risk is severe near escape speed. TPS: an ablative shield, because convective heating $\propto v^3$ and radiative heating (steeper still) are both extreme at $12\ \text{km/s}$; reusable tiles could not survive it. This is a scaled-up Apollo problem.

7.19 $E = m\,\tfrac12 v^2 = 5{,}000 \times \tfrac12(7{,}800)^2 = 5{,}000 \times 3.042\times10^7 = 1.52\times10^{11}\ \text{J} = 152\ \text{GJ}$. Water needs ~$2.6\ \text{MJ/kg}$ to heat from room temperature and boil, so $1.52\times10^{11}/2.6\times10^6 \approx 58{,}500\ \text{kg} \approx 58\ \text{t}$ of water. The capsule's re-entry energy could boil about 12 times its own mass in water — and it must shed all of it as heat in minutes. This is why the shield, not the structure, is the star of re-entry.

7.21 Two failures at the shallow extreme: (1) past the shallow corridor wall the vehicle skips back out of the atmosphere entirely — it fails to decelerate enough, its path arcs up, and it is flung back to space (for a capsule with no propulsion and limited battery/consumables, this can mean it never lands or lands uncontrolled); (2) even short of a skip, a very shallow entry prolongs the heat soak, raising the total heat load and demanding a heavier shield. "As shallow as possible" is not "as safe as possible" — safety is inside the corridor, steep enough to be captured, shallow enough to stay under the g and peak-heating limits.

7.22 By vis-viva, $v = \sqrt{\mu(2/r - 1/a)}$ is larger where $r$ is smaller, so the vehicle is faster at perigee (the low point) than at apogee. Heating scales as $\rho v^3$, and perigee is simultaneously the fastest point (largest $v$) and the deepest point (largest $\rho$), so both factors are maximized there: the heating peaks near perigee. This is exactly why an aerobraking or grazing-entry pass does its thermal work at the low point of the orbit.

7.23 Both $\rho v^2$ (drag) and $\rho v^3$ (heating) contain the density $\rho$, which falls off exponentially with altitude, and a speed $v$ that behaves oppositely: high up, $v$ is enormous but $\rho \to 0$; low down, $\rho$ is large but the vehicle has already been decelerated so $v$ is small. The product is small at both ends and peaks at a middle altitude, where there is simultaneously enough air and enough speed. On ascent, the mirror phenomenon is max-Q — the peak of dynamic pressure $q = \tfrac12\rho v^2$ (Ch. 4/5) — for exactly the same reason.

7.24 Mass ratio $= e^{\Delta v/v_e} = e^{7800/3400} = e^{2.294} = 9.9$. Propellant fraction $= 1 - 1/9.9 = 0.90$, so ~90% of the returning vehicle would have to be propellant just to brake — as hard as launching all over again. This is why no vehicle brakes from orbit with rockets: the rocket equation makes it as expensive as the launch, so we use the atmosphere (a ~100–130 m/s deorbit burn plus a heat shield) instead.

7.25 A capsule returning from the Moon falls into Earth's gravity well and arrives near escape speed, which by the Ch. 2 relation is $v_{\text{esc}} = \sqrt{2}\,v_{\text{circ}} \approx 1.41\times7.8 \approx 11\ \text{km/s}$. Energy scales as $v^2$: $(11/7.8)^2 \approx 1.99$, so ~2× the energy per kilogram of a LEO entry. Heating scales as $v^3$: $(11/7.8)^3 \approx 2.8$, so ~2.8× the peak convective flux (and radiative heating, steeper still, adds more). So a lunar return carries about twice the energy and suffers about three times the peak heating of a LEO entry — the quantitative reason it demands an ablative shield and a lifting entry, and the reason the $\sqrt{2}$ factor of Chapter 2 is felt so sharply on the way home.

Chapter 8 — Kepler's Laws and the Two-Body Problem (solutions to † and odd-numbered exercises)

Uses $\mu_\oplus = 3.986\times10^{5}\ \text{km}^3/\text{s}^2$, $R_\oplus = 6{,}371\ \text{km}$ unless noted.

8.1 (a) First law (ellipses): every orbit is an ellipse with the primary at one focus — this is the conic solution $r = p/(1+e\cos\nu)$ of the inverse-square equation of motion. (b) Second law (equal areas): the primary–body line sweeps equal areas in equal times — a statement of conservation of angular momentum ($dA/dt = h/2$, constant). (c) Third law (periods): $T^2 \propto a^3$ — obtained by equating the ellipse's area $\pi ab$ to the steady area rate times the period, $(h/2)T$, giving $T^2 = (4\pi^2/\mu)a^3$.

8.3 $a = (r_p + r_a)/2 = (6{,}678 + 9{,}922)/2 = 8{,}300\ \text{km}$. $e = (r_a - r_p)/(r_a + r_p) = (9{,}922 - 6{,}678)/(9{,}922 + 6{,}678) = 3{,}244/16{,}600 = 0.195$. A low-eccentricity ellipse (bound).

8.5 Period scales as $T \propto a^{3/2}$. If $a_B = 4a_A$, then $T_B/T_A = 4^{3/2} = (4^{1/2})^3 = 2^3 = 8$. Orbit B's period is 8 times longer. (Four times the size, eight times the time.)

8.7 GPS: $a = 26{,}560\ \text{km}$. (a) $T = 2\pi\sqrt{a^3/\mu} = 2\pi\sqrt{(26{,}560)^3/3.986\times10^{5}} = 2\pi\sqrt{1.8736\times10^{13}/ 3.986\times10^{5}} = 2\pi\sqrt{4.7010\times10^{7}} = 2\pi(6{,}857.9) = 43{,}090\ \text{s} = 11.97\ \text{h} \approx 11\ \text{h}\ 58\ \text{min}$. (b) Two periods: $2T = 86{,}180\ \text{s}$, essentially the sidereal day ($86{,}164\ \text{s}$; the $\sim 16\ \text{s}$ difference is rounding in $a$). So a GPS satellite makes exactly two revolutions per sidereal day and repeats its ground track daily — which is why the constellation's geometry over any site recurs each day.

8.9 Orbit equation $r = p/(1+e\cos\nu)$ with $p = a(1-e^2)$. Set $r = a$: $a = a(1-e^2)/(1+e\cos\nu) \Rightarrow 1+e\cos\nu = 1-e^2 \Rightarrow \cos\nu = -e$. So the radius equals the semi-major axis exactly where $\cos\nu = -e$. For $e = 0.2$: $\cos\nu = -0.2 \Rightarrow \nu = 101.5^\circ$ (and by symmetry $258.5^\circ$). (Neat general result: $r = a$ at $\cos\nu = -e$ — the ends of the minor axis of the ellipse.)

8.11 $e = 0.6$, $E = 60^\circ = 1.0472\ \text{rad}$, $a = 12{,}000\ \text{km}$. (a) Kepler's equation: $M = E - e\sin E = 1.0472 - 0.6\sin(60^\circ) = 1.0472 - 0.6(0.8660) = 1.0472 - 0.5196 = 0.5276\ \text{rad} = 30.2^\circ$. (b) $\tan(\nu/2) = \sqrt{(1+e)/(1-e)}\tan(E/2) = \sqrt{1.6/0.4}\tan(30^\circ) = 2(0.5774) = 1.1547 \Rightarrow \nu/2 = 49.1^\circ \Rightarrow \nu = 98.2^\circ$. (Check: $\cos\nu = (\cos E - e)/(1-e\cos E) = (0.5-0.6)/(1-0.3) = -0.1/0.7 = -0.143 \Rightarrow \nu = 98.2^\circ$. ✓) (c) $r = a(1 - e\cos E) = 12{,}000(1 - 0.6\times0.5) = 12{,}000(0.7) = 8{,}400\ \text{km}$. Note $M < E < \nu$ ($30.2^\circ < 60^\circ < 98.2^\circ$), as required.

8.13 period(mu, a) = 2*pi*sqrt(a**3/mu), SI. period(3.986e14, 6.771e6): $a^3 = (6.771\times10^6)^3 = 3.1043\times10^{20}$; $/\mu = 7.788\times10^{5}$; $\sqrt{} = 882.5$; $\times 2\pi = 5{,}545\ \text{s}$. # Expected output: 5545. period(3.986e14, 42.164e6): $a^3 = 7.4959\times10^{22}$; $/\mu = 1.8806\times10^{8}$; $\sqrt{} = 13{,}713$; $\times 2\pi = 86{,}164\ \text{s}$. # Expected output: 86164.

8.15 solve_kepler(1.1877, 0.723) by Newton from $E_0 = M + e\sin M$: $E_0 = 1.1877 + 0.723\sin(1.1877) = 1.1877 + 0.723(0.92752) = 1.85829$. Iter 1: $f = 1.85829 - 0.723\sin(1.85829) - 1.1877 = -0.02273$; $f' = 1 - 0.723\cos(1.85829) = 1.20501$; $E_1 = 1.85829 + 0.02273/1.20501 = 1.87716$. Iter 2: $f = 1.87716 - 0.723\sin(1.87716) - 1.1877 = +0.00012$; $f' = 1.21805$; $E_2 = 1.87716 - 0.00012/1.21805 = 1.87706$. Converged. `# Expected output: 1.87706` (rad) $= 107.55^\circ$.

8.17 At $t = T/4$ the mean anomaly is $M = n\,(T/4) = (2\pi/T)(T/4) = \pi/2 = 90^\circ$ — but $M$ is not the true anomaly. Solving $M = E - e\sin E$ with $M = 1.5708$, $e = 0.5$: $E_0 = 1.5708 + 0.5(1) = 2.0708$; two Newton steps give $E \approx 2.0210\ \text{rad} = 115.8^\circ$; then $\cos\nu = (\cos E - e)/(1 - e\cos E) = (-0.4361 - 0.5)/(1 + 0.2180) = -0.7685 \Rightarrow \nu \approx 140.2^\circ$. So at a quarter period the true anomaly is about $140^\circ$, larger than $90^\circ$ — because the satellite raced through the fast perigee arc first (Kepler's second law), covering more than a quarter of the angle in the first quarter of the time. The claim $\nu = 90^\circ$ at $t = T/4$ holds only for a circle ($e = 0$).

8.19 Averaging "perigee and apogee circular periods" is conceptually wrong: there is no such thing as a circular period at perigee or apogee for an elliptical orbit, and Kepler's third law says the period depends only on the semi-major axis, not on $e$ or on the apsidal radii separately. Correct period with $a = 20{,}000\ \text{km}$: $T = 2\pi\sqrt{a^3/\mu} = 2\pi\sqrt{(20{,}000)^3/3.986\times10^{5}} = 2\pi\sqrt{2.0070\times10^{7}} = 2\pi(4{,}480) = 28{,}150\ \text{s} = 7.82\ \text{h}$. (The wrong "average" method would give circular periods $T(r_p{=}10{,}000) = 9{,}952\ \text{s}$ and $T(r_a{=}30{,}000) = 51{,}710\ \text{s}$, averaging to $30{,}830\ \text{s}$ — about $10\%$ too high.) Eccentricity contributes nothing to the period.

8.21 Molniya design. $T = 43{,}082\ \text{s}$ (half a sidereal day). (a) $a = (\mu T^2/4\pi^2)^{1/3} = ((3.986\times10^{5})(43{,}082)^2/39.478)^{1/3} = (1.874\times10^{13})^{1/3} = 26{,}562\ \text{km}$. (b) $r_p = R_\oplus + 500 = 6{,}871\ \text{km}$, so $e = 1 - r_p/a = 1 - 6{,}871/26{,}562 = 0.741$. $r_a = a(1+e) = 26{,}562(1.741) = 46{,}244\ \text{km}$, apogee altitude $= 46{,}244 - 6{,}371 = 39{,}873\ \text{km}$ (just above GEO altitude). (c) Set the argument of periapsis $\omega = 270^\circ$: this places perigee at the southernmost point and apogee at the northernmost, so the slow, high-loitering half of the orbit hangs over the northern hemisphere. (Set $i = 63.4^\circ$, the critical inclination, so this stays put — Ch. 12.)

8.23 (Mission project — pattern shown for Track A, GEO comsat.) Six elements: $a = 42{,}164\ \text{km}$, $e \approx 0$, $i = 0^\circ$ (equatorial), $\Omega$ and $\omega$ undefined for a circular equatorial orbit (a real, slightly-elliptical design fixes them), $\nu$ = wherever the satellite is stationed. Period $T = 2\pi\sqrt{a^3/\mu} = 86{,}164\ \text{s}$; mean motion $n = 2\pi/T = 7.29\times10^{-5}\ \text{rad/s} = 1.00\ \text{rev/day}$. Constrained by the mission: $a$ and $i$ (GEO must be $42{,}164\ \text{km}$, equatorial); free to choose: $\Omega$/longitude slot. Track C (Mars orbiter) records the science-orbit elements about Mars using $\mu_{\text{Mars}} = 4.283\times10^{4}\ \text{km}^3/\text{s}^2$; Track B (lunar) uses $\mu_{\text{Moon}} = 4.903\times10^{3}$. Grade on: correct primary $\mu$, a sensible full element set, correct period/mean motion, and a clear split of constrained vs. free elements.

8.25 Why can't you read $\nu$ off a clock? Because Kepler's second law makes the angular rate non-uniform: the body sweeps the true anomaly quickly near perigee and slowly near apogee, so equal times do not correspond to equal angles. What does advance uniformly with the clock is the mean anomaly $M = n(t-t_p)$ — the "fraction of the period $\times 360^\circ$" quantity — but $M$ equals the true anomaly $\nu$ only at perigee and apogee. To recover $\nu$ you must solve Kepler's equation for $E$ and convert. The error is worst for highly eccentric orbits (large $M$–$\nu$ gap; e.g. a Molniya or a comet). It works exactly only for a circular orbit ($e = 0$), where $M = E = \nu$ and motion really is uniform.

8.27 Solar orbits obey $T^2 \propto a^3$ with Earth as reference ($T = 1\ \text{yr}$ at $a = 1\ \text{AU}$), so $T_{\text{Mars}} = a^{3/2} = (1.52)^{3/2}\ \text{yr}$. $(1.52)^{3/2} = 1.52\sqrt{1.52} = 1.52(1.2329) = 1.874\ \text{yr}$. In days: $1.874 \times 365.25 = 684\ \text{days}$, within about $0.4\%$ of the true $687\ \text{days}$ (the small gap comes from using $1.52$ rather than Mars's exact $1.524\ \text{AU}$). Kepler's third law, no dynamics required.

8.29 GTO ($a = 24{,}468\ \text{km}$, $e = 0.723$) at $\nu = 147.3^\circ$, $r = 29{,}802\ \text{km}$. Speed (vis-viva, Ch. 6): $v = \sqrt{\mu(2/r - 1/a)} = \sqrt{3.986\times10^{5}(2/29{,}802 - 1/24{,}468)} = \sqrt{3.986\times10^{5}(6.711\times10^{-5} - 4.087\times10^{-5})} = \sqrt{10.46} = 3.23\ \text{km/s}$. Energy two ways: $\varepsilon = -\mu/(2a) = -3.986\times10^{5}/(2\times24{,}468) = -8.145\ \text{MJ/kg}$; and $\varepsilon = v^2/2 - \mu/r = 10.46/2 - 3.986\times10^{5}/29{,}802 = 5.23 - 13.37 = -8.14\ \text{MJ/kg}$. They agree — as they must, since $\varepsilon$ is conserved and the two expressions are the definition and the size-relation of the same orbit.

Chapter 9

Chapter 9 — Solutions to odd-numbered and †-marked exercises

Constants: $\mu_\oplus = 3.986\times10^{5}\ \text{km}^3/\text{s}^2$, $R_{\text{mean}}=6{,}371\ \text{km}$, $R_{\text{eq}}=6{,}378\ \text{km}$, sidereal day $86{,}164\ \text{s}$, solar day $86{,}400\ \text{s}$, $g_0=9.81\ \text{m/s}^2$. All arithmetic hand-computed (code never executed).

1. (⭐)† $r = 6{,}371 + 550 = 6{,}921\ \text{km}$. (a) $v = \sqrt{\mu/r} = \sqrt{3.986\times10^{5}/6{,}921} = \sqrt{57.59} = 7.59\ \text{km/s}$. (b) $T = 2\pi\sqrt{r^3/\mu} = 2\pi\sqrt{(6{,}921)^3/3.986\times10^{5}} = 2\pi(911.7\ \text{s}) = 5{,}728\ \text{s} = 95.5\ \text{min}$. (c) Orbits/day $= 1{,}440/95.5 = 15.1$. Sanity check: $95.5\ \text{min}$ is squarely in the $90$–$100\ \text{min}$ LEO band, and $15.1$ orbits/day matches the "~16 sunrises a day" folklore. ✓

3. (⭐⭐)† From one sidereal day $T = 86{,}164\ \text{s}$: $a = \left(\mu T^2/4\pi^2\right)^{1/3} = \left((3.986\times10^{5})(86{,}164)^2/39.478\right)^{1/3} = (7.496\times10^{13})^{1/3} = 42{,}164\ \text{km}$. Altitude $= a - R_{\text{eq}} = 42{,}164 - 6{,}378 = 35{,}786\ \text{km}$. Speed $= \sqrt{\mu/a} = \sqrt{3.986\times10^{5}/42{,}164} = \sqrt{9.454} = 3.07\ \text{km/s}$. Reproduces the canonical numbers. ✓

5. (⭐⭐⭐)† Half a sidereal day, $T = 43{,}082\ \text{s}$. (a) $a = \left(\mu T^2/4\pi^2\right)^{1/3} = (1.874\times10^{13})^{1/3} = 26{,}562\ \text{km}$. (b) $r_p = 6{,}371 + 500 = 6{,}871\ \text{km}$; $r_a = 2a - r_p = 53{,}124 - 6{,}871 = 46{,}253\ \text{km}$; apogee altitude $= 46{,}253 - 6{,}371 = 39{,}882\ \text{km}$. (c) $e = (r_a - r_p)/(r_a + r_p) = 39{,}382/53{,}124 = 0.741$. (d) $v_p = \sqrt{\mu(2/r_p - 1/a)} = \sqrt{3.986\times10^{5}(2.9107\times10^{-4} - 3.7648\times10^{-5})} = \sqrt{101.0} = 10.05\ \text{km/s}$; $v_a = \sqrt{\mu(2/r_a - 1/a)} = \sqrt{3.986\times10^{5}(4.3240\times10^{-5} - 3.7648\times10^{-5})} = \sqrt{2.23} = 1.49\ \text{km/s}$. The satellite is $10.05/1.49 = 6.7\times$ slower at apogee than at perigee. By Kepler's second law (equal areas in equal times), moving that slowly at apogee means it spends most of the 12-hour orbit up high over the north — the loiter that makes the orbit useful. ✓ (Sanity: apogee altitude ~$39{,}900\ \text{km}$ is above GEO, as expected for a Molniya.)

7. (⭐)† Footprint area (flat-disk approximation) $= \pi(2{,}560)^2 = 2.06\times10^{7}\ \text{km}^2$. Naive count $N \approx (5.1\times10^{8})/(2.06\times10^{7}) \approx 25$ satellites. The true Starlink number is in the thousands because: (1) footprints must overlap, and useful coverage demands a minimum elevation angle (say $\ge 25^\circ$), which shrinks the effective footprint far below the horizon-to-horizon disk; (2) real systems build in redundancy, spare capacity, and per-user bandwidth (many satellites over a busy region), and the orbital geometry cannot pack satellites at the ideal density everywhere at once. The back-of-envelope gives the right floor and the right story (LEO needs many), not the operational count.

9. (⭐⭐)† A GEO satellite's signal reaches to an Earth-central half-angle of ~$81^\circ$ around its sub-satellite point on the equator. Three satellites $120^\circ$ apart in longitude therefore blanket every longitude at low and middle latitudes with heavy overlap — nowhere on the equator is more than $60^\circ$ of longitude from a satellite, well within reach. But because each satellite sits on the equator, its coverage in latitude also tops out near $\pm 81^\circ$, and the useful elevation angle drops toward the horizon well before that. Above ~$70^\circ$–$75^\circ$ latitude the satellites are too low in the sky to be reliable, and past ~$81^\circ$ they are below the horizon. The poles are left unserved — the exact gap that Molniya/HEO orbits (apogee over the pole) are designed to fill.

11. (⭐)† Three distinct reasons not to put everything in GEO: (i) Delta-v: reaching GEO costs several km/s more than LEO, so a given rocket lofts far less mass — GEO is expensive to reach. (ii) Distance: at $35{,}786\ \text{km}$ a camera's resolution is poor and a radio round trip takes ~$0.25\ \text{s}$, a latency you notice on calls and cannot tolerate for fast interactive data. (iii) Scarcity: every GEO satellite must share one thin equatorial ring without interfering with neighbors, so usable slots are a genuinely limited, internationally regulated resource. GEO is superb for hemispheric broadcast and useless for high-resolution imaging or low-latency links.

13. (⭐⭐)† For the lighting angle to stay constant, the orbital plane must precess eastward at $0.9856^\circ$/day (one turn per year, following the Earth–Sun line). Earth's equatorial bulge (the J2 effect) makes an orbit's node drift, but the direction depends on inclination: a prograde orbit ($i<90^\circ$) drifts westward — the wrong way. Only a slightly retrograde orbit ($i>90^\circ$) precesses eastward, and matching exactly $0.9856^\circ$/day at typical SSO altitudes ($\sim 600$–$800\ \text{km}$) requires $i \approx 98^\circ$. An exactly polar ($90^\circ$) orbit has zero J2 nodal drift, so its lighting would slowly slip — it cannot be sun-synchronous. (The J2 formula that pins down the exact inclination is Chapter 12's.)

15. (⭐⭐⭐)† MEO is the navigation sweet spot from a two-sided argument: Why not lower (LEO)? Each LEO satellite sees only a small patch of Earth (~$23^\circ$ half-angle at $550\ \text{km}$), so guaranteeing four-in-view everywhere, always would take hundreds of satellites — each an expensive, atomic-clocked, precisely tracked machine. The constellation cost explodes. Why not higher (GEO)? Equatorial GEO satellites give poor geometry at high latitudes (they crowd the southern horizon) and cannot cover the poles at all; a navigation fix also degrades when all satellites lie in nearly the same direction. MEO (~$20{,}200\ \text{km}$) threads the needle: a footprint nearly as wide as GEO's (~$76^\circ$) means just ~$24$ satellites keep four-plus in view worldwide, and a $55^\circ$ inclination spreads them across the sky for strong geometry including high latitudes. The altitude minimizes total system cost for the coverage required.

17. (⭐⭐) Correct (sidereal): $a_{\text{sid}} = \left(\mu(86{,}164)^2/4\pi^2\right)^{1/3} = 42{,}164\ \text{km}$. Wrong (solar): $a_{\text{sol}} = \left(\mu(86{,}400)^2/4\pi^2\right)^{1/3} = 42{,}241\ \text{km}$. Difference $= 42{,}241 - 42{,}164 = 77\ \text{km}$. Physically: a satellite placed at the solar-day radius has a period ~4 minutes too long, so it falls behind Earth's rotation by ~$1^\circ$ of longitude per day and slowly drifts westward out of its assigned slot — it is not stationary. The four-minute sidereal/solar distinction is the difference between a satellite that stays put and one that wanders. Expected code output: correct $42164$, wrong $42241$, error $77$ km.

18. (⭐⭐)† $\lambda = \arccos\!\big(R_{\text{eq}}/(R_{\text{eq}}+h)\big)$ with $R_{\text{eq}}=6{,}378\ \text{km}$: - $h=400$: $\arccos(6378/6778) = \arccos(0.9410) = 19.8^\circ$. - $h=800$: $\arccos(6378/7178) = \arccos(0.8886) = 27.3^\circ$. - $h=20{,}200$: $\arccos(6378/26578) = \arccos(0.2400) = 76.1^\circ$. - $h=35{,}786$: $\arccos(6378/42164) = \arccos(0.1513) = 81.3^\circ$. The half-angle grows steeply with altitude and then saturates toward $90^\circ$: coverage improves fast at first (LEO→MEO) and only marginally from MEO→GEO. This is the geometric root of why few high satellites can see the whole globe while many low ones cannot — the coverage half of the coverage-vs-resolution trade.

19. (⭐⭐⭐) With $a = 26{,}562\ \text{km}$ (half-sidereal-day Molniya): - Perigee $600\ \text{km}$: $r_p = 6{,}971$, $r_a = 2a - r_p = 46{,}153$, $e = 39{,}182/53{,}124 = 0.738$, apogee altitude $= 39{,}781\ \text{km}$. - Perigee $1{,}000\ \text{km}$: $r_p = 7{,}371$, $r_a = 45{,}753$, $e = 38{,}382/53{,}124 = 0.722$, apogee altitude $= 39{,}381\ \text{km}$. Raising the perigee lowers the eccentricity because the semi-major axis is fixed by the period ($a=26{,}562$): lifting $r_p$ forces $r_a$ down by the same amount ($r_a = 2a - r_p$), so the orbit becomes less lopsided ($e$ falls) and the apogee sinks slightly. Expected output: perigee 600 e 0.738 apoalt 39781; perigee 1000 e 0.722 apoalt 39381.

21. (⭐⭐)† Low latency (<$50\ \text{ms}$ round trip) rules out GEO immediately: a single GEO round trip is already ~$0.25\ \text{s} = 250\ \text{ms}$. The only regime with short enough light-travel is LEO (a few ms up-and-down). Global, continuous LEO coverage requires a large constellation — on the order of hundreds to thousands of satellites (Starlink is the real example). The single hardest operational consequence: because every LEO satellite races overhead in minutes and sets, the system must continuously hand over each user from satellite to satellite and manage a huge, dynamic constellation (collision avoidance, phasing, replenishment, ground-station networks) — an enormous operations and coordination burden that a single stationary GEO satellite entirely avoids. The trade is latency and capacity (LEO wins) against simplicity and satellite count (GEO wins).

23. (⭐⭐⭐)† Rubric (your mission). A complete answer records altitude/apsides, inclination, period, and a one-line requirement→orbit justification, with the period computed from the chosen altitude using the correct body's $\mu$. Examples: - Track A (comsat → GEO): $35{,}786\ \text{km}$, $i=0^\circ$, period $23\ \text{h}\ 56\ \text{min}$. "A fixed customer dish needs a satellite motionless in the sky." Also note the GTO on-ramp and ~$11\ \text{m/s}$ graveyard reserve. - Track B (lunar lander → LLO): ~$100\ \text{km}$ circular, inclination over the landing site, $v=\sqrt{4{,}903/1{,}837}=1.63\ \text{km/s}$, $T=2\pi\sqrt{1{,}837^3/4{,}903}\approx118\ \text{min}$. "A low parking orbit to stage the descent from." - Track C (Mars orbiter → low Mars orbit): ~$400\ \text{km}$ near-polar, $v=\sqrt{4.283\times10^{4}/3{,}790}=3.36\ \text{km/s}$, $T\approx118\ \text{min}$. "Near-polar for global mapping; low for resolution." - Track D (asteroid rendezvous → heliocentric match): semi-major axis $\approx$ the target's (~$1.1$–$1.2\ \text{AU}$), essentially zero relative velocity for proximity ops. "You match the Sun-orbit and fly formation; the body is too small to orbit." Full credit requires the justification to trace from the requirement, not just name an orbit.

25. (⭐⭐)† The claim is not necessarily true. A polar orbit ($i\approx90^\circ$) guarantees global reach — the satellite overflies every latitude — but says nothing about the local time of those passes. As the seasons and orbital plane drift relative to the Sun, a plain polar orbit crosses a given city at different local times through the year. The missing condition is sun-synchronicity: the orbital plane must precess once per year (via J2) so the Sun angle stays fixed. The orbit the brochure is actually describing (if the claim is true) is a sun-synchronous orbit — near-polar (~$98^\circ$) and precessing to hold constant local time. Polar buys where; sun-synchronous additionally buys when/under what light.

27. (⭐⭐)† $v_e = I_{sp}\,g_0 = 320 \times 9.81 = 3{,}139\ \text{m/s} = 3.139\ \text{km/s}$. Mass ratio $= e^{\Delta v/v_e} = e^{1.47/3.139} = e^{0.4683} = 1.597$. With dry mass $m_f = 1{,}800\ \text{kg}$: $m_0 = 1.597 \times 1{,}800 = 2{,}875\ \text{kg}$, so propellant $m_p = m_0 - m_f = 1{,}075\ \text{kg}$. That propellant is $1{,}075/2{,}875 = 37\%$ of the fueled satellite's mass — for a single $1.47\ \text{km/s}$ burn. It is a vivid reminder of the rocket equation's bite (Ch. 3): even the "small" last step to GEO demands over a third of the launched mass be propellant.

29. (⭐⭐⭐)† (a) $a=26{,}562\ \text{km}$, $e=0.74$, $i=63.4^\circ$ → a Molniya orbit (half-sidereal-day HEO at the critical inclination). (b) $T = 2\pi\sqrt{a^3/\mu} = 2\pi\sqrt{(26{,}562)^3/3.986\times10^{5}} = 2\pi(6{,}857\ \text{s}) = 43{,}083\ \text{s} = 11\ \text{h}\ 58\ \text{min}$ (half a sidereal day). ✓ (c) $r_p = a(1-e) = 26{,}562(0.26) = 6{,}906\ \text{km}$ (perigee altitude ~$535\ \text{km}$); $r_a = a(1+e) = 26{,}562(1.74) = 46{,}218\ \text{km}$ (apogee altitude ~$39{,}847\ \text{km}$). (d) $i=63.4^\circ$ is the critical inclination: at this tilt the J2 perturbation from Earth's equatorial bulge produces zero apsidal precession, so the argument of perigee — and hence the carefully placed northern apogee — stays frozen for years instead of rotating away. (The J2 mechanism is Chapter 12's.)

Chapter 10 — Full solutions to † and odd-numbered exercises

Earth: $\mu = 3.986\times10^{5}\ \text{km}^3/\text{s}^2$, $R = 6{,}371\ \text{km}$. Speeds in km/s.

Part A

10.1 † (Hohmann 300→2000 km). $r_1=6{,}671$, $r_2=8{,}371$, $a_t=(6{,}671+8{,}371)/2=7{,}521\ \text{km}$. $v_{c1}=\sqrt{398600/6671}=7.730$, $v_{c2}=\sqrt{398600/8371}=6.901$. $v_p=\sqrt{398600(2/6671-1/7521)}=\sqrt{66.50}=8.155$, so $\Delta v_1=8.155-7.730=0.425\ \text{km/s}$. $v_a=\sqrt{398600(2/8371-1/7521)}=\sqrt{42.24}=6.499$, so $\Delta v_2=6.901-6.499=0.402\ \text{km/s}$. Total $0.827\ \text{km/s}$. Both prograde; a modest change, so a modest cost.

10.3 † (retrograde 50 m/s at 400 km). $v_{\text{new}}=7.673-0.050=7.623$. From vis-viva, $1/a = 2/6771 - 7.623^2/398600 = 2.9538\times10^{-4} - 1.4579\times10^{-4} = 1.4959\times10^{-4}$, so $a=6{,}685\ \text{km}$. A retrograde burn removes energy and lowers the opposite side, so the burn point ($6{,}771\ \text{km}$) becomes the new apoapsis. The opposite end (new perigee) is $2a-r=2(6{,}685)-6{,}771=6{,}598\ \text{km}$, an altitude of about $\mathbf{227\ \text{km}}$ (down from 400).

10.5 † (ratio 25 → bi-elliptic?). Yes — consider it. The rule: the Hohmann transfer is provably optimal only for $r_2/r_1 < 11.94$; for $r_2/r_1 > 15.58$ a bi-elliptic transfer with a high enough intermediate apoapsis $r_b$ always beats it (and between 11.94 and 15.58 it depends on $r_b$). A ratio of 25 is well above 15.58, so a bi-elliptic transfer will save some delta-v — at the cost of a much longer trip.

Part B

10.7 † (Hohmann 400 km LEO → 20,200 km MEO). $r_1=6{,}771$, $r_2=26{,}571$, $a_t=16{,}671\ \text{km}$. $v_{c1}=7.673$, $v_{c2}=\sqrt{398600/26571}=3.873$. $v_p=\sqrt{398600(2/6771-1/16671)}=\sqrt{93.83}=9.686$, $\Delta v_1=9.686-7.673=2.014\ \text{km/s}$. $v_a=\sqrt{398600(2/26571-1/16671)}=\sqrt{6.096}=2.469$, $\Delta v_2=3.873-2.469=1.404\ \text{km/s}$. Total $3.418\ \text{km/s}$. This matches the energy-only reconstruction of Chapter 6, Exercise 6.16.

10.9 † (transfer time for 10.7). $t=\pi\sqrt{a_t^3/\mu}=\pi\sqrt{16{,}671^3/3.986\times10^{5}} =\pi\sqrt{1.1624\times10^{7}}=\pi(3{,}409\ \text{s})=10{,}712\ \text{s}\approx \mathbf{2.98\ \text{hours}}$ (half the transfer-ellipse period).

Part C

10.10 † (hohmann). hohmann(3.986e5, 6771, 42164) → (2.399, 1.457, 3.857) km/s (see code/exercise-solutions.py). These are the LEO→GEO burns and total from §10.2.

10.11 (plane_change). plane_change(7.673, 28.5) $=2(7.673)\sin(14.25^\circ)=3.778\ \text{km/s}$. It represents a $28.5^\circ$ plane change performed in LEO — e.g., rotating a Cape-Canaveral orbit toward the equator, and a warning of how ruinous plane changes are down low.

10.12 † (dv_budget). dv_budget([("perigee",2.40),("apogee",1.46),("plane",1.51)]) $=5.37\ \text{km/s}$. Section 10.6 eliminates the separate "plane" leg by folding the plane change into the apogee burn (law of cosines), cutting the real total to about $4.23\ \text{km/s}$.

Part D

10.13 (find the error). The colleague is right ($\approx 1.83\ \text{km/s}$). The student added the two burns as scalars, but delta-vs are vectors: the circularization ($+1.46$, in-plane) and the plane change ($1.51$, out-of-plane) are (nearly) perpendicular contributions to a single velocity change, so the true cost is the magnitude of their vector sum, $\sqrt{v_a^2+v_{c2}^2-2v_av_{c2}\cos\Delta i}=1.825\ \text{km/s}$, not the arithmetic sum $2.97$. The student overcharged the mission by more than a kilometer per second.

10.14 † (find the error). The physics is backwards. Plane-change cost is $2v\sin(\Delta i/2)$, proportional to orbital speed $v$, so it is cheapest where you move slowest — high up, at apoapsis — not in fast, low LEO. Doing the $28.5^\circ$ turn in LEO ($v=7.673$) costs $3.78\ \text{km/s}$; doing the same turn at GTO apogee ($v\approx 1.6$–$3.1$) costs far less and, combined with the circularization, $\approx 1.83\ \text{km/s}$. Correct advice: defer the plane change to the highest, slowest point.

Part E

10.15 † (design the GEO apogee burn). Separate: circularize $\Delta v = 3.075-1.617=1.458$, then change plane $2(3.075)\sin(14.25^\circ)=1.514$; total $2.972\ \text{km/s}$. Combined: $\sqrt{1.617^2+3.075^2-2(1.617)(3.075)\cos 28.5^\circ}=\sqrt{3.331}=1.825\ \text{km/s}$. Saving $= 2.972-1.825 = 1.147\ \text{km/s}$ — nearly 40%, from one well-aimed burn.

10.17 † (ratio 20 bi-elliptic). (a) Yes — $20 > 15.58$, so a bi-elliptic transfer (with a suitably high $r_b$) will cost less total delta-v than the Hohmann. (b) The two costs: (i) a much longer transfer time (the detour out to $r_b$ and back can take days to weeks versus hours); and (ii) an extra, precise burn far from the planet, where tracking is weaker and a mistimed or mis-sized burn is harder to correct — added operational risk for a small delta-v saving.

Part F

10.18 † (why can't you thrust straight at a target ahead?). Pointing at the target and firing is a prograde burn. It raises your orbit, lengthens your period, and — because a higher orbit is a slower orbit — makes you drift behind the target, not toward it. To actually catch a target ahead you must burn retrograde, dropping into a lower orbit that is faster and shorter-period; you gain on the target, and once alongside you burn prograde to raise back and match orbits. Slow down to catch up.

10.19 (90° plane change in LEO). $\Delta v = 2(7.673)\sin 45^\circ = 15.346\times0.7071 = 10.85\ \text{km/s}$ (equal to $\sqrt{2}\,v$, the escape-speed factor). That is more than the $\approx 9.4\ \text{km/s}$ it took to reach orbit in the first place, and more than three times the $\approx 3.2\ \text{km/s}$ that escapes Earth from LEO. The lesson is stark: a large plane change is the single most expensive routine maneuver in spaceflight, so you never launch into the wrong plane and fix it later — you launch into the right plane.

Part G

10.21 † (vis-viva at $r=10{,}000$, $a=15{,}000$). $v=\sqrt{398600(2/10000-1/15000)}=\sqrt{398600(2.0\times10^{-4}-6.667\times10^{-5})}=\sqrt{53.15}=7.290\ \text{km/s}$. You would use this speed to price a burn made at $r=10{,}000\ \text{km}$ on this transfer orbit — for example, the difference between it and the local circular speed, or between two transfer orbits sharing that point.

10.23 (Kourou vs Cape). Kourou ($5.2^\circ$) is better because a launch cannot reach an inclination below its site latitude, so Kourou starts nearly in the equatorial plane a GEO satellite needs, while Cape Canaveral ($28.5^\circ$) starts $28.5^\circ$ off. Since plane-change cost is $2v\sin(\Delta i/2)$ and GEO is equatorial, the standalone plane change at GEO speed ($3.075$) is $2(3.075)\sin(14.25^\circ)=1.514\ \text{km/s}$ from the Cape but only $2(3.075)\sin(3^\circ)=0.322\ \text{km/s}$ from Kourou (taking $\Delta i\approx 6^\circ$) — a difference of about $1.2\ \text{km/s}$ standalone (and, folded into the apogee burn, $\approx 0.35\ \text{km/s}$; see Case Study 1). That saving becomes propellant, hence payload or years of life.

10.25 (energy ledger of the LEO→GEO Hohmann). Using $\varepsilon=-\mu/(2a)$: parking $\varepsilon_1=-398600/(2\cdot6{,}771)=-29.44\ \text{MJ/kg}$; transfer ellipse $\varepsilon_t=-398600/(2\cdot24{,}467.5)=-8.15\ \text{MJ/kg}$; GEO $\varepsilon_2=-398600/(2\cdot42{,}164)=-4.73\ \text{MJ/kg}$. The energy jumps only at the two burns ($-29.44\to-8.15$ at perigee, $-8.15\to-4.73$ at apogee) and is unchanged during the coast across the transfer ellipse (where the speed falls from $10.07$ to $1.62\ \text{km/s}$ but $\varepsilon$ holds at $-8.15$). Total energy added $=-4.73-(-29.44)=\mathbf{24.71\ \text{MJ/kg}}$.

10.24 † (rocket equation for the transfer). $\Delta v=4.23\ \text{km/s}$, $v_e=3.14\ \text{km/s}$. Mass ratio $m_0/m_f=e^{\Delta v/v_e}=e^{4.23/3.14}=e^{1.347}=3.85$. Propellant fraction $=1-1/3.85=0.740=\mathbf{74\%}$. Nearly three-quarters of the spacecraft (at the start of the transfer) is propellant just to fly this one LEO→GEO trip — a vivid reminder of the tyranny of the rocket equation, and of why the $1.15\ \text{km/s}$ saved by the combined maneuver (§10.6) matters so much.

Chapter 11 — Interplanetary Trajectories (solutions to † and odd-numbered exercises)

Constants: $\mu_\odot = 1.327\times10^{11}$, $\mu_\oplus = 3.986\times10^{5}$, $\mu_{\text{Mars}} = 4.283\times10^{4}$, $\mu_{\text{Venus}} = 3.249\times10^{5}$, $\mu_{\text{Jup}} = 1.267\times10^{8}$ (all km³/s²); $1\ \text{AU} = 1.496\times10^{8}\ \text{km}$; $R_\oplus = 6{,}371\ \text{km}$.

11.1 $v_\infty = \sqrt{C_3} = \sqrt{12} = 3.46\ \text{km/s}$.

11.3 $924{,}000 / 6{,}371 = 145$ Earth radii; $924{,}000 / 384{,}400 = 2.40$ Earth–Moon distances (i.e., Earth's SOI reaches about 2.4× the Moon's distance).

11.5 $\dfrac{1}{T_{\text{syn}}} = \dfrac{1}{224.70} - \dfrac{1}{365.25} = 4.4504\times10^{-3} - 2.7379\times10^{-3} = 1.7126\times10^{-3}\ \text{day}^{-1}$, so $T_{\text{syn}} = 583.9\ \text{days} \approx 1.60\ \text{yr} \approx 19.2\ \text{months}$. (Venus's synodic period is famously ~584 days.)

11.7 Inward transfer, $r_1 = 1.496\times10^8$ (Earth), $r_2 = 0.723\ \text{AU} = 1.0816\times10^8\ \text{km}$ (Venus). $a_t = (r_1+r_2)/2 = 1.2888\times10^8\ \text{km}$. - $v_1 = \sqrt{\mu_\odot/r_1} = 29.78\ \text{km/s}$; $v_2 = \sqrt{\mu_\odot/r_2} = \sqrt{1226.9} = 35.03\ \text{km/s}$. - Earth is at the transfer aphelion: $v_{a} = \sqrt{\mu_\odot(2/r_1 - 1/a_t)} = \sqrt{744.2} = 27.28\ \text{km/s}$. $v_{\infty,\oplus} = |27.28 - 29.78| = 2.50\ \text{km/s}$ (the probe must be slower than Earth). - Venus is at the transfer perihelion: $v_{p} = \sqrt{\mu_\odot(2/r_2 - 1/a_t)} = \sqrt{1423.9} = 37.73\ \text{km/s}$. $v_{\infty,\text{Venus}} = |35.03 - 37.73| = 2.70\ \text{km/s}$. - Transfer time $= \pi\sqrt{a_t^3/\mu_\odot} = \pi\sqrt{2.1407\times10^{24}/1.327\times10^{11}} = \pi\times4.0165\times10^{6} = 1.262\times10^{7}\ \text{s} = 146\ \text{days}$ (~4.8 months). Sanity: real Venus transfers run ~150 days. ✓

11.9 $a_t = (1 + 2.7)/2 = 1.85\ \text{AU} = 2.7676\times10^8\ \text{km}$. $t = \pi\sqrt{a_t^3/\mu_\odot} = \pi\sqrt{2.1200\times10^{25}/1.327\times10^{11}} = \pi\sqrt{1.5976\times10^{14}} = \pi\times1.2639\times10^{7} = 3.971\times10^{7}\ \text{s} = 460\ \text{days}$ (~1.26 yr).

11.11 $\dfrac{1}{T_{\text{syn}}} = \dfrac{1}{365.25} - \dfrac{1}{10{,}759} = 2.7379\times10^{-3} - 9.295\times10^{-5} = 2.6449\times10^{-3}$, so $T_{\text{syn}} = 378\ \text{days} \approx 1.03\ \text{yr}$. It is barely over a year because Saturn is so slow ($1/T_2$ is tiny) that Earth almost laps it in a single year — the synodic period approaches Earth's own year as the outer planet's period $\to\infty$.

11.13 Venus sweeps $360^\circ \times 146/224.70 = 234^\circ$ during the 146-day cruise; the probe sweeps $180^\circ$. Required phase $= 180^\circ - 234^\circ = -54^\circ$: Venus must be about $54^\circ$ behind Earth at departure (a negative lead, because the faster inner planet, arriving $180^\circ$ around, started behind).

11.14 $r_{\text{park}} = 6{,}371 + 200 = 6{,}571\ \text{km}$. $v_{\text{circ}} = \sqrt{3.986\times10^5/6571} = 7.789\ \text{km/s}$; $v_{\text{esc}} = \sqrt{2\times3.986\times10^5/6571} = 11.015\ \text{km/s}$; $v_{\text{peri}} = \sqrt{2.95^2 + 11.015^2} = \sqrt{130.02} = 11.403\ \text{km/s}$; $\Delta v = 11.403 - 7.789 = 3.61\ \text{km/s}$. This is slightly larger than the $300\ \text{km}$ result ($3.59$). It does not contradict Oberth: Oberth compares burning deep and fast (at perigee) versus high and slow (at the SOI edge), where deep wins hugely. Among different circular parking altitudes the injection burn is nearly flat and marginally rises as you go lower; the real advantage of a low orbit is that you spent less delta-v reaching it, so the total from the surface still favors the lower parking orbit.

11.15 Capture ellipse $r_p = 3{,}790$, $r_a = 33{,}390\ \text{km}$, $a_{\text{cap}} = 18{,}590\ \text{km}$. - Arrival hyperbola periapsis speed: $v_{\text{hyp}} = \sqrt{2.65^2 + 2(4.283\times10^4)/3790} = \sqrt{29.62} = 5.44\ \text{km/s}$. - Capture-ellipse periapsis speed: $v_{\text{ell}} = \sqrt{\mu_{\text{Mars}}(2/3790 - 1/18590)} = \sqrt{20.30} = 4.51\ \text{km/s}$. - $\Delta v_{\text{MOI}} = 5.44 - 4.51 = 0.94\ \text{km/s}$ — less than half the $2.08\ \text{km/s}$ for direct low-circular capture. This is why real orbiters capture into a loose ellipse, then aerobrake.

11.17 $C_3 = 3.8^2 = 14.44\ \text{km}^2/\text{s}^2$. From $300\ \text{km}$: $v_{\text{esc}} = 10.932$, $v_{\text{peri}} = \sqrt{3.8^2 + 10.932^2} = \sqrt{133.95} = 11.574\ \text{km/s}$; $\Delta v_{\text{TMI}} = 11.574 - 7.730 = 3.84\ \text{km/s}$. Versus the Hohmann ($v_\infty = 2.95$, $\Delta v = 3.59$): the faster trip costs an extra $\approx 0.25\ \text{km/s}$ at departure (and its $C_3$ jumped from $8.7$ to $14.4$) — plus a faster, more expensive arrival. Speed costs energy.

11.18 hohmann_transfer(1.327e11, 1.496e8, 2.2799e8): $v_1 = 29.78$, $v_2 = 24.13$, $a = 1.888\times10^8$, $v_p = 32.73$, $v_a = 21.48$; returns $(|32.73-29.78|,\ |24.13-21.48|,\ \pi\sqrt{a^3/\mu}) = (2.95,\ 2.65,\ 2.237\times10^7\ \text{s})$; $2.237\times10^7/86400 = 259$ d. # Expected output: v_inf_dep=2.95, v_inf_arr=2.65, tof=259 d.

11.19 synodic_period(365.25, 4332.59) $= 1/|1/365.25 - 1/4332.59| = 1/(2.7379\times10^{-3} - 2.3081\times10^{-4}) = 1/2.5070\times10^{-3} = 398.9\ \text{days}$. # Expected output: 398.9 days (Earth–Jupiter synodic ~399 days).

11.20 injection_dv(3.986e5, 6671, 2.95) $= \sqrt{2.95^2 + 2(3.986\times10^5)/6671} - \sqrt{3.986\times10^5/6671} = \sqrt{8.70 + 119.50} - \sqrt{59.75} = 11.323 - 7.730 = 3.59\ \text{km/s}$. # Expected output: 3.59.

11.21 Before: heliocentric $= 13.06 - 8 = 5.06\ \text{km/s}$ (probe moving with Jupiter but slower, $v_\infty$ pointed backward). After a $180^\circ$ turn $v_\infty$ points forward: heliocentric $= 13.06 + 8 = 21.06\ \text{km/s}$. Gain $= 21.06 - 5.06 = 16.0\ \text{km/s} = 2\times8 = 2v_\infty$. This is the theoretical maximum boost from a single flyby (a full reversal — unachievable in practice, but the ceiling).

11.23 A closer flyby (smaller $r_p$) makes $e = 1 + r_p v_\infty^2/\mu$ smaller (closer to 1), so $1/e = \sin(\delta/2)$ is larger, so the turn angle $\delta$ is larger. A bigger turn rotates $v_\infty$ more, producing a larger change when it is added to the planet's velocity. Physically: passing closer means diving deeper into the planet's gravity, which deflects the path more strongly.

11.25 The $32.73\ \text{km/s}$ is the speed only at perihelion (Earth's orbit). On the transfer ellipse the spacecraft slows as it climbs the Sun's gravity well (vis-viva, Ch. 6): by Mars's orbit it is doing only $21.48\ \text{km/s}$. Speed is not constant on the ellipse — the student assumed a circular orbit's constant speed. At Mars the probe is slower than Mars ($24.13$), which is why Mars overtakes it.

11.27 $\varepsilon = -\mu_\odot/(2a_t) = -1.327\times10^{11}/(2\times1.888\times10^8) = -351.4\ \text{km}^2/\text{s}^2$. - At perihelion ($r_1 = 1.496\times10^8$): $v = \sqrt{2(\varepsilon + \mu_\odot/r_1)} = \sqrt{2(-351.4 + 887.0)} = \sqrt{2\times535.6} = \sqrt{1071.2} = 32.73\ \text{km/s}$. ✓ - At aphelion ($r_2 = 2.280\times10^8$): $v = \sqrt{2(-351.4 + 582.0)} = \sqrt{2\times230.6} = \sqrt{461.2} = 21.48\ \text{km/s}$. ✓ Both match §11.2's vis-viva results — because vis-viva is energy conservation, $v^2/2 - \mu/r = \varepsilon = -\mu/(2a)$.

11.29 More $C_3$. A $\approx 180$-day trip is a higher-energy, non-tangential transfer — a bigger, faster ellipse (or a Type-II arc), so the perihelion speed at departure is higher, giving a larger $v_\infty$ and hence a larger $C_3$. Cutting ~30% off the cruise time roughly doubles the departure $C_3$ (from ~8.7 into the ~14–20 range) and raises the departure burn (and a faster, costlier arrival). Speed costs energy — the tyranny of the rocket equation applied to trajectories.

Chapter 12

Chapter 12 — Solutions to Selected Exercises (Perturbations)

Full solutions to the daggered (†) and odd-numbered problems. Constants: $\mu_\oplus = 3.986\times10^5\ \text{km}^3/\text{s}^2$, $R_\oplus = 6{,}378\ \text{km}$, $J_2 = 1.0826\times10^{-3}$, $g_0 = 9.81\ \text{m/s}^2$. All arithmetic hand-verified.

12.1 † $a = 6{,}971\ \text{km}$, $i = 40^\circ$, $e = 0$. $n = \sqrt{\mu/a^3} = \sqrt{3.986\times10^5/(6{,}971)^3} = 1.085\times10^{-3}\ \text{rad/s}$; $(R_\oplus/a)^2 = (6{,}378/6{,}971)^2 = 0.837$; $\cos 40^\circ = 0.766$. $$\dot\Omega = -\tfrac{3}{2}(1.085\times10^{-3})(1.0826\times10^{-3})(0.837)(0.766) = -1.13\times10^{-6}\ \text{rad/s}.$$ $\times 86{,}400\times(180/\pi) = \mathbf{-5.59^\circ/\text{day}}$ — the node drifts west (prograde orbit). Sanity: a touch larger in magnitude than the ISS's $-5.02^\circ$/day because $\cos 40^\circ > \cos 51.6^\circ$ outweighs the slightly higher altitude.

12.3 † $a = 7{,}071\ \text{km}$, $i = 45^\circ$. $n = 1.062\times10^{-3}\ \text{rad/s}$; $(R_\oplus/a)^2 = 0.814$; $5\cos^2 45^\circ - 1 = 5(0.5) - 1 = 1.5$. $$\dot\omega = \tfrac{3}{4}(1.062\times10^{-3})(1.0826\times10^{-3})(0.814)(1.5) = 1.05\times10^{-6}\ \text{rad/s} = \mathbf{+5.21^\circ/\text{day}}.$$ Positive, so perigee advances. (At $i = 45^\circ$ the bracket $5\cos^2 i - 1$ is positive — we are below the $63.4^\circ$ critical inclination.)

12.5 † Absorbed: $P = S/c = 1{,}361/(2.998\times10^8) = \mathbf{4.54\ \mu\text{Pa}}$. Reflecting: $\mathbf{9.08\ \mu\text{Pa}}$ (twice). A reflector reverses each photon's momentum ($\Delta p = 2p$ instead of $p$ for absorption), so it feels twice the force. A real surface (reflectivity $r$) is in between: $P = (S/c)(1 + r)$.

12.7 † $a = 7{,}071\ \text{km}$, require $\dot\Omega = +1.991\times10^{-7}\ \text{rad/s}$. $n = 1.062\times10^{-3}$; denominator $\tfrac{3}{2}nJ_2(R_\oplus/a)^2 = 1.403\times10^{-6}$. $$\cos i = -\frac{1.991\times10^{-7}}{1.403\times10^{-6}} = -0.1419 \Rightarrow i = \mathbf{98.2^\circ}.$$ Slightly less than the $98.6^\circ$ at $800\ \text{km}$, as expected (see 12.8).

12.9 † For a circular orbit, $\dot\Omega \propto n\,(R_\oplus/a)^2 \cos i$. Now $n = \sqrt{\mu/a^3} = \sqrt{\mu}\,a^{-3/2}$ and $(R_\oplus/a)^2 \propto a^{-2}$, so $$\dot\Omega \propto a^{-3/2}\cdot a^{-2} = \mathbf{a^{-7/2}}\quad(\text{at fixed } i, e). \blacksquare$$ Because the exponent is steep and negative, low orbits regress far faster. LEO ($a = 6{,}771$) vs. GEO ($a = 42{,}164$): the ratio is $(42{,}164/6{,}771)^{7/2} = (6.23)^{3.5} \approx 600$. A LEO orbit's plane swivels about 600 times faster than a GEO orbit's at the same inclination — the reason nodal regression rules LEO mission design and is a footnote at GEO.

12.10 † $\beta = m/(C_D A) = 1.3/(2.2\times0.01) = \mathbf{59.1\ \text{kg/m}^2}$. $$\dot a = -\frac{\rho v a}{\beta} = -\frac{(3\times10^{-12})(7{,}670)(6.771\times10^6)}{59.1} = -2.64\times10^{-3}\ \text{m/s} = \mathbf{-228\ \text{m/day}}.$$ The CubeSat's $\beta$ ($59$) is about half the ISS's ($105$), and decay $\propto 1/\beta$, so it sinks about $1.8\times$ faster ($228$ vs. $128\ \text{m/day}$). A 1U CubeSat dropped at $400\ \text{km}$ has a life of only months — which is exactly why most rideshare CubeSats are deployed a bit higher, and why $400\ \text{km}$ CubeSats are treated as short-duration missions.

12.11 Drag force $\propto \rho v^2$, and both $\rho$ (air density) and $v$ (orbital speed) are largest at perigee, the lowest, fastest point. So almost all the braking happens in the brief perigee passes, and almost none at apogee. Removing energy at perigee lowers the orbit's energy, which shows up as a lower apogee on the far side (the perigee altitude barely moves, since you are not braking there). Orbit after orbit the apogee is pulled down toward the perigee altitude while perigee holds, so the eccentricity shrinks and the ellipse rounds out. Only once the orbit is nearly circular does drag act all the way around and spiral the whole thing inward to re-entry.

12.13 † $a_{\text{SRP}} = (S/c)(1 + r)(A/m) = (4.54\times10^{-6})(1.9)(20) = \mathbf{1.73\times10^{-4}\ \text{m/s}^2}$. Compared with the Sun's gravity at $1\ \text{AU}$ ($\approx 5.9\times10^{-3}\ \text{m/s}^2$), that is a fraction $1.73\times10^{-4}/5.9\times10^{-3} \approx \mathbf{0.029}$, i.e. about 3% of solar gravity, directed continuously outward. Three percent sounds small, but it never stops and costs no propellant; over months it can spiral the sail outward through the solar system. This "lightness number" near $0.03$ is representative of real demonstrated sails (IKAROS), and it is why solar sailing is a genuine propulsion option (Ch. 21), not a curiosity.

12.15 † (See code/exercise-solutions.py.) j2_nodal_rate(7078, 0, 98.2): $n = 1.060\times10^{-3}$, $(R_\oplus/a)^2 = 0.812$, $\cos 98.2^\circ = -0.1426$, giving $\dot\Omega = +1.99\times10^{-7}\ \text{rad/s} = \mathbf{+0.987^\circ/\text{day}}$ — essentially the sun-synchronous rate ($+0.9856$), confirming that $98.2^\circ$ is sun-synchronous at $700\ \text{km}$. # Expected output: 0.987.

12.17 † decay_rate(3e-12, 7670, 6.771e6, 105) $= -\rho v a/\beta = \mathbf{-1.48\times10^{-3}\ \text{m/s}}$; per day $\times 86{,}400 = \mathbf{-128\ \text{m/day}}$. Matches the §12.3 worked ISS value. # Expected output: -1.48 mm/s -> -128 m/day.

12.18 † Rotating the plane $0.9856^\circ$/day by thrust costs, per day, $\Delta v \approx v\,\theta_{\text{rad}} = 7{,}500\times(0.9856\times\pi/180) = 129\ \text{m/s per day}$. Over a year: $129\times365 \approx \mathbf{47{,}000\ \text{m/s} = 47\ \text{km/s per year}}$ — about five times the cost of reaching orbit in the first place, every year, and utterly impossible. J2 provides the identical $0.9856^\circ$/day drift for $0\ \text{m/s}$, purely by choosing $i \approx 98^\circ$. That enormous gap — 47 km/s/yr versus free — is the entire reason §12.2 matters and the entire reason sun-synchronous orbits are ubiquitous.

12.19 At $i = 90^\circ$, $\cos i = 0$, so $\dot\Omega = 0$: the plane does not precess at all. Over a year the Earth–Sun line sweeps a full $360^\circ$ while the orbital plane stays fixed in inertial space, so the local solar time of each pass drifts through all 24 hours — the opposite of sun-synchronous. To make the plane precess eastward at $+0.9856^\circ$/day you need $\cos i < 0$, i.e. $i > 90^\circ$; the retrograde tilt (near $98^\circ$) is not optional but essential.

12.20 † At $\sim 128\ \text{m/day}$, three years of decay is $128\times365\times3 \approx 140{,}000\ \text{m} = \mathbf{140\ \text{km}}$ — dropping from $400$ to $\sim 260\ \text{km}$. But this uses a moderate density; at solar maximum the density (and decay) can be several times larger, and as the satellite sinks into denser air the decay accelerates, so it would likely re-enter well before three years. "Moderate solar activity" hides a factor-of-ten uncertainty in density, which means you cannot responsibly ignore drag at $400\ \text{km}$: either budget reboost delta-v or accept a real risk of losing the mission early. The period is exact; the lifetime is a gamble on the Sun.

12.21 (Project — guidance.) A complete answer names the dominant perturbation and gives a numeric budget: Track A (GEO): luni-solar N–S $\approx 48$ + E–W $\approx 3\ \text{m/s/yr}$, $\times$ life $\to \sim 0.8\ \text{km/s}$ over 15 yr. Track B (low lunar orbit): unstable mascon field + Earth third body $\to$ tens of m/s per year. Track C (low Mars orbit): thin-atmosphere drag + Mars $J_2$ ($1.96\times10^{-3}$) $\to$ modest drag make-up. Track D (heliocentric asteroid match): SRP and solar third-body dominate $\to$ formation-keeping corrections. Full credit records the number, its dominant cause, and that it feeds the delta-v budget and vehicle sizing.

12.23 $a = 7{,}000\ \text{km}$, $i = 98^\circ$: $n = 1.078\times10^{-3}$, $(R_\oplus/a)^2 = 0.830$, $\cos 98^\circ = -0.1392$, giving $\dot\Omega = +2.02\times10^{-7}\ \text{rad/s} = \mathbf{+1.00^\circ/\text{day}}$. This is close to but not exactly sun-synchronous ($+0.9856$) — it is a hair too fast. To slow the eastward drift you need $|\cos i|$ smaller, i.e. $i$ closer to $90^\circ$: decrease the inclination slightly (to about $97.9^\circ$). This is exactly the fine-tuning real mission designers do to lock the local time.

12.25 † By vis-viva $v = \sqrt{\mu(2/r - 1/a)}$, speed is greatest at perigee (smallest $r$), and perigee is also the lowest altitude ($300\ \text{km}$), where the air is densest. Drag $\propto \rho v^2$ is therefore overwhelmingly a perigee effect. Braking at perigee removes energy that lowers the apogee (from $3{,}000\ \text{km}$) on the opposite side, while the perigee altitude ($300\ \text{km}$) stays roughly fixed. So apogee drops first, the eccentricity falls, and the orbit circularizes toward $\sim 300\ \text{km}$ before spiraling in — the same mechanism as 12.11, now with concrete apsides.

12.27 † At $1{,}000\ \text{km}$ the §12.3 table gives a lifetime of centuries — effectively permanent. A dead satellite there will not self-clean on any useful timescale, so debris accumulates: every spent stage and defunct spacecraft stays put for generations, raising collision risk and feeding the Kessler syndrome (Chapter 35). This is why $\sim 800$–$1{,}200\ \text{km}$ is the most debris-sensitive band, why the "25-year rule" for post-mission disposal exists, and why constellations are deliberately placed low enough that drag will clear them. Drag is the only free garbage collector, and above $\sim 600\ \text{km}$ it clocks out.

Chapter 13 — Orbit Determination (solutions to † and odd-numbered exercises)

13.1 Six. A state is position (3 components) and velocity (3 components) — six independent numbers, equivalent to the six orbital elements. Fewer than six independent measurements leave the orbit under-determined. (Measurements taken at different times still count toward the six, because the dynamics link them onto one orbit.)

13.3 $n = 14.00\ \text{rev/day} = 14.00\times2\pi/86{,}400 = 1.0181\times10^{-3}\ \text{rad/s}$. Period $T = 2\pi/n = 6{,}171\ \text{s} = 102.9\ \text{min}$. Semi-major axis $a = (\mu/n^2)^{1/3} = (3.986\times10^5/1.0366\times10^{-6})^{1/3} = (3.845\times10^{11})^{1/3} \approx 7{,}272\ \text{km}$, i.e. altitude $7{,}272 - 6{,}371 = 901\ \text{km}$. (Sanity: larger $a$ and longer period than the ISS's $\sim 6{,}795\ \text{km}$/$92.9\ \text{min}$, as a lower mean motion demands.)

13.5 $a = (r_1 + r_2)/2 = (6{,}671 + 20{,}000)/2 = 13{,}335.5\ \text{km}$. Time of flight $= \pi\sqrt{a^3/\mu} = \pi\sqrt{(13{,}335.5)^3/3.986\times10^5} = \pi\sqrt{5.950\times10^6} = \pi(2{,}439) = 7{,}663\ \text{s} = 2.13\ \text{h}$. (Sanity: shorter than the LEO→GEO $5.29\ \text{h}$, as this transfer reaches a smaller radius.)

13.7 Two position vectors fix only the endpoints; through them passes a one-parameter family of Keplerian orbits — a lazy low-energy ellipse, a taut high-energy one, up to a parabola/hyperbola. They differ in energy/size, hence in how fast they cross between the points. Specifying the time of flight selects the single family member that takes exactly that long (for a given transfer direction and revolution count). Time is the missing constraint.

13.8 tle_semimajor(15.0): $n = 15.0\times2\pi/86{,}400 = 1.0908\times10^{-3}\ \text{rad/s}$; $a = (3.986\times10^5/(1.0908\times10^{-3})^2)^{1/3} = (3.350\times10^{11})^{1/3} \approx 6{,}945\ \text{km}$. # Expected output: 6945. (Altitude $\approx 574\ \text{km}$.)

13.9 kalman_gain(9.0, 3.0) $= 9/(9+3) = 0.75$. # Expected output: 0.75. The prediction variance (9) is three times the measurement variance (3), so the filter leans toward the (sharper) measurement, weight $0.75$.

13.10 ls_constant([10.2, 9.8, 10.1, 9.9, 10.0]): the constant minimizing $\sum r_i^2$ is the mean $= (10.2+9.8+10.1+9.9+10.0)/5 = 50.0/5 = 10.0$. # Expected output: 10.0.

13.11 The student confused the mean anomaly with the true anomaly. $M = 45^\circ$ is the fictitious uniform-time angle; the physical angle $\nu$ equals $M$ only at perigee and apogee. You must solve Kepler's equation $M = E - e\sin E$ for $E$, then convert to $\nu$. With $e = 0.5$, iterating from $E_0 = M + e\sin M = 1.139$ gives $E \approx 1.262\ \text{rad} = 72.3^\circ$, and $\cos\nu = (\cos E - e)/(1 - e\cos E) = (0.305 - 0.5)/(1 - 0.153) = -0.230$, so $\nu \approx 103^\circ$. The true anomaly ($103^\circ$) is much larger than $45^\circ$ — the body runs ahead of the uniform schedule after perigee.

13.13 In a $550\ \text{km}$ orbit, an onboard GNSS receiver is the natural primary source: it delivers a near-direct state (position to metres, velocity from carrier phase) continuously and autonomously, exactly where GNSS works best (below the navigation constellation). Back it with ground tracking for independence and safety: radar and optical observations from the surveillance network produce the public TLE used for conjunction screening, and serve as a cross-check or fallback if the receiver fails. So: GNSS primary (accuracy, autonomy, cost), ground tracking secondary (independence, catalog/conjunction, redundancy). Optical is unnecessary for routine ops at this altitude but is how the catalog sees you.

13.15 Inverse-variance fusion: $1/\sigma^2 = 1/\sigma_1^2 + 1/\sigma_2^2 = 1/1.5^2 + 1/2.0^2 = 0.4444 + 0.2500 = 0.6944\ \text{km}^{-2}$, so $\sigma^2 = 1.44$ and $\sigma = 1.2\ \text{km}$. The combined estimate ($1.2\ \text{km}$) is sharper than either input ($1.5$ and $2.0\ \text{km}$) — two imperfect measurements of one quantity beat both.

13.16 A single image supplies only two numbers — a direction (two angles) — out of the six the state needs, and it contains no range and no rates. Precision cannot manufacture the four missing dimensions; a perfectly measured direction is still just a direction. You need at least three observations (six angles) so the orbital dynamics can supply the unknown ranges — which is exactly Gauss's method (§13.4).

13.17 Received skin-track power $\propto 1/\rho^4$, so the ratio is $(36{,}000/2{,}000)^4 = 18^4 = 104{,}976 \approx 1.05\times10^5$. The same target at GEO returns about $100{,}000$ times less power than at $2{,}000\ \text{km}$ — which is why skin-track radar is a LEO tool, and high orbits are tracked optically (or with cooperative transponders whose link falls only as $1/\rho^2$ each way, §13.2 and Ch. 26).

13.18 The GNSS constellation orbits at Earth's MEO; its signals point down at Earth and do not usefully reach Mars, and in any case a Mars spacecraft is far outside (above) the constellation, where there is no GNSS at all. Deep-space missions instead use DSN radiometric tracking — ranging, Doppler, and delta-DOR (Ch. 26) — plus optical navigation against the target and stars on approach (Case Study 2).

13.19 $M = 0.5236\ \text{rad}$, $e = 0.6$. Start $E_0 = M + e\sin M = 0.5236 + 0.6(0.5000) = 0.8236$. Newton, $f(E) = E - e\sin E - M$, $f'(E) = 1 - e\cos E$: - $E_0 = 0.8236$: $f = 0.8236 - 0.6(0.7326) - 0.5236 = -0.1396$; $f' = 1 - 0.6(0.6806) = 0.5916$; $E_1 = 0.8236 + 0.1396/0.5916 = 1.0595$. - $E_1 = 1.0595$: $f = 1.0595 - 0.6(0.8722) - 0.5236 = +0.0125$; $f' = 1 - 0.6(0.4891) = 0.7065$; $E_2 = 1.0595 - 0.0125/0.7065 = 1.0418$. Converged $E \approx 1.042\ \text{rad} = 59.7^\circ$. Then $\cos\nu = (\cos E - e)/(1 - e\cos E) = (0.505 - 0.6)/(1 - 0.6(0.505)) = -0.095/0.697 = -0.136$, so $\nu \approx 97.8^\circ$. Ordering $M(30^\circ) < E(59.7^\circ) < \nu(97.8^\circ)$: the true anomaly is larger than the mean anomaly between perigee and apogee.

13.21 Two plausible causes of a growing along-track residual trend: (1) mis-modeled atmospheric drag — a wrong density or ballistic coefficient makes the predicted mean motion slightly off, so the along-track error accumulates smoothly and grows with time; it correlates with solar/geomagnetic activity and with the orbit's perigee altitude. (2) An unmodeled maneuver or small unbalanced thrust (momentum dumping, a propellant leak) — this shows a step or slope change beginning at a specific epoch. Distinguish them by the residual shape (smooth accelerating drift vs. a discontinuity/kink) and by whether the onset correlates with a known thruster event in telemetry; a drag error also scales with atmospheric conditions, a maneuver does not.

13.23 Higher. The minimum-energy (Hohmann) transfer is already the slowest reasonable half-revolution path; arriving sooner than its $5.29\ \text{h}$ requires a higher-energy transfer ellipse (larger $a$ on the fast branch), which needs a bigger departure burn. You buy arrival time with propellant — the fundamental Lambert trade.

13.25 Pipeline: (1) IOD — run Gauss's method on the first few observations to get a rough first orbit (a lead). (2) Batch least squares — feed that a-priori orbit and all 200 radar observations into a differential-correction fit using a propagator that includes J2, drag, and third-body perturbations (Ch. 12); iterate until the residuals stop shrinking, yielding a precise orbit and its covariance. Inspect the residuals for structure. (3) Maintain — as new observations arrive, either re-run the batch periodically (producing fresh TLEs) or run a Kalman/EKF that folds each new measurement in recursively; monitor residuals continuously to catch model errors, maneuvers, or bad data. IOD bootstraps, batch refines, the filter keeps it current.

Chapter 14 — Spacecraft Attitude Dynamics (solutions to † / odd exercises)

14.1 The three translational degrees of freedom (the position of the center of mass) are the orbit, driven by force (gravity and thrust). The three rotational degrees of freedom (orientation about the center of mass) are the attitude, driven by torque. The two sets are dynamically independent.

14.3 Norm$^2 = 0.60^2 + 0 + 0.80^2 + 0 = 0.36 + 0.64 = 1.00$, so $\mathbf{q}$ is a valid unit quaternion. The scalar part gives the angle: $q_0 = \cos(\theta/2) = 0.60 \Rightarrow \theta/2 = 53.13^\circ \Rightarrow \theta = 106.3^\circ$. The vector part gives the axis: $(q_1,q_2,q_3) = \hat{\mathbf{e}}\sin(\theta/2)$ with $\sin(53.13^\circ) = 0.80$, so $\hat{\mathbf{e}} = (0, 0.80, 0)/0.80 = (0, 1, 0)$ — the body $y$-axis. The rotation is $106.3^\circ$ about $+y$.

14.5 (a) Versus Euler angles: the quaternion has no singularity (no gimbal lock) — it is well behaved at every orientation. (b) Versus the DCM: the quaternion has minimal redundancy — four numbers with one constraint, versus nine numbers with six constraints — so it is far cheaper to store, transmit, and keep normalized.

14.7 With $\mathbf{q}_x = (0.7071, 0.7071, 0, 0)$, the Hamilton product $\mathbf{q}_x \otimes \mathbf{q}_x$: $w = (0.7071)^2 - (0.7071)^2 - 0 - 0 = 0$; $x = (0.7071)(0.7071) + (0.7071)(0.7071) + 0 - 0 = 1$; $y = z = 0$. Result $(0, 1, 0, 0)$: a $180^\circ$ rotation about $+x$ ($q_0 = \cos(\theta/2) = 0 \Rightarrow \theta = 180^\circ$, axis $(1,0,0)$). This makes sense — two $90^\circ$ turns about the same axis add to a single $180^\circ$ turn.

14.9 $\mathbf{q}^{*} = (0.7071, -0.7071, 0, 0)$. Then $\mathbf{q} \otimes \mathbf{q}^{*}$: $w = (0.7071)^2 - (0.7071)(-0.7071) - 0 - 0 = 0.5 + 0.5 = 1$; $x = (0.7071)(-0.7071) + (0.7071)(0.7071) = 0$; $y = z = 0$. Result $(1, 0, 0, 0)$ — the identity (zero) rotation. You must get the identity, because the conjugate is the inverse rotation, and composing any rotation with its inverse leaves the orientation unchanged.

14.11 A DCM uses 9 numbers subject to 6 orthonormality constraints (rows are mutually perpendicular unit vectors), leaving $9 - 6 = 3$ DOF and 6 redundant numbers. A quaternion uses 4 numbers subject to 1 unit-norm constraint, leaving $4 - 1 = 3$ DOF and only 1 redundant number. The quaternion is cheaper to keep valid in flight: renormalizing is one square root and four divides (divide by the norm), versus re-orthonormalizing a $3\times3$ matrix (a Gram–Schmidt pass) for the DCM.

14.13 quat_multiply([0.7071,0,0,0.7071], [0.7071,0,0,0.7071]) composes two $90^\circ$ rotations about $z$: $w = (0.7071)^2 - (0.7071)^2 = 0$, $z = (0.7071)^2 + (0.7071)^2 = 1$, $x = y = 0$. Result $(0,0,0,1)$ — a $180^\circ$ rotation about $z$. # Expected output: [0. 0. 0. 1.]

14.15 dcm_from_quat([1, 0, 0, 0]) is the identity quaternion (zero rotation): every off-diagonal term has a factor of $0$, and each diagonal is $1 - 2(0) = 1$. Result is the $3\times3$ identity matrix. # Expected output: [[1 0 0] [0 1 0] [0 0 1]]

14.17 The wheels saturate. Holding attitude against a steady one-directional disturbance forces the (internal) wheel to keep absorbing momentum; it reaches capacity after $t_{\text{sat}} = H_{\max}/M_d = 4/(1\times10^{-4}) = 4\times10^4\ \text{s} \approx 11\ \text{hours}$, after which it is at maximum speed and provides no control torque. The total angular momentum obeys $\dot{\mathbf{H}}_{\text{total}} = \mathbf{M}_{\text{external}}$, so the accumulated momentum can only be removed by an external actuator. The corrected memo: the spacecraft does need thrusters (or magnetorquers in LEO) for periodic momentum dumping — "no propellant" does not mean "no external torque ever required."

14.19 (Model answer, Track A — GEO comsat.) (a) Pointing requirement: beam-pointing accuracy of about $0.05$–$0.1^\circ$, set by keeping the antenna footprint on the service region. (b) Stabilization: three-axis — a GEO comsat must hold a fixed Earth-pointing attitude continuously, which spin cannot do without a despun platform. (c) Suite: reaction wheels for fine control; thrusters for momentum dumping and station-keeping (Earth's magnetic field is far too weak at GEO for magnetorquers); star tracker plus rate gyros for determination, with coarse Sun sensors for safe mode. Momentum dumped roughly daily, folded into station-keeping burns (see §14.5 worked example: $t_{\text{sat}}\sim 14\ \text{h}$ for $H_{\max}=5\ \text{N·m·s}$, $M_d\sim 10^{-4}\ \text{N·m}$). (Other tracks: Track B/C use star tracker + gyros + thrusters, no magnetorquers off-Earth; Track D similar, with SRP the dominant disturbance.)

14.21 Requirement $0.03^\circ = 108\ \text{arcsec}$. RSS: $\sqrt{30^2 + 20^2 + x^2 + 60^2} \le 108$, so $900 + 400 + x^2 + 3600 \le 108^2 = 11{,}664$, giving $x^2 \le 6{,}764$ and $x \le 82.2\ \text{arcsec}$. So wheel-quantization error up to about $82\ \text{arcsec}$ can be tolerated. Regardless of $x$, the term to fix first is the alignment/thermal $60\ \text{arcsec}$ contribution — its square, $3{,}600$, is the largest in the sum, so reducing it lowers the RSS total the most.

14.23 At $r = 6.871\times10^6\ \text{m}$, $\dfrac{3\mu}{2r^3} = \dfrac{3(3.986\times10^{14})} {2(6.871\times10^6)^3} = 1.84\times10^{-6}\ \text{s}^{-2}$, so $M_{gg} = (1.84\times10^{-6})(0.1) \approx 1.8\times10^{-7}\ \text{N·m}$. This is about 500 times smaller than the $\sim 1\times10^{-4}\ \text{N·m}$ magnetorquer torque, so the magnetorquer easily overpowers the gravity-gradient disturbance — the cubesat is comfortably controllable. (Small satellites are so light that even weak magnetorquers dominate their tiny disturbance torques.)

14.25 A reaction wheel is an internal actuator, so the vehicle's total angular momentum changes only under an external torque: $\dot{\mathbf{H}}_{\text{total}} = \mathbf{M}_{\text{external}}$. Spinning a wheel merely trades momentum between body and wheel; the sum is fixed. A steady disturbance keeps injecting momentum, which accumulates in the wheel until it saturates. Only an external actuator (thrusters, or magnetorquers pushing on Earth's field) can carry that momentum back out. Hence you can never hold attitude against a persistent disturbance on wheels alone — an external actuator for momentum dumping is mandatory.

14.27 A nadir-pointing satellite keeps one body axis aimed at Earth's center, so its body frame stays aligned with the local orbit frame — which itself rotates once per orbit relative to the stars. Therefore, in inertial (ECI) space the satellite is rotating at the orbital rate, i.e. at the mean motion $n = 2\pi/T = \sqrt{\mu/a^3}$ (Chapter 8). For a low orbit $n \approx 1.13\times10^{-3}\ \text{rad/s} \approx 0.065^\circ/\text{s}$. It looks steady to someone on the ground but is turning at $n$ against the stars — which is exactly why holding "nadir" is an active attitude task, not a static one.

14.29 For a circular orbit $a = r$, so the mean motion satisfies $n^2 = \mu/a^3 = \mu/r^3$, hence $3\mu/r^3 = 3n^2$. Evaluate for a $400\ \text{km}$ orbit: with $n = 1.13\times10^{-3}\ \text{rad/s}$, $3n^2 = 3(1.13\times10^{-3})^2 = 3(1.277\times10^{-6}) = 3.83\times10^{-6}\ \text{s}^{-2}$. Direct from $\mu$ and $r = 6.771\times10^6\ \text{m}$: $3\mu/r^3 = 3(3.986\times10^{14})/(6.771\times10^6)^3 = 1.196\times10^{15}/3.104\times10^{20} = 3.85\times10^{-6}\ \text{s}^{-2}$. The two agree to rounding (the small difference is only from rounding $n$), confirming $3\mu/r^3 = 3n^2$ — the gravity-gradient stiffness is three times the squared orbital rate.

Chapter 15 — The Three-Body Problem and Lagrange Points

Solutions to daggered (†) and odd-numbered exercises. Constants from Appendix B.

15.1 † $\mu^{}_{\text{SE}} = 5.972\times10^{24}/(1.989\times10^{30}+5.972\times10^{24}) = 3.00\times10^{-6}$. $\mu^{}_{\text{EM}} = 7.342\times10^{22}/(5.972\times10^{24}+7.342\times10^{22}) = 0.0121$. The Sun–Earth system is far closer to the lopsided limit $\mu^{*}\to 0$ (by four orders of magnitude) — the Sun so dominates that the Earth is very nearly a test particle.

15.3 † $r_{\text{H}} = 384{,}400\,(7.342\times10^{22}/(3\times5.972\times10^{24}))^{1/3} = 384{,}400\,(4.098\times10^{-3})^{1/3} = 384{,}400\times0.1600 = 61{,}500\ \text{km}$. As a fraction of the Earth–Moon distance: $61{,}500/384{,}400 = 0.16$, i.e. 16%. (True L1 $\approx 58{,}000$ km, L2 $\approx 64{,}500$ km — the leading-order Hill estimate sits between them.)

15.5 † In lunar distances: $1.5\times10^6/384{,}400 = 3.9$. As a fraction of an AU: $1.5\times10^6/1.496\times10^8 = 0.0100 = 1.0\%$. So Sun–Earth L1/L2 are about four times the Moon's distance, yet only one percent of the way to the Sun.

15.7 † $R = 5.203\ \text{AU} = 7.784\times10^8\ \text{km}$; $m_2/m_1 = 1.898\times10^{27}/1.989\times 10^{30} = 9.54\times10^{-4}$. $r_{\text{H}} = 7.784\times10^8\,(9.54\times10^{-4}/3)^{1/3} = 7.784\times10^8\times0.0683 = 5.3\times10^7\ \text{km}$ (about 53 million km, $\approx 740$ Jupiter radii). Callisto orbits at $1.88\times10^6\ \text{km}$ — only $\sim 3.5\%$ of the Hill radius — so all four Galilean moons sit deep inside Jupiter's gravitational territory and are securely bound.

15.9 † With $x = R(0.5-\mu^{*})$ and $y = R\sqrt3/2$, the barycentric distance is $d^2 = x^2+y^2 = R^2[(0.5-\mu^{*})^2 + 3/4] = R^2[0.25 - \mu^{*} + \mu^{*2} + 0.75] = R^2[1-\mu^{*}+\mu^{*2}]$, so $d = R\sqrt{1-\mu^{*}+\mu^{*2}}$. For Earth–Moon: $d = 384{,}400\sqrt{1-0.0121+0.000147} = 384{,}400\times0.9940 = 382{,}100\ \text{km}$. It is close to $R$ because $\mu^{*}$ is small (the correction is $\approx 1-\mu^{*}/2$), so the triangular point sits almost exactly one Earth–Moon distance from the barycenter.

15.11 † (a) Sun–Neptune $m_1/m_2 \approx 1.9\times10^4 > 24.96$ → stable. (b) Earth–Moon $81.3 > 24.96$ → stable. (c) $20 < 24.96$ → unstable — a binary that even, its triangular points cannot trap material.

15.13 † $\tfrac12(25+\sqrt{621}) = \tfrac12(25+24.920) = \tfrac12(49.920) = 24.96$. As a mass parameter: $m_2/m_1 = 1/24.96 = 0.04006$, so $\mu^{*}_{\text{crit}} = 0.04006/(1+0.04006) = 0.0385$. Stability requires $\mu^{*} < 0.0385$.

15.14 † $C_4 = 3 - \mu^{} + \mu^{2} = 3 - 0.0121 + (0.0121)^2 = 3 - 0.0121 + 0.000147 = 2.988$.

15.15 Earth at $(-0.0121, 0)$, Moon at $(0.9879, 0)$. At $(0.5, 0)$: $r_1 = |0.5+0.0121| = 0.5121$, $r_2 = |0.5-0.9879| = 0.4879$. $\Omega = \tfrac12(0.5^2) + 0.9879/0.5121 + 0.0121/0.4879 = 0.125 + 1.9291 + 0.0248 = 2.0789$. With $v = 0$, $C_J = 2\Omega = 4.158$. (Compare $C_4 = 2.988$: this point has a higher Jacobi constant, so it sits "deeper" in the forbidden-region sense — a spacecraft released here at rest is more tightly hemmed in than one at the triangular-point energy.)

15.17 † hill_radius(1.989e30, 5.972e24, 1.496e8) $= 1.496\times10^8\times(5.972\times10^{24}/ (3\times1.989\times10^{30}))^{1/3} = 1.496\times10^8\times(1.0008\times10^{-6})^{1/3} = 1.496\times10^8\times0.01000 = 1{,}496{,}418$. # Expected output: 1496418.

15.19 † jacobi_C4(0.0121) $= 3 - 0.0121 + 0.0121^2 = 2.988$. # Expected output: 2.988.

15.20 † A low-energy (weak-stability-boundary) transfer threads the cargo tug gently through the Earth–Moon L1/L2 necks and lets multi-body dynamics do the capture, so it needs almost no insertion burn — propellant that a direct Hohmann would spend braking into lunar orbit. The one resource it spends lavishly in return is time: months of cruise instead of days. For a non-urgent cargo run, propellant is the scarce resource and time is cheap, so the trade favors the low-energy route (exactly the logic that rescued Hiten and delivered GRAIL).

15.21 Place the telescope at L2. From L2 the Sun, Earth, and Moon all lie in nearly one direction (sunward), so a single sunshield blocks every warm, infrared-bright body at once and the optics face cold deep space, cooling passively to tens of kelvin. At L1 the Sun and Earth are on opposite sides of the spacecraft, so one shield cannot hide both — the Earth would shine into the cold field of view. One shield, one hot direction: that is L2.

15.22 † Total station-keeping $= 3\ \text{m/s/yr}\times20\ \text{yr} = 60\ \text{m/s}$. Compared with a single LEO→GTO leg ($\approx 2{,}500\ \text{m/s}$), that is only $60/2500 = 2.4\%$ — a rounding error. Living at an unstable point for two decades costs less delta-v than a minor orbital maneuver, because the instability is slow and predictable and each correction is tiny. Instability is cheap to manage, not expensive.

15.23 † Two errors. (1) The gravitational pulls do not cancel at L1: the Sun outweighs the Earth enormously, and at Sun–Earth L1 the Sun's pull is far stronger than Earth's — they are nowhere near equal. (2) The net force is not zero, and an object there does not "just float." The correct picture: in the rotating frame that co-orbits with the Earth, equilibrium is a balance of net gravity plus the centrifugal force. The net gravity is large and points sunward, and it supplies exactly the centripetal force needed to carry the spacecraft around the Sun once per year in lockstep with Earth. The Lagrange point is a rotating-frame force balance, not a gravitational null.

15.25 † Sphere of influence: $r_{\text{SOI}} \approx a(m_2/m_1)^{2/5}$ — a two-body force-ratio comparison used for patched-conic flybys; for Earth, $\approx 924{,}000\ \text{km}$. Hill radius: $r_{\text{H}} = R(m_2/3m_1)^{1/3}$ — a tidal / rotating-frame comparison of the planet's pull to the Sun's tidal stretch; for Earth, $\approx 1.5\times10^6\ \text{km}$. They differ (exponent $2/5$ vs $1/3$) because they answer different questions, but both mark "how far the planet stays in charge." The Moon, at $384{,}400\ \text{km}$, lies well inside both boundaries, which is exactly why it is gravitationally bound to Earth.

15.27 † The tyranny of the rocket equation (Ch. 3) is that required propellant grows exponentially with delta-v — so the thing to minimize, above all, is delta-v. A low-energy transfer is not a way around physics (energy and momentum are conserved throughout); it is the purest possible response to the rocket equation: it refuses to spend delta-v, substituting the free motion of the multi-body gravity field (and a great deal of patience) for propulsive maneuvers. Where a Hohmann pays delta-v to force the trajectory, the superhighway lets the natural dynamics carry the spacecraft for nearly nothing — spending the one currency the rocket equation does not tax: time.

Chapter 16 — Rocket Propulsion Fundamentals (solutions to † and odd problems)

Use $g_0 = 9.80665\ \text{m/s}^2$ (or $9.81$ where noted).

16.1 † $F = \dot m\, c = 280 \times 3{,}000 = 840{,}000\ \text{N} = 840\ \text{kN}$.

16.3 † $c = I_{sp}\,g_0 = 350 \times 9.80665 = 3{,}432\ \text{m/s} \approx 3.43\ \text{km/s}$.

16.5 † $T/W = F/(mg) = 4{,}000{,}000 / (300{,}000 \times 9.81) = 4{,}000{,}000/2{,}943{,}000 = 1.36$. Since $T/W > 1$, it lifts off, with initial acceleration $a = g(T/W - 1) = 9.81 \times 0.36 \approx 3.5\ \text{m/s}^2$ (about $0.36\,g$).

16.7 † Only the pressure term changes with altitude, so $F_{\text{vac}} = F_{\text{SL}} + p_{a,\text{SL}}A_e = 800{,}000 + 101{,}325 \times 1.0 = 901{,}325\ \text{N} \approx 901\ \text{kN}$. (For completeness: $c_{\text{SL}} = 290 \times 9.80665 = 2{,}844\ \text{m/s}$ and $\dot m = 800{,}000/2{,}844 \approx 281\ \text{kg/s}$, though these are not needed for the answer.)

16.9 † $c = v_{\text{ex}} + \dfrac{p_e A_e}{\dot m} = 3{,}200 + \dfrac{40{,}000}{250} = 3{,}200 + 160 = 3{,}360\ \text{m/s}$. Then $I_{sp} = c/g_0 = 3{,}360/9.80665 \approx 343\ \text{s}$.

16.11 †

def thrust(mdot, ve, pe, pa, ae):
    return mdot * ve + (pe - pa) * ae
print(round(thrust(300, 2900, 70000, 101325, 0.68)))
# Expected output: 848699

Momentum term $300\times2900 = 870{,}000$; pressure term $(70{,}000 - 101{,}325)\times0.68 = -21{,}301$; sum $= 848{,}699\ \text{N} \approx 848.7\ \text{kN}$.

16.13 †

def total_impulse(F, tb):  return F * tb
def burn_time(mp, mdot):   return mp / mdot
print(round(total_impulse(845000, 162)))   # 136890000
print(round(burn_time(43900, 306), 1))     # 143.5

$I_t = 845{,}000 \times 162 = 1.369\times10^{8}\ \text{N·s}$; $t_b = 43{,}900/306 = 143.5\ \text{s}$ (the per-engine burn, using per-engine propellant $\approx 43.9\ \text{t}$).

16.15 † Two errors. (1) Wrong mass: an upper stage fires after staging, so it must lift only itself plus payload, not the entire launch vehicle. With a realistic stage+payload mass of, say, $115\ \text{t}$, $T/W = 900{,}000/(115{,}000\times9.81) \approx 0.80$. **(2)** The liftoff test $T/W > 1$ applies to leaving a surface; an upper stage ignites in space, where any positive thrust accelerates it. A $T/W$ below 1 in orbit is completely normal and fine. The stage fires without difficulty.

16.17 † From $I_t = I_{sp}\,g_0\,m_p$, $m_p = \dfrac{I_t}{I_{sp}\,g_0} = \dfrac{3.0\times10^{8}}{320 \times 9.80665} = \dfrac{3.0\times10^{8}}{3{,}138} \approx 9.56\times10^{4}\ \text{kg} = 95.6\ \text{t}$.

16.19 † Jet power $P = \tfrac12 \dot m_{\text{total}}\, c^2$ with $\dot m_{\text{total}} = 9\times306 = 2{,}754\ \text{kg/s}$ and $c = 2{,}765\ \text{m/s}$: $P = 0.5 \times 2{,}754 \times (2{,}765)^2 = 0.5 \times 2{,}754 \times 7.645\times10^{6} \approx 1.05\times10^{10}\ \text{W} = 10.5\ \text{GW}$. That is roughly ten large ($1\ \text{GW}$) power stations' worth of kinetic energy per second. (Cross-check: $P = \tfrac12 F c = 0.5\times7.605\times10^{6}\times2{,}765 \approx 1.05\times10^{10}\ \text{W}$ ✓.)

16.21 † For a $500\ \text{kg}$ vehicle, weight on Earth is $W = 500\times9.81 = 4{,}905\ \text{N}$, while an ion engine's thrust is about $0.1\ \text{N}$. So $T/W = 0.1/4{,}905 \approx 2\times10^{-5}$ — about fifty-thousand times too small to lift off. Its enormous $I_{sp}$ is irrelevant to launch: without $T/W > 1$ the vehicle never leaves the pad. (This is §16.4's threshold concept — delta-v and $T/W$ are independent tests, and electric propulsion fails the second one spectacularly.)

16.22 † $\Delta v = c\,\ln(m_0/m_f) = 3{,}050 \times \ln(120/30) = 3{,}050 \times \ln 4 = 3{,}050 \times 1.386 = 4{,}228\ \text{m/s} \approx 4.23\ \text{km/s}$. The effective exhaust velocity $c$ is set by the engine (propellant + nozzle); the masses $m_0$ and $m_f$ — the mass ratio — are set by the vehicle's construction (how much propellant and how light the structure).

16.23 Thrust does not appear because a fixed propellant load at a fixed exhaust velocity carries a fixed total impulse $I_t = c\, m_p$ and a fixed delta-v (through the mass ratio). Thrust only sets the burn time $t_b = I_t/F$ — how fast that fixed impulse is delivered. High thrust burns it fast, low thrust slowly; the delta-v is identical either way, which is exactly why the rocket equation omits it.

16.24 † Higher liftoff $T/W$ means the vehicle accelerates and climbs faster, so it spends less time fighting gravity during ascent — smaller gravity losses (Ch. 4). But faster acceleration low in the atmosphere means it hits high speed while the air is still dense, raising dynamic pressure (max-Q), drag, and structural/aerodynamic loads; a higher $T/W$ also demands more (heavier, costlier) engine. Real first stages cluster near $T/W \approx 1.2$–$1.5$ because that band gives most of the gravity-loss benefit without paying excessively in aero loads or engine mass. (A powered landing, by contrast, sometimes wants a high $T/W$ "suicide burn" to minimize gravity losses while hovering is expensive — Ch. 22.)

16.25 The exhaust velocity rises by $c_{\text{vac}}/c_{\text{SL}} = 311/282 = 1.103$, i.e. +10.3%. For a stage flying a fixed mass ratio, delta-v is directly proportional to $c$ ($\Delta v = c\ln(m_0/m_f)$), so the delta-v also rises by +10.3% — the same percentage, because delta-v is linear in $c$ at fixed mass ratio. (Note: the propellant saving for a fixed target delta-v would be larger than 10.3%, because there $c$ sits in the exponent of $e^{\Delta v/c}$; but the question fixes the mass ratio, so the two percentages match.)

16.26 † Both versions have the same delta-v: $\Delta v = c\ln(m_0/m_f)$ contains no thrust, and the two builds share the same $c$, $m_0$, and $m_f$. But engine A, with ten times the thrust at the same $c$, has ten times the mass flow ($\dot m = F/c$), so its burn time $t_b = m_p/\dot m$ is one-tenth that of engine B. Same delta-v, delivered in one-tenth the time (and with ten times the acceleration and thus smaller gravity losses on a launch).

Chapter 17 — Chemical Rocket Engines (solutions to † and odd-numbered exercises)

17.1 LOX / liquid hydrogen (hydrolox), ~450–465 s. Hydrogen's exhaust — water plus leftover $\text{H}_2$ from running fuel-rich — has the lowest mean molecular weight of any chemical rocket exhaust, and lighter molecules leave the nozzle faster at a given temperature, so hydrolox has the highest exhaust velocity and thus the highest specific impulse. (Rigorous in Ch. 19.)

17.3 The gas-generator (open) cycle.

17.5 Merlin = (b) gas-generator; RS-25 = (d) fuel-rich staged combustion; RL10 = (a) expander; Raptor = (c) full-flow staged combustion.

17.7 Hydrolox: $V = 200{,}000/360 = 555.6\ \text{m}^3$. Kerolox: $V = 200{,}000/1020 = 196.1\ \text{m}^3$. Ratio $= 555.6/196.1 = 2.83$. The hydrogen tank is ~2.8× larger for the same propellant mass; putting hydrogen on the upper stage (which is smaller and carries less propellant) limits the tank-mass and drag penalty while still capturing hydrogen's high $I_{sp}$, whose benefit is carried all the way to orbit.

17.9 Methane (and oxygen) can be manufactured on Mars from atmospheric $\text{CO}_2$ and subsurface water ice (the Sabatier reaction, $\text{CO}_2 + 4\text{H}_2 \to \text{CH}_4 + 2\text{H}_2\text{O}$, plus electrolysis for $\text{O}_2$). A methalox Mars-return vehicle can therefore refuel in situ instead of hauling its entire return propellant from Earth — an enormous saving via the rocket equation, since that return propellant would otherwise have to be carried all the way to Mars. Two further reasons methane is attractive: (1) it burns cleanly with no coking/soot, easing engine reuse; (2) it is only mildly cryogenic ($-162\,^\circ\text{C}$) and far denser than hydrogen, so tanks are smaller and can share thermal and structural architecture with the LOX.

17.10 Effective $I_{sp} \approx (1 - 0.04)\times 340 = 0.96 \times 340 = 326.4\ \text{s}$; loss $\approx 340 - 326.4 = 13.6\ \text{s}$ (~4%). Small in seconds, but because exhaust velocity sits in the exponent of the mass ratio (Ch. 3), it is worth recovering on a high-performance stage — which staged combustion does by feeding the turbine gas into the main chamber instead of dumping it.

17.11 The expander turbine is driven only by heat picked up from the chamber and nozzle walls. Wall (heat-transfer) area scales roughly as the square of engine size ($\sim L^2$), while thrust and propellant flow scale roughly as chamber volume / throat area ($\sim L^3$). So as an engine is scaled up, the heat available to drive the pumps grows more slowly than the flow the pumps must move — the cycle "runs out of heat." Pure closed expander cycles are therefore practical only at modest, upper-stage thrust (RL10 ~110 kN, Vinci ~180 kN).

17.12 (1) Both propellants enter the chamber as hot gas, so the injector mixes gas with gas → better combustion efficiency and much better stability. (2) Each turbine handles only one propellant type, eliminating the interpropellant seal between the fuel-side and oxidizer-side gases on a shared shaft (historically a major failure point). (3) Passing the full flow of a propellant through its turbine lets the turbine make its power at a lower temperature, aiding engine life and enabling the very high chamber pressure that reuse rewards.

17.13 Burn rate rises by a factor $1.15^{0.4} = e^{0.4\ln 1.15} = e^{0.4\times 0.1398} = e^{0.0559} = 1.0575$ — about **+5.75%**. Because $n = 0.4 < 1$, a 15% pressure rise produces only a ~5.75% rise in gas generation (less than proportional), so the pressure feedback loop is self-limiting and any disturbance dies away. If $n \ge 1$, gas would be generated faster than pressure rose, and the motor would run away to detonation — which is why propellant chemists keep $n$ comfortably below 1.

17.15 $f_{ox} = 6/7 = 0.857$, $f_{fuel} = 1/7 = 0.143$. Volume per kg $= 0.857/1141 + 0.143/71 = 0.000751 + 0.002012 = 0.002763\ \text{m}^3/\text{kg}$, so $\rho = 1/0.002763 \approx 362\ \text{kg/m}^3$. # Expected output: 362. (Matches the ~360 kg/m³ bulk density used for hydrolox in §17.1.)

17.17 $\dot m = \rho_p A_b r = 1800 \times 25 \times 0.009 = 405\ \text{kg/s}$; $F = \dot m\, I_{sp}\, g_0 = 405 \times 265 \times 9.80665 = 1{,}052{,}499\ \text{N} \approx 1.05\ \text{MN}$. `# Expected output: 1052499`. (Using $g_0 = 9.81$ instead gives $1{,}052{,}858\ \text{N}$ — the same ~1.05 MN.)

17.18 Total $\approx 33 \times 650\ \text{kg/s} \times 150\ \text{s} = 33 \times 97{,}500 = 3{,}217{,}500\ \text{kg} \approx 3.2\times10^{6}\ \text{kg} \approx 3{,}200\ \text{t}$. Compared with the quoted ~3,400 t capacity, this is the right order of magnitude (within ~6%). Assumptions: all 33 engines run the full 150 s at a steady 650 kg/s — an over-estimate, since some engines throttle or shut down early and flow varies — but a fair order-of-magnitude check. Takeaway: Super Heavy consumes on the order of 3,000 tonnes of propellant in ~2.5 minutes.

17.19 A liquid engine throttles by cutting propellant flow through a valve. A solid has no feed valve: its "flow" is propellant converted to gas at the burning surface, whose rate is fixed by the exposed surface area (grain geometry, set at casting) and by chamber pressure through $r = a\,p_c^{\,n}$ — not by anything you can adjust in flight. To throttle you would have to change the burning area or pressure on command, which a sealed, burning grain does not permit. So a solid cannot be throttled for a gentle landing; its thrust is pre-programmed and can only be terminated violently (blow-out ports).

17.20 The flaw: hydrogen's high $I_{sp}$ comes with extremely low density (~71 kg/m³) and deep cryogenic temperature. For a first stage, the resulting huge, heavy tanks — plus their drag and reduced thrust-to-weight — often cost more delta-v than the $I_{sp}$ advantage buys, so a denser propellant gives the whole vehicle more delta-v (the density–$I_{sp}$ trade of §17.1 and Chapter 3's second case study). Real first stages that deliberately avoid hydrogen: Falcon 9 (kerolox), Super Heavy (methalox), and the Soyuz and Atlas V first stages (kerolox). Hydrogen is used instead on upper stages, where its $I_{sp}$ is carried all the way to orbit and its bulk is less punishing.

17.21 (Rubric: named propellant + cycle + reason tied to thrust/restart/throttle/storability.) Model answers by track: - Track A (GEO comsat): apogee/orbit-insertion and station-keeping → storable hypergolic bipropellant, pressure-fed, because it must fire reliably after a coast, restart many times over a ~15-year life, and stay liquid for years (station-keeping may instead use electric propulsion, Ch. 20). - Track B (lunar lander): deeply throttleable, restartable pressure-fed hypergolic (Apollo-LM style) or methalox — it must hover and set down (deep throttle) and restart across descent phases; solids are disqualified (no throttle/shutdown). - Track C (Mars orbiter): storable hypergolic for Mars-orbit insertion after the 6–9-month cruise (cryogens would boil off); one large, reliable burn with restart for trims. - Track D (asteroid rendezvous): storable hypergolic or, better, electric propulsion (Ch. 20) for the low-thrust, long-duration matching; chemistry only if a fast or large burn is required.

17.23 Momentum thrust $= \dot m\, c = 300 \times 3{,}300 = 990{,}000\ \text{N} = 990\ \text{kN}$. With the nozzle pressure term adding 5%: total $F = 990{,}000 \times 1.05 = 1{,}039{,}500\ \text{N} \approx 1.04\ \text{MN}$. (This is the Chapter 16 thrust equation $F = \dot m\,c + (p_e - p_a)A_e$, with the pressure term contributing 5% of the momentum term here.)

17.25 The RS-25 has the higher specific impulse, but it is enormously expensive, complex, and slow to refurbish — "reusability 1.0." Merlin's lower $I_{sp}$ is more than offset by low unit cost and mass production (nine per first stage, giving engine-out reliability), mechanical simplicity and robustness (old gas-generator cycle, dense easy-to-handle kerolox), and deep throttling that enables propulsive landing and rapid, low-refurbishment reuse. That combination cut launch cost by roughly an order of magnitude (Appendix H, §H.3) — something no single high-$I_{sp}$ expendable engine could achieve. Themes: #5 (reusability changes everything — cheap, rapid reuse, not peak $I_{sp}$, sets cost per kilogram) and #6 (what an engineer optimizes for is a historical and economic choice). The vehicle and program, not the engine's $I_{sp}$, decide success — echoing Chapter 3's lesson that the true currency (there delta-v, here cost per kilogram) is what matters, with $I_{sp}$ only one input to it.

Chapter 18 — Combustion and Propellants

18.1 Balance methane combustion: $\text{CH}_4 + 2\,\text{O}_2 \rightarrow \text{CO}_2 + 2\,\text{H}_2\text{O}$ (1 C, 4 H, 4 O on each side). Stoichiometric mass ratio $r = \dfrac{2 \times 32}{16} = \dfrac{64}{16} = 4.0$. So $8$ for hydrogen but only $4$ for methane — methane needs half as much oxygen per unit fuel mass.

18.3 Fuel mass fraction of the stoichiometric mixture is $1/(1+r) = 1/(1+4) = 0.20$. Energy per kg of mixture $= 50\ \text{MJ/kg fuel} \times 0.20 = 10.0\ \text{MJ/kg}$. Comment: this is ~25% below hydrogen–oxygen's $13.4\ \text{MJ/kg}$ but essentially tied with kerosene's $\approx 9.8$. All chemical propellants land in the $\sim 10\text{–}13\ \text{MJ/kg}$ band — the narrow energy window that caps chemical specific impulse.

18.5 Using $v_e = \sqrt{\frac{2\gamma}{\gamma-1}\frac{R_u T_c}{\mathcal{M}}[1-(p_e/p_c)^{(\gamma-1)/\gamma}]}$: $\frac{2\gamma}{\gamma-1} = \frac{2.4}{0.2} = 12$; $\frac{R_u T_c}{\mathcal{M}} = \frac{8.314\times3500}{0.020} = 1.455\times10^6\ \text{J/kg}$; product $= 1.746\times10^7$. Bracket: $(0.1/100)^{0.1667} = (0.001)^{0.1667} = 10^{-0.5} = 0.316$, so $[\,1-0.316\,] = 0.684$. Then $v_e = \sqrt{1.746\times10^7 \times 0.684} = \sqrt{1.194\times10^7} \approx 3{,}455\ \text{m/s}$, and $I_{sp} = 3455/9.81 \approx 352\ \text{s}$. With $T_c = 3{,}500\ \text{K}$ and $\mathcal{M} = 20$, this is LOX/CH$_4$ (methalox).

18.7 $\sqrt{T_c/\mathcal{M}}$: LH$_2$ $= \sqrt{3300/13} = \sqrt{253.8} = 15.93$; CH$_4$ $= \sqrt{3500/20} = \sqrt{175} = 13.23$. Ratio $= 15.93/13.23 = 1.20$: hydrogen delivers about 20% more exhaust velocity. The molecular-weight knob is responsible — hydrogen's exhaust ($\mathcal{M}=13$) is lighter than methane's ($\mathcal{M}=20$), and hydrogen actually burns cooler here (3,300 vs 3,500 K), so temperature works slightly against it. Lightness wins.

18.9 Work per gram of H$_2$ (take $\mathcal{M}_{\text{H}_2} \approx 2\ \text{g/mol}$, so $1\ \text{g} = 0.5\ \text{mol}$). Oxygen mass $= r$ grams $= r/32\ \text{mol}$ O$_2$. In $2\,\text{H}_2 + \text{O}_2 \rightarrow 2\,\text{H}_2\text{O}$ each mole of O$_2$ burns 2 mol H$_2$ into 2 mol H$_2$O. So H$_2$ consumed $= 2(r/32) = r/16\ \text{mol}$; water produced $= r/16\ \text{mol}$; leftover H$_2$ $= 0.5 - r/16\ \text{mol}$. Total product moles $= (r/16) + (0.5 - r/16) = 0.5\ \text{mol}$ — *constant* for any $r \le 8$. Total product mass equals total reactant mass $= (1 + r)\ \text{g}$. Therefore mean molar mass $= \dfrac{1+r}{0.5} = 2(1+r)\ \text{g/mol}$. $\blacksquare$ (Check: $r=8 \Rightarrow 18$, pure water; $r=4 \Rightarrow 10$.)

18.10 exit_velocity(1.20, 3300, 0.013, 0.1, 200): term $= 12 \times (8.314\times3300/0.013) = 12 \times 2.110\times10^6 = 2.533\times10^7$; bracket $= 1 - (0.0005)^{0.1667} = 1 - 0.282 = 0.718$; $v_e = \sqrt{2.533\times10^7 \times 0.718} = 4{,}265\ \text{m/s}$; $I_{sp} = 4265/9.80665 = 435\ \text{s}$. # Expected output: 4265 435.

18.11 mean_molar_mass(6) = 2(1+6) = 14.0; knob(3300,6) = sqrt(3300/14) = sqrt(235.7) = 15.35, rounded to 15.4. # Expected output: 14.0 15.4.

18.12 specific_energy(-241.8, 2, 0.03603): energy $= -2\times(-241.8) = 483.6\ \text{kJ} = 0.4836\ \text{MJ}$; $0.4836/0.03603 = 13.4\ \text{MJ/kg}$. # Expected output: 13.4.

18.13 The flaw: "hotter → higher $I_{sp}$" uses only one of the two knobs and ignores exhaust molecular weight. $v_e \propto \sqrt{T_c/\mathcal{M}}$. Kerosene exhaust ($\mathcal{M}\approx 23$) is nearly twice as heavy as hydrogen's ($\mathcal{M}\approx 13$): $\sqrt{3600/23} = 12.5$ versus $\sqrt{3300/13} = 15.9$. Hydrogen wins by ~27% despite its cooler flame. Correct conclusion: hydrogen has the higher specific impulse; the $350\ \text{K}$ temperature edge cannot overcome kerosene's molecular-weight penalty.

18.15 Choose N$_2$O$_4$/MMH (hypergolic) — or a green monopropellant (ASCENT, LMP-103S) for a lower-handling-cost modern alternative. Dominant properties: (1) storability for the multi-year mission with no boiloff, and (2) restart reliability — hypergolics ignite on contact, so station-keeping burns light every time with no igniter to fail. What you sacrifice: specific impulse (~320 s versus ~450 for LOX/LH$_2$) and, for hypergolics, toxic handling — acceptable because a comsat's total impulse is modest, so tank size and efficiency matter far less than reliability and storability.

18.17 At $r \approx 3.5$ the knob $\sqrt{T_c/\mathcal{M}}$ (and thus $I_{sp}$) is maximized, but running that fuel-rich means burning enormous volumes of very-low-density liquid hydrogen. That demands larger, heavier tanks, whose extra structural (tank) mass eats into the mass ratio the higher $I_{sp}$ was meant to improve. Flying at $r \approx 6$ (oxidizer-ward of the chamber optimum) gives up a few seconds of $I_{sp}$ to shrink the hydrogen tank, lowering vehicle mass. The chamber optimum and the vehicle optimum differ — Theme 4, mass is the enemy.

18.19 Two reasons, one performance and one hardware. (1) Performance: the knob $\sqrt{T_c/\mathcal{M}}$ does not peak at stoichiometric — running fuel-rich lowers exhaust molecular weight faster than it lowers temperature, so peak $I_{sp}$ occurs fuel-rich; stoichiometric is not even the efficiency optimum. (2) Hardware: the stoichiometric flame is the hottest (~3,500+ K), beyond what chamber walls, throat, and cooling can survive; a fuel-rich mixture runs cooler and often leaves a protective fuel-rich boundary layer. Both point the same way — away from stoichiometric.

18.20 Two distinct reasons a designer picks a lower-$I_{sp}$ propellant: (1) Density — LOX/LH$_2$ has a bulk density near $0.36\ \text{g/cm}^3$ versus ~$1.0$ for kerosene, so hydrogen needs roughly triple the tank volume; on a first stage this bulky, low-thrust-density propellant is a poor choice despite its efficiency. (2) Storability/handling — liquid hydrogen boils off fast (20 K), leaks through seals, and embrittles metal, making it unsuitable for long pad stays or long-duration missions; storable or mildly-cryogenic propellants win where the vehicle must wait. (Also acceptable: low liftoff thrust-to-weight from hydrogen's low density.)

18.21 $v_e = I_{sp}\,g_0 = 452 \times 9.81 \approx 4{,}434\ \text{m/s}$ (using $9.80665$: $4{,}433\ \text{m/s}$). Mass ratio $= e^{\Delta v/v_e} = e^{4100/4433} = e^{0.925} \approx 2.52$. So a hydrogen upper stage needs a mass ratio of about 2.5 for the $4.1\ \text{km/s}$ TLI burn — modest, because hydrogen's high $I_{sp}$ makes the required mass ratio small (contrast a storable stage at $I_{sp}=320$: $e^{4100/3139} = e^{1.306} = 3.7$).

18.23 Running the preburner extremely fuel-rich means only a small fraction of the fuel is actually oxidized, so relatively little heat is released — and that heat is spread over a large mass of mostly-unburned propellant, so the drive gas stays cool (~800 K) instead of reaching the ~3,500 K of a stoichiometric burn. This is necessary because turbine blades cannot survive full combustion temperature; the fuel-rich preburner deliberately makes cool, high-pressure gas to spin the turbopump without melting it. In a staged-combustion cycle the still-unburned fuel is then routed to the main chamber and finished there, so no propellant (and no energy) is wasted — which is exactly why staged combustion outperforms an open gas-generator cycle.

Chapter 19 — Full solutions to daggered (†) and odd-numbered exercises

All arithmetic hand-checked. Use $g_0 = 9.81\ \text{m/s}^2$, $R_u = 8.314\ \text{J/(mol·K)}$.

Part A

19.1 † $p^*/p_c = \left(\frac{2}{\gamma+1}\right)^{\gamma/(\gamma-1)} = \left(\frac{2}{2.2}\right)^{1.2/0.2} = (0.9091)^{6} = 0.564$. Since $0.564 > 0.5$, the gas is above half the chamber pressure when it reaches the throat (it has dropped to about 56%). Throat temperature ratio for reference: $T^*/T_c = 2/2.2 = 0.909$.

19.3 † $a = \sqrt{\gamma R T} = \sqrt{1.20 \times 520 \times 2000} = \sqrt{1{,}248{,}000} = 1{,}117\ \text{m/s}$.

19.5 † $I_{sp} = v_e/g_0 = 4200/9.81 = 428\ \text{s}$. This is closest to LOX/LH2 (hydrogen, ~360–460 s), toward its lower end — a good but not top-tier hydrogen engine, or a very strong methane one.

Part B

19.7 † Isentropic relation: $\frac{T_c}{T} = 1 + \frac{\gamma-1}{2}M^2 = 1 + 0.1(2)^2 = 1.4$. So $T = T_c/1.4 = 3600/1.4 = 2{,}571\ \text{K}$. The gas has cooled by ~1000 K by the time it reaches Mach 2, and that thermal energy became kinetic energy.

19.9 † Once the throat is sonic ($M=1$), the gas leaves the throat at the local speed of sound. Pressure disturbances (which is how "there is now lower pressure downstream, send more gas" would be communicated) travel at the speed of sound relative to the gas. Since the gas is already moving downstream at exactly that speed, a disturbance from downstream cannot propagate back upstream through the throat to the chamber. The chamber therefore never "hears" that the back pressure dropped, and keeps delivering the same mass flow, set only by $p_c$, $A_t$, and $T_c$. Physically: the throat is passing the maximum mass flux the gas can carry at those chamber conditions.

Part C

19.10 † $R = 8.314/0.018 = 461.9\ \text{J/(kg·K)}$. Front factor $\frac{2\gamma}{\gamma-1} = 12$. Bracket: $(0.01)^{(0.2/1.2)} = (0.01)^{0.1667} = 0.464$, so $1 - 0.464 = 0.536$. Then $v_e = \sqrt{12 \times 461.9 \times 3500 \times 0.536} = \sqrt{1.040\times10^{7}} = 3{,}224\ \text{m/s}$. $I_{sp} = 3224/9.81 = 329\ \text{s}$.

19.11 $\sqrt{T_c/\mathcal{M}}$: for A, $\sqrt{3600/18} = \sqrt{200} = 14.14$; for B, $\sqrt{3200/10} = \sqrt{320} = 17.89$. Ratio $17.89/14.14 = 1.265$, so B has ~26% higher exhaust velocity — even though B is cooler (3200 K vs 3600 K). Lesson: "hotter is always better" is false; the lighter exhaust (molar mass 10 vs 18) wins because $\mathcal{M}$ sits in the denominator under the square root.

19.12 † exit_velocity(1.20, 3600, 0.013, 1e4, 1e7): $R = 8.314/0.013 = 639.5$; $\frac{2\gamma}{\gamma-1} = 12$; bracket $= 1 - (0.001)^{0.1667} = 1 - 0.316 = 0.684$; $v_e = \sqrt{12 \times 639.5 \times 3600 \times 0.684} = \sqrt{1.889\times10^{7}} = 4{,}346\ \text{m/s}$. # Expected output: 4346.

Part D

19.13 † A nozzle is perfectly expanded when $p_e = p_a$, so this one is perfectly expanded at ambient pressure $p_a = 0.05\ \text{bar}$ — about 20 km altitude, where the atmosphere is roughly 5 kPa. At sea level $p_a = 1.0\ \text{bar} > p_e = 0.05\ \text{bar}$, so $p_e < p_a$: it is over-expanded (and, this severely, would separate). It belongs on an upper stage (or a high-altitude stage), never a sea-level first stage.

19.15 † expansion_ratio(1.20, 0.5 bar, 200 bar): $p_c/p_e = 400$. Numerator $= (2/2.2)^{1/0.2}(400)^{1/1.2} = (0.9091)^{5}(400)^{0.833} = 0.621 \times 147.4 = 91.5$. Denominator $= \sqrt{(2.2/0.2)[1 - (1/400)^{0.1667}]} = \sqrt{11 \times (1 - 0.368)} = \sqrt{6.95} = 2.636$. So $\epsilon = 91.5/2.636 = \mathbf{34.7}$. (Cross-check: $p_c/p_e=400$ gives $M_e \approx 4.14$, and the area–Mach relation gives $\epsilon \approx 34.7$ — agrees.)

Part E

19.16 † $p_c/p_e = 90/0.06 = 1500$. Exit Mach: $(1 + 0.1 M_e^2)^{6} = 1500 \Rightarrow 1 + 0.1M_e^2 = 1500^{1/6} = 3.383 \Rightarrow M_e^2 = 23.83 \Rightarrow M_e = 4.88$. Expansion ratio (area-ratio relation, $\gamma=1.2$): numerator $= (0.9091)^{5}(1500)^{0.833} = 0.621 \times 443 = 275$; denominator $= \sqrt{11[1 - (1/1500)^{0.1667}]} = \sqrt{11 \times 0.704} = 2.78$; $\epsilon = 275/2.78 = \mathbf{99}$. A $\epsilon \approx 99$ bell — large, appropriate for a vacuum stage; the exit is nearly 100× the throat area, a deep skirt.

19.17 (Project — no single answer.) Rubric: correct isentropic set-up; a defensible propellant with the chapter's $T_c$, $\mathcal{M}$, $\gamma$; $p_e$ chosen for the firing altitude (near-vacuum → low $p_e$, sea level → $\sim$ambient); $v_e$ and $I_{sp}$ computed with units and a sanity check against the real-engine ranges (LH2 ~450 s, CH4 ~370–380 s, RP-1 ~330–350 s). Record in the MDR.

19.18 † Choked-flow, solved for $A_t$: bracket $b = (2/2.2)^{5.5} = 0.592$; $\sqrt{\gamma/(R T_c)} = \sqrt{1.2/(500 \times 3400)} = \sqrt{7.06\times10^{-7}} = 8.40\times10^{-4}$. Denominator $= p_c \cdot 8.40\times10^{-4} \cdot 0.592 = 8\times10^{6} \times 4.97\times10^{-4} = 3{,}979$. $A_t = 50/3979 = 0.01257\ \text{m}^2$. Throat diameter $d_t = 2\sqrt{A_t/\pi} = 2\sqrt{0.00400} = 0.126\ \text{m}$ (12.6 cm).

Part F

19.19 † The error is "always longer is better." A longer nozzle means a larger $\epsilon$ and a lower $p_e$. In vacuum that helps (more $v_e$), but at sea level, once $p_e$ falls below ambient the nozzle is over-expanded; push $\epsilon$ too far and the flow separates from the wall, causing large, unsteady, asymmetric side loads that can destroy the engine — and the extra bell is dead mass regardless. The mistake is worst on the first stage, which fires at sea level (where separation bites) and must carry the bell's mass through liftoff and max dynamic pressure. First-stage bells are deliberately kept short enough to stay attached at $p_a \approx 1\ \text{bar}$.

19.21 † From the area–velocity relation $\frac{dA}{A} = (M^2-1)\frac{dV}{V}$: a converging duct ($dA<0$) accelerates subsonic flow only until $M=1$, which it can reach at most at its minimum-area exit. To go beyond $M=1$ the gas must keep accelerating, which for $M>1$ requires $dA>0$ — a diverging section. A purely converging nozzle has no diverging section, so it caps at $M=1$. Raising the chamber pressure only increases the (choked) mass flow; it cannot make the exit supersonic without a diverging bell.

Part G

19.22 † $F = \dot m v_e + (p_e - p_a)A_e$. Momentum term $\dot m v_e = 300 \times 3200 = 960{,}000\ \text{N}$. Sea level: $(p_e - p_a)A_e = (0.5 - 1.013)\times10^{5} \times 1.2 = -6.16\times10^{4}\ \text{N}$, so $F_{\text{SL}} = 960{,}000 - 61{,}560 = 898{,}440\ \text{N} \approx \mathbf{898\ kN}$. Vacuum: $(0.5\times10^{5})(1.2) = 60{,}000\ \text{N}$, so $F_{\text{vac}} = 960{,}000 + 60{,}000 = 1{,}020{,}000\ \text{N} \approx \mathbf{1{,}020\ kN}$. Thrust rises by $122\ \text{kN}$ from sea level to vacuum; the pressure term $(p_e-p_a)A_e$ is responsible — it swings by $p_a A_e = 1.013\times10^{5} \times 1.2 = 121{,}560\ \text{N}$ as $p_a$ goes to zero.

19.23 $v_e \propto \sqrt{T_c/\mathcal{M}}$. Kerosene's higher flame temperature $T_c$ raises the numerator only under a square root, while its much heavier exhaust (CO2, CO → $\mathcal{M} \approx 22$) raises the denominator; the molar-mass penalty outweighs the temperature gain, so $\sqrt{T_c/\mathcal{M}}$ — and hence $I_{sp}$ — is lower than hydrogen's ($\mathcal{M} \approx 13$). Molar mass dominates.

19.24 † $v_{e,\text{H2}} = 450 \times 9.81 = 4{,}415\ \text{m/s}$; $\Delta v_{\text{H2}} = 4415 \ln 4 = 4415 \times 1.386 = 6{,}120\ \text{m/s} = 6.12\ \text{km/s}$. $v_{e,\text{RP}} = 340 \times 9.81 = 3{,}335\ \text{m/s}$; $\Delta v_{\text{RP}} = 3335 \times 1.386 = 4{,}624\ \text{m/s} = 4.62\ \text{km/s}$. Difference $= \mathbf{1.50\ km/s}$ from the identical mass ratio. Tie to the tyranny of the rocket equation: because $v_e$ multiplies $\ln(m_0/m_f)$ directly, a higher-$I_{sp}$ propellant buys delta-v for free relative to a fixed mass ratio — the cheapest way to gain delta-v is a better exhaust, which is exactly why hydrogen's molar-mass advantage matters.

19.25 Two reasons a first stage may prefer RP-1 or methane despite LH2's higher $I_{sp}$: (1) Density / mass ratio. LH2 is ~70 kg/m³ vs RP-1 ~810; carrying the same propellant mass needs a ~10× larger tank → more structural mass → a worse mass ratio, and first stages are acutely mass-ratio sensitive (Chapter 3's tyranny). (2) Operability and thrust density. LH2 demands 20 K cryogenics, insulation, boil-off management, and a physically larger (higher-drag) vehicle; dense RP-1/methane give higher thrust per unit engine size, are storable/cheaper, and (methane) support reuse. The first stage gains more from low tank mass and thrust density than from specific impulse.

19.26 † $c = F/\dot m$. Sea level: $c = 898{,}440/300 = 2{,}995\ \text{m/s}$, which is less than $v_e = 3{,}200\ \text{m/s}$ — the nozzle is over-expanded at sea level, so the negative pressure term makes the effective exhaust velocity below the gas velocity. Vacuum: $c = 1{,}020{,}000/300 = 3{,}400\ \text{m/s}$, greater than $v_e$ — under-expanded in vacuum, so the positive pressure term boosts $c$ above the gas velocity. They would be equal only if the nozzle were perfectly expanded ($p_e = p_a$).

Chapter 20 — Electric Propulsion (solutions to † and odd-numbered exercises)

Use $g_0 = 9.81\ \text{m/s}^2$, $\mu_\oplus = 3.986\times10^5\ \text{km}^3/\text{s}^2$, xenon $q = 1.602\times10^{-19}\ \text{C}$, $m = 2.18\times10^{-25}\ \text{kg}$.

20.1 $v_e = I_{sp}\,g_0 = 3{,}000 \times 9.81 = 29{,}430\ \text{m/s} \approx 29.4\ \text{km/s}$. Against a chemical engine's $3.3\ \text{km/s}$, that is $29.4/3.3 \approx 8.9$, so about 9 times larger.

20.3 $F = \dfrac{2\eta P}{v_e} = \dfrac{2 \times 0.60 \times 5{,}000}{20{,}000} = \dfrac{6{,}000}{20{,}000} = 0.30\ \text{N} = \mathbf{300\ \text{mN}}$.

20.5 $v_e = 2{,}000 \times 9.81 = 19{,}620\ \text{m/s}$, so $P/F = v_e/(2\eta) = 19{,}620/(2 \times 0.50) = \mathbf{19{,}620\ \text{W/N} \approx 19.6\ \text{kW/N}}$. It means the thruster must draw about $19.6\ \text{kW}$ of electrical power for every newton of thrust — the quantitative face of "low thrust": useful (chemical-scale) thrust would demand impossibly large power.

20.7 (a) $F = 2\eta P/v_e$: with $P,\eta$ fixed, $F \propto 1/v_e$, so doubling $v_e$ halves the thrust. (b) $\dot m = F/v_e$: the numerator halves and the denominator doubles, so $\dot m$ falls to one quarter. (Physically: pushing each ion twice as fast, at fixed power, means you can push only a quarter as many per second — kinetic energy scales as $v_e^2$.)

20.9 Ion: $v_e = 3{,}300 \times 9.81 = 32{,}373\ \text{m/s}$, $F = 2(0.5)(5{,}000)/32{,}373 = 0.154\ \text{N} = 154\ \text{mN}$. Hall: $v_e = 1{,}700 \times 9.81 = 16{,}677\ \text{m/s}$, $F = 5{,}000/16{,}677 = 0.300\ \text{N} = 300\ \text{mN}$. The Hall thruster makes about $300/154 \approx 1.9\times$ more thrust. But the ion engine uses less propellant for a given delta-v, because its higher $I_{sp}$ (larger $v_e$) makes the mass ratio $e^{\Delta v/v_e}$ smaller.

20.10 ep_thrust(2300, 3100, 0.61): $v_e = 3{,}100 \times 9.80665 = 30{,}401\ \text{m/s}$; $F = 2(0.61)(2{,}300)/30{,}401 = 0.0923\ \text{N}$. # Expected output: 92.3 (mN).

20.11 $\dot m = F/(I_{sp}g_0) = 0.28/(1{,}800 \times 9.80665) = 0.28/17{,}652 = 1.586\times10^{-5}\ \text{kg/s}$. $t = 356/1.586\times10^{-5} = 2.244\times10^{7}\ \text{s} = 2.244\times10^{7}/86{,}400 \approx \mathbf{260\ \text{days}}$. # Expected output: 260.

20.12 power_to_thrust(3100, 0.61) $= I_{sp}g_0/(2\eta) = 3{,}100 \times 9.80665/(2 \times 0.61) = 30{,}401/1.22 = \mathbf{24{,}919\ \text{W/N}}$ ($\approx 24.9\ \text{kW/N}$). # Expected output: 24919.

20.13 The flaw conflates efficiency with thrust. Efficiency measures how much electrical power becomes beam kinetic energy; it says nothing about the force produced. Thrust is $F = 2\eta P/v_e$, and because an ion engine's $v_e$ is enormous, its thrust is only milli-newtons even at 60% efficiency. To launch you need thrust to exceed weight ($T/W > 1$); a chemical engine, though it "wastes" much of its energy as heat, produces meganewtons of thrust by flowing hundreds of kg/s of propellant. Electric propulsion's $T/W \ll 1$ makes launch physically impossible regardless of efficiency.

20.15 Electric ($I_{sp}=2{,}000$, $v_e = 19{,}613\ \text{m/s}$): $m_p = 800\,(e^{4000/19613} - 1) = 800\,(e^{0.2040} - 1) = 800\,(0.2262) = \mathbf{181\ \text{kg}}$. Chemical ($I_{sp}=320$, $v_e = 3{,}138\ \text{m/s}$): $m_p = 800\,(e^{4000/3138} - 1) = 800\,(e^{1.275} - 1) = 800\,(2.578) = \mathbf{2{,}062\ \text{kg}}$. Ratio $2{,}062/181 \approx \mathbf{11.4\times}$ less propellant for the electric option.

20.17 Dry mass 500 kg, $\Delta v = 8{,}000\ \text{m/s}$. - $I_{sp}=2{,}000$ ($v_e=19{,}613$): $m_p = 500\,(e^{8000/19613}-1) = 500\,(e^{0.4079}-1) = 500\,(0.5036) = \mathbf{252\ \text{kg}}$. - $I_{sp}=4{,}000$ ($v_e=39{,}227$): $m_p = 500\,(e^{8000/39227}-1) = 500\,(e^{0.2039}-1) = 500\,(0.2262) = \mathbf{113\ \text{kg}}$. At $P=5\ \text{kW}$, $\eta=0.6$: $F = 2(0.6)(5000)/v_e$, so $F_{2000} = 6000/19613 = 0.306\ \text{N}$ and $F_{4000} = 6000/39227 = 0.153\ \text{N}$. Burn times $t = m_p v_e/F$: $t_{2000} = 252\times19613/0.306 \approx 1.62\times10^7\ \text{s} \approx 187\ \text{days}$; $t_{4000} = 113\times39227/0.153 \approx 2.90\times10^7\ \text{s} \approx 335\ \text{days}$. **Trade:** the higher $I_{sp}$ saves ~139 kg of xenon (252→113) but nearly doubles the transfer time (187→335 days) and halves the thrust. Choose the higher $I_{sp}$ if propellant/launch mass is the binding constraint; the lower one if schedule (or radiation dose, or boiloff) is.

20.19 To hold or lower a $500\ \text{kg}$ lander against the Moon's gravity, thrust must at least equal its lunar weight: $F \ge m\,g_\text{Moon} = 500 \times 1.62 = 810\ \text{N}$. An ion engine provides $\sim 0.1\ \text{N}$, so $T/W = 0.1/810 \approx 1.2\times10^{-4} \ll 1$: the thruster cannot support the lander at all — it would simply fall. Landing (like launch) needs high thrust, which is chemical territory; electric propulsion can deliver a lander to lunar orbit but cannot perform the descent.

20.21 Chemical ($v_e = 3.14\ \text{km/s}$): $m_0/m_f = e^{6000/3138} = e^{1.912} = 6.77$, propellant fraction $1 - 1/6.77 = \mathbf{85\%}$. Electric ($v_e = 29.4\ \text{km/s}$): $m_0/m_f = e^{6000/29420} = e^{0.204} = 1.23$, propellant fraction $1 - 1/1.23 = \mathbf{18\%}$. The tyranny of the rocket equation lives in the exponent $\Delta v/v_e$: the chemical engine sits high on the exponential curve (85% propellant), while the electric engine's ~9× larger $v_e$ shrinks the exponent and drops it to the gentle part of the curve (18%). Raising $v_e$ is how you escape the tyranny.

20.23 $v_1 = \sqrt{398{,}600/6{,}778} = 7.669\ \text{km/s}$; $v_2 = \sqrt{398{,}600/10{,}000} = \sqrt{39.86} = 6.313\ \text{km/s}$. Spiral delta-v $= v_1 - v_2 = 7.669 - 6.313 = \mathbf{1.355\ \text{km/s}}$. This is larger than the equivalent Hohmann transfer would be, because the low-thrust spiral thrusts continuously — including at points where an impulsive burn would spend its delta-v more efficiently — so it is not delta-v-optimal. (The two-burn Hohmann is the delta-v minimum between coplanar circular orbits.)

20.25 $\Delta v = 50 \times 15 = 750\ \text{m/s}$, satellite $3{,}000\ \text{kg}$. Electric ($I_{sp}=1{,}500$, $v_e = 14{,}710\ \text{m/s}$): $m_p = 3{,}000\,(e^{750/14710} - 1) = 3{,}000\,(e^{0.0510} - 1) = 3{,}000\,(0.0523) = \mathbf{157\ \text{kg}}$. Chemical ($I_{sp}=300$, $v_e = 2{,}942\ \text{m/s}$): $m_p = 3{,}000\,(e^{750/2942} - 1) = 3{,}000\,(e^{0.2549} - 1) = 3{,}000\,(0.2904) = \mathbf{871\ \text{kg}}$. Electric station-keeping saves $871 - 157 = \mathbf{714\ \text{kg}}$ over the 15-year life — mass that becomes payload or years of extra propellant. This propellant saving is exactly why GEO station-keeping was electric propulsion's first operational application.

Chapter 21

Chapter 21 — Solutions to selected exercises (starred † and odd-numbered)

Constants used: $g_0 = 9.81\ \text{m/s}^2$, $c = 2.998\times10^{8}\ \text{m/s}$, $R_u = 8.314\ \text{J/(mol·K)}$, $S = 1361\ \text{W/m}^2$, $g_{\odot} = 5.9\times10^{-3}\ \text{m/s}^2$.

21.1 † Hot hydrogen, $\gamma=1.4$ (so $2\gamma/(\gamma-1)=7$), $T_c=2500$ K, $\mathcal{M}=0.002$ kg/mol: $v_e = \sqrt{7 \times 8.314\times2500/0.002} = \sqrt{7\times1.039\times10^{7}} = \sqrt{7.27\times10^{7}} \approx 8{,}530\ \text{m/s}$; $I_{sp}=8530/9.81 \approx 869\ \text{s}$. This is an upper bound: the formula drops the pressure-ratio term (which is $\le 1$) and assumes complete, uniform heating, so a real engine (finite expansion, losses) does worse — NERVA demonstrated ~825 s.

21.3 † Reflector, $A=400\ \text{m}^2$ at 1 AU: $F = 2SA/c = 2\times1361\times400/2.998\times10^{8} = 1.089\times10^{6}/2.998\times10^{8} \approx 3.63\times10^{-3}\ \text{N} = 3.63\ \text{mN}$. (About the weight of a grain of rice — tiny, but continuous and propellantless.)

21.5 † Ratio $= 8\times10^{13}/1.3\times10^{7} \approx 6.2\times10^{6}$ — about 6 million times more energy per kilogram. It does not give a 6-million-fold $I_{sp}$ because specific impulse depends on $v_e \propto \sqrt{T_c/\mathcal{M}}$ — the temperature and molecular weight of the exhaust — not directly on the fuel's energy density. The reactor's temperature is capped by the melting point of its materials (~2,500–2,700 K), so almost all of that vast energy reserve cannot be turned into exhaust speed; it only means the reactor can heat a great deal of propellant, not that each kilogram leaves faster. The $I_{sp}$ gain over chemical is only ~2×, set by hydrogen's low molar mass.

21.7 † A photon of energy $E$ has momentum $p=E/c$. A beam of power $P$ delivers energy $P$ per second, hence momentum $P/c$ per second onto an absorbing surface — and momentum per second is force, so $F_{\text{abs}} = P/c$. If the surface reflects the light straight back, each photon arrives with momentum $+p$ and leaves with $-p$, a change of $2p$; summed over the beam the surface receives momentum at rate $2P/c$, so $F_{\text{refl}} = 2P/c$. The factor of 2 is the reversal of the photons' momentum. $\blacksquare$

21.9 † $\beta = a_c/g_{\odot} = 0.5\times10^{-3}/5.9\times10^{-3} \approx 0.085$. Since $\beta<1$, the sail cannot hover against solar gravity — the Sun's pull exceeds the light's push, so the sail cannot simply hold station or fly straight outward against gravity; it can only spiral out slowly. $\beta$ is the distance-independent ratio of light-push to solar gravity: $\beta=1$ is the break-even sail, and near-term sails sit well below it.

21.11 † ideal_ve(1.4, 2700, 0.002) $= \sqrt{7 \times 8.314\times2700/0.002} = \sqrt{7.857\times10^{7}} \approx 8{,}864$. # Expected output: 8864 (m/s).

21.13 † thrust_from_power(1e6, 5000, 0.6): $v_e = 5000\times9.80665 = 49{,}033\ \text{m/s}$; $F = 2\times0.6\times10^{6}/49{,}033 = 1.2\times10^{6}/49{,}033 \approx 24.5\ \text{N}$. # Expected output: 24.5 (N).

21.14 † The article confuses the solar wind (a stream of charged particles) with sunlight (photons). A solar sail is pushed by radiation pressure — reflected photons — not by particles. At 1 AU the momentum flux carried by sunlight is roughly a thousand times greater than that in the solar wind, so the sail is a light-mill, not a windmill. (Devices that ride the actual solar wind — electric sails, magnetic sails — exist as separate concepts with different physics.)

21.15 Both clauses are wrong. (a) A NERVA-class reactor runs cooler than a chemical flame (~2,500–2,700 K vs ~3,300 K), not hotter — the solid fuel elements would melt otherwise. (b) The higher $I_{sp}$ comes from the low molecular weight of pure hydrogen exhaust ($\mathcal{M}=2$ vs ~10–18 for chemical exhaust), via $v_e \propto \sqrt{T_c/\mathcal{M}}$ — despite the lower temperature, not because of a higher one.

21.16 † Propellant fraction $= 1 - e^{-\Delta v/v_e}$, $\Delta v = 6000\ \text{m/s}$. (a) Chemical, $v_e = 450\times9.81 = 4{,}414\ \text{m/s}$: $1 - e^{-6000/4414} = 1 - e^{-1.359} = 1 - 0.257 = 0.743$, i.e. 74.3% propellant. (b) NTP, $v_e = 850\times9.81 = 8{,}339\ \text{m/s}$: $1 - e^{-6000/8339} = 1 - e^{-0.719} = 1 - 0.487 = 0.513$, i.e. 51.3% propellant. Consequence: for a heavy crewed vehicle, dropping from ~74% to ~51% propellant is tens of tonnes of propellant not launched (or the same propellant spent on a faster, lower-radiation-dose trajectory). This is the entire physical case for nuclear thermal propulsion. (Caveat: hydrogen's bulk and the reactor's shielding mass trim the raw advantage.)

21.17 (Mission project — guidance.) A complete answer names the actual propulsion (Track A: chemical apogee motor + solar-electric station-keeping; Track B: chemical; Track C: chemical, optionally solar-electric cruise; Track D: solar-electric) and states one requirement change that would bring an exotic engine into the trade — e.g., "if Track C became a crewed Mars mission, nuclear thermal enters the trade because transit time becomes a crew-radiation driver," or "if Track D were relocated to a Jupiter Trojan, nuclear-electric enters because solar arrays fail that far out." The graded point is the reasoning — matching the engine to the mission's hardest constraint — not the specific engine.

21.18 † NEP cargo tug, $P=2\ \text{MW}$, $\eta=0.55$, $I_{sp}=6000\ \text{s}$, $\alpha=20\ \text{kg/kW}$. (a) $v_e = 6000\times9.81 = 58{,}860\ \text{m/s}$; $F = 2\times0.55\times2\times10^{6}/58{,}860 = 2.2\times10^{6}/58{,}860 \approx 37.4\ \text{N}$. (b) Power-system mass $= \alpha P = 20\ \text{kg/kW} \times 2{,}000\ \text{kW} = 40{,}000\ \text{kg} = 40\ \text{t}$. (c) $a = F/m = 37.4/60{,}000 = 6.2\times10^{-4}\ \text{m/s}^2 = 0.62\ \text{mm/s}^2$. (d) The power system dominates: 40 t of reactor/converter/radiator on a 60 t dry vehicle — the payload (and everything else) is a minority of the mass. A nuclear-electric vehicle is essentially a power plant with cargo attached.

21.19 † Two decisive reasons: (1) fallout — launching from the ground means detonating nuclear devices in the atmosphere, scattering radioactive debris (an environmental/ethical problem); (2) the 1963 Partial Test Ban Treaty, which prohibits nuclear detonations in the atmosphere and space, making the launch itself illegal. Neither is a physics problem — the pusher-plate mechanism works (it was tested with chemical explosives). Orion is the textbook case of "possible but not permitted": the physics says yes, law and ethics say no.

21.21 † $t = 10\ \text{yr} \times 3.156\times10^{7}\ \text{s/yr} = 3.156\times10^{8}\ \text{s}$; $\Delta v = a_c\,t = 0.5\times10^{-3}\times3.156\times10^{8} = 1.58\times10^{5}\ \text{m/s} = 158\ \text{km/s}$. This is an overestimate for an outbound cruise because (i) sunlight — and therefore the acceleration — falls off as $1/r^2$ as the sail recedes from the Sun, and (ii) steering the sail off Sun-facing to shape the trajectory reduces the useful thrust component. Still, even a fraction of 158 km/s dwarfs a good chemical stage (~a few km/s), and it costs no propellant — the point of a sail.

21.22 † $\Delta v = v_e\ln(R) = 8{,}340 \times \ln 2 = 8{,}340\times0.693 \approx 5{,}780\ \text{m/s} = 5.78\ \text{km/s}$. A chemical stage with the *same* mass ratio ($R=2$) but $v_e=4{,}400\ \text{m/s}$ delivers $4{,}400\times0.693 \approx 3{,}050\ \text{m/s} = 3.05\ \text{km/s}$. The NTP stage gives ~1.9× the delta-v from an identical mass ratio, purely because its exhaust velocity is ~1.9× higher — a direct illustration that $v_e$ multiplies the whole rocket equation.

21.23 High $I_{sp}$ is an asset for an asteroid cargo mission: the trip is not time-critical, so the tiny-thrust/long-burn penalty is acceptable, and the propellant savings (from very high $v_e$) are large, letting a small tank do a big-$\Delta v$ job. It is a liability for a fast crewed Mars mission: at fixed power, higher $I_{sp}$ means lower thrust ($F=2\eta P/v_e$), which means a longer transit — and a longer transit means more crew radiation exposure, more consumables, and more risk. Crews want speed (high thrust), which favors nuclear-thermal or chemical over ultra-high-$I_{sp}$ electric.

21.25 † "Infinite $I_{sp}$" is literally defensible: $I_{sp}$ is impulse per unit weight of propellant consumed, and a solar sail consumes no propellant, so the ratio is formally infinite. But it is a meaningless figure of merit here, because $I_{sp}$ measures how efficiently a vehicle uses the reaction mass it carries — and a sail carries none. The sail does not "use propellant efficiently"; it sidesteps the rocket equation entirely, taking its momentum from external sunlight. The right figures of merit for a sail are its characteristic acceleration and lightness number $\beta$, not $I_{sp}$. Saying a sail has "infinite $I_{sp}$" correctly signals "not bound by the mass ratio" but wrongly implies the metric still applies.

Chapter 22

Chapter 22 — Solutions to selected exercises († and odd)

Use $g_0 = 9.81\ \text{m/s}^2$. Structural coefficient $\varepsilon = m_s/(m_s+m_p)$; payload ratio $\pi = (1-\varepsilon R)/[R(1-\varepsilon)]$; stage mass $m_{\text{stage}} = m_{\text{above}}(R-1)/(1-\varepsilon R)$.

22.1 † Ceiling $\Delta v_{\max} = v_e\ln(1/\varepsilon) = 3.2\ \text{km/s}\times\ln(1/0.09) = 3.2\times\ln(11.11) = 3.2\times2.408 = 7.71\ \text{km/s}$. That is below the ~9.4 km/s to orbit — and it assumes zero payload — so no, a single such stage cannot reach orbit with anything aboard.

22.3 † Each stage should provide half: $4\ \text{km/s}$. For identical stages ($v_e$, $\varepsilon$ equal), holding the total delta-v fixes the product of mass ratios; by the AM–GM inequality the payload fraction $\prod\pi_i$ is largest when the mass ratios — hence the delta-vs — are equal.

22.5 † $\dot m = Q/h_{fg} = 50\ \text{W} / 213{,}000\ \text{J/kg} = 2.35\times10^{-4}\ \text{kg/s}$. Times $86{,}400\ \text{s/day}$: $\approx 20.3\ \text{kg/day}$.

22.6 † Even split: $\Delta v_i = 4.5\ \text{km/s}$. $R = e^{4500/3300} = e^{1.364} = 3.91$ (below $1/0.07 = 14.3$, OK). $\pi = (1-0.07\times3.91)/(3.91\times0.93) = 0.7263/3.636 = 0.1997$. Overall payload fraction $\pi^2 = 0.0399$. Lift-off mass $m_0 = 500/0.0399 \approx 12{,}530\ \text{kg} \approx 12.5\ \text{t}$.

22.7 Let stage mass $S = m_s + m_p$, so $\varepsilon = m_s/S$ and $\pi = m_L/m_0$. With $m_0 = m_L + S$ and $m_f = m_L + m_s = m_L + \varepsilon S$: $R = m_0/m_f = (m_L+S)/(m_L+\varepsilon S)$. Divide top and bottom by $m_0$; since $S/m_0 = (m_0-m_L)/m_0 = 1-\pi$, we get $R = (\pi + (1-\pi))/(\pi + \varepsilon(1-\pi)) = 1/[\varepsilon + \pi(1-\varepsilon)]$. $\blacksquare$

22.8 † $N=2$: each $\Delta v = 5\ \text{km/s}$, $R = e^{5/3.5} = e^{1.4286} = 4.173$, $\pi = (1-0.08\times4.173)/(4.173\times0.92) = 0.6662/3.839 = 0.1735$, overall $\pi^2 = 0.0301$ (3.01%). $N=3$: each $3.333\ \text{km/s}$, $R = e^{0.9524} = 2.592$, $\pi = (1-0.2074)/(2.592\times0.92) = 0.7926/2.385 = 0.3324$, overall $\pi^3 = 0.0367$ (3.67%). The third stage improves payload fraction by only ~0.66 percentage points, while adding a whole stage's engines, structure, and a separation event that must work perfectly — cost and risk the numbers alone don't show.

22.9 † In $R_i = (v_{e,i}-\mu)/(v_{e,i}\varepsilon_i) = (1 - \mu/v_{e,i})/\varepsilon_i$, a larger $v_{e,i}$ makes $\mu/v_{e,i}$ smaller, so the numerator $1-\mu/v_{e,i}$ is larger and $R_i$ is larger. Since $\Delta v_i = v_{e,i}\ln R_i$, a larger mass ratio (and larger $v_{e,i}$) means more delta-v: the more efficient stage should do more of the work, because delta-v spent through a high-$v_e$ engine costs less mass.

22.10 † payload_fraction(9000, 3300, 0.07, 2): $R = e^{(9000/2)/3300} = e^{1.364} = 3.910 < 1/0.07 = 14.3$; $\pi = (1-0.07\times3.910)/(3.910\times0.93) = 0.1997$; return $\pi^2 = $ 0.0399.

22.11 stage_mass(1000, 4500, 3400, 0.08): $R = e^{4500/3400} = e^{1.3235} = 3.757$; $m = 1000\times(3.757-1)/(1-0.08\times3.757) = 2756.6/0.6995 = $ 3941 kg.

22.12 † can_hover(338000, 25600): weight $= 25600\times9.81 = 251{,}136\ \text{N}$. Minimum thrust $338{,}000\ \text{N} > 251{,}136\ \text{N}$, so thrust_min <= weight is false; return False — it cannot throttle low enough to hover, hence the hoverslam.

22.13 † More engines multiply the chances that some engine fails; they only yield engine-out capability if the vehicle can detect a failure fast, isolate it so it doesn't damage neighbors, and re-plan the burn. The N1 had 30 engines but an engine-management system that could not do this reliably, so a single failure could cascade — all four flights failed. Engine count is raw material for graceful degradation, not graceful degradation itself.

22.15 † Even split: $\Delta v_i = 4.7\ \text{km/s}$, $R = e^{4700/3300} = e^{1.4242} = 4.155$ (below $1/0.08 = 12.5$, OK). Top-down. Upper stage (carries payload $2000\ \text{kg}$): $m_{\text{stage2}} = 2000\times(4.155-1)/(1-0.08\times4.155) = 2000\times3.155/0.6676 = 9452\ \text{kg}$; gross $= 11{,}452\ \text{kg}$. Lower stage (carries that stack): $m_{\text{stage1}} = 11{,}452\times3.155/0.6676 = 54{,}120\ \text{kg}$. Lift-off mass $= 11{,}452 + 54{,}120 = 65{,}570\ \text{kg} \approx 65.6\ \text{t}$ (payload fraction $\approx 3.0\%$).

22.17 † The ceiling is $1/\varepsilon$: at $\varepsilon = 0.08$ it is $12.5$; at $\varepsilon = 0.10$ it drops to $10.0$. The reusable stage can achieve a smaller maximum mass ratio, so for the same delta-v its payload ratio $\pi = (1-\varepsilon R)/[R(1-\varepsilon)]$ falls (the $\varepsilon R$ term grows). Concretely, at $R = 4.155$: $\pi$ drops from $(1-0.08\times4.155)/(4.155\times0.92) = 0.1749$ to $(1-0.10\times4.155)/(4.155\times0.90) = 0.1563$ — roughly an 11% cut in that stage's payload ratio. The recovery hardware and reserved landing propellant behave like added dead mass, and the rocket equation taxes it directly.

22.19 † Three reasons parachutes are worse: (1) mass — parachutes large enough for a heavy booster are themselves heavy, and they cannot achieve a precise, upright landing; (2) precision — parachute descent lands imprecisely over a wide area, not on a pad or droneship; (3) damage — an uncontrolled or ocean/saltwater landing corrodes and shocks the engines, defeating the point of recovering flightworthy hardware. Propulsive landing needs one thing parachutes don't: an engine that can restart and throttle to null the velocity on command.

22.21 † $\Delta v = v_e\ln(m_0/m_f) = 2.9\ \text{km/s}\times\ln(500/150) = 2.9\times\ln(3.333) = 2.9\times1.204 = 3.49\ \text{km/s}$. No — the number of stages does not appear; this is a single stage's delta-v, a property of its own $v_e$ and mass ratio alone. Staging matters only when you sum stages.

22.23 † A liquid bipropellant engine can be shut down, restarted, and deeply throttled by controlling propellant flow, which is exactly what the boostback/entry/landing burns and the timed hoverslam require. A solid motor burns its cast grain to completion once ignited — it cannot be throttled, shut off, or relit — so it can perform none of the controlled burns propulsive landing needs.

22.25 † During the coast the stage faces: (1) ullage — the propellant floats away from the outlet in weightlessness, threatening the restart; the fix is an ullage (settling) burn just before ignition so the pumps draw liquid. (2) Boiloff — heat leaking in vaporizes cryogenic propellant, steadily losing mass over six hours; the fix is insulation (multi-layer insulation), shading the tanks, and, for long coasts, active cooling.

Chapter 23 — Structures and Materials (solutions to † and odd-numbered exercises)

Use $g_0 = 9.81\ \text{m/s}^2$ unless a problem specifies $g_0 = 9.80665$.

23.1 † Axial force $F = m\,n\,g_0 = 1{,}500\times5\times9.81 = 73{,}575\ \text{N} \approx 73.6\ \text{kN}$. As an equivalent hanging mass, $m\,n = 1{,}500\times5 = 7{,}500\ \text{kg} = 7.5\ \text{t}$ (equivalently $F/g_0 = 73{,}575/9.81 = 7{,}500\ \text{kg}$). Sanity: a 1.5-tonne spacecraft at 5 g presses on its adapter with the weight of a 7.5-tonne object — modest structurally, but real.

23.3 † $\sigma = F/A = 8{,}000\ \text{N} / (20\times10^{-6}\ \text{m}^2) = 8{,}000/2\times10^{-5} = 4\times10^{8}\ \text{Pa} = 400\ \text{MPa}$. An aluminum rod (yield ~350 MPa) is not safe: 400 MPa exceeds the yield strength, so the rod would deform permanently even before any factor of safety is applied. (Note $20\ \text{mm}^2 = 20\times10^{-6}\ \text{m}^2$ — the unit conversion is where errors hide.)

23.5 † Hooke's law: $\sigma = E\varepsilon = 72\times10^{9}\ \text{Pa}\times6\times10^{-4} = 4.32\times10^{7}\ \text{Pa} = 43.2\ \text{MPa}$. Well below aluminum's yield (~350 MPa), so this strain is elastic — the material springs back.

23.7 † Specific strength (UTS/$\rho$): titanium $= 950/4.43 = 214$; aluminum $= 470/2.7 = 174$. Titanium is higher by $(214-174)/174 = 23\%$. Yet aluminum is more common because (i) by specific strength the gap is only ~23% (raw strength overstated the difference), and (ii) aluminum is far cheaper, easier to machine and weld, and has vast heritage. Titanium is reserved for parts where its properties (heat, compatibility, high specific strength) are genuinely needed and worth the cost.

23.9 † A buckling-limited wall's mass scales as $\rho/E$ (inverse specific stiffness). Specific stiffness: steel $E/\rho = 193\times10^{9}/7{,}900 = 2.44\times10^{7}$; aluminum $= 70\times10^{9}/2{,}700 = 2.59\times10^{7}\ \text{m}^2/\text{s}^2$. Mass ratio steel/aluminum $= (\rho/E){\text{steel}}/(\rho/E) = (E/\rho)}{\text{Al}}/(E/\rho) = 2.59/2.44 = 1.06$. So a buckling-limited steel wall is only ~6% heavier — }essentially equal, not triple. Steel's 3× density is offset by its ~2.7× stiffness (the wall can be ~3× thinner). This is the hidden foundation under the Starship steel decision (§23.6).

23.11 † hoop_stress(3.5e5, 1.83, 0.004) $= 3.5\times10^{5}\times1.83/0.004 = 640{,}500/0.004 = 1.60\times10^{8}\ \text{Pa} = 160\ \text{MPa}$. Reported in MPa: # Expected output: 160.

23.13 † kg_per_kg_structure(6000, 348): $v_e = 348\times9.80665 = 3{,}413\ \text{m/s}$; mass ratio $R = e^{6000/3413} = e^{1.758} = 5.80$. So 1 kg of structure adds 5.80 kg to the fueled stage (1 kg of itself + 4.80 kg of propellant). # Expected output: 5.8.

23.14 † The student wrote $\sigma = pr\,t$ (multiplying by the wall thickness) instead of $\sigma = pr/t$ (dividing). Correct: $\sigma_{\text{hoop}} = pr/t = 3.4\times10^{5}\times1.83/0.004 = 1.56\times10^{8}\ \text{Pa} = 156\ \text{MPa}$. The tell-tale: their answer, $2{,}489\ \text{Pa} \approx 0.0025\ \text{MPa}$, is absurdly tiny — a real tank wall is stressed to tens or hundreds of MPa. Dividing by a 4 mm wall (not multiplying) is what turns a small pressure into a large wall stress; a factor of $t^2 = (0.004)^2$ separates the wrong answer from the right one.

23.15 (odd) A margin of safety of $+2.5$ means the bracket is $3.5\times$ (i.e. $1+2.5$) its required factored load — grossly overbuilt for a mass-limited spacecraft, dragging mass the rocket equation punishes. The chief wants it thinned to a small positive MS (near zero) at the required factor of safety. The distinction the lead is drawing: a factor of safety is a requirement you must meet (strength ÷ limit load $\ge$ ~1.4); a margin of safety is how much you exceed that requirement (MS $= \sigma_{\text{allow}}/ (\text{FoS}\cdot\sigma_{\text{limit}}) - 1 \ge 0$). A large MS is not a virtue on a rocket — it is wasted mass. The goal is FoS met, MS near zero.

23.16 † (a) Minimum ultimate strength $= \text{FoS}\times\sigma_{\text{limit}} = 1.4\times300 = 420\ \text{MPa}$. (b) With 7075-T6 (UTS 570 MPa): $\text{MS} = 570/(1.4\times300) - 1 = 570/420 - 1 = 0.357 = +36\%$. (c) The part is overweight — a +36% margin means it carries far more than required; thin it toward $\text{MS}\approx0$ to save mass (until another limit — buckling, stiffness, minimum-gauge, or handling — takes over). On a rocket, a comfortable margin is an invitation to lighten, not a reason to relax.

23.17 (odd, mission project) No single numeric answer. A complete response provides: a subsystem mass table (structure, power, payload, propulsion-dry, ADCS/avionics, thermal, comms, harness) with first-pass masses sanity-checked against §23.5's representative shares; a dry-mass sum; a 25–30% mass margin at this early stage; the wet mass computed from the delta-v budget (Chapter 3) via the mass ratio $R = e^{\Delta v/v_e}$; and a check that the wet mass fits under the launch vehicle's payload cap (to be confirmed in Chapter 30).

23.18 † (a) From $\sigma_{\text{hoop}} = pr/t$, the minimum wall is $t = pr/\sigma_{\text{allow}} = (5\times10^{5}\times2.25)/(300\times10^{6}) = 1.125\times10^{6}/3\times10^{8} = 3.75\times10^{-3}\ \text{m} = 3.75\ \text{mm}$. (b) Mass per metre of length $= \rho\times(2\pi r\,t) = 2{,}700\times(2\pi\times2.25\times 0.00375) = 2{,}700\times0.0530 = 143\ \text{kg/m}$. (c) If the wall were buckling-limited, its mass would scale as $\rho/E$, and steel and aluminum have nearly identical $E/\rho$ (Ex. 23.9), so a steel wall would be about the same mass (roughly 3× thinner but 3× denser) — not triple. Density alone would dominate only if the wall were purely strength-limited; for the compression/buckling case it does not.

23.19 † You cannot slap a factor of safety of 3 on a rocket "to be safe" because roughly tripling the structural mass is taxed exponentially by the rocket equation (Chapter 3) — a 3×-heavy structure could push the mass ratio out of reach and prevent the vehicle from reaching orbit at all. A building can afford FoS 2–5 because its "launch cost" is zero. Aerospace instead keeps FoS near 1.4 and buys down uncertainty by other means: rigorous analysis, extensive ground testing (static, vibration, acoustic, proof-pressure), tight manufacturing control, and material-allowable databases. It earns the thin margin rather than padding it.

23.21 † With two stages each of mass ratio $R\approx3$: adding 1 kg of payload adds ~$R = 3\ \text{kg}$ to the upper stage's initial mass, and that 3 kg is in turn payload to the booster, multiplied again by $R\approx3$, for ~$3\times3 = 9\ \text{kg}$ (order 10 kg) at lift-off. So one payload kilogram "costs" roughly ten kilograms of rocket. Because the multiplication compounds per stage, a kilogram saved highest in the stack (payload, upper stage) is worth the most at lift-off — which is exactly why structures engineers fight hardest for mass at the top (and why the Shuttle external tank of Case Study 1, dropped near orbital speed, traded ~1:1 with payload).

23.23 (odd) The max-Q dynamic-pressure force is a quasi-static (steady) aerodynamic load — it acts along the axis as drag (a component of the axial load) and, through $q\alpha$, sideways as a lateral load. The $q\alpha$ bending is the lateral quasi-static load. The primary structure — the monocoque tank/body — must carry that bending moment distributed along its length (a slender tube is weak in bending, §23.2), which is why holding $\alpha\approx0$ (Chapter 5) protects it. (Transonic buffeting near max-Q also feeds the vibration environment.)

23.24 † Reusability adds dead structural mass — landing legs, grid fins, reinforced structure, and thermal protection — that must be carried but does no payload work, so it reduces payload via the rocket equation (theme #4): a structural "tax." It is worth paying because the recovered stage's cost is amortized over many flights, dropping the cost per flight by up to 10–100× (theme #5). The mass penalty is a one-time performance cost paid every flight; the cost saving repeats and compounds every flight. For a high-flight-rate vehicle, the economics win decisively — the logic that also drives the Starship steel choice (§23.6).

23.25 (odd) By Newton's second law $F = ma$, the acceleration is $a = F/m$. During a burn the thrust $F$ is roughly constant while the mass $m$ falls as propellant is consumed, so $a$ rises through the burn and peaks near burnout, when $m$ is smallest. The peak axial launch load on the payload, $m_{\text{payload}}\,a$, peaks with it. To keep that load (and the acceleration on crew or delicate payload) within limits, engines often throttle down near the end of the burn — reducing $F$ to cap $a$ — the same "throttle to manage a load" logic seen at max-Q in Chapter 5, now driven by g rather than dynamic pressure.

Chapter 24 — Full solutions to starred (†) and odd-numbered exercises

Constants: $S = 1361\ \text{W/m}^2$, $\sigma = 5.670\times10^{-8}\ \text{W m}^{-2}\text{K}^{-4}$. All temperatures in kelvin; convert to °C by subtracting 273.15. Numbers hand-computed.

Part A

A1 (†). $S(d) = 1361/d^2$. - Venus ($0.723$ AU): $1361/0.723^2 = 1361/0.5227 = 2{,}604\ \text{W/m}^2$. - Saturn ($9.537$ AU): $1361/9.537^2 = 1361/90.95 = 15.0\ \text{W/m}^2$. - Ratio $= 2604/15.0 = 174$ (equivalently $(9.537/0.723)^2 = 13.19^2 = 174$). Venus receives ~174× the sunlight Saturn does — the inner solar system is a fundamentally hotter thermal problem.

A3 (†). One-sided plate: $A_{\text{sun}} = A_{\text{rad}}$, so $T = (\alpha S/\varepsilon\sigma)^{1/4}$. - White ($\alpha=0.20,\varepsilon=0.90$): $T = (0.20\times1361/(0.90\times5.67\times10^{-8}))^{1/4} = (5.334\times10^{9})^{1/4} = 270.3\ \text{K} = -2.9\ ^\circ\text{C}$. - Polished aluminum ($\alpha=0.15,\varepsilon=0.05$): $T = (0.15\times1361/(0.05\times5.67\times10^{-8}))^{1/4} = (7.201\times10^{10})^{1/4} = 518.0\ \text{K} = +245\ ^\circ\text{C}$. - For a sun-facing surface you want cool, use white paint (low $\alpha/\varepsilon = 0.22$). Bare aluminum, despite looking "reflective," has such low $\varepsilon$ it cannot dump what it absorbs and bakes at $+245\ ^\circ\text{C}$. Choose by $\alpha/\varepsilon$, not by appearance.

A5 (†). Heater power $= \varepsilon\sigma A T^4$ at $T=278\ \text{K}$, $A=0.15\ \text{m}^2$; $278^4 = 5.973\times10^{9}$. - With MLI ($\varepsilon^{*}=0.04$): $0.04\times5.67\times10^{-8}\times0.15\times5.973\times10^{9} = 2.0\ \text{W}$. - Bare anodized ($\varepsilon=0.76$): $0.76\times5.67\times10^{-8}\times0.15\times5.973\times10^{9} = 38.6\ \text{W}$. - The blanket cuts the heater by the emissivity ratio $0.76/0.04 = 19\times$. MLI and heaters are partners: insulation is what makes the heater affordable.

A7 (†). Absorbed from Earth $= (\alpha\, q_{\text{alb}} + \varepsilon\, q_{\text{IR}})\,A = (0.20\times400 + 0.88\times237)\times0.5 = (80 + 208.6)\times0.5 = 144.3\ \text{W}$. Equilibrium (one face emitting): $\varepsilon\sigma A T^4 = 144.3$, so $T^4 = 144.3/(0.88\times5.67\times10^{-8}\times0.5) = 5.78\times10^{9}$ and $T = 275.8\ \text{K} = +2.7\ ^\circ\text{C}$. Even with no direct sunlight, Earth's albedo and IR alone hold a nadir panel near $+3\ ^\circ\text{C}$ — Earth fills its field of view and loads it. Lesson: the nadir face is a poor place for a radiator; radiators belong where they see cold space (anti-nadir / anti-sun).

Part B

B1 (†). See code/exercise-solutions.py. t_eq(alpha, eps, a_sun, a_rad, flux, q_int) = ((alpha*flux*a_sun + q_int)/(eps*SIGMA*a_rad))**0.25. Sphere (1,1,1,4) → $278.3\ \text{K} = +5.2\ ^\circ\text{C}$; plate (1,1,1,1) → $393.6\ \text{K} = +120.5\ ^\circ\text{C}$.

B3 (†). Grey sphere $T = (S(d)/4\sigma)^{1/4}$, i.e. $278.3\,d^{-1/2}$:

The $d^{-1/2}$ scaling is gentle (fourth root of an inverse square): a 30-fold distance change moves $T$ by only ~9×. Even so it spans habitable-warm to cryogenic across the solar system.

Part C

C1 (†). Model answer (Track A, GEO comsat). (a) Location: GEO is at ~1 AU, so $S = 1361\ \text{W/m}^2$; near the equinoxes the satellite passes through up to ~72-min eclipses. (b) Cases for a representative radiator ($\alpha=0.2,\varepsilon=0.85$, area $0.6\ \text{m}^2$, one-sided, internal $150\ \text{W}$): hot (full sun on panel + internal, EOL $\alpha\to0.35$): $T=((0.35\times1361\times0.6+150)/(0.85\times5.67\times10^{-8}\times0.6))^{1/4}$. Numerator $= 285.8+150 = 435.8$; denominator $=2.892\times10^{-8}$; $T=(1.507\times10^{10})^{1/4}=350\ \text{K}=+77\ ^\circ\text{C}$. cold (eclipse, internal only): $T=(150/2.892\times10^{-8})^{1/4}=(5.187\times10^{9})^{1/4}=268\ \text{K}=-5\ ^\circ\text{C}$. (c) Demand: hot case exceeds a $+60\ ^\circ\text{C}$ limit → enlarge the radiator or lower EOL $\alpha$ (use OSR); cold case needs a modest heater to hold electronics above 0 °C. Report the heater watts to the power budget. (Accept any internally consistent set with stated assumptions; the method is graded, not the exact numbers.)

C3. Two ways to cut the $25\ \text{W}$ eclipse heater demand without adding heater power: 1. Reduce the heat loss. The loss is $\varepsilon^{*}\sigma A T^4$; adding MLI over the zone (lowering $\varepsilon^{*}$) or shrinking the exposed radiating area cuts it proportionally. Quantitatively: dropping the effective emittance from $0.05$ to $0.03$ cuts the radiative loss by $0.03/0.05 = 40\%$ — a $25\ \text{W}$ heater becomes ~$15\ \text{W}$. 2. Add a louver over the radiator so its effective emittance falls in eclipse (open ~0.7 → shut ~0.1), roughly a $7\times$ reduction in cold-case loss — largely eliminating the heater, at the cost of mass and a moving part. Both work by attacking the loss, not by pushing more power in. (A third option: raise thermal mass so the zone coasts through eclipse — but that fights mass is the enemy.)

Part D

D1 (†). A fan cools by convection — it carries heat away in moving air. In vacuum there is no air (or any fluid) to carry heat, so convection simply does not exist; blowing a (nonexistent) gas accomplishes nothing. The only way heat leaves a spacecraft to its surroundings is radiation. Inside the vehicle you can still move heat by conduction (straps, heat pipes) to a radiator, but the final hand-off to the environment is always radiative. The absent ingredient is a fluid medium.

D3 (†). "Space is cold" is misleading: space is a near-perfect vacuum, an empty sink — it does not actively cool a sunlit surface; only radiation to it does, and a sunlit surface is simultaneously being heated by over $1\ \text{kW/m}^2$ of sun. A black radiator ($\alpha=0.95$) at $300\ \text{K}$ emits $\varepsilon\sigma T^4 \approx 0.85\times459 = 390\ \text{W/m}^2$ but, facing the Sun, absorbs $\alpha S = 0.95\times1361 = 1293\ \text{W/m}^2$. Net flux is $390 - 1293 = -903\ \text{W/m}^2$ — it gains ~900 W for every square meter, the opposite of a radiator. It would heat until $\varepsilon\sigma T^4$ caught the absorbed sun, near $405\ \text{K}$ ($+132\ ^\circ\text{C}$) with no waste heat rejected at all. That is why radiators are kept edge-on to the Sun or on the shaded side and finished in low-$\alpha$ OSR.

D5 (†). Temperature depends on $\alpha/\varepsilon$, not on $\alpha$ alone. Gold foil: $\alpha/\varepsilon = 0.25/0.04 = 6.3$. Black paint: $0.95/0.85 = 1.1$. Gold's ratio is nearly six times higher, so a gold sphere at 1 AU sits at $278.3\times6.3^{1/4} = 441\ \text{K}$ ($+168\ ^\circ\text{C}$) versus a black one at $286\ \text{K}$ ($+13\ ^\circ\text{C}$). Gold reflects far more sunlight, but its emissivity is so tiny ($0.04$) that it can barely radiate away even the little it absorbs, so it bakes. Gold foil's real job is as an infrared reflector/insulator — the low-$\varepsilon$ outer layer of MLI blankets and thermal shields — not as a cool sunlit finish.

Part E

E1 (†). Grey sphere at $2.5\ \text{AU}$: $T = 278.3/\sqrt{2.5} = 278.3/1.581 = 176\ \text{K} \approx -97\ ^\circ\text{C}$. Far below the freezing point of water (273 K / 0 °C) — as expected for the asteroid belt, where sunlight is $1361/2.5^2 = 218\ \text{W/m}^2$, about a sixth of Earth's.

E3 (†). LOX boiloff: $\dot m = Q/h_{fg} = 10/213{,}000 = 4.7\times10^{-5}\ \text{kg/s} = 4.06\ \text{kg/day}$. Over a week: $\approx 28\ \text{kg}$ — more than half of a $50\ \text{kg}$ tank. A week-long loiter is marginally feasible but wasteful; you would either accept the loss, add insulation to cut the leak, or add active cooling (a cryocooler / vapor-cooled shield). This is why long cryogenic coast phases are a serious design driver.

Part F

F1 (†). The error is using the full surface area $4\pi r^2$ as the absorbing area. A sphere intercepts sunlight only on its projected cross-section (its shadow), $A_{\text{sun}} = \pi r^2$; its back half is self-shaded. The correct area ratio is $\pi r^2/4\pi r^2 = 1/4$, giving $T = (S/4\sigma)^{1/4} = 278\ \text{K} = +5\ ^\circ\text{C}$, not $394\ \text{K}$. The mistake inflated the temperature by $116\ ^\circ\text{C}$. Rule: $A_{\text{sun}}$ is the projected area facing the Sun; $A_{\text{rad}}$ is the total area seeing cold space.

F3 (†). The formula's $\varepsilon^{*}=1/[(N+1)(2/\varepsilon-1)] = 0.0012$ is a laboratory ideal for perfect, uncompressed, uninterrupted layers. Flown MLI is stitched, taped, and pierced by struts and wires, and compressed at seams where layers touch and conduct — each a thermal short. Real blankets achieve $\varepsilon^{*}\approx0.01$–$0.05$, i.e. 10–40× leakier. Designing the heaters on $0.0012$ would badly under-size them (predict far less heat loss than real), so the spacecraft could run too cold in eclipse with no heater margin. Design with $\varepsilon^{*}\approx0.03$ (a defensible mid-range value), and confirm by test.

Part G

G1 (†). Re-entry vs. on-orbit heat: - Source: re-entry heat comes from compression of air in the shock layer (kinetic energy of the vehicle converted to gas enthalpy) — see Ch. 7; on-orbit heat comes from absorbed sunlight, albedo, planetary IR, and internal electronics. - Magnitude: re-entry deposits enough to reach thousands of kelvin and must dispose of ~$30\ \text{MJ/kg}$; on-orbit loads are $\sim 10^3\ \text{W/m}^2$ at most and hold surfaces within roughly $\pm 150\ ^\circ\text{C}$. - Timescale: re-entry lasts minutes; on-orbit thermal balance is permanent (the whole mission). - Hence ablative shields / tiles (sacrificial or refractory, for a brief inferno) versus paint, MLI, and radiators (steady management of a lopsided trickle). Same word "heat," opposite engineering.

G3 (†). Delta-v from the rocket equation $\Delta v = v_e\ln(m_0/m_f)$ with $v_e=4400\ \text{m/s}$, $m_f=2000\ \text{kg}$ (dry+payload), $m_0=2000+m_p$. - Full ($m_p=120$): $\Delta v = 4400\ln(2120/2000) = 4400\times0.05827 = 256.4\ \text{m/s}$. - After one day (boiloff $0.97\ \text{kg}$, $m_p=119.03$): $\Delta v = 4400\ln(2119.03/2000) = 4400\times0.05781 = 254.4\ \text{m/s}$. - Loss $\approx 2.0\ \text{m/s}$ per day of loiter. Small per day, but a two-week hold in orbit costs ~$28\ \text{m/s}$ — real margin gone to a thermal problem, which is why cryogenic upper stages minimize coast time or insulate aggressively (Ch. 22 boiloff; Ch. 3 rocket equation).

G5 (†). (a) Heater energy per eclipse $= 25\ \text{W}\times35\ \text{min} = 25\times(35/60)\ \text{Wh} = 14.6\ \text{Wh}$ (= $52.5\ \text{kJ}$). (b) Averaged over the $90$-min orbit: $25\times(35/90) = 9.7\ \text{W}$ (equivalently $52{,}500\ \text{J}/5{,}400\ \text{s} = 9.7\ \text{W}$). (c) The battery must deliver that $14.6\ \text{Wh}$ each eclipse on top of all other eclipse loads, deepening its discharge; because batteries are sized by eclipse energy and allowable depth-of-discharge (Ch. 25), the survival heaters directly enlarge the battery — a thermal requirement paid for in mass by the power subsystem.

Chapter 25

Chapter 25 — Solutions to starred (†) and odd-numbered exercises

Units carried throughout; every final number sanity-checked. $S = 1361\ \text{W/m}^2$, $\eta = 0.30$, $X_d = 0.85$, $X_e = 0.65$ unless stated.

25.1 † Saturn at $9.54\ \text{AU}$: $S = 1361/9.54^2 = 1361/91.0 \approx 15\ \text{W/m}^2$. Only ~1.1% of Earth's flux — an array would need ~91× the Earth area for the same power. Solar power is impractical at Saturn; missions there use RTGs.

25.3 † Fraction $= \cos 40^\circ = 0.766 \approx 77\%$. The array loses ~23% purely to being 40° off Sun-pointing — the cosine loss.

25.5 † $\text{DoD} = 150/500 = 0.30 = 30\%$.

25.7 † $P_{sa} = \dfrac{1500\times60/0.85 + 1500\times36/0.65}{60} = \dfrac{105{,}882 + 83{,}077}{60} = \dfrac{188{,}959}{60} \approx 3{,}149\ \text{W}$. (~2.1× the 1,500 W load — the eclipse-plus-storage penalty, as expected.)

25.9 † $L_d = (1 - 0.03)^{10} = 0.97^{10} = 0.737$. From the BOL area of 25.8 ($A_{\text{BOL}} = 3149/(1361\times0.30\times0.90) = 3149/367.5 = 8.57\ \text{m}^2$), the array you must build is $A = 8.57/0.737 \approx 11.6\ \text{m}^2$ — about 36% larger, bought to pay for a decade of radiation damage.

25.11 † Eclipse energy $E = 1200 \times (72/60) = 1{,}440\ \text{W·h}$. Capacity $C = 1440/(0.70\times0.90) = 1440/0.63 = 2{,}286\ \text{W·h}$. Mass $= 2286/150 \approx 15.2\ \text{kg}$.

25.13 † A LEO satellite is eclipsed once per ~95-min orbit — ~16 times a day, ~5,500 times a year, ~40,000 cycles over a life. A GEO satellite is eclipsed only in two ~45-day seasons around the equinoxes, ~90 times a year, ~1,350 cycles over 15 years. Because lithium-ion cycle life falls steeply as DoD rises, the many-cycle LEO battery must be run shallow (~25%) while the few-cycle GEO battery tolerates deep (~70%) discharge — the reason a GEO battery is ~⅓ the mass for the same eclipse energy.

25.14 † Thermal decay: $110 \times 0.5^{14/87.7} = 110 \times 0.895 = 98.5\ \text{W}$. The real output is lower because the thermocouples themselves degrade over the 14 years on top of the plutonium's decay; a real MMRTG delivers somewhat below this thermal-only figure.

25.15 $P_{\text{th}} = 300/0.07 = 4{,}286\ \text{W}$; Pu-238 mass $= 4286/0.54 = 7{,}936\ \text{g} \approx 7.9\ \text{kg}$ (pure isotope). The GPHS-RTG carries ~10.9 kg because the fuel is plutonium dioxide (oxygen dilutes the heat-per-gram), is not 100% Pu-238, and includes margin — the practical fuel mass always exceeds the ideal isotope mass.

25.16 † (a) $P_{\text{th}} = 5000/0.06 = 83{,}333\ \text{W}$; mass $= 83{,}333/0.54 = 154{,}321\ \text{g} \approx 154\ \text{kg}$ of Pu-238. (b) At ~1.5 kg/yr production that is ~100 years of the entire national output for one spacecraft — infeasible. The alternative is a fission reactor (Kilopower class), which makes kilowatts from a few kilograms of uranium-235 (far more available than Pu-238) at ~30% efficiency.

25.17 † solar_flux(1.52) $= 1361/1.52^2 = 1361/2.3104 = 589.1 \to$ 589 W/m². (See code/exercise-solutions.py.)

25.19 † array_area(3149, 1361, 0.30, 0.90, 0.7374) $= 3149/(1361\times0.30\times0.90\times0.7374) = 3149/270.97 = 11.62\ \text{m}^2$. (Same as 25.9 — the code reproduces the hand calculation.)

25.20 † The estimate ignores: (1) cosine loss — the array is rarely perfectly Sun-pointed (×~0.9 or worse); (2) the array works only in sunlight yet must power the whole orbit and recharge the battery through lossy paths — required output is ~2× the average load; (3) end-of-life degradation (~20% over years); (4) high cell temperature and packing/inherent losses (~0.77). Stacked, the real array is roughly 2–3× the naive $5\ \text{m}^2$ — on the order of $10$–$15\ \text{m}^2$.

25.21 Flux at $5.2\ \text{AU} = 1361/27.0 = 50.3\ \text{W/m}^2$. Ideal area for 400 W: $A = 400/(50.3\times0.30) = 400/15.1 \approx 26.5\ \text{m}^2$. Juno's real ~60 m² panels are larger because they must also cover eclipse/pointing, the severe radiation degradation at Jupiter, spin-averaged cosine loss, packing factor, and margin — pushing the physical area well above the ~27 m² ideal.

25.22 † A battery only stores energy; it generates none. At Saturn the flux (~15 W/m²) is far too weak to recharge it, so however large the battery, it discharges once and is dead. You need a power source that works without sunlight — an RTG or reactor — which makes electricity from decay or fission regardless of the Sun.

25.23 (Project — open.) Rubric: a two-mode load table with harness (+5%) and margin (+25%); correct source selection (solar for Sun-lit tracks, RTG/fission for deep-space or long-dark); battery sized from the worst eclipse with a DoD justified by cycle count; a chosen bus voltage; all recorded in the MDR with units and a sanity check.

25.24 † The 60 kg is 60 kg of dry (structural) mass in the Ch. 23 mass budget; by the rocket equation (Ch. 3) every kilogram of dry mass demands propellant to accelerate, and propellant demands tanks — more dry mass — compounding (theme 4). If the launcher delivers only 2.4% of lift-off mass as payload, the power system's share alone implies a lift-off mass of $60/0.024 = 2{,}500\ \text{kg}$. Every 6 kg trimmed from the battery saves ~250 kg of rocket.

25.25 The other 70% of the absorbed sunlight becomes heat in the array. The thermal control subsystem (Ch. 24) handles it: the array radiates it to space (running hot, ~60 °C, which itself lowers cell voltage), and that heat is part of the spacecraft's overall thermal balance. Power and thermal are two accountings of the same energy flow.

25.26 † A 5 kW ion thruster (Ch. 20) is easily fed near Earth, where a few square meters of array make kilowatts (solar-electric propulsion, as on Dawn/BepiColombo). But flux falls as $1/d^2$: at Jupiter you'd need ~27× the array for the same 5 kW, and past Saturn solar collapses. For sustained kilowatts of electric propulsion in the outer solar system, only a fission reactor (nuclear-electric propulsion, Ch. 21) scales to the job.

Chapter 26 — Communications, Navigation, and Data Handling

(Full solutions to daggered (†) and odd-numbered exercises. Constants: $c = 2.998\times10^8\ \text{m/s}$, $k \to -228.6\ \text{dBW/K/Hz}$, $1\ \text{AU} = 1.496\times10^{11}\ \text{m}$, X-band $\lambda = 0.0357\ \text{m}$.)

26.1 $P = 10\log_{10}(20) = +13.0\ \text{dBW}$.

26.3 $\lambda = c/f = (2.998\times10^8)/(32\times10^9) = 9.37\times10^{-3}\ \text{m} = 9.37\ \text{mm}$.

26.5 $\lambda = (2.998\times10^8)/(2.2\times10^9) = 0.136\ \text{m}$. Path loss $L_{\text{fs}} = 20\log_{10}(4\pi(3.84\times10^8)/0.136) = 20\log_{10}(3.54\times10^{10}) = 20(10.55) = 211.0\ \text{dB}$.

26.6 $10\log_{10}(2\times10^{-19}) = 10(-18.70) = -187.0\ \text{dBW}$.

26.7 $G_t = 10\log_{10}[0.6(\pi(2.0)/0.0357)^2] = +42.7\ \text{dBi}$; $P_t = 10\log_{10}(15) = +11.8\ \text{dBW}$; $\text{EIRP} = 54.5\ \text{dBW}$. At $d = 2.5\ \text{AU} = 3.74\times10^{11}\ \text{m}$, $L_{\text{fs}} = 20\log_{10}(4\pi(3.74\times10^{11})/0.0357) = +282.4\ \text{dB}$. With $G_r(70\ \text{m}) = +74.2\ \text{dBi}$ and $L_{\text{other}} = 2\ \text{dB}$: $P_r = 54.5 + 74.2 - 282.4 - 2 = -155.7\ \text{dBW}$. $N_0 = -228.6 + 10\log_{10}(30) = -213.8\ \text{dBW/Hz}$, so $P_r/N_0 = -155.7 + 213.8 = 58.1\ \text{dB-Hz}$.

26.8 $R_{\max} = 10^{(58.1 - 2.3)/10} = 10^{5.58} \approx 3.8\times10^5\ \text{bit/s} = 380\ \text{kbit/s}$. At $256\ \text{kbit/s}$: $E_b/N_0 = 58.1 - 10\log_{10}(256000) = 58.1 - 54.1 = 4.0\ \text{dB} > 2.3\ \text{dB}$ → margin $+1.7\ \text{dB}$, so yes, the link closes at 256 kbit/s.

26.9 Doubling a dish quadruples its area, so each gain rises by $10\log_{10}(4) = +6.02\ \text{dB}$; doing it at both ends adds $+12\ \text{dB}$. Path loss is unchanged (it depends on $d$ and $\lambda$, not dish size). So received power rises by 12 dB, a factor of ~16.

26.11 fspl_db(3.84e8, 2.2e9) $= 20\log_{10}(4\pi(3.84\times10^8)/0.1363) = 211.0$. # Expected output: 211.0.

26.13 link_budget(15, 42.7, 74.2, 8.4e9, 3.74e11, 2.0): EIRP $= 10\log_{10}(15) + 42.7 = 54.46$; $L_{\text{fs}} = 282.4$; $P_r = 54.46 - 282.4 + 74.2 - 2 = -155.7$. # Expected output: -155.7.

26.14 Free-space path loss is $L_{\text{fs}} = 20\log_{10}(4\pi d/\lambda)$, not $10\log_{10}$, because the loss factor is $(4\pi d/\lambda)^2$ — a squared quantity — and $10\log_{10}(x^2) = 20\log_{10}(x)$. The student's $10\log_{10}$ gives exactly half the correct decibels: their $152\ \text{dB}$ should be $\approx 305\ \text{dB}$. They have understated the loss by ~153 dB, i.e. by a factor of about $10^{15}$ — which is why their signal looked absurdly strong.

26.15 Omitting $\eta = 0.6$ replaces $G = 0.6(\pi D/\lambda)^2$ with $(\pi D/\lambda)^2$, overstating the gain by $10\log_{10}(1/0.6) = +2.2\ \text{dB}$ (at each dish where it is dropped). The predicted link looks better (more optimistic) than reality — a dangerous error, since it can make a link that does not close appear to close.

26.16 Required $P_r/N_0 = (E_b/N_0)_{\text{req}} + 10\log_{10}R = 2 + 10\log_{10}(5\times10^5) = 59.0\ \text{dB-Hz}$. $N_0 = -228.6 + 10\log_{10}(25) = -214.6\ \text{dBW/Hz}$, so required $P_r = 59.0 - 214.6 = -155.6\ \text{dBW}$. At $d = 5\ \text{AU} = 7.48\times10^{11}\ \text{m}$, $L_{\text{fs}} = +288.4\ \text{dB}$; with $P_t = 10\log_{10}(30) = 14.8\ \text{dBW}$, $G_r(70\ \text{m}) = 74.2\ \text{dBi}$, $L_{\text{other}} = 2\ \text{dB}$: $$G_t = P_r + L_{\text{fs}} + L_{\text{other}} - P_t - G_r = -155.6 + 288.4 + 2 - 14.8 - 74.2 = +45.8\ \text{dBi}.$$ Invert the dish-gain formula: $10^{4.58} = 0.6(\pi D/\lambda)^2 \Rightarrow (\pi D/\lambda)^2 = 6.3\times10^4 \Rightarrow \pi D/\lambda = 251 \Rightarrow D = 251(0.0357)/\pi \approx \boxed{2.85\ \text{m}}$.

26.17 (Mission project — no single numeric answer.) A complete note states: the band (S/X/Ka) matched to distance and required rate; the data rate and daily volume from the science payload; the ground network (commercial near-Earth for A/B, the DSN for C/D); and the antenna diameter and transmitter power sized so the worst-case link closes, with the transmitter's electrical draw taken from the Chapter 25 power budget. Rubric rewards a decibel-column link budget, explicit units, and a check that the received power is near $10^{-13}$–$10^{-18}\ \text{W}$ depending on range.

26.18 (a) Ka-band. A lunar relay is close (~$3.8\times10^8\ \text{m}$), so path loss is small and margin is large; use the highest practical band to get 100 Mbit/s, since the $f^2$ advantage is available and pointing at that short range is manageable (keep an X-band fallback for weather). (b) X-band. At 45 AU with only ~200 W, Ka's narrower beam demands pointing accuracy that is hard to guarantee, and the extra data rate cannot be exploited with so little power that far out; X-band's robustness and mature DSN support make it the right primary link.

26.19 Received power falls as $1/d^2$; the data rate a link supports is proportional to $P_r/N_0$, hence to $P_r$, so rate $\propto 1/d^2$. From Jupiter ($5\ \text{AU}$) to the Kuiper Belt ($45\ \text{AU}$) the distance grows ×9, so the rate falls by $9^2 = 81$: $40\ \text{kbit/s} \to 40{,}000/81 \approx 490\ \text{bit/s} \approx 0.5\ \text{kbit/s}$.

26.21 Round-trip light time $= 2d/c = 2(1.5)(1.496\times10^{11})/(2.998\times10^8) = 1{,}500\ \text{s} = 25\ \text{minutes}$. You cannot steer the rover in real time — a hazard seen on Earth is already 12.5 min old, and any command takes 12.5 min to arrive. Rovers therefore drive autonomously between daily command loads, with the ground planning a sol's activities and reviewing the results a cycle later.

26.23 Electrical draw $= 25/0.35 = 71.4\ \text{W}$ (the other $\sim 46\ \text{W}$ is waste heat). As a fraction of a 200 W RTG, that is $71.4/200 = 36\%$ — over a third of the entire power budget. Implication: communications is a headline load, so data rate is limited by power as much as by distance; the power system (and the thermal system that rejects the waste heat) must be sized around the transmitter, and downlink is scheduled when other loads are quiet.

26.25 The 40 kg comms subsystem is 40 kg of payload lifted at the rocket equation's exponential cost (Ch. 3) — mass is the enemy (theme #4). A bigger dish buys data rate, since gain $\propto D^2$, but also costs mass, stowed volume, structure, and tighter pointing. The tension: data return pushes the dish larger; the mass budget and launch cost push it smaller. The design settles where the next kilogram of dish no longer buys enough extra science to justify its launch and propellant cost — the same marginal trade that governs every subsystem in Part IV.

Chapter 27 — Solutions to selected exercises (Guidance, Navigation, and Control)

Solutions to the daggered (†) and odd-numbered problems. All arithmetic hand-computed; units carried and sanity-checked throughout.

Part A

27.1 † $K = \sigma_{\text{pred}}^2/(\sigma_{\text{pred}}^2+\sigma_{\text{meas}}^2) = 9/(9+3) = 0.75$. Update: $\hat{x}^+ = 100 + 0.75(112-100) = 100 + 9 = 109\ \text{m}$. Variance: $(\sigma^+)^2 = (1-0.75)(9) = 2.25$, so $\sigma^+ = 1.5\ \text{m}$. Sanity: $\sigma^+ = 1.5 < \sqrt{3} = 1.73$ (measurement) $< 3$ (prediction) — fused beats both, as required.

27.3 † $v_{\text{go}} = 8{,}000 - 7{,}100 = 900\ \text{m/s}$. Current accel $a = F/m = 800{,}000/40{,}000 = 20\ \text{m/s}^2$, so naive $t = 900/20 = 45\ \text{s}$. Flow $\dot m = 800{,}000/3{,}400 = 235.3\ \text{kg/s}$; rocket-equation mass at cutoff $m_f = 40{,}000/e^{900/3400} = 40{,}000/e^{0.2647} = 40{,}000/1.3031 = 30{,}697\ \text{kg}$, so $\Delta m = 9{,}303\ \text{kg}$ and $t = 9{,}303/235.3 = 39.5\ \text{s}$. The rocket-equation time is shorter than the naive 45 s because the stage's acceleration rises as it lightens, so it gains the last of the $900\ \text{m/s}$ faster than a constant-accel estimate.

27.5 † One-way $t = D/c = 3.0\times10^{8}/3.0\times10^{5} = 1{,}000\ \text{s} = 16.7\ \text{min}$; round-trip $= 33.3\ \text{min}$. Since even the one-way time (16.7 min) exceeds the 7-min event, the signal announcing the event's start reaches the ground about 16.7 min after it began — roughly 10 min after the event already ended. The ground cannot react in time; the vehicle must be autonomous.

Part B

27.7 † $\theta_{n+1} = \theta_n + (0.4\,e_n + 1)$ with $e_n = -\theta_n$, from $\theta_0 = 0$: $\theta_1 = 0 + (0+1) = 1$; $\theta_2 = 1 + (-0.4+1) = 1.6$; $\theta_3 = 1.6 + (-0.64+1) = 1.96$; $\theta_4 = 1.96 + (-0.784+1) = 2.176$. Steady state: $u + w = 0 \Rightarrow 0.4\,e_{ss} + 1 = 0 \Rightarrow e_{ss} = -2.5$, so $\theta_{ss} = 2.5^{\circ}$ (the droop; the sequence converges to it).

27.9 † The map $\theta_{n+1} = (1-K_p)\theta_n + 1$ is a linear recursion with multiplier $(1-K_p)$; it converges iff $|1-K_p| < 1$, i.e. $0 < K_p < 2$. Fastest non-oscillatory settling is at $K_p = 1$ (multiplier $0$: it jumps to the fixed point $\theta^* = 1/K_p = 1$ in one step — "deadbeat"). For $1 < K_p < 2$ the multiplier is negative, giving decaying oscillation. At $K_p = 2$ the multiplier is $-1$: undamped, sustained oscillation (marginal stability). For $K_p > 2$, $|1-K_p| > 1$: the error grows without bound — unstable.

27.11 † Running integral $S_n = \sum e\,\Delta t$ and unclipped $u_n = K_p e + K_i S_n = 3 + S_n$:

The output is pinned at $u_{\max}=5$ the whole time, but the integral keeps charging to $S = 12$. That is integral windup: when the error finally clears, the term $K_i S = 12$ is still commanding a large output, so the controller massively overshoots in the opposite direction until the integral discharges. The cure is anti-windup: stop accumulating the integral (or clamp it) whenever the actuator is saturated.

Part C

27.12 † Predict: $\text{var}_{\text{pred}} = 16 + 9 = 25$. Gain $K = 25/(25+4) = 25/29 = 0.862$. Update: $x = 500 + 0.862(530-500) = 500 + 25.86 = 525.86$; variance $= (1-0.862)(25) = (4/29)(25) = 3.448$. Expected output: 525.86 3.448 0.862.

27.13 Proportional-only ($K_p=0.5$, $+1$/step disturbance, start 0): angles [1.0, 1.5, 1.75, 1.875, 1.9375], marching toward the $2^{\circ}$ droop.

27.14 † $v_{\text{go}} = 600$; $m_f = 30{,}000/e^{600/3400} = 30{,}000/1.193 = 25{,}150$; $\Delta m = 4{,}850$; $\dot m = 600{,}000/3{,}400 = 176.5$; $t = 4{,}850/176.5 = 27.5\ \text{s}$. Expected output: 27.5.

Part D

27.15 † With $K_p = 5$ the loop is $\theta_{n+1} = (1-5)\theta_n + 1 = -4\theta_n + 1$. The multiplier $-4$ has magnitude $> 1$, so any error is multiplied fourfold (and sign-flipped) each step: the response diverges, oscillating with growing amplitude — the vehicle is driven unstable. High $K_p$ does not remove droop; it destabilizes the loop (stable only for $0 < K_p < 2$). The correct cure for steady-state droop is the integral term, which drives the error to zero without runaway gain (add a derivative term to damp the transient).

Part E

27.17 † (Model answer, Track A GEO comsat — accept a coherent note for any track.) Navigation: two star trackers (redundant, non-parallel) + coarse Sun sensors + rate gyros for attitude; ground ranging/Doppler (Ch. 26) plus GNSS while in GTO for orbit; an extended Kalman filter fuses them. Guidance: apogee-kick and station-keeping targeting via velocity-to-be-gained; Lambert if repositioning. Control: three-axis, four reaction wheels (skewed, redundant) for pointing to the ~0.05–0.1° comsat requirement, thrusters and/or magnetorquers for momentum dumping. Autonomy: ground-supervised for maneuver planning (sub-second round-trip light-time), but attitude control fully onboard and continuous, with an autonomous Sun-pointing safe mode. Rubric: names N/G/C boxes with specific hardware, ties control to the Ch. 14 pointing budget, and argues the autonomy level from light-time.

27.19 † Choose a star tracker (few-arcsec absolute attitude — it sets the pointing knowledge to meet the 2-arcsec requirement), paired with low-drift rate gyros (smooth high-rate attitude between star fixes, blended by the Kalman filter). For actuation, use reaction wheels (or CMGs): they are smooth and propellant-free, avoiding the impulsive jitter that thruster firings would inject into a telescope's line of sight — thrusters and/or magnetorquers are relegated to occasional momentum dumping between science observations. The star tracker sets knowledge; the reaction-wheel family avoids thruster jitter for stability.

Part F

27.20 † An IMU measures accelerations and angular rates, not position and attitude. Recovering position requires integrating acceleration twice, and attitude requires integrating rate once; each integration turns the sensor's small, ever-present bias and noise into an error that grows without bound (metres, then kilometres; degrees, then tumbling). So the IMU is an excellent short-term motion memory but a hopeless long-term navigator, and it must be reset by absolute sensors (star trackers, GNSS, radar, ground tracking) through the Kalman filter.

27.21 Proportional control's two failings: (1) steady-state droop — against a constant disturbance it holds a nonzero error; fixed by the integral term. (2) overshoot / oscillation when the gain is raised for speed; fixed (damped) by the derivative term, which also lets you keep $K_p$ modest and inside the stability limit.

27.22 † To reject a $1\ \text{Hz}$ slosh disturbance the loop should run at least $\sim 10\times$ its frequency, i.e. $\ge 10\ \text{Hz}$. But running the control action out at the $3\ \text{Hz}$ structural bending mode is dangerous: the controller can inject energy into that lightly-damped mode (control-structure interaction) and destabilize it, so the loop must be band-limited or notch-filtered to avoid commanding at the bending frequency. Too slow → cannot reject slosh; too fast (into the bending mode) → excites a structural resonance. The loop rate must sit above the disturbance but the control effort must be filtered out of the structural modes.

27.23 Open-loop ascent cannot handle: (1) wind/gusts and wind shear, (2) engine underperformance or thrust dispersion (an engine running hot/cold), and (3) mass, aerodynamic, and center-of-mass uncertainties (plus propellant slosh). The one thing that handles all of them is closed-loop feedback: measuring the actual state and correcting the steering and cutoff in real time (PEG for guidance, a PID attitude loop for control).

Part G

27.24 † Treat $100\ (\text{var }16)$ as the prediction and $106\ (\text{var }9)$ as the measurement. $K = 16/(16+9) = 16/25 = 0.64$. Fused estimate $= 100 + 0.64(106-100) = 100 + 3.84 = 103.84\ \text{km}$. Fused variance $= (1-0.64)(16) = 0.36 \times 16 = 5.76$ (so $\sigma = 2.4\ \text{km}$). Inverse-variance check: $1/16 + 1/9 = 0.0625 + 0.1111 = 0.1736$, and $1/0.1736 = 5.76$ ✓ — the fused variance (5.76) is below both inputs (16 and 9).

27.25 Time to saturation $t_{\text{sat}} = H_{\max}/M_d = 4/(1\times10^{-4}) = 4\times10^{4}\ \text{s} = 40{,}000\ \text{s} \approx 11.1\ \text{hours}$. Before then the vehicle must dump momentum (desaturate) the wheels using an external torque, because internal wheels can only redistribute momentum, not remove it (§14.5). The actuator that performs it — thrusters or (in LEO) magnetorquers — is from Chapter 14.

27.26 † $v_{\text{go}} = 500$; $m_f = 35{,}000/e^{500/3200} = 35{,}000/e^{0.15625} = 35{,}000/1.1691 = 29{,}938\ \text{kg}$; $\Delta m = 5{,}062\ \text{kg}$; $\dot m = 700{,}000/3{,}200 = 218.75\ \text{kg/s}$; $t = 5{,}062/218.75 = 23.1\ \text{s}$. Thrust $F$ does not appear in the rocket equation itself ($\Delta v = v_e \ln(m_0/m_f)$ contains only exhaust velocity and masses). $F$ (with the flow $\dot m = F/v_e$) sets the rate at which delta-v is delivered, and therefore when cutoff occurs — not how much delta-v the stage carries. This is the Chapter 3 point that thrust governs the rate of spending delta-v, not the amount.

Chapter 28 — Life Support and Human Spaceflight

(Solutions to daggered † and odd-numbered exercises. Planning figures: O$_2$ 0.84, potable water 2.5, food 1.8 kg/crew-day → supplied 5.14; CO$_2$ out 1.0 kg/crew-day. $g_0 = 9.81\ \text{m/s}^2$.)

28.1 Crew 3, 30 days: $3 \times 30 \times 0.84 = 75.6\ \text{kg}$ of oxygen.

28.3 Per person, 500 days: $500 \times 5.14 = 2{,}570\ \text{kg}$ — about $2.57\ \text{t}$ of consumables for one crew member, which is why long crewed missions are mass-dominated by life support.

28.5 Crew 4, one week: $4 \times 7 \times 1.0 = 28\ \text{kg}$ of CO$_2$. The scrubber's workload is set by the production rate (a fixed mass per day), not the cabin volume, because at steady state the scrubber must remove exactly what the crew produces to hold the partial pressure constant. A larger cabin only delays the buildup; it does not reduce how much must be removed per day.

28.7 Balanced: $\text{CO}_2 + 4\,\text{H}_2 \rightarrow \text{CH}_4 + 2\,\text{H}_2\text{O}$ — four H$_2$ per CO$_2$. Electrolysis supplies only about two H$_2$ per O$_2$ (and the crew makes ~1 CO$_2$ per O$_2$), so only about half the CO$_2$ can be reduced — hence ~50% oxygen recovery; the rest of the CO$_2$, with its oxygen, is vented.

28.9 Person-days $= 4 \times 500 = 2{,}000$. Makeup water at 90% recovery: $0.10 \times 2.5 = 0.25\ \text{kg/crew-day} \Rightarrow 500\ \text{kg}$. At 98%: $0.02 \times 2.5 = 0.05\ \text{kg/crew-day} \Rightarrow 100\ \text{kg}$. The upgrade saves $500 - 100 = 400\ \text{kg}$ of makeup water over the mission.

28.11 consumables(crew, days, rate=5.14): return crew*days*rate. Trace consumables(4, 180) $= 4 \times 180 \times 5.14 = 3{,}700.8$. # Expected output: 3701 (rounded).

28.13 ag_rpm(50): $\omega = \sqrt{9.81/50} = \sqrt{0.1962} = 0.4429\ \text{rad/s}$; $\text{rpm} = 0.4429 \times 60/2\pi = 4.23$. # Expected output: 4.23.

28.14 The claim confuses consuming the CO$_2$ with recovering the oxygen. The reaction is limited not by CO$_2$ but by hydrogen: electrolysis yields ~2 H$_2$ per O$_2$ delivered, the crew exhales ~1 CO$_2$ per O$_2$ consumed, and Sabatier needs 4 H$_2$ per CO$_2$ — so with only half the hydrogen it needs, the system reclaims oxygen from only about half the CO$_2$. Correct recovery is $\approx 50\%$; the missing hydrogen leaves, chemically bound, in the vented methane.

28.15 For $1\,g$ at $r = 5\ \text{m}$: $\omega = \sqrt{9.81/5} = 1.40\ \text{rad/s}$, $\text{rpm} = 1.40 \times 9.549 = 13.4\ \text{rpm}$. Coriolis on a hand at $1\ \text{m/s}$: $a_{\text{Cor}} = 2\omega v = 2 \times 1.40 \times 1 = 2.80\ \text{m/s}^2 \approx 0.29\,g$ sideways. Gravity gradient over $1.8\ \text{m}$: head at $r = 3.2\ \text{m}$ feels $3.2/5 = 0.64$ of the foot value — $36\%$ lighter at the head. The claim is not credible: $13\ \text{rpm}$ is far above any comfort limit, the sideways Coriolis nudge is nearly a third of a gravity, and a $36\%$ head-to-foot gradient is grossly disorienting. Comfortable $1\,g$ needs a radius of hundreds of meters, not five.

28.16 Crew 6, 400 days ($2{,}400$ crew-days). (a) Open-loop: $2{,}400 \times 5.14 = 12{,}336\ \text{kg} \approx 12.3\ \text{t}$. (b) Closed-loop resupply per crew-day $= 1.8 + 0.10(2.5) + 0.50(0.84) = 1.8 + 0.25 + 0.42 = 2.47$; $\times 2{,}400 = 5{,}928\ \text{kg}$, plus $2{,}500\ \text{kg}$ hardware $= 8{,}428\ \text{kg}$. (c) Closed-loop is lighter by $12{,}336 - 8{,}428 = 3{,}908\ \text{kg} \approx 3.9\ \text{t}$. (d) Food dominates the closed-loop mass ($2{,}400 \times 1.8 = 4{,}320\ \text{kg}$, about half of the $8{,}428$). It implies future gains must come from closing the food loop (growing food / bioregenerative systems) — the last, hardest loop.

28.17 (Model answer.) For an uncrewed track (A/B/C/D): "ECLSS — N/A (uncrewed). A crewed variant would require a pressurized habitable volume and structure (Ch. 23), an ECLSS power load large enough to move the power budget (Ch. 25), cabin heat/humidity rejection (Ch. 24), radiation-shielding and storm-shelter mass (§28.3), a launch-escape system, and human-rated reliability — reshaping every MDR budget." For a crewed variant, size consumables from §28.1 for your crew and duration and record the open- vs closed-loop decision from §28.2.

28.18 (a) $20\ \text{g/cm}^2 = 200\ \text{kg/m}^2$; over $8\ \text{m}^2$: $200 \times 8 = 1{,}600\ \text{kg}$ of water. (b) A Mars-class mission carries and recycles far more than $1{,}600\ \text{kg}$ of water (its drinking-water throughput alone is several tonnes), so arranging existing water tanks as the shelter wall adds essentially no new launch mass — the shielding piggybacks on water carried anyway. (c) Food thins as it is eaten: the food wall is thickest at launch and thinnest on the return leg, exactly when a late solar particle event would still threaten the crew. Water (topped up by the recycler) and stored waste are steadier shielding.

28.19 Shielding mass $\approx$ (surface area) $\times$ (areal density). The whole habitat has a large surface area, so raising it to storm-shelter areal density costs enormous mass; a shelter walls off only the small volume of a huddled crew. The physics compounds the problem: galactic cosmic rays are so energetic that even heavy shielding only modestly cuts the dose and can create secondary particles (spallation), so there are diminishing — even negative — returns. You cannot practically wall off GCR; the shelter helps against the lower-energy protons of an SPE, and the real GCR defenses are speed and (someday) active magnetic deflection.

28.21 Two reasons. (1) Fire: a pure-oxygen atmosphere, especially above one atmosphere, is violently flammable — materials that merely smolder in air burn fiercely. (2) Oxygen toxicity: sustained breathing of high $p_{\text{O}_2}$ damages the lungs and nervous system. The settling event was the Apollo 1 fire (January 1967), which killed three astronauts during a ground test in a pure-oxygen cabin pressurized above one atmosphere and ended the use of high-pressure pure oxygen.

28.22 Energy $= 1.0\ \text{kW} \times (180 \times 24)\ \text{h} = 1.0 \times 4{,}320 = 4{,}320\ \text{kWh}$. It feeds the power budget because this is a continuous, non-negotiable load that must be supplied at all times — including in eclipse from batteries — so it directly sizes the solar array and battery of Chapter 25. Life support is one of the largest steady power consumers on a crewed vehicle.

28.23 $A = \dfrac{P}{\varepsilon \sigma T^4} = \dfrac{2000}{0.85 \times 5.67\times10^{-8} \times (300)^4}$. Now $(300)^4 = 8.1\times10^{9}$, so $\sigma T^4 = 5.67\times10^{-8} \times 8.1\times10^{9} = 459\ \text{W/m}^2$ and $\varepsilon\sigma T^4 = 0.85 \times 459 = 390\ \text{W/m}^2$. Then $A = 2000/390 \approx 5.1\ \text{m}^2$ (less if both faces radiate; deep space at ~3 K contributes negligibly). It ties life support to Chapter 24 because the crew and ECLSS are heat sources whose load the thermal system must reject by radiator — the only heat path available in vacuum.

28.24 Consumables are payload that must be accelerated to escape velocity, and by the rocket equation (Ch. 3) each kilogram of payload to trans-Mars injection demands several kilograms of propellant. For a TMI $\Delta v \approx 3.6\ \text{km/s}$ with a hydrogen upper stage ($v_e \approx 4.4\ \text{km/s}$), the mass ratio is $e^{3600/4400} = e^{0.82} \approx 2.3$, so each kg of habitat mass needs ~$2.3\ \text{kg}$ in LEO — and getting that to LEO carries its own rocket-equation multiplier. Removing $7.5\ \text{t}$ of consumables therefore removes on the order of $7.5 \times 2.3 \approx 17\ \text{t}$ of in-LEO mass, and far more at lift-off. Mass saved in the crew cabin is propellant saved on the pad — the tyranny of the rocket equation reaching all the way inside.

28.25 Radiation: a storm shelter (plus shielding and faster transits) — but it does not fully solve the problem, because galactic cosmic rays are chronic and nearly unshieldable, and a Mars mission still approaches or exceeds the career dose limit. Microgravity: exercise (~2 h/day, resistive) — but it only slows the losses; crews still demineralize bone, shed plasma, and risk the vision syndrome (SANS). Only artificial gravity would remove the cause, and it has never flown.

Chapter 29 — Mission Design (solutions to † and odd-numbered exercises)

All sizing uses $g_0 = 9.80665$ m/s²; $v_e = I_{sp} g_0$; mass ratio $R = e^{\Delta v/v_e}$; $m_\text{wet} = R\,m_\text{dry}$; $m_\text{prop} = m_\text{wet} - m_\text{dry}$.

29.1 (a) functional (what it does); (b) constraint (a fence on the solution — a mass limit); (c) performance (a function with a number attached).

29.3 Ideal budget $= 1{,}500 + 600 + 40 = 2{,}140\ \text{m/s}$.

29.5 MDR confirms the requirements, orbit, delta-v budget, and architecture are consistent — that the mission closes. PDR confirms the preliminary subsystem designs meet their requirements with adequate margin (freezing the architecture and interfaces). CDR confirms the detailed build-to design is complete and verified, authorizing flight-hardware fabrication.

29.7 (a) $v_e = 320 \times 9.80665 = 3{,}138\ \text{m/s}$. (b) $R = e^{2726/3138} = e^{0.8687} = 2.384$. (c) $m_\text{wet} = 2.384 \times 1{,}500 = 3{,}576\ \text{kg}$. (d) $m_\text{prop} = 3{,}576 - 1{,}500 = 2{,}076\ \text{kg}$. Sanity check: matches the chapter's Checkpoint (2,076 kg propellant, 3,576 kg wet). ✓

29.9 $v_e = 3{,}138\ \text{m/s}$; $R = e^{5980/3138} = e^{1.906} = 6.72$; $m_\text{wet} = 6.72 \times 2{,}200 = 14{,}791\ \text{kg}$. The mass ratio (6.72) is far larger than the comsat's 2.38 because the delta-v is more than double AND the mass ratio is exponential in delta-v: $e^{5980/3138}$ vs $e^{2726/3138}$. Physically, a 6-km/s mission with a chemical engine must be almost entirely propellant.

29.11 Margined budget $= 1{,}470(1.05) + 750(1.10) + 11 = 1{,}543.5 + 825 + 11 = 2{,}379.5\ \text{m/s}$. $R = e^{2379.5/3138} = e^{0.7583} = 2.135$; $m_\text{wet} = 2.135 \times 1{,}500 = 3{,}202\ \text{kg}$; $m_\text{prop} = 1{,}702\ \text{kg}$. Saving vs. the 28.5° case: $3{,}576 - 3{,}202 = 374\ \text{kg}$ of wet mass. The equatorial launch site removes the plane change, lowering the transfer delta-v and hence the mass ratio — a launch-site choice that ripples all the way to launch mass (theme: everything is connected).

29.12 roll_up_dv([("insertion",1500,0.05),("SK",600,0.10),("disposal",40,0.0)]): ideal $= 1500+600+40 = 2140$; margined $= 1500(1.05)+600(1.10)+40 = 1575+660+40 = 2275.0$. # Expected output: (2140, 2275.0).

29.13 size_vehicle(2726, 320, 1500): $v_e = 3138.1$, $R = e^{2726/3138.1} = 2.3837$, $m_\text{wet} = 3575.6$, $m_\text{prop} = 2075.6$. # Expected output: (2075.6, 3575.6) (≈ 2076, 3576 kg).

29.14 amplifier(5980,320) = $e^{5980/3138.1} = 6.72$; amplifier(2726,1800) = $e^{2726/17652.0} = 1.17$. # Expected output: 6.72 1.17. The amplifier is the mass ratio: wet-mass-per-dry-mass. The lander's 6.72 vs the electric comsat's 1.17 shows why high-delta-v missions punish mass so much harder.

29.15 close_check(1750,5000): $\lfloor 5000/1750 \rfloor = 2$, fits → (2, True). close_check(3576,5000): $\lfloor 5000/3576 \rfloor = 1$, fits → (1, True). The electric comsat (1,750 kg) fits two under a 5,000 kg launcher; the chemical (3,576 kg) only one — the §29.5 / CS-01 result.

29.16 The 2,561 m/s is a point estimate of a genuinely uncertain quantity: masses grow, engines underperform, and the budget is likely missing small legs (finite-burn/pointing losses, extra trajectory corrections). Designing to the exact estimate guarantees that any adverse deviation makes the mission fall short. Margin is not padding on top of the right answer — it is the honest acknowledgment that the estimate is a distribution, and that a lost mission costs far more than carried reserve. The right move is to reduce uncertainty so margin can shrink with justification, never to delete it.

29.17 The launch leg (~11.9 km/s) and the spacecraft-own budget (2.7 km/s) are provided by different, staged vehicles: the launch vehicle burns its share and is discarded (or flies home) before the spacecraft ever fires its own engine. Summing them and sizing one vehicle for 14.6 km/s double-counts by making the spacecraft carry the launch vehicle's job into its own mass ratio — exactly the dead-mass error staging exists to avoid (Ch. 3). Size the launch vehicle for the wet mass it must lift; size the spacecraft for its own 2.7 km/s.

29.18 Margin exists to cover uncertainty, and uncertainty falls as estimates are replaced by built-and-measured hardware. Holding 25% margin to launch means either you never bought down the uncertainty (you don't actually know your mass at launch — alarming) or you are carrying 25% of dead capability you no longer need (wasting payload/launch mass). Healthy programs buy down margin as they buy knowledge; a margin that isn't shrinking as the design matures is itself a warning sign.

29.19 (Project / rubric.) A complete answer has: every leg listed with its ideal delta-v traced to the delta-v map / Appendix G; a justified per-leg margin (larger for less-certain or less-forgiving legs — e.g., descent, capture — smaller for well-characterized burns); and both an ideal and a margined total. Common tracks: A (comsat) GTO→GEO + station-keeping + disposal ≈ 2.6 km/s ideal; B (lunar lander) TLI + LOI + descent ≈ 5.5 km/s; C (Mars orbiter) TMI (launcher) + MOI ~2.1 + corrections ~0.3; D (asteroid) departure + rendezvous + proximity ops. Reward units on every number and a sanity check.

29.21 (Project / rubric.) A complete trade study: names ≥ 2 concrete architectures; lists ≥ 4 weighted criteria (mass, cost, transfer time, risk, heritage, power); scores each option per criterion; and states the winner together with the weighting that produced it ("electric wins because we weighted launch mass above time-to-station"). The key learning is that the answer is conditional on the weighting — answers that declare one option "objectively best" miss the point.

29.23 Budget $= 2{,}100 + 300 = 2{,}400\ \text{m/s}$ (ideal; a back-of-envelope estimate). Amplifier $R = e^{2400/3138} = e^{0.765} = 2.15$. Wet mass $= 2.15 \times 900 = 1{,}934\ \text{kg} \approx 2{,}000\ \text{kg}$ (one sig fig). So a ~900 kg Mars orbiter needs roughly a two-tonne wet mass — the insertion burn (which the spacecraft pays propulsively) roughly doubles the vehicle.

29.25 Subsystems are coupled (§29.5): the lightest camera might draw more power, forcing a bigger, heavier array and more thermal hardware, whose added dry mass — multiplied by the rocket-equation amplifier — adds more propellant and wet mass than the camera saved. Optimizing each part in isolation ignores the interfaces and shared budgets through which a local win becomes a system loss. Only optimizing the whole (systems engineering) — watching all the budgets and the coupling at once — reliably yields the lightest system.

29.27 From the rocket equation directly: $m_\text{wet} = R\,m_\text{dry} = 2.38 \times 1{,}500 = 3{,}570\ \text{kg}$; $m_\text{prop} = 3{,}570 - 1{,}500 = 2{,}070\ \text{kg}$ (rounding $R$ to 2.38). Using the exact $R = e^{2726/3138} = 2.384$ gives 3,576 kg wet and 2,076 kg propellant — identical to size_vehicle(2726, 320, 1500), because size_vehicle is the rocket equation inverted. ✓

29.29 The amplifier is the mass ratio, so a dry-mass increase is multiplied by it: $\Delta m_\text{wet} = R\,\Delta m_\text{dry} = 3.0 \times \Delta m_\text{dry}$ — the wet mass overruns by 3× the dry overrun (in kilograms). To the structures engineer: your 8%-of-dry overrun adds $3.0 \times 8\% = 24\%$ of the dry mass in wet mass; if the launch vehicle's margin was under that, the design no longer closes. Their 8% is a 24% problem for the vehicle because every dry kilogram drags three wet kilograms behind it through the rocket equation.

Chapter 30

Chapter 30 — Solutions to daggered (†) and odd-numbered exercises

Numbers use $g_0 = 9.81\ \text{m/s}^2$, equatorial rotation $0.465\ \text{km/s}$, and (for the GEO maneuver) $v_a = 1.60$, $v_{\text{GEO}} = 3.07\ \text{km/s}$. Vehicle numbers are Tier 2 (Appendix H).

30.1 † (a) GTO payload is smaller because reaching GTO demands ~2.4 km/s more delta-v than LEO, and by the rocket equation ($m_0/m_f = e^{\Delta v/v_e}$) that extra delta-v must be paid for in payload traded for propellant. (b) The "missing" 15 t was converted into velocity (extra propellant expended), not mass delivered — the vehicle spent capacity on delta-v instead of on payload.

30.3 † $0.465\cos 19.6^\circ = 0.465 \times 0.9418 = 0.438\ \text{km/s}$ (Wenchang). Cape Canaveral: $0.465\cos 28.5^\circ = 0.409\ \text{km/s}$. Wenchang's lower latitude gives about $0.029\ \text{km/s}$ (~29 m/s) more free eastward velocity — small, but real payload near the margin.

30.5 † Any two of: (i) to which orbit — LEO, GTO, or escape, and at what inclination (fewer kg to a higher-energy orbit); (ii) price or internal cost — an advertised price is not the operator's cost; (iii) how full the rocket flew — a dedicated launch of a light payload has a far worse $/kg than a full one, and a rideshare seat differs from a dedicated flight.

30.7 † $\cos i = \cos\phi\sin\beta \Rightarrow \sin\beta = \cos i / \cos\phi = \cos 60^\circ / \cos 28.5^\circ = 0.5 / 0.8788 = 0.5689$. So $\beta = 34.7^\circ$ (northeasterly, ascending) or $\beta = 180^\circ - 34.7^\circ = 145.3^\circ$ (southeasterly, descending). The $34.7^\circ$ solution is the northeasterly launch.

30.9 † With $\Delta v = \sqrt{v_a^2 + v_{\text{GEO}}^2 - 2 v_a v_{\text{GEO}}\cos i}$: $i = 7^\circ$: $\sqrt{2.56 + 9.42 - 9.82\cos 7^\circ} = \sqrt{11.98 - 9.75} = \sqrt{2.23} = 1.49\ \text{km/s}$. $i = 28.5^\circ$: $\sqrt{11.98 - 8.63} = \sqrt{3.35} = 1.83\ \text{km/s}$. The higher-latitude launch costs about $1.83 - 1.49 = 0.34\ \text{km/s}$ more of the satellite's own delta-v.

30.11 † A supersynchronous transfer orbit has its apogee above GEO, so by vis-viva the satellite is moving slower at that apogee — the $v_a$ term in the combined-burn formula decreases. Because a plane change costs $\Delta v \approx 2v\sin(i/2)$, doing it at a lower speed $v_a$ makes it cheaper; the inclination-removal cost drops. The satellite then lowers its apogee to GEO. The trade: the launch vehicle must inject to a higher, more energetic apogee (it does more work so the satellite does less).

30.13 inclination(28.5, 90) $= \deg(\arccos(\cos 28.5^\circ \cdot \sin 90^\circ)) = \arccos(0.8788) = 28.5^\circ$ (minimum inclination = latitude). `inclination(28.5, 45)` $= \arccos(0.8788 \times 0.7071) = \arccos(0.6214) = 51.6^\circ$ (a northeasterly launch reaches the ISS-class plane). Expected output: 28.5 then 51.6.

30.15 windows_per_day(28.5, 51.6) $= 2$ (target inclination $>$ latitude → site crosses the plane twice/day); windows_per_day(28.5, 28.5) $= 1$ (equal → tangent, one crossing); windows_per_day(28.5, 20.0) $= 0$ (target inclination $<$ latitude → plane unreachable directly). Expected output: 2 1 0.

30.17 The minimum inclination equals the launch-site latitude, so from Kourou it is $5.2^\circ$, not $0^\circ$. Kourou cannot launch directly into an equatorial ($0^\circ$) GEO; a small $5.2^\circ$ plane change is still required. Its low latitude makes that plane change small and cheap — the whole point of the site — but not zero.

30.19 † Need $\geq 5{,}200 \times 1.10 = 5{,}720\ \text{kg}$ to GTO. From Appendix H, vehicles clearing it: Falcon 9 (expendable, ~8.3 t GTO), Falcon Heavy (~27 t), Vulcan Centaur (~7–14 t), Ariane 6 (~5–11.5 t, upper configs), New Glenn (~13 t), Long March 5 (~14 t), Proton-M (~6.3 t), Atlas V (~8.9 t in high config). Screened out: LVM3 (~4 t), Electron. The single deciding factor depends on the mission: for a GEO comsat, launch-site latitude (an equatorial Ariane launch minimizes the satellite's own plane change) or price (a reused Falcon 9) is usually decisive; for a flagship, reliability/heritage. A defended answer naming one factor and tying it to the mission earns full marks.

30.21 † Margin $= 2{,}300 / 1{,}700 = 1.35$ (35%), which clears the 10% rule ($1.35 > 1.10$). [Using the case study's 1,705 kg gives $2{,}300/1{,}705 = 1.35$ as well.] One non-performance factor to check: fairing volume (does the folded orbiter fit?), launch-site reachability of the departure asymptote's declination, injection accuracy, price, schedule, or reliability/heritage.

30.23 † At least three reasons: (1) Geographic/range safety — very few nations own equatorial land with a safe eastward over-water launch corridor; you must launch where spent stages fall harmlessly. (2) Political/logistical — sovereignty, existing infrastructure, workforce, and the cost of transporting large stages mean nations use their own territory (the US flies from Florida/California, not a foreign equator). (3) Mission-dependent — polar and sun-synchronous missions launch north/south and gain nothing from an equatorial site, and rendezvous missions must match a target's existing inclination regardless of site. The equatorial credit itself is modest (~0.14 km/s Kourou vs Cape), so it rarely outweighs the above.

30.25 † From the Cape ($i = 28.5^\circ$, $\Delta v = 1.83$): propellant $= 3{,}000(e^{1.83/3.1} - 1) = 3{,}000(1.805 - 1) = 2{,}414\ \text{kg}$. From Kourou ($i = 5.2^\circ$, $\Delta v = 1.48$): $3{,}000 (e^{1.48/3.1} - 1) = 3{,}000(1.612 - 1) = 1{,}836\ \text{kg}$. Extra $\approx 2{,}414 - 1{,}836 = 578\ \text{kg}$ — order a few hundred kilograms. That is a design driver, not a rounding error: ~580 kg on a 3-tonne satellite is roughly a fifth of its on-station mass, equal to many years of station-keeping propellant or a large payload increment. (Illustrative, Tier 3.)

30.27 † A reusable first stage reserves propellant for boostback, entry, and landing burns. That reserved propellant raises the stage's effective burnout mass $m_f$ (it is not all spent on thrust for the payload), lowering the stage's mass ratio $m_0/m_f$ and hence the delta-v it delivers to the stack — so the payload it can push to orbit falls (Falcon 9: ~22.8 t → ~17 t). It is still usually worth it because the recovered stage, costing tens of millions of dollars, can fly a dozen or more times; the hardware cost amortized over many flights is far below the value of the ~25% payload given up on any single flight.

30.29 † The vehicle needs ~9.4 km/s to LEO and ~11.8 km/s (surface-referenced) to GTO. For the same vehicle, deliverable mass scales roughly as $e^{-\Delta v/v_e}$, so the LEO-to-GTO payload ratio is about $e^{(11.8 - 9.4)/3.3} = e^{2.4/3.3} = e^{0.727} \approx 2.1$. This sits at the low end of the "~2–3×" rule of thumb; the real ratio often runs a bit higher (~2.5–3×) because GTO delivery also carries the upper stage farther and pays extra losses, effects this simple estimate omits.

30.31 † Fixing launch-commit criteria before launch day means the go/no-go limits are set calmly, with full engineering analysis, rather than being rationalized under launch-day schedule pressure ("go fever"). Giving every console a veto in the GO/NO-GO poll means any single engineer can halt the count without being overruled or pressured. Together they institutionalize "any unresolved doubt stops the launch" — precisely the discipline whose absence destroyed Challenger, where an engineering concern (cold O-rings) was overruled by management under schedule pressure.

Chapter 31 — Ground Operations and Mission Control

Solutions to the daggered (†) and odd-numbered exercises. Constants: $c = 2.998\times10^{5}\ \text{km/s}$, $1\ \text{AU} = 1.496\times10^{8}\ \text{km}$, $R_\oplus = 6378\ \text{km}$, $\mu_\oplus = 3.986\times10^{5}\ \text{km}^3/\text{s}^2$.

31.1 GEO one-way latency $= 35{,}786 / 2.998\times10^{5} = 0.119\ \text{s} \approx 0.12\ \text{s}$. Round trip $= 2 \times 0.119 = 0.239\ \text{s} \approx 0.24\ \text{s}$. (This "quarter-second each way" is the delay you hear on a call routed through a geostationary relay.)

31.3 $d = 1.5\ \text{AU} = 1.5 \times 1.496\times10^{8} = 2.244\times10^{8}\ \text{km}$. One-way $= 2.244\times10^{8} / 2.998\times10^{5} = 748.5\ \text{s} = 12.5\ \text{min}$. This distance sits almost exactly midway between the opposition (~$0.52\ \text{AU}$) and conjunction (~$2.52\ \text{AU}$) extremes — their average is $1.52\ \text{AU}$ — so it represents an intermediate geometry, neither closest approach nor farthest. It is marginally nearer the opposition distance but is best described as a typical mid-cycle range.

31.5 Sun at $1\ \text{AU}$: $1.496\times10^{8} / 2.998\times10^{5} = 499\ \text{s} = 8.3\ \text{min}$. So one AU corresponds to about 8.3 light-minutes — a handy conversion: multiply a distance in AU by $\sim 8.3$ to get its one-way light time in minutes.

31.7 The flight-rules book is a collection of pre-decided responses to anticipated contingencies ("if $X$, then do $Y$"), written by the experts who know the systems, calmly, long before launch. Its purpose is to remove improvisation — and therefore panic-driven error — from the vast majority of in-flight decisions, so they are made correctly and instantly by prior agreement. The flight director's authority to override a rule exists only for the rare situation the rules did not foresee; even then, FLIGHT is breaking a known rule knowingly and accountably, not guessing in a vacuum. Pre-deciding is how you make good decisions under conditions (time pressure, fear, fatigue) that otherwise produce bad ones.

31.9 Telemetry is the spacecraft's downlinked stream of measurements about its own state — its vital signs (temperatures, pressures, voltages, valve positions, attitude). Commanding is the uplink of instructions to the vehicle (individual commands or stored sequences). To "verify a command in the telemetry" is to watch the downlink for the specific state change the command was meant to cause (a valve now reads open, a mode has changed). It is never skipped because commanding is open-loop until telemetry closes the loop: an unverified command may have been lost, rejected, or misexecuted, and on a distant vehicle the telemetry confirmation is the only proof you will ever get that it worked.

31.10 light_time(3.78e8) $= 3.78\times10^{8} / 2.998\times10^{5} = 1260.8\ \text{s}$, which round() gives as 1261; in minutes, $1260.8/60 = 21.0\ \text{min}$. So:

# Expected output:
# 1261
# 21.0

This is Mars near conjunction — a $21$-minute one-way delay, $42$-minute round trip.

31.11 For the Moon, round_trip_light_time(384_400) $= 2 \times (384{,}400/2.998\times10^{5}) = 2 \times 1.282 = 2.56\ \text{s}$. Since $1 \le 2.56 < 10$, autonomy_level returns "supervised". Output:

# Expected output:
# 2.56 supervised

A $2.6$-second round trip is sluggish but human-in-the-loop still works (as Apollo showed); critical events are time-tagged.

31.12 pass_duration_min(550): $r = 6378 + 550 = 6928\ \text{km}$. Period $T = 2\pi\sqrt{6928^3/3.986\times10^{5}} \approx 5{,}738\ \text{s} = 95.6\ \text{min}$. Half-angle $\lambda = \arccos(6378/6928) = \arccos(0.9206) \approx 23.0^\circ$. Contact $= (2\times23.0/360)\times95.6 = (46.0/360)\times95.6 \approx 12.2\ \text{min}$.

# Expected output:
# 12.2

Slightly longer than the $400\ \text{km}$ case's $10.2\ \text{min}$, because a higher orbit sees farther — but still only a few usable minutes per pass.

31.13 Near opposition the one-way light time is $\approx 4.4\ \text{min}$ and the round trip $\approx 8.7\ \text{min}$; the landing lasts $\approx 7\ \text{min}$. Trace the timeline from atmospheric entry ($t=0$): telemetry showing a problem does not reach Earth until $t \approx 4.4\ \text{min}$; if the team instantly commanded an abort, it would not arrive at the vehicle until $t \approx 8.7\ \text{min}$ — by which point the landing (over at $t \approx 7\ \text{min}$) has already succeeded or failed. A ground-commanded abort is physically impossible: the round trip is longer than the event. The abort logic — every fault response — must live onboard and execute autonomously. The plan confuses LEO-style real-time operations with deep-space reality.

31.15 (Mission project — model answer for Track C, a Mars orbiter; adapt to your track.) Ground segment: track via the Deep Space Network, a shared resource; assume ~one $8$-hour pass per day, so commanding and downlink are batched, not continuous. Latency: one-way $4.4\ \text{min}$ (opposition) to $21\ \text{min}$ (conjunction); round trip $8.7$–$42\ \text{min}$; plus a ~2-week solar-conjunction blackout each synodic cycle. Autonomy level: fully autonomous — sequence-based commanding, no real-time control; orbit insertion and all fault responses execute onboard. Safe-mode concept: on any fault, halt the science plan, turn arrays to the Sun, point the high-gain antenna at Earth (or beacon), shed nonessential loads, and wait for the next pass; a command-loss timer triggers autonomous recovery if contact is lost for $N$ days. Rubric: an explicit latency computation with units, an autonomy level that follows from it, and a safe-mode/anomaly plan honest about how little the ground can do in the moment.

31.17 A $400\ \text{km}$ satellite has period $\approx 92.6\ \text{min}$ and a usable pass of only $\sim 6\ \text{min}$ over any one station — about $6/92.6 \approx 6.5\%$ of an orbit. Naively, covering one ground track continuously would need $92.6/6 \approx 15$ stations strung along it. But it is worse than that: Earth rotates $\sim 23^\circ$ under the orbit each revolution, so the ground track marches around the globe and you would need a dense, world-spanning grid (dozens of stations, many over inaccessible ocean) to catch every pass — which is why no one does it this way. A single GEO relay ($35{,}786\ \text{km}$ up) can see almost the entire LEO shell at once, so it maintains contact with the satellite over most of its orbit; two or three relays (like NASA's TDRS) give near-continuous coverage. One well-placed relay replaces a global ground network — the operational reason relays exist.

31.18 Data per day $= 500\ \text{kbit/s} \times 8\ \text{h} = 0.5\times10^{6}\ \text{bit/s} \times (8\times3600)\ \text{s} = 0.5\times10^{6} \times 28{,}800 = 1.44\times10^{10}\ \text{bit}$. Convert to bytes: $/8 = 1.8\times10^{9}\ \text{byte} = 1.8\ \text{GB/day}$. The camera makes $10\ \text{GB/day}$, so there is an $\approx 8.2\ \text{GB/day}$ deficit. The operations concept must therefore include onboard mass storage, data compression, and prioritization (downlink the most valuable data first), plus scheduling data-heavy campaigns for when the range — and thus the link rate — is most favorable. You cannot return everything; operations decides what.

31.19 Because the satellite physically drops below the horizon. The pass geometry of §31.3 limits an overhead pass to $\sim 10\ \text{min}$ maximum; after that the satellite is on the far side of the Earth, and the planet itself blocks the line of sight. No dish, however large, can see through the Earth. Antenna size improves the link (data rate, sensitivity) but cannot change the geometry — so a single station gets only brief, periodic contacts no matter what.

31.20 Round-trip light time to Mars is $8.7\ \text{min}$ at opposition and $42\ \text{min}$ at conjunction (one-way $4.4$–$21\ \text{min}$). A "stop" command sent from Earth answers a hazard the rover's cameras saw $4$–$21\ \text{min}$ ago; in between, a joysticked rover would drive blind. So Mars rovers move slowly, in short pre-planned drives, using onboard autonomous hazard detection and avoidance — not a human joystick. A drone on Earth has millisecond latency, which is why real-time piloting works there and cannot work at Mars. The difference is entirely the speed of light over the distance.

31.21 A spacecraft's state has six unknowns (three position, three velocity — equivalently the six orbital elements). A single short tracking pass observes range and range-rate over only a small arc of the orbit — too little independent geometry to solve for all six, and too short to reveal the orbit's shape. Several passes spread across the orbit (and across time) sample different geometry and different points in the orbit, providing the independent observations needed to fit all six elements and to average down measurement noise in a least-squares or Kalman solution (Ch. 13). Tracking over time is also what lets flight dynamics see the perturbations (Ch. 12) that make the real orbit drift from a simple ellipse.

31.23 The Deep Space Network has only a few dozen large antennas, at three sites spread in longitude (so Earth's rotation always keeps one facing deep space), yet it must serve dozens of active missions across the Solar System. There are far more missions than dishes, so antenna time is scheduled: each mission is allocated tracking passes, prioritized around its critical events, and competes for the rest. Between its scheduled passes a spacecraft is out of contact, so it copes by recording telemetry and science to onboard storage and replaying it when its pass comes (store-and-forward), executing autonomous stored sequences in the meantime, and safing itself if a fault occurs while no one is listening.

31.25 The lunar module Aquarius, built for two crew for ~two days, had to sustain three for ~four. The binding consumables were electrical power (finite battery amp-hours — the crew powered down to roughly a quarter of normal load), water (used to cool the electronics by boiling it to space, rationed to dangerous lows), and carbon-dioxide removal (the lithium-hydroxide scrubber capacity). It was not primarily oxygen — the LM carried far more than the crew needed. The CO₂ fix was the improvised "mailbox": because the command module's square LiOH canisters would not fit the lunar module's round scrubber sockets, a ground team designed an adapter from only the materials aboard (plastic stowage bags, cardboard, tape, a sock), tested it, and read the assembly procedure up to the crew, who built it in orbit. The CO₂ level fell and the crew came home.

Chapter 32 — Reliability, Testing, and Why Rockets Fail

(Solutions to the daggered (†) and odd-numbered exercises. All reliabilities treated as independent unless the problem says otherwise.)

32.1 † Series: $R = 0.99 \times 0.95 \times 0.98 = 0.9217$. Mission-loss probability $= 1 - 0.9217 = 0.0783 \approx 7.8\%$. (Note it lands below the weakest link, $0.95$ — the series product always does.)

32.3 † Parallel pair: $R = 1 - (1 - 0.90)^2 = 1 - (0.10)^2 = 1 - 0.01 = 0.99$. Two units each failing 10% of the time fail together only 1% of the time.

32.5 † $R = 1 - 0.08 = 0.92$. The number is meaningless without: (1) the function (what counts as success — full payload? basic bus survival?), (2) the conditions (the environment it must survive), and (3) the duration (over one launch? 15 years? one re-entry?). A reliability without all three is an adjective, not a requirement.

32.7 † (a) Series $= 0.97 \times 0.92 \times 0.99 \times 0.95 = 0.8393$. (b) The weakest link is the $0.92$ function. (c) Add one redundant copy of it: block $= 1 - (1 - 0.92)^2 = 1 - (0.08)^2 = 0.9936$; new system $= 0.97 \times 0.9936 \times 0.99 \times 0.95 = 0.9064$. (d) Mission-loss dropped from $1 - 0.8393 = 16.1\%$ to $1 - 0.9064 = 9.4\%$ — a fall of about $6.7$ points, or ~42% relative, from one backup on the weakest link.

32.9 † Require $1 - (0.15)^n \ge 0.95 \Rightarrow (0.15)^n \le 0.05 \Rightarrow n \ge \dfrac{\ln 0.05}{\ln 0.15} = \dfrac{-2.996}{-1.897} = 1.58$. So $n = 2$ strings. Check: $n=2$ gives $1 - (0.15)^2 = 0.9775 \ge 0.95$ ✓ (and $n=1$ gives only $0.85$).

32.10 † series([0.99, 0.98, 0.97, 0.96]) $= 0.99 \times 0.98 \times 0.97 \times 0.96 = 0.90345$. # Expected output: 0.9035 (rounded to 4 dp).

32.11 redundant(0.85, 3) $= 1 - (1 - 0.85)^3 = 1 - (0.15)^3 = 1 - 0.003375 = 0.996625$. # Expected output: 0.9966.

32.12 † k_of_n(0.98, 4, 3) $= P(4) + P(3) = 0.98^4 + \binom{4}{3}(0.98)^3(0.02) = 0.922368 + 4(0.941192)(0.02) = 0.922368 + 0.075295 = 0.99766$. # Expected output: 0.9977. (Four units that tolerate one failure.)

32.13 † The student computed the reliability of two computers in series — $0.99 \times 0.99 = 0.9801$ — i.e. treated the design as if both had to work. A backup is in parallel: the pair fails only if both fail, so $R = 1 - (1 - 0.99)^2 = 1 - (0.01)^2 = 1 - 0.0001 = 0.9999$. The pair is $0.9999$, not $0.9801$. The error is modeling a redundant unit with the series (multiply) rule instead of the parallel (multiply-the-unreliabilities) rule.

32.15 † (a) Four equal series subsystems: each $R = 0.90^{1/4} = 0.9740$. (b) If one is fixed at $0.95$, the other three must supply $0.90 / 0.95 = 0.94737$ between them, so each $= 0.94737^{1/3} = 0.9821$. (Weakening one subsystem forces the others up — the coupling of a reliability allocation.)

32.17 † Rank by relative system gain per kg (new-block/old-block ÷ mass), since series coupling means each redundancy multiplies the whole system reliability:

Ranking per kg: valve (1.32) ≈ computer (1.25) > wheels (0.58). Total mass $8 + 12 + 3 = 23\ \text{kg} \le 60\ \text{kg}$ margin, so you can afford all three — but if forced to choose, the valve and computer buy far more reliability per kilogram than the (heavy) wheel set.

32.18 † Treat the 200 events as an independent series: $r_{\text{avg}}^{200} = 0.95 \Rightarrow r_{\text{avg}} = 0.95^{1/200} = e^{\ln 0.95 / 200} = e^{-0.0002565} = 0.99974$. Each of the 200 events must average about 99.974% reliable. For contrast, "three nines" per event ($0.999$) gives only $0.999^{200} = 0.819$ — 82%, not 95%. Reaching 95% over 200 must-work events demands nearly four nines on every single step: workmanship on each ignition, separation, and deployment is not optional.

32.19 You cannot make everything redundant because every redundant unit is dry mass, and dry mass is punished by the rocket equation (Ch. 3): $m_0 = (m_0/m_f)\,m_f$, so each kilogram of dry mass drags a mass-ratio's worth of wet mass behind it, and propellant grows exponentially with delta-v. Doubling subsystems could more than double the launch mass and cost. So redundancy is spent, not sprinkled: put it where the reliability math says it pays most (the weakest links), and make the rest reliable by margin, derating, and testing.

32.20 † Testing consumes life: vibration, thermal cycling, and hot-fires accumulate fatigue, wear, and cycles. A unit tested "until it never fails" has spent much of its design life and moved toward the wear-out end of the bathtub curve — flying that worn article risks a wear-out failure. That is exactly why the campaign is split: qualification over-stresses a dedicated article (proving the design has margin) and does not fly it, while acceptance testing screens each fresh flight unit at mild flight levels to catch build defects without using up its life. The flight article must be new. (Protoflight is the compromise when you cannot afford a separate qual unit.)

32.21 † Margin buy-down is the reduction of uncertainty. Early in a program the mass and performance figures are estimates with wide uncertainty, so you hold a large reserve (25–30% mass at MDR) to cover wherever the true value lands. As the design matures, estimates are replaced by built-and-measured hardware; the uncertainty shrinks, so reserve can be safely removed — you "buy down margin by buying knowledge." Holding margin constant would be wrong both ways: it would carry needless mass at launch (when you know the numbers) and, if you set it low, too little reserve at the start (when you don't). It is the same instinct as screening out the uncertain, high-failure-rate region before committing.

32.23 † An upper stage's mass is "payload" to every stage beneath it, so it is carried through the whole ascent and amplified by each lower stage's mass ratio (Ch. 3): a kilogram added high in the stack can cost several kilograms of extra lift-off mass. So $400\ \text{kg}$ of redundant engine on the upper stage propagates into a far larger wet-mass and delta-v penalty than the same $400\ \text{kg}$ low in the stack. Buying reliability by redundancy is therefore most expensive high in the stack — which is why upper stages are often single-engine and made reliable by margin and testing instead.

32.25 $P(\text{20 clean} \mid R = 0.90) = 0.90^{20} = 0.122$; $P(\text{20 clean} \mid R = 0.99) = 0.99^{20} = 0.818$. Twenty straight successes is *likely* for the $0.99$ rocket (82%) but not rare for the $0.90$ rocket (12%), so 20 flights cannot cleanly tell them apart. To demonstrate $R = 0.99$ at 90% confidence from flights alone needs about $\ln(0.10)/\ln(0.99) \approx 229$ consecutive successes. This is why reliability is argued mostly from analysis plus test, not flight history — especially for new vehicles, which have too few flights to prove a high number statistically.

Chapter 33

Chapter 33 — Solutions to selected exercises (Small Satellites, CubeSats, Constellations)

Full solutions to the daggered (†) and odd-numbered problems. Use $R_E = 6{,}371\ \text{km}$, solar flux $1{,}361\ \text{W/m}^2$, $c = 299{,}792\ \text{km/s}$, $g_0 = 9.80665\ \text{m/s}^2$. Coverage geometry: $\sin\eta = \frac{R_E}{R_E+h}\cos\varepsilon$, $\lambda = 90^\circ - \varepsilon - \eta$, $f = \frac{1-\cos\lambda}{2}$, $N_{\min} \gtrsim 1/f$.

33.1 † (classify by mass)

(a) $4\ \text{kg}$ → nanosatellite ($1$–$10\ \text{kg}$); (b) $60\ \text{kg}$ → microsatellite ($10$–$100\ \text{kg}$); (c) $300\ \text{kg}$ → minisatellite ($100$–$500\ \text{kg}$); (d) $0.3\ \text{kg}$ → picosatellite ($0.1$–$1\ \text{kg}$).

33.3 † (power of a 3U face)

Area $A = 0.10 \times 0.30 = 0.030\ \text{m}^2$. Peak power $P = 1361 \times 0.030 \times 0.30 \times 0.8 = 9.8\ \text{W}$. A CubeSat is power-limited: its mass is fixed by the standard, but every subsystem (radio, computer, actuators, payload) competes for these few watts, and the orbit-average (tumbling + eclipse) is lower still — so you run out of power before you run out of mass.

33.5 † (horizon-limited footprint at 500 km)

With $\varepsilon = 0$, $\cos\lambda_{\max} = R_E/(R_E+h) = 6371/6871 = 0.92723$, so $\lambda_{\max} = \arccos(0.92723) = 22.0^\circ$. (This is the absolute largest footprint half-angle — a satellite just on the horizon; any usable elevation shrinks it.)

33.7 † (1000 km, 25°)

$\sin\eta = \frac{6371}{7371}\cos 25^\circ = 0.86433 \times 0.90631 = 0.78335$, so $\eta = 51.57^\circ$; $\lambda = 90 - 25 - 51.57 = 13.43^\circ$; $f = (1 - \cos 13.43^\circ)/2 = (1-0.97266)/2 = 0.01367$ ($1.37\%$); $N_{\min} = 1/0.01367 = 73.1 \to \mathbf{74}$. Going higher helps a lot: from $\sim 184$ at $550\ \text{km}$ to $\sim 74$ here — a factor of about $2.5\times$ fewer satellites, because each higher satellite sees a larger footprint. (The cost is longer latency and slower orbital decay.)

33.9 † (LEO vs GEO latency)

One-way ground→sat→ground delay $= 2h/c$ for an overhead satellite. (a) LEO: $2 \times 550 / 299{,}792 = 0.00367\ \text{s} = 3.67\ \text{ms}$. (b) GEO: $2 \times 35{,}786 / 299{,}792 = 0.2387\ \text{s} = 238.7\ \text{ms}$. Ratio $\approx 238.7 / 3.67 \approx \mathbf{65\times}$. Interactive applications (calls, gaming, trading) are ruined by a quarter-second lag but comfortable at a few milliseconds — so LEO's tiny latency is the business case that justifies flying thousands of satellites to do what three could do from GEO.

33.10 † (coverage_fraction code)

coverage_fraction(550, 25) → $\eta = 56.54^\circ$, $\lambda = 8.46^\circ$, $f = 0.005439$. round(..., 4) = 0.0054.

33.11 (min_satellites_for_global code)

min_satellites_for_global(700, 20): $\sin\eta = \frac{6371}{7071}\cos 20^\circ = 0.90100 \times 0.93969 = 0.84667$, $\eta = 57.85^\circ$, $\lambda = 12.15^\circ$, $f = (1-\cos 12.15^\circ)/2 = 0.011195$, $1/f = 89.3 \to \lceil\cdot\rceil = \mathbf{90}$. Fewer than the chapter's $184$ because two knobs changed in the fleet-reducing direction: the satellite is higher ($700$ vs $550\ \text{km}$, so a bigger footprint) and the elevation requirement is lower ($20^\circ$ vs $25^\circ$, so the footprint is allowed to reach closer to the horizon) — both enlarge $\lambda$, enlarge $f$, and shrink $N_{\min}$.

33.12 † (cost_per_kg code)

cost_per_kg(7.5e6, 300) = $25{,}000$; cost_per_kg(1.0e6, 200) = $5{,}000$. Ratio $25{,}000/5{,}000 = \mathbf{5.0}$ — the dedicated small launcher is five times more expensive per kilogram than the rideshare slot, the premium you pay for control over orbit and schedule.

33.13 † (the "Starlink only needs 184" error)

$N_{\min} \approx 184$ leaves out two things. (1) Overlap: it assumes footprints tile the sphere perfectly with no overlap, which is geometrically impossible — guaranteeing no gaps at every latitude and every instant, as satellites and Earth rotate, takes $2$–$3\times$ the floor. (2) Capacity: $N_{\min}$ is a coverage number, but a broadband constellation is sized by bandwidth — each satellite serves only so many users at once, so densely populated regions need many satellites overhead simultaneously. Starlink's thousands are driven mainly by capacity, far above any coverage floor.

33.15 † (550 vs 1200 km trade)

$550\ \text{km}$ needs $\sim 184$ satellites; $1{,}200\ \text{km}$ needs only $\sim 57$. So the higher shell needs fewer satellites (each sees a bigger footprint). But the lower shell is more sustainable: at $550\ \text{km}$ a dead satellite re-enters within a few years (atmospheric drag self-cleans the orbit), while at $1{,}200\ \text{km}$ it persists for centuries, so a high constellation must add propulsion to every satellite to force a deorbit. The trade: fewer satellites and launches (favoring high) versus lower latency and self-disposal (favoring low). A responsible design often accepts the larger low fleet precisely because the satellites are cheap and the orbit disposes of them for free.

33.17 † (launch campaign)

$600$ satellites $\div\ 60$ per launch $= \mathbf{10}$ launches. Total launch cost $= 10 \times \$60\ \text{M} = \$600\ \text{M}$. Launch cost per satellite $= \$600\ \text{M}/600 = \$1\ \text{M}$. This is only the launch cost — the total cost of a satellite on orbit also includes the satellite hardware itself, the ground segment (stations, network), operations, insurance, and the replacements needed over the constellation's life.

33.18 † (why 550 km decays in years, 1200 km in centuries)

Decay time scales roughly inversely with atmospheric density, and density falls quasi-exponentially with a scale height of $\sim 50$–$60\ \text{km}$ in this regime. From $550$ to $1{,}200\ \text{km}$ is $\sim 650\ \text{km}$, i.e. roughly $650/55 \approx 12$ scale heights, so the density drops by a factor of about $e^{12} \sim 10^{5}$. If a satellite at $550\ \text{km}$ decays in a few years, one at $1{,}200\ \text{km}$ — in air $\sim 10^{5}$ times thinner — takes $\sim 10^{5}$ times longer, i.e. effectively centuries to millennia. (Order-of-magnitude only; real decay also depends on the ballistic coefficient and the solar cycle, which swells the upper atmosphere.)

33.19 (why not GEO broadband)

Three GEO satellites do cover the globe, but the one-way latency is $\sim 0.24\ \text{s}$ (from 33.9), a round-trip of nearly half a second, which ruins interactive uses — voice calls, video conferencing, gaming, and financial trading all degrade badly. GEO also cannot serve high latitudes and the poles, where a geostationary satellite sinks to (or below) the horizon (Chapter 9). So GEO trades away latency and polar coverage for a small satellite count — the wrong trade for a modern low-latency broadband service, which is why the constellations chose LEO despite needing thousands of satellites.

33.20 † (why not give every CubeSat propulsion)

On a $4\ \text{kg}$ satellite the mass, volume, and power budgets are all tiny, and a propulsion system — tank, propellant, valves, and the power to run it — consumes a large fraction of all three, plus cost and new failure modes. Many science and education CubeSats simply cannot afford it, which is exactly why so many fly with no propulsion at all and rely on atmospheric drag to bring them down. That is the root of the §33.6 tension: propulsion would let every satellite deorbit responsibly and dodge collisions, but it is unaffordable on the very cheapest satellites — so the sustainable answer is instead to choose low orbits that dispose of a dead satellite passively.

33.21 (dedicated vs rideshare for one CubeSat)

Rideshare: $4\ \text{kg} \times \$5{,}000/\text{kg} = \$20{,}000$. Dedicated Electron: $\$7.5\ \text{M}$. Factor $= \$7.5\ \text{M}/\$20{,}000 = \mathbf{375\times}$ more expensive. The dedicated launch buys one thing rideshare cannot: control — your exact orbit, on your schedule, launched when you are ready rather than whenever the rideshare provider flies to its standard orbit.

33.22 † (inclination for 70° N)

A satellite's sub-point latitude never exceeds its orbital inclination, so to place satellites over (and give high-elevation coverage to) $70^\circ\text{N}$ you need an inclination of at least $\sim 70^\circ$ (near-polar, $\sim 87^\circ$, is the usual choice for truly global/polar coverage). A $53^\circ$-inclination shell's satellites never go north of $53^\circ$ sub-point latitude, so from $70^\circ\text{N}$ they appear only very low on the southern horizon — or not at all — and coverage there is poor. Inclination sets which latitudes a constellation can serve well.

33.23 (requirements flow-down for global broadband)

• Mission requirement: "provide continuous global broadband."
• System requirement (orbit): operate a LEO constellation giving continuous coverage of all latitudes at a minimum elevation $\varepsilon \ge 25^\circ$ → near-polar inclination, chosen altitude $h$.
• Subsystem/architecture requirement (constellation size): fly $N$ satellites in $P$ planes, where $N$ is the coverage floor $N_{\min}(h,\varepsilon)$ from §33.5 times a margin (overlap + capacity + spares).
• Component/program requirement (launch & disposal): a launch campaign of $\lceil N/\text{per-launch}\rceil$ flights plus on-orbit spares, and a disposal plan meeting the 5-year rule. Each level is derived from the one above by the coverage geometry (orbit → $N$) and launch capacity ($N$ → launches), exactly the Chapter 29 flow-down.

33.24 † (deorbit propellant)

$v_e = I_{sp}\,g_0 = 220 \times 9.80665 = 2{,}157.5\ \text{m/s}$. Mass ratio $m_0/m_f = e^{\Delta v/v_e} = e^{150/2157.5} = e^{0.06953} = 1.0720$. Propellant mass $m_p = m_f(m_0/m_f - 1) = 260 \times 0.0720 = \mathbf{18.7\ \text{kg}}$. That is $18.7/(260+18.7) = 6.7\%$ of the wet mass — a modest, affordable fraction, which is why a satellite that carries any propulsion can usually afford the small budget needed to deorbit itself. (The problem, per 33.20, is the satellites that carry none at all.)

33.25 (why small antenna + low power → low data rate)

A small antenna has low gain and a few watts gives low transmit power. In the link budget (Chapter 26) the received signal-to-noise ratio scales with (transmit power × transmit gain × receive gain)/(path loss), so low power and low gain shrink the SNR — and the achievable data rate is bounded by SNR (Shannon's limit). A weak link therefore forces a low rate. Two constellation-scale workarounds: (1) inter-satellite links plus ground-station networks to relay data so a satellite is always connected; (2) store-and-forward — buffer data on board and dump it in a high-rate burst during a ground-station pass. (Better deployable/ steerable high-gain antennas and stronger error-correcting codes also help at the satellite level.)

33.26 † (constellation reliability)

Expected number still working after one year $= 100 \times 0.95 = \mathbf{95}$. The service can still be effectively $100\%$ available because the constellation is designed with coverage redundancy (the §33.5 margin of $2$–$3\times$ the floor, plus on-orbit spares), so losing $\sim 5$ of $100$ satellites does not open a permanent coverage gap, and failed satellites are replaced on a routine schedule. Reliability lives in the system, not the unit — the threshold concept of §33.5: a flagship must make every part perfect, but a constellation is engineered to keep working while individual satellites fail and are replaced.

Chapter 34

Chapter 34 — Solutions to selected exercises (daggered † and odd-numbered)

Constants: $\mu_\oplus = 3.986\times10^5$, $\mu_{\text{Mars}} = 4.283\times10^4\ \text{km}^3/\text{s}^2$; Earth radius $6{,}371$, Mars radius $3{,}390\ \text{km}$; $g_0 = 9.80665\ \text{m/s}^2$. All numbers hand-traced; no code executed. (All daggers fall on odd numbers, so this set = every odd exercise.)

34.1 † Synodic period. $\dfrac{1}{T_{\text{syn}}} = \left|\dfrac{1}{365.25} - \dfrac{1}{686.98}\right| = (2.7379 - 1.4557)\times10^{-3} = 1.2822\times10^{-3}\ \text{day}^{-1}$, so $T_{\text{syn}} = 780\ \text{days} = 780/30.44 \approx 25.6 \approx 26\ \text{months}$. It is longer than either planet's year because it measures the time for the faster body (Earth) to gain a full lap ($360^\circ$) on the slower (Mars); their per-day angular rates differ only slightly, so lapping takes far longer than one orbit of either.

34.3 † Conjunction-class timeline. $259 + 500 + 259 = 1{,}018\ \text{days} = 1{,}018/365.25 \approx 2.8\ \text{years}$. The surface-stay length ($\sim 500$ days) is set by waiting for the return launch window — Earth and Mars must re-phase for a minimum-energy Mars→Earth transfer, which does not recur until roughly $500$ days after arrival. (The stay is orbital mechanics, not a choice.)

34.5 † ISRU chemistry. Sabatier: $\text{CO}_2 + 4\,\text{H}_2 \rightarrow \text{CH}_4 + 2\,\text{H}_2\text{O}$. Electrolysis: $2\,\text{H}_2\text{O} \rightarrow 2\,\text{H}_2 + \text{O}_2$. The pair manufactures methane (fuel) and oxygen (oxidizer) — methalox — the same propellant Starship's Raptor burns and the natural product of Mars ISRU on the $96\%$-CO$_2$ atmosphere.

34.7 † TMI from a $400\ \text{km}$ parking orbit, $r = 6{,}771\ \text{km}$, $v_\infty = 2.95\ \text{km/s}$. $v_{\text{circ}} = \sqrt{3.986\times10^5/6{,}771} = \sqrt{58.87} = 7.673\ \text{km/s}$; $2\mu_\oplus/r = 117.74$; $v_{\text{peri}} = \sqrt{2.95^2 + 117.74} = \sqrt{126.44} = 11.24\ \text{km/s}$; $\Delta v_{\text{TMI}} = 11.24 - 7.67 = 3.57\ \text{km/s}$. This is slightly smaller than the $300\ \text{km}$ result ($3.59\ \text{km/s}$): a higher parking orbit gives a marginally smaller injection burn (the burn is nearly flat in altitude and rises slightly as you go lower). This does not contradict the Oberth effect, which is about burning deep-and-fast versus slowly at the SOI edge, not the mild parking-altitude dependence.

34.9 † Entry speed at $r_{\text{atm}} = 3{,}490\ \text{km}$, $v_{\infty,\text{Mars}} = 2.65\ \text{km/s}$. $v_{\text{entry}} = \sqrt{2.65^2 + 2(4.283\times10^4)/3{,}490} = \sqrt{7.02 + 24.55} = \sqrt{31.57} = 5.62\ \text{km/s} \approx 5.6\ \text{km/s}$ — matches the chapter. (This is relative to Mars, not the Sun's $21.5\ \text{km/s}$.)

34.11 † TEI from low Mars orbit, $r = 3{,}540\ \text{km}$, $v_\infty = 2.65\ \text{km/s}$. $2\mu_{\text{Mars}}/r = 24.20$; $v_{\text{peri}} = \sqrt{2.65^2 + 24.20} = \sqrt{31.22} = 5.588\ \text{km/s}$; $v_{\text{circ}} = \sqrt{4.283\times10^4/3{,}540} = \sqrt{12.10} = 3.478\ \text{km/s}$; $\Delta v_{\text{TEI}} = 5.588 - 3.478 = 2.11\ \text{km/s}$. Ascent ($\sim 4.1$) is bigger than TEI ($\sim 2.1$); total "surface → Earth-bound" $= 4.1 + 2.1 = 6.2\ \text{km/s}$ — all of it made on Mars by ISRU.

34.13 † Terminal velocity, $3{,}000\ \text{kg}$ under a $314\ \text{m}^2$ chute, $C_d = 0.5$. Mars ($\rho = 0.020$, $g = 3.71$): $v = \sqrt{2(3000)(3.71)/((0.020)(0.5)(314))} = \sqrt{22{,}260/3.14} = \sqrt{7{,}089} = 84\ \text{m/s}$. Earth ($\rho = 1.225$, $g = 9.81$): $v = \sqrt{58{,}860/192.3} = \sqrt{306} = 17\ \text{m/s}$. Factor $\approx 84/17 = 4.8\times$ worse. Implication: even a huge parachute leaves the vehicle at $\sim 84\ \text{m/s}$ ($\sim 300\ \text{km/h}$) — lethal — so parachutes alone can never land a spacecraft on Mars; a powered/airbag/sky-crane terminal phase is mandatory.

34.15 † (Conceptual.) The heat shield removes the bulk of the speed — from $\sim 5.6\ \text{km/s}$ ($\sim$ Mach 25) down to roughly Mach 2 ($\sim 0.5\ \text{km/s}$), i.e. on the order of $90\%$ — as heat, paid for by shield mass rather than propellant. The last few hundred m/s cannot be removed by a parachute because (Exercise 34.13) the thin air limits even a giant chute to $\sim 84\ \text{m/s}$, far too fast to land; so the terminal phase must use retropropulsion, a sky crane, or airbags. Mars gives you the braking for the fast part (a shock forms) but not for the slow part (too little air) — the "cruel middle."

34.17 † MAV sizing. $v_e = 360 \times 9.80665 = 3{,}530\ \text{m/s} = 3.53\ \text{km/s}$; $m_0/m_f = e^{4.1/3.53} = e^{1.161} = 3.19$; $m_p = 5\,(3.19 - 1) = 5 \times 2.19 = 11.0\ \text{t}$. Propellant fraction $= 11.0/16.0 = 69\%$. Matches the chapter's $\sim 11$–$12\ \text{t}$ (the Checkpoint's margined $4.305\ \text{km/s}$ gives $11.9\ \text{t}$; this unmargined $4.1$ gives $11.0\ \text{t}$).

34.19 † Methalox split of $\sim 12\ \text{t}$ at O/F $= 3.5$: oxygen $= 12 \times 3.5/4.5 = 9.3\ \text{t}$; methane $= 12 \times 1/4.5 = 2.7\ \text{t}$. Mars's $96\%$-CO$_2$ atmosphere supplies the carbon and oxygen directly (as CO$_2$) — so it most directly furnishes the oxygen (and the carbon skeleton of the methane); the scarce ingredient is hydrogen, which must be brought from Earth or mined from Martian water ice and electrolyzed.

34.21 † injection_prop(3.6, 380, 220). $v_e = 380 \times 9.80665 = 3{,}727\ \text{m/s}$; mass ratio $= e^{3600/3727} = e^{0.966} = 2.628$; propellant $= 220\,(2.628 - 1) = 220 \times 1.628 = 358\ \text{t}$.

# Expected output:
# 358

34.23 † synodic_period(365.25, 686.98) $= 1/|1/365.25 - 1/686.98| = 780\ \text{days}$. synodic_period(365.25, 400.0) $= 1/|0.0027379 - 0.0025| = 1/0.00023785 = 4{,}204\ \text{days} \approx 11.5\ \text{years}$ — huge, because a period close to Earth's makes the rate difference tiny, so the two bodies take an age to re-align.

# Expected output:
# 780
# 4204

34.25 † Poor window, $C_3 = 16\ \text{km}^2/\text{s}^2$. $v_\infty = \sqrt{16} = 4.0\ \text{km/s}$. TMI from $300\ \text{km}$ ($r = 6{,}671$): $v_{\text{peri}} = \sqrt{4.0^2 + 119.50} = \sqrt{135.50} = 11.64\ \text{km/s}$; $\Delta v_{\text{TMI}} = 11.64 - 7.73 = 3.91\ \text{km/s}$, up from $3.59$ at the good window. The higher $C_3$ hits the injection — the launch vehicle's payload-to-$C_3$ capability and, in a refuel-in-orbit architecture where the vehicle performs its own TMI, the vehicle's own injection propellant load. It does not change the arrival or return legs (those depend on the arrival $v_\infty$).

34.27 † (Conceptual.) Opposition-class missions are usually judged worse for the crew despite being shorter because of two competing effects: (1) total mission duration — shorter for opposition; but (2) time spent in the harsh deep-space radiation environment — longer for opposition, because it trades the long surface stay (where Mars's atmosphere and regolith offer partial shielding and a storm shelter is easy) for more time in transit, plus a fast trajectory and a Venus swingby that runs closer to the Sun. Since dose accumulates fastest in deep space, a mission that spends more of its time in transit can deliver a higher total dose than the longer conjunction mission — shorter overall, but more irradiated. Radiation, not calendar time, is the crew's binding constraint.

34.29 † (Why can't you just bring the return propellant?) Because return propellant carried from Earth is multiplied by every leg it rides through: to have it on the Martian surface you must (1) land it — through the mass-limited EDL of §34.4; which means (2) inject it toward Mars — paying the TMI mass ratio; which means (3) launch it to LEO — paying the ascent mass ratio. That stack of rocket equations turns $\sim 11\ \text{t}$ on the Mars pad into many tens of tonnes in LEO and hundreds on the Earth pad. ISRU deletes the entire chain by manufacturing the propellant at the destination from local CO$_2$ (plus hydrogen), so it is never launched, injected, or landed at all — the return propellant simply never has to be accelerated across the solar system.

Chapter 35 — Solutions to Selected Exercises († and odd-numbered)

Constants: $\mu = 3.986\times10^{5}\ \text{km}^3/\text{s}^2$, $R_\oplus = 6{,}371\ \text{km}$, $g_0 = 9.81\ \text{m/s}^2$, $3.156\times10^{7}\ \text{s/yr}$. Densities are flagged as solar-cycle-dependent.

35.1 † — energy of a fragment

$E = \tfrac12 m v^2 = \tfrac12(0.002)(9{,}000)^2 = \tfrac12(0.002)(8.1\times10^{7}) = 8.1\times10^{4}\ \text{J} = 81\ \text{kJ}$. Rifle bullet: $\tfrac12(0.010)(900)^2 = 4{,}050\ \text{J} = 4.05\ \text{kJ}$. The 2 g orbital fragment carries about 20 times the energy of the bullet — from one-fifth the mass — because orbital closing speed ($9\ \text{km/s}$) is ten times a bullet's and energy goes as the square of speed.

35.3 † — the three defensive regimes

• Shield it: below $\sim 1\ \text{cm}$ — a Whipple shield fragments/vaporizes the impactor.
• Track and dodge it: above $\sim 10\ \text{cm}$ — cataloged individually, avoided by maneuver.
• Neither: $1$–$10\ \text{cm}$ — too energetic to shield, too small to track. The "lethal non-trackable" band falls here, which is why it is the most dangerous.

35.5 † — closing speed vs. crossing angle

$v_{\text{rel}} = 2v\sin(\theta/2)$. At $90^\circ$: $2(7.6)\sin45^\circ = 15.2(0.7071) = 10.7\ \text{km/s}$. At $20^\circ$: $2(7.6)\sin10^\circ = 15.2(0.1736) = 2.64\ \text{km/s}$. A steep crossing meets near the full sum of both speeds; a shallow one (nearly co-moving) has a small relative speed. Since impact energy $\propto v_{\text{rel}}^2$, the $90^\circ$ case is $(10.7/2.64)^2 \approx 16$ times more energetic — steep crossings dominate the hazard.

35.6 † — collision probability

$N = \Phi A t = (2\times10^{-5})(8)(7) = 1.12\times10^{-3}$. $P = 1 - e^{-N} = 1 - e^{-0.00112} = 1.12\times10^{-3} \approx 0.11\%$ (about 1 in 893).

35.7 (odd) — fleet risk

Expected strikes $= 8{,}000 \times 1.12\times10^{-3} = 8.96 \approx 9$ over 7 years. $P(\geq 1\ \text{lost}) = 1 - e^{-8{,}000 N} = 1 - e^{-8.96} = 1 - 1.3\times10^{-4} \approx 0.9999$ — essentially certain the fleet loses at least one satellite to debris. The per-satellite risk is tiny; the fleet rolls the dice 8,000 times.

35.8 † — the linear approximation

$P = 1 - e^{-N} = N - \tfrac{N^2}{2} + \tfrac{N^3}{6} - \cdots$, so for small $N$, $P \approx N$. The leading dropped term is $N^2/2$, a fractional error $\approx N/2$ relative to $N$. Setting $N/2 = 0.05$ gives $N \approx 0.1$. Check: at $N = 0.1$, exact $P = 1 - e^{-0.1} = 0.0952$ versus linear $0.1$ — a $5\%$ error. So the approximation is good to $5\%$ up to about $N \approx 0.1$.

35.9 † — GEO graveyard vs. de-orbit

$\Delta v \approx \tfrac{v}{2}\tfrac{\Delta h}{a} = \tfrac{3{,}070}{2}\tfrac{285}{42{,}164} = 1{,}535 \times 0.00676 = 10.4\ \text{m/s}$. De-orbiting instead means lowering the perigee from $42{,}164\ \text{km}$ radius down to $\sim 6{,}500\ \text{km}$ (into the atmosphere) — a full descending Hohmann costing $\sim 1{,}500\ \text{m/s}$, over a hundred times more, because you must shed a large fraction of the orbital speed rather than nudge a near-circular orbit up a few hundred km.

35.11 † — natural lifetime at 700 km

$a = 7{,}071\ \text{km} = 7.071\times10^{6}\ \text{m}$; $v = \sqrt{\mu/a} = 7{,}508\ \text{m/s}$; $H = 7.5\times10^{4}\ \text{m}$; $\beta = 80$; $\rho_0 = 4\times10^{-14}\ \text{kg/m}^3$ (flag: solar-cycle-dependent). $$\tau \approx \frac{(7.5\times10^{4})(80)}{(4\times10^{-14})(7.071\times10^{6})(7{,}508)} = \frac{6.0\times10^{6}}{2.12\times10^{-3}} = 2.83\times10^{9}\ \text{s} \approx 90\ \text{years}.$$ This fails the 25-year rule (90 > 25) and badly fails the 5-year rule. Even allowing the density to swing by an order of magnitude (lifetime ~45–180 yr), it fails both — active de-orbit or a drag device is required.

35.13 † — implement collision_probability

collision_probability(1e-5, 10, 5): $N = 5\times10^{-4}$; $1 - e^{-0.0005} = 0.00049988$. # Expected output: 0.0005 (rounded to 6 places).

35.15 † — implement graveyard_dv

graveyard_dv(3070, 300, 42164) $= 0.5 \times 3070 \times 300/42164 = 10.92$. # Expected output: 10.9.

35.16 † — disposal plan (model answer, Track A GEO comsat)

Destination: graveyard $\sim 300\ \text{km}$ above GEO. Disposal $\Delta v \approx 11\ \text{m/s}$, reserved (not "leftover"). No natural decay (that is why we vacate the belt). Governing rule: IADC GEO disposal guideline (re-orbit altitude $\geq 235 + 1000\,C_r(A/m)\ \text{km}$ above GEO); a $300\ \text{km}$ raise satisfies it for a typical $A/m$. Passivate: vent propellant, safe batteries, relieve tank pressure. (Other tracks: B controlled lunar impact/graveyard + passivation; C long-quarantine Mars orbit for planetary protection + passivation; D stable heliocentric disposal or documented target impact + passivation.)

35.17 (odd) — active de-orbit at 800 km

Lower perigee to $150\ \text{km}$ ($r_p = 6{,}521\ \text{km}$). $v_1 = \sqrt{398{,}600/7{,}171} = 7.456\ \text{km/s}$; $a_t = (7{,}171+6{,}521)/2 = 6{,}846\ \text{km}$; $v_{\text{apo}} = \sqrt{398{,}600(2/7{,}171 - 1/6{,}846)} = \sqrt{52.95} = 7.277\ \text{km/s}$; $\Delta v = 7.456 - 7.277 = 0.179\ \text{km/s} = 179\ \text{m/s}$. Propellant: $v_e = 220(9.81) = 2{,}158\ \text{m/s}$; $m_p = 120(1 - e^{-179/2158}) = 120(1 - 0.9204) = 9.6\ \text{kg}$. Disposal propellant fraction $= 9.6/120 = 8.0\%$.

35.19 † — why not "vacuum up" small debris

The $1$–$10\ \text{cm}$ population numbers $\sim 10^{6}$ objects spread through orbital shells the size of a planet, all at $\sim 10\ \text{km/s}$ closing speeds. A collector is itself moving at orbital speed, so it meets debris at hypervelocity — a net or scoop would be shattered, not gently filled; the flux is so low (mean free paths of centuries) that sweeping a meaningful fraction by chance is hopeless; and a large collector is itself a huge collision cross-section, adding risk. You cannot catch hypervelocity confetti dispersed across a volume that vast.

35.21 † — why not leave a defunct satellite

It becomes a permanent collision target, and if struck it yields thousands of fragments. Because collision rate $\propto N^2$, each added derelict raises everyone's risk super-linearly. And a shell has a finite carrying capacity: leave enough derelicts and the region crosses the supercritical threshold where debris self-amplifies (Kessler syndrome), degrading the orbit for generations. "Empty enough" is a property of staying below carrying capacity, not a licence to keep adding.

35.23 † — space-law reasoning

(a) OST Article VI: states bear international responsibility for national space activities, including private ones, which they must "authorize and continually supervise" — so a company must be licensed and overseen by its government for the state to meet its treaty duty (in the US, FCC/FAA/NOAA). (b) OST Article VIII: the registering state keeps jurisdiction, control, and ownership of its object indefinitely, even after abandonment; the derelict is not res nullius, so removing it without the owner state's consent violates its retained rights.

35.25 † — design for demise (synthesis with Ch. 7)

Re-entry converts orbital kinetic energy to heat (compression heating, Ch. 7), melting and ablating most structure. Survivors are dense, high-melting-point parts — steel/tungsten reaction wheels, titanium tanks, glass optics — because their high ballistic coefficient and melting temperature let them reach the ground before fully ablating. A design change that helps demise: swap titanium tanks for aluminum (lower melting point → melts higher up), or fragment dense components / choose lower-melting materials, keeping casualty risk below 1 in 10,000.

35.27 † — reserved disposal propellant (synthesis with Ch. 3)

$v_e = 300(9.81) = 2{,}943\ \text{m/s}$; $m_p = 3{,}000(1 - e^{-11/2943}) = 3{,}000(1 - 0.99627) = 11.2\ \text{kg}$. So reserve $\sim 11\ \text{kg}$. Planning on "leftover" propellant is a mistake because disposal must be guaranteed: consumption varies (station-keeping overruns, leaks, anomalies), and a satellite that runs its tanks dry — or fails outright — cannot dispose of itself, stranding a derelict in a scarce, regulated slot. Reserve the disposal propellant explicitly and lock it out of operational use.

Chapter 36

Chapter 36 — Solutions to selected exercises († and odd)

Uses $g_0 = 9.81\ \text{m/s}^2$. All vehicle figures are approximate/illustrative (Tier 2/3).

36.1 † $v_e = I_{sp}\,g_0 = 80 \times 9.81 = 785\ \text{m/s} \approx 0.79\ \text{km/s}$. (Black powder is a poor propellant — barely a tenth the exhaust velocity of a good chemical engine.)

36.3 † With $v_e = 2.0\ \text{km/s}$ and mass ratio $3.2$: $\Delta v = 2{,}000 \times \ln(3.2) = 2{,}000 \times 1.163 = 2{,}326\ \text{m/s} \approx 2.3\ \text{km/s}$. (Using the precise $I_{sp} = 203$ s → $v_e = 1{,}991$ m/s gives ~2,319 m/s; both round to 2.3 km/s.)

36.5 † The Soviet Union, in 1957, on the R-7 Semyorka. The R-7 used parallel staging — a core plus four strap-on boosters that dropped away after launch — which the single-stage V-2 lacked; shedding that dead mass is what let it reach the ~9.4 km/s of orbital delta-v.

36.6 † Mass ratio $= e^{\Delta v/v_e} = e^{7800/800} = e^{9.75} \approx 17{,}150$. To reach orbital speed alone, a gunpowder rocket would need to be ~17,000-to-1 propellant — utterly unbuildable. (And that ignores gravity/drag losses; against the full ~9.4 km/s to orbit the ratio is ~160,000.) The lesson: a low $v_e$ cannot be rescued by any realistic mass ratio, because $v_e$ sits in the exponent.

36.7 The missing ~0.7 km/s went to gravity losses (thrust spent supporting the rocket's weight during the climb, not adding speed) and drag losses (air resistance on the way up) — the two ascent losses of Chapter 4. The rocket equation gives the ideal delta-v (~2.3 km/s); losses reduce the achieved burnout to ~1.6 km/s.

36.8 † $m_0/m_f = e^{9400/3000} = e^{3.13} \approx 22.9 \approx 23$. The R-7 needed an effective mass ratio near 23 — versus the V-2's 3.2. No single stage can reach 23 (the structural ceiling $1/\varepsilon$ is ~10–12), so the R-7 used parallel staging to reach it. Note also that its better engine mattered: at the V-2's $v_e = 2.0\ \text{km/s}$ the required ratio would be $e^{9400/2000} = e^{4.7} \approx 110$, far worse. Better $v_e$ and staging were both needed.

36.9 Single V-2 stage: $\Delta v = 2.0 \times \ln(3.2) = 2.33\ \text{km/s}$. Two idealized stages (each $v_e = 2.0$, mass ratio 3.2, ignoring the upper stage's own mass): $2 \times 2.33 = 4.66\ \text{km/s}$. Staging roughly doubles the capability — a vivid illustration of §36.4. But 4.66 km/s is still well short of orbit's 9.4, which is why 1950s designers also needed better exhaust velocities (and, in the R-7, clustering); see Case Study 2. (Real staging gives a bit less than double, because the first stage must accelerate the fueled upper stage.)

36.10 †

def v2_delta_v():
    return delta_v(isp_to_ve(203), 12_500, 3_900)
# isp_to_ve(203) = 203 * 9.80665 = 1990.75 m/s
# ln(12500/3900) = ln(3.205) = 1.1648
# 1990.75 * 1.1648 = 2318.7
# Expected output:
# 2319   (m/s)

36.11

def gunpowder_mass_ratio(target_dv, ve=785):
    return math.exp(target_dv / ve)
# gunpowder_mass_ratio(9400) = e^(9400/785) = e^11.9745 = 158,650
# Expected output (rounded to the nearest 10,000):
# 160,000

A gunpowder rocket would have to be ~160,000-to-1 propellant to orbit — physically impossible.

36.12 †

def timeline(events):
    ordered = sorted(events, key=lambda e: e[1])
    span = ordered[-1][1] - ordered[0][1]
    return ordered, span
# Input: [("Tsiolkovsky",1903), ("Apollo 11",1969), ("Sputnik",1957)]
# ordered = [("Tsiolkovsky",1903), ("Sputnik",1957), ("Apollo 11",1969)]
# span    = 1969 - 1903 = 66
# Expected output:
# ([("Tsiolkovsky",1903), ("Sputnik",1957), ("Apollo 11",1969)], 66)

36.13 † Two flaws. (1) Altitude is not orbit. Reaching ~100 km means touching space on a ballistic arc; staying in space needs ~7.8 km/s of sideways speed. The V-2's ideal ~2.3 km/s (burnout ~1.6 km/s) is nowhere near that, so it fell straight back. (2) You cannot orbit a single stage by adding fuel. The mass ratio orbit requires (~23 even with a good engine, ~110 at the V-2's $v_e$) exceeds the structural ceiling $1/\varepsilon$; the extra tankage outweighs the gain, and a bigger warhead makes the mass ratio worse. Orbit needs staging and a higher $v_e$ — not more propellant.

36.15 † Even split, $\Delta v_1 = \Delta v_2 = 4.7\ \text{km/s}$: $R_1 = e^{4700/2000} = e^{2.35} = 10.49$ and $R_2 = e^{4700/2400} = e^{1.958} = 7.09$. The ceiling for $\varepsilon = 0.10$ is a mass ratio of $1/0.10 = 10$. Stage 1 needs $R_1 = 10.49 > 10$ ($R_1\varepsilon = 1.05 > 1$) — infeasible; no first stage with V-2-class engines and 1940s structure can deliver 4.7 km/s. Stage 2 ($R_2 = 7.09$, $R_2\varepsilon = 0.71$) is feasible. Conclusion: orbit had to wait for better engines (higher $v_e$ lowers the required $R$) and/or lighter structures (lower $\varepsilon$ raises the ceiling); both matured by the mid-1950s, enabling the R-7 in 1957.

36.17 † Factor $= 12.9 / 2.3 \approx 5.6$. U.S. delta-v capability had to grow roughly five- to sixfold in under a decade — from a V-2-class suborbital hop to a three-stage Moon rocket. (A staggering pace, driven by the Space Race and Apollo funding.)

36.19 No. Crossing the ~100 km Kármán line means reaching space briefly, on a ballistic trajectory that comes straight back down. A "spacecraft" implies staying in space — achieving orbit (or escape), which needs ~7.8 km/s of sideways speed. The V-2 had ~1.6 km/s at burnout and fell back to Earth within minutes. Altitude is not orbit (Ch. 1, Ch. 4).

36.18 † Because gunpowder's exhaust velocity ($v_e \approx 785\ \text{m/s}$) is far too low, and the required mass ratio grows exponentially with $\Delta v / v_e$: to orbit, $m_0/m_f = e^{9400/785} = e^{11.97} \approx 160{,}000$. A rocket that is 160,000-to-1 propellant cannot be built — the tanks alone would outweigh everything. The ancients had the principle of reaction propulsion (Newton's third law), but not a propellant energetic enough to make the exponent manageable. Space waited for high-energy liquid propellants, not a bigger idea.

36.20 † Assume the first-stage booster is ~60% of the vehicle's build cost, and refurbishing it costs ~10% of a new booster, i.e. $0.10 \times 0.60 = 0.06$ of a vehicle's cost. Each reflight avoids building a new booster (saving $0.60$) but costs refurbishment ($0.06$), for a net saving of about $0.54$ of a vehicle's cost — per reuse. So even a single reuse recovers roughly half a vehicle's cost, and reuse pays off after just 1–2 flights on these assumptions. Ignored: the payload penalty from reserving landing propellant (Ch. 22), the R&D cost to develop reuse, recovery/refurbishment infrastructure, and booster life limits — all of which push the true break-even to more flights, but not by a huge factor.

36.21 † Sputnik's ~7.8 km/s sideways speed keeps it up because orbit is sideways: gravity is still pulling it down (it is constantly falling), but it moves sideways so fast that the Earth's surface curves away beneath it just as fast as it falls — it perpetually "misses the ground." The V-2, with far less sideways speed, simply fell back to the surface. Orbit is not about height; it is about horizontal velocity (Ch. 1, Ch. 4).

36.23 † The common problem is re-entry: safely shedding the ~7.8 km/s of orbital kinetic energy, which converts to enormous heat, without burning up or decelerating too hard. It is covered in Chapter 7 (compression heating, heat shields, the re-entry corridor). Gagarin, every Apollo crew, and every returning astronaut faced it.

Chapter 37 — The Space Shuttle (solutions to † and odd-numbered exercises)

All figures Tier 2 (historical/rounded) unless noted; the lessons do not depend on a third significant figure. Arithmetic verified by hand.

37.1 † Cost per flight $= \$209\text{B}/135 \approx \$1.55\ \text{billion}$. Cost per kilogram $= \$1.55\times10^{9} / 27{,}500\ \text{kg} \approx \$56{,}000/\text{kg}$. Against the promised $\sim\$220/\text{kg}$, the miss factor is $56{,}000/220 \approx 255$ — more than two orders of magnitude. (Even using the marginal per-flight cost of ~\$450M gives ~\$16,000/kg, still ~70× the promise.)

37.3 † Solid fraction $= \dfrac{2\times 12.5}{2\times 12.5 + 3\times 1.8} = \dfrac{25}{30.4} = 0.822$, i.e. ~82% of liftoff thrust. The crew could throttle and shut down the SSMEs (the ~18% they controlled), but had no control over the boosters supplying the other ~82% once lit — the majority of liftoff thrust came from motors that could not be commanded.

37.5 † $2/135 = 0.0148 = 1.48\%$, or about 1 in 68 ($135/2 = 67.5$). This is the same order of magnitude as the engineers' ~1-in-100 estimate, and roughly 1,500× worse than management's claimed 1-in-100,000. Reality landed on the engineers' side of Feynman's factor-of-1,000 disagreement.

37.7 † $c = m + F/n$ with $m = 0.45$, $F = 4.0$ (\$B). $n=4$: $0.45 + 1.00 = \$1.45\text{B}$. $n=10$: $0.45 + 0.40 = \$0.85\text{B}$. $n=40$: $0.45 + 0.10 = \$0.55\text{B}$. The flight rate $n$ does almost all the work: the fixed term $F/n$ collapses from \$1.00B to \$0.10B as $n$ rises, while the marginal cost stays put. Cost per flight is governed by how often you fly, not by the hardware.

37.9 † Reuse saves money only if (1) refurbishment is cheaper than building new and (2) the flight rate is high enough to spread the fixed program cost. Exercise 37.7 shows condition (2) is decisive: at low flight rate the fixed cost per flight ($F/n$) swamps everything, so even free hardware would be expensive to fly. Exercise 37.8 shows condition (1): if refurbishment is a large fraction of build cost, reuse's savings are capped no matter how many flights you fly. The Shuttle failed both — expensive refurbishment (months of work, engine overhaul, 24,000 tiles, ocean-recovered boosters) and a low flight rate (~4–5/yr). Its hardware was reusable; its operations were not cheap — hence "reusability is an operations problem, not a hardware problem."

37.10 † A gas-generator cycle burns a little propellant to drive the turbopumps and then dumps that exhaust overboard, throwing away a few percent of the propellant's energy and lowering $I_{sp}$. Staged combustion routes the turbine-drive (preburner) exhaust into the main chamber and burns it completely, recovering that energy — so, same propellants, higher $I_{sp}$. What it makes harder: the preburner gas is hot and at very high pressure, and it must pass through the turbopumps and plumbing, so every component lives in an environment that wants to melt or burst it. This is worse for a reusable engine because a single-use engine must survive that environment once, while a reusable one must survive it dozens of times — accumulating thermal fatigue, cracking, and wear — and be inspected and certified fit to fly again each time. The reuse requirement multiplies an already-brutal difficulty.

37.11 (odd) Once ignited, each SRB burned to completion (~124 s) with no throttle and no shutdown, so for the first ~2 minutes the crew was committed: there was no way to stop the boosters and separate to safety, hence no effective abort during SRB burn. The SSMEs, by contrast, could be throttled (67–109%) and commanded to shut down, which is why the liquid-only phases (after SRB separation) supported abort modes (RTLS, TAL, ATO, AOA). The uncontrollable 82% of liftoff thrust defined the most dangerous window.

37.12 † Max-Q is the peak of dynamic pressure $q = \tfrac12 \rho v^2$ (Ch. 5), where aerodynamic loads on the structure are greatest. Throttling the SSMEs down there reduces acceleration and speed through that region, holding $q$ and the loads within structural limits, then throttling back up after. A solid motor cannot be throttled on command — its thrust is fixed by the propellant grain geometry and burning area, so any thrust "dip" must be cast into the grain in advance and cannot adapt to the day's actual winds or air density. Less in-flight control means less margin exactly where loads peak.

37.13 †

def cost_per_flight(marginal, fixed_annual, flights_per_year):
    return marginal + fixed_annual / flights_per_year

print(round(cost_per_flight(0.45, 4.0, 4), 2))
print(round(cost_per_flight(0.45, 4.0, 40), 2))
# Expected output:
# 1.45
# 0.55

Trace: $0.45 + 4.0/4 = 1.45$; $0.45 + 4.0/40 = 0.55$.

37.15 †

def prob_no_loss(p, n):
    return (1 - p) ** n

print(round(prob_no_loss(0.015, 135), 2))
# Expected output:
# 0.13

Trace: $(1-0.015)^{135} = 0.985^{135}$. $\ln 0.985 \approx -0.01511$; $\times 135 \approx -2.040$; $e^{-2.040} \approx 0.13$. Interpretation: at a 1.5% per-flight loss rate there was only about a 13% chance of flying all 135 missions with zero losses — so two losses is entirely consistent with the rate. A loss-free program would have been the statistical surprise, not the losses.

37.16 † $R = r^{k} = 0.999995^{2000}$. $\ln R = 2000 \times \ln(0.999995) \approx 2000\times(-5.0\times 10^{-6}) = -0.01$; $R = e^{-0.01} \approx 0.9900$. Probability of catastrophic failure $1 - R \approx 0.0099 \approx 1\%$, i.e. ~1 in 100 — the same order as the engineers' estimate and within a factor of ~1.5 of the demonstrated 1-in-68. What "independent" hides: real Criticality-1 items are not independent — a single common cause (a cold snap, a management decision to fly, a bad manufacturing batch) can defeat many "independent" items at once, so the true reliability is usually worse than the product formula suggests.

37.17 (odd) The primary and secondary O-rings shared a common cause: the same joint rotation (the gap opening at ignition) and the same low temperature acted on both. Redundancy multiplies reliability only when the redundant elements fail independently; here one event — cold plus rotation — could unseat both, so the secondary was not true independent redundancy and the joint behaved as a single point of failure despite its nominal backup. On Challenger the cold stiffened both seals and the rotation opened the gap for both simultaneously.

37.18 † The failure mode is normalization of deviance. The rule the team should hold to: judge the anomaly against the design requirement (it should not be occurring at all; if the item is Criticality 1, any anomaly is a stop condition), not against "we survived it the last six times." Each survival is not evidence of safety — it may simply be luck that has not yet run out. The Shuttle's history says what happens if they do not hold that line: a normalized anomaly eventually recurs under worse conditions and the latent failure turns catastrophic. O-ring erosion (Challenger) and foam shedding (Columbia) were both normalized anomalies that each, in the end, killed a crew.

37.19 (odd, mission project) No single numeric answer. A strong reflection: (1) states the reuse posture honestly and ties it to the projected flight rate (per 37.7/37.9, low-flight-rate missions may rationally choose an expendable vehicle); (2) names the mission's Criticality-1 single points of failure carried over from the Chapter 32 risk assessment; (3) names one specific, plausible warning sign the program might be tempted to normalize and the concrete rule it will hold instead. Reward specificity over generality.

37.21 † Reusing the hardest hardware — a crewed spaceplane, high-pressure staged-combustion engines, and 24,000 fragile hand-fitted tiles — meant refurbishment was extraordinarily expensive and slow (months of inspection, engine removal and overhaul, tile-by-tile checks, boosters recovered from salt water and rebuilt), and the low flight rate could not amortize the huge fixed workforce. So "reuse" did not lower cost. For reuse to win, refurbishment must be cheap (a small fraction of build cost and quick) and the flight rate high. Reusability by itself is not the prize; cheap, frequent reuse is.

37.23 † The leading edge has a small nose radius $R_n$, and the stagnation heat flux scales as $\dot q \propto 1/\sqrt{R_n}$, so a small radius gives a high flux; the shock also stands closest to a sharp edge. Highest flux forces the use of reinforced carbon-carbon (good to ~$1{,}650\,^\circ\text{C}$), beyond what silica tiles survive. A hole there is catastrophic — not merely damaging — because behind the RCC is ordinary aluminum with no thermal protection (none was ever supposed to be needed): once the ~$10{,}000\ \text{K}$ shock-layer gas enters through the breach it melts the internal structure, and on the way down there is no repair and no abort (Ch. 7). This is exactly the Columbia failure.

37.25 † (a) Criticality 1 — an SRB field-joint O-ring during first-stage burn: its failure leads to burn-through and loss of vehicle, and the "redundant" secondary is defeated by common cause (37.17), so effectively no backup. (b) Not a single Criticality-1 loss in the same sense — after SRB separation the Shuttle had engine-out abort modes (ATO/AOA/TAL/RTLS) designed to tolerate the loss of one SSME, so one engine out is (intended to be) survivable, though serious. (c) A single belly tile is generally not Criticality 1 on its own — the mosaic and margins give some tolerance, and consequence is location-dependent; loss of a tile in a specially hot/critical spot could approach Criticality 1. (d) An RCC wing-leading-edge panel during re-entry is Criticality 1: a breach admits shock-layer gas to unprotected structure (the Columbia mechanism) with no backup.

37.27 † With one SSME out: thrust $= 2\times 1.8 + 2\times 12.5 = 3.6 + 25 = 28.6\ \text{MN}$. Weight $= 2{,}030{,}000\ \text{kg}\times 9.81 = 19.9\ \text{MN}$. $T/W = 28.6/19.9 = 1.44 > 1$, so yes, it still leaves the pad. The reveal: the two SRBs (25 MN) dominate liftoff thrust, so losing one SSME (1.8 MN) drops $T/W$ only from 1.53 to 1.44. The liftoff margin lives in the solids — precisely the elements the crew could neither throttle nor shut down, which is the safety irony at the heart of the design.

Chapter 38 — SpaceX and the Reusability Revolution

Solutions to the daggered (†) and odd-numbered exercises. All figures Tier 2/3, consistent with the chapter; $g_0 = 9.81\ \text{m/s}^2$.

38.1 Nine Merlins × $845\ \text{kN} = 7{,}605\ \text{kN} \approx 7.6\ \text{MN}$ of sea-level thrust.

38.3 $62\times10^{6}/22{,}800 \approx \$2{,}719/\text{kg}$. The Shuttle's ~$\$54{,}500/\text{kg}$ is about $20\times$ higher — Falcon 9 is roughly an order of magnitude cheaper per kilogram (Tier 2).

38.5 (a) Merlin — kerolox, gas-generator cycle. (b) Raptor — methalox, full-flow staged combustion. (c) RS-25 — hydrolox, (fuel-rich) staged combustion. Raptor is the full-flow one, and it was the first full-flow staged combustion engine ever to fly (2019).

38.7 (a) All propellant burned: $m_0 = 120+25+400 = 545\ \text{t}$, $m_f = 120+25 = 145\ \text{t}$, so $\Delta v = 2{,}850\ln(545/145) = 2{,}850\times1.324 = 3{,}774\ \text{m/s}$. (b) Reserve $12\%$ of $400 = 48\ \text{t}$: $m_f' = 145+48 = 193\ \text{t}$, so $\Delta v' = 2{,}850\ln(545/193) = 2{,}850\times1.038 = 2{,}959\ \text{m/s}$. (c) Recovery cost $= 3{,}774 - 2{,}959 = 815\ \text{m/s}$ of ascent delta-v.

38.9 Cheaper when $30/N + 1 + 18 < 30 + 18$, i.e. $30/N < 29$, i.e. $N > 30/29 = 1.03$. So reuse first wins essentially by the second flight ($N \approx 1.03$). (This matches Chapter 22's "break-even in ~1–2 flights.")

38.11 $m_0 = 120+1{,}200+100 = 1{,}420\ \text{t}$, $m_f = 120+100 = 220\ \text{t}$, so $\Delta v = 3{,}700\ln(1{,}420/220) = 3{,}700\times1.865 = 6{,}900\ \text{m/s} = 6.9\ \text{km/s}$. Yes — that comfortably exceeds trans-Mars injection's ~$3.6\ \text{km/s}$ from LEO (with margin left over).

38.12 payload_penalty(3862, 3140) returns $3862 - 3140 = 722$. # Expected output: 722.

38.13 cost_per_kg(62, 22800) $= 62\times10^{6}/22{,}800 = 2719$; cost_per_kg(1500, 27500) $= 1500\times10^{6}/27{,}500 = 54545$. # Expected output: 2719 then 54545.

38.14 reuse_cost(30, 1, 18, N) = 30/N + 19: $N{=}1\to 49.0$; $N{=}2\to 34.0$; $N{=}5\to 25.0$; $N{=}10\to 22.0$.

38.15 The flaw ignores the fixed per-flight cost $F$ that reuse never recovers. In the model $C_{\text{reuse}}(N) = M/N + R + F \to R + F$ as $N \to \infty$ — a floor far above propellant cost. The dominant ignored term is $F$, the expended second stage (built new every flight) plus operations and range costs. Recovering the first stage removes the first-stage manufacturing term $M/N$, not the whole cost; propellant is cheap, but you keep buying a fresh upper stage.

38.17 Droneship (reserve $10\% = 40\ \text{t}$): $m_0 = 545\ \text{t}$, $m_f = 120+25+40 = 185\ \text{t}$, $\Delta v = 2{,}850\ln(545/185) = 2{,}850\times1.081 = 3{,}079\ \text{m/s}$. RTLS (reserve $18\% = 72\ \text{t}$): $m_f = 120+25+72 = 217\ \text{t}$, $\Delta v = 2{,}850\ln(545/217) = 2{,}850\times0.921 = 2{,}624\ \text{m/s}$. RTLS produces $455\ \text{m/s}$ less ascent delta-v, so it loses more payload — the boostback burn is pure overhead. You still choose RTLS when the faster, cheaper turnaround of a booster back at the pad (lower operations cost $F$) outweighs the payload cost, or when no droneship is positioned downrange.

38.19 (a) $1{,}200/120 = 10$ tanker flights. (b) $10 \times \$25\ \text{M} = \$250\ \text{M}$ to deliver the propellant (plus the mission Ship itself → 11 launches total). (c) Eleven launches per mission is only affordable if each launch is cheap and the vehicles are reused; at Shuttle-era ~$\$1.5\ \text{B}$ per flight the campaign would cost ~$\$16\ \text{billion}$ and be absurd. Orbital refueling is a child of reuse economics.

38.20 A reusable second stage must carry a heat shield, control surfaces, and landing propellant to orbit — dead mass. This hurts most on an upper stage because that mass is accelerated all the way to orbital velocity (the mass ratio it ruins is the one that matters most), whereas first-stage recovery mass is dropped at a lower speed. Falcon 9's small, highly optimized second stage would lose most of its payload if made reusable, and it cannot cheaply be refueled — so partial reuse of the first stage alone is the sweet spot for Falcon 9. Crossing to full reuse only pays as a purpose-built, refuelable vehicle (Starship), where the whole architecture is redesigned around the penalty.

38.21 $4{,}000/50 = 80$ launches (order of magnitude: ~$10^2$). At Shuttle-era prices (~$\$1.5\ \text{B}$/flight), 80 launches would cost ~$\$120\ \text{billion}$ in launch alone — economically impossible for a commercial constellation. At reusable prices (~$\$60\ \text{M}$/flight), ~$\$5\ \text{billion}$ of launch (and fewer flights with higher payload) makes the constellation feasible — which is exactly why mega-constellations appeared only after reusable launch (ties to Chapter 33).

38.22 Steel wins for a reusable, re-entering vehicle for two reasons: (1) stainless steel keeps its strength to well over $1{,}000\ \text{K}$, where carbon composite's resin degrades (~$450$–$600\ \text{K}$), so the steel structure can double as the heat shield and the vehicle needs far less thermal protection — the heat-shield mass saved offsets steel's ~$3$–$5\times$ density penalty; (2) steel is roughly one-fiftieth the cost of composite per kilogram, weldable in the open air (no autoclave), and actually strengthens at cryogenic temperature — ideal for a mass-produced, frequently reflown vehicle. Carbon fiber is lighter per unit strength but cannot survive re-entry heat without a full (heavy, fragile) heat shield, and is expensive and hard to inspect and repair.

38.23 A solid motor cannot be shut down, relit, or throttled once ignited (Chapter 17). Propulsive landing requires an engine that can relight in flight (for the entry and landing burns) and throttle deeply (to time the hoverslam so velocity reaches zero exactly at the ground). A solid has neither capability, so it cannot perform a controlled landing.

38.24 $\Delta(\ln R) = \Delta v / v_e = 722/2{,}840 = 0.254$, so the mass ratio was reduced by a factor $e^{0.254} = 1.29$. Reserving landing propellant cut the first stage's mass ratio by about $29\%$ (check: $3.90/1.29 = 3.02$, the reusable ratio in §38.3). Because $\Delta v$ scales with $\ln R$, a modest-looking mass-ratio cut translates directly into the $722\ \text{m/s}$ ascent loss.

38.25 In a gas-generator engine (Merlin) a small preburner drives the turbine and its exhaust is dumped overboard — that dumped flow is the "wasted" propellant (a small $I_{sp}$ loss), and the open cycle limits chamber pressure to ~$100\ \text{bar}$. In full-flow staged combustion (Raptor) all the turbine gas is routed into the main chamber (nothing dumped → the waste is recovered), and the two-preburner, gas–gas, no-interpropellant-seal, cooler-turbine design lets it reach ~$300\ \text{bar}$ while remaining reusable. The dumped-overboard turbine flow is the waste; closing the cycle recovers it and enables higher pressure.

38.26 Same equation $C_{\text{reuse}} = M/N + R + F$, opposite outcomes. The Shuttle had a huge refurbishment cost $R$ (tiles inspected/replaced, engines torn down between flights) and huge fixed program cost $F$ (standing army, infrastructure), spread over a low flight rate $N$ — so the cost per flight stayed enormous despite technical reusability. Falcon 9 keeps $R$ small by designing for reuse (Block 5), has a lower $F$, and flies at a high $N$ that amortizes the build cost $M$ — so its cost per flight actually fell. Reuse only cuts cost when $R$ and $F$ are small and $N$ is large.

38.27 (Answers will vary; example for Starship.) Chapter 3 — the rocket equation and the mass-ratio penalty that full reuse must pay. Chapter 17 — full-flow staged combustion and the Raptor engine cycle. Chapter 22 — staging, propulsive landing, reuse economics, autogenous pressurization, and the tower catch. Chapter 23 — stainless steel as a structural material. Chapters 7 and 24 — re-entry heating and thermal control, explaining why steel needs less thermal protection. Chapter 34 — Mars, methane ISRU, and orbital refueling. Chapter 38 braids all of these into one vehicle, each choice a response to the tyranny of the rocket equation and the economics of throwing hardware away.

Chapter 39 — The Future

39.1 Artemis is a program — NASA's international effort to return humans to the Moon to stay. The Gateway is a station — a small crewed outpost in a near-rectilinear halo orbit around the Moon. The Human Landing System (HLS) is a lander — the crewed vehicle that carries astronauts from lunar orbit to the surface and back, bought from commercial providers.

39.3 Suborbital means going up into space but not sideways fast enough to stay there — you cross the edge of space and fall back. Orbital means reaching the sideways speed (~7.8 km/s) at which you keep missing the ground as you fall, circling the planet. The difference is horizontal velocity, not altitude.

39.5 The near-term prize is water, because it can be split into hydrogen and oxygen for propellant — made in space, it never has to be launched up from Earth through ~9.4 km/s of delta-v. Water is therefore worth far more where it is mined (in orbit or on a small body) than on Earth's surface, where it is nearly free; the value is in avoiding the rocket equation's launch cost.

39.7 Specific kinetic energy $\tfrac{1}{2}v^2$: suborbital $= \tfrac{1}{2}(1000)^2 = 0.5\ \text{MJ/kg}$; orbital $= \tfrac{1}{2}(7800)^2 = 30.4\ \text{MJ/kg}$. Ratio $= 30.4/0.5 \approx 61$. The vehicle difference is larger still because delta-v enters the mass ratio exponentially: the hop needs ~35% propellant, orbit ~94% — a reusable rocket plane versus an orbital-class staged rocket. A 61× energy gap becomes a change of vehicle species.

39.9 For 1 kg of $\mathrm{CH_4}$ ($16\ \text{g/mol} \Rightarrow 62.5\ \text{mol}$): the reaction needs $4 \times 62.5 = 250\ \text{mol}$ of $\mathrm{H_2}$, i.e. $250 \times 2 = 500\ \text{g} = \mathbf{0.5\ kg\ H_2}$ imported from Earth. It also yields $2 \times 62.5 = 125\ \text{mol}$ of water $= 125 \times 18 = 2250\ \text{g} = \mathbf{2.25\ kg\ H_2O}$ as a bonus (electrolyze it for oxygen). The $\mathrm{CO_2}$ ($2.75\ \text{kg}$) comes free from the Martian atmosphere.

39.11 From the delta-v map, LEO → trans-Mars injection is ~3.6 km/s and LEO → Earth escape is ~3.2 km/s — nearly the same. So once you are in LEO, reaching Mars costs barely more than simply leaving Earth's gravity. The expensive part (~9.4 km/s) was getting to LEO in the first place, which you pay whether you are going to Mars or to GEO. Distance adds almost nothing; "too far" misdiagnoses the problem — the barrier is climbing out of Earth's gravity well and, for a crew, surviving the round trip.

39.12 launch_cost(6200, 2700) $= 6200 \times 2700 = 16{,}740{,}000$. # Expected output: 16740000.

39.13 energy_ratio(7800, 1000) $= (7800/1000)^2 = 7.8^2 = 60.84 \to 60.8$. # Expected output: 60.8.

39.14 hydrogen_for_methane(10000) $= 0.5 \times 10000 = 5000.0$. # Expected output: 5000.0. (To make 10 t of methane you must import 5 t of hydrogen — which is why mining Martian water ice for hydrogen, rather than shipping it, is so attractive.)

39.15 economy_2040(500, 0.06, 17) $= 500 \times 1.06^{17} = 500 \times 2.6928 \approx 1346$. # Expected output: 1346. Tier 2 — a projection, not a measurement; the answer depends entirely on the assumed growth rate (4%/yr gives ~$974 B, 8%/yr gives ~$1,850 B).

39.16 Two independent errors. (1) Delivery: the platinum is worth thousands of dollars per kg on Earth's surface, but the rocket equation charges enormously to return bulk mass down a gravity well and land it — the return cost can exceed the commodity value, which is why the real business is using resources in space. (2) Market: "today's prices" assume scarcity; delivering even a fraction of an asteroid's platinum would collapse the price, so the headline valuation evaporates the instant you succeed. Ignoring what physics charges to deliver and what economics does to the price is the classic double error.

39.17 The rocket equation makes suborbital and orbital different species, not different sizes: a hop needs ~1.5 km/s (mass ratio ~1.5, ~35% propellant) and orbit needs ~9.4 km/s (mass ratio ~16, ~94% propellant). They sit on opposite sides of the exponential, so hundreds of suborbital passengers tell you almost nothing about when an orbital seat becomes affordable — that depends on the separate, harder problem of cheap orbital-class reusable launch.

39.18 The ">$1 trillion by 2040" projection (Tier 2) reaches that figure mainly by compounding the existing satellite-services economy — broadband constellations, Earth observation, navigation — not by assuming asteroid mines or Mars cities. Those exotic sectors are the models' upside, not the base case. So the conclusion "most of that money will come from mining and colonies" inverts the projection's own assumptions; the load-bearing growth is boringly terrestrial-facing satellites.

39.19 (Mission project — model answers.) Track A (GEO comsat): the trend that would change the design is in-space servicing/refuelling (life-extension changes the end-of-life disposal plan) and LEO-constellation competition; the physics that stays untouched is the GTO→GEO delta-v and station-keeping budget. Track B (lunar lander): ISRU-produced lunar propellant and commercial cargo-to-surface services could reshape sourcing; the descent delta-v itself is unchanged. Track C (Mars orbiter): becoming a relay node in a "Mars network" changes the comms concept; the trans-Mars trajectory and orbit-insertion burn are unchanged. Track D (asteroid rendezvous): your mission becomes prospecting for a future resource venture; the rendezvous delta-v and small-body operations are unchanged. Full credit for naming one changed and one untouched element and saving it to the MDR.

39.21 One reason it could help: a libration-point (or high-orbit) staging station near Mars — analogous to the Gateway at Earth–Moon L2 — could be a fuel-cheap waypoint and a communications relay (e.g., for the far side or for landers out of direct Earth view), reusing the Chapter 15 halo-orbit logic. One reason the analogy may not carry: the Mars case has a much longer, window-constrained transit (Ch. 11), different Sun–Mars and Mars–Phobos geometries, and a crew that wants to reach the surface quickly, so a staging station may add delay and mass without the continuous-relay payoff that makes the lunar Gateway attractive. Verdict: worth studying, not obviously worth building — the honest posture of the whole chapter.

39.22 Delta-v: returning bulk mass from a heliocentric or asteroid orbit down to Earth's surface means matching orbits, hauling the mass back, and paying delta-v the whole way down the gravity well — the rocket equation charges as much (or more) to return mass as to launch it. Economics: platinum's high price assumes scarcity; delivering even a fraction of an asteroid's worth would flood the market and crash the price. So the maneuver is both physically expensive and self-defeating: you spend a fortune in propellant to destroy the very scarcity that made the cargo valuable. The sound business is water sold as propellant, in space.

39.23 A Mars launch window opens only about every 26 months (Ch. 11), and a minimum-energy transit takes ~6–9 months, so a "resupply" package ordered today might arrive well over a year later — and cannot be sent at all for most of that 26-month cycle. Resupply-on-demand, the model that keeps the ISS stocked every few months, is therefore impossible at Mars. This forces the life-support system (Ch. 28) to be nearly closed-loop — recycling air and water (and ideally growing some food) with years of reliability and no lifeboat — which is exactly why closed-loop ECLSS is one of the two "not-yet-demonstrated" Mars challenges.

39.24 For the Gateway: it is a fuel-cheap, self-maintaining waypoint (a few m/s/yr, Ch. 15) that stays in continuous view of Earth and the lunar south pole, serves as a communications relay, an international docking node, and an aggregation point where reusable landers and crews can meet — infrastructure that makes repeated missions cheaper. Against: it forces the lander to supply extra delta-v to descend from and return to the high NRHO (versus a direct or low-lunar-orbit approach), and it is an added system to build and operate. The tie-breaker is the mission goal: a single Apollo-style sortie is better served by flying straight to the surface, but a sustained, reusable program — Artemis's stated aim — is better served by the persistent infrastructure the Gateway provides.

39.25 All three follow from the rocket equation's exponential link between delta-v and propellant mass: Mars missions must use ISRU because bringing return propellant compounds that exponential across landing, cruise, and launch; asteroid platinum should stay in space because returning mass down Earth's gravity well pays the same exponential in reverse; and orbital tourism is far harder than suborbital because orbit's ~9.4 km/s versus a hop's ~1.5 km/s turns, exponentially, into ~94% versus ~35% propellant — a different vehicle class. One equation, three consequences.

39.27 The master variable is the falling cost of launch, driven by reusability (Ch. 38, theme 5). It is not new physics — the rocket equation is unrepealed and unchanged — but for sixty years the price of obeying it sat frozen near the top of the curve, and that price (not the physics) is what kept the Moon empty, stations governmental, and passengers professional. Drop the price by an order of magnitude and activities that were always physically possible become economically worth doing: "possible" migrates into "probable." The binding constraint was economic all along.

Chapter 40 — Solutions to starred (†) and odd-numbered exercises

All arithmetic hand-traced; $g_0 = 9.80665\ \text{m/s}^2$ where a Checkpoint value must match, else 9.81. Design problems (40.19–40.23, 40.29, 40.31, 40.33) give a model answer / rubric, since the deliverable is the reader's own mission.

Part A

40.1 † The ten MDR rows (§40.1), in order: (1) objective & requirements — what must it do? (2) orbit / destination — where must it fly? (3) delta-v budget — what's the velocity price? (4) propulsion selection — what engine pays it? (5) vehicle mass sizing — how big is it? (6) trajectory & window — when can it go? (7) spacecraft systems — what keeps it useful? (8) launch vehicle — what lifts it? (9) operations & timeline — how is it flown? (10) risk assessment — what can kill it?

40.3 † The blank is the delta-v budget. It is the master constraint because it is the one budget that converts, through the exponential rocket equation $m_0/m_f = e^{\Delta v/v_e}$, into propellant mass — which dominates wet mass, which picks the launch vehicle, which sets the cost. An error there is amplified exponentially; an error in any other budget is not.

40.5 † A design defense leads with (1) the driving requirement (the one requirement that sizes the mission), then (2) the margins (mass/power/delta-v, and why each is sized as it is), then (3) the single points of failure (the un-abortable events, named before the board finds them).

Part B

40.6 † $v_e = 320 \times 9.80665 = 3{,}138\ \text{m/s}$. Mass ratio $= e^{2726/3138} = e^{0.8687} = 2.38$. Wet mass $= 2.38 \times 1{,}500 = 3{,}576\ \text{kg}$; propellant $= 3{,}576 - 1{,}500 = 2{,}076\ \text{kg}$. Matches §40.2. Sanity: mass ratio 2.38 → 58% propellant, reasonable for a chemical apogee stage.

40.7 (odd) Electric $v_e = 1{,}800 \times 9.80665 = 17{,}652\ \text{m/s}$; mass ratio $= e^{2726/17652} = e^{0.1544} = 1.17$; wet $= 1{,}750\ \text{kg}$; propellant $= 250\ \text{kg}$. Propellant drops by a factor $2{,}076 / 250 \approx 8.3$. Two electric comsats ($2 \times 1{,}750 = 3{,}500\ \text{kg}$) fit under 5,500 kg to GTO; the chemical pair (7,152 kg) does not.

40.8 † Mass ratio 6.72 → propellant fraction $1 - 1/6.72 = 0.851$, i.e. 85%. The comsat's 2.38 → 58%. The delta-v roughly doubled (2,726 → 5,980 m/s), but because it sits in the exponent of $e^{\Delta v/v_e}$, doubling the exponent squares the ratio ($2.38^2 \approx 5.7$, and the extra bit of delta-v pushes it to 6.72). The exponential punishes, so ~2× delta-v gives ~3× mass ratio.

40.9 (odd) At $r = 3{,}790\ \text{km}$, $\mu = 4.283\times10^{4}$: $v_{\text{hyp}} = \sqrt{2.65^2 + 2(42830)/3790} = \sqrt{7.02 + 22.60} = \sqrt{29.62} = 5.44\ \text{km/s}$; $v_c = \sqrt{42830/3790} = \sqrt{11.30} = 3.36\ \text{km/s}$; MOI $= 5.44 - 3.36 = 2.08\ \text{km/s}$. Confirmed.

40.10 † Ion: $v_e = 29{,}420\ \text{m/s}$, ratio $e^{5300/29420} = e^{0.180} = 1.20$, wet $838\ \text{kg}$, xenon $138\ \text{kg}$. Chemical: $v_e = 3{,}138\ \text{m/s}$, ratio $e^{5300/3138} = e^{1.689} = 5.41$, wet $3{,}789\ \text{kg}$, propellant $3{,}089\ \text{kg}$. Propellant factor $3089/138 \approx 22\times$. Mandatory because at chemical $I_{sp}$ the 3-tonne propellant load turns an ~840 kg probe into a ~3,800 kg one, blowing the mass (and launch) budget; only the high $v_e$ collapses the mass ratio toward 1.

Part C (see code/exercise-solutions.py for the traced code)

40.11 † roll_up_dv([("GTO->GEO",1800,0.05),("SK",750,0.10),("disposal",11,0.0)]) → ideal $1800+750+11 = 2561$; margined $1890+825+11 = 2726.0$. Output (2561, 2726.0).

40.13 † closes(3576, 5500) → $3576 < 5500$ True, $5500//3576 = 1$ → (True, 1). closes(1750, 5500) → True, $5500//1750 = 3$ → (True, 3). Caveat: 3 is the mass-only count; operationally two are stacked (fairing volume + dispenser). Mass closure is necessary, not sufficient.

40.15 † capture_burn(2.65, 42830, 3790, "circ"): $v_{\text{hyp}} = 5.443$, $v_c = 3.362$ → $2.081\ \text{km/s}$. "ellipse": $v_{\text{esc}} = \sqrt{2(42830)/3790} = 4.754$ → $5.443 - 4.754 = 0.689\ \text{km/s}$. Output 2.081 0.689.

Part D

40.16 † Adding 5,980 + 2,900 = 8,880 m/s and sizing once treats the whole vehicle (descent structure included) as if it flew the ascent leg too — but the descent stage is left on the Moon, so its dry mass must not be carried up. Onion sizing sizes the ascent stage first (its wet mass becomes the descent stage's payload), so the descent propellant only ever pushes the fueled ascent stack, not a phantom combined vehicle. Adding the budgets over-counts what must be accelerated on the return and mis-sizes both stages.

40.17 (odd) They double-counted the launch vehicle's job. The ~11.9 km/s is paid by the launch vehicle, which stages away (or flies home) before the spacecraft's own engine ever fires; the spacecraft only ever provides its own ~2.7 km/s. Merging them violates the staging principle of Ch. 3: you never carry the launch vehicle's delta-v into your own mass ratio.

40.18 † The 2,726 m/s is a point estimate of an uncertain quantity — masses grow, engines underperform, small legs (finite-burn, extra TCMs) get missed. Designing to it exactly guarantees that any adverse deviation makes the mission fall short, and there is always at least one. The correct move is to reduce the uncertainty (better mass tracking, analysis) so the margin can shrink with justification — never to delete the margin and hope (Ch. 29).

Part E (model answers / rubric)

40.19 † Full MDR for the reader's track. Rubric: all ten rows present with a defensible number and source chapter; delta-v budget margined per-leg; vehicle sized via the rocket equation; closure shown (wet mass < launch capacity, with margin); driving requirement named. Model (Track A): 15-yr GEO comsat, GEO orbit, 2,726 m/s margined, chemical Isp 320 s → 3,576 kg wet, one Falcon 9 to GTO, ~12 m² array, 3-axis GN&C, disposal to graveyard, dominant risk = 15-yr endurance; driving requirement = lifetime.

40.21 † Sensitivity proof of the driving requirement. Rubric: pick the candidate, perturb it a little (e.g. Track A lifetime 15 → 20 yr adds ~250 m/s station-keeping → re-size shows wet-mass jump), then perturb another requirement by a comparable fraction and show a smaller wet-mass/closure change. The driving requirement is the one with the largest sensitivity. Track A: lifetime; Track B/CS: landed/returned mass; Track C: science-orbit altitude (→ capture Δv); Track D: target inclination.

40.23 † Harder-requirement redesign. Rubric: state the change, re-run the spine, report what broke (usually: wet mass exceeds launcher, or a subsystem exceeds mass budget) and the fix (better engine, staging, aerobraking, relaxed elsewhere, bigger launcher). E.g. Track B cargo 2,000 → 4,000 kg roughly doubles landed mass → wet ~29,600 kg single-stage → now needs Falcon Heavy/Starship or staging.

Part F

40.24 † The 46:1 ratio was for a 200 kg capsule; the ratio is (to first order) independent of capsule mass because the rocket equation scales linearly in payload for fixed delta-v and stage dry-mass fractions. So a 100 kg capsule → roughly $46 \times 100 \approx 4{,}600\ \text{kg}$ in LEO, ≈ 5,000 kg to one significant figure. (In reality the ratio drifts because fixed stage-structure masses do not scale perfectly with payload; the linear estimate is the back-of-envelope answer.)

40.25 (odd) A single stage carries all its dry structure (engines, empty tanks) through every leg, so its propellant fights dead mass the whole trip — worst on the return, where it hauls empty descent tanks back up. Staging sets down mass the moment it stops being useful (spent descent stage stays on the Moon), so the ascent propellant pushes only the small ascent stage. Same lesson as Falcon 9 (Ch. 3): staging buys delta-v not with more fuel but by not accelerating dead structure.

40.26 † Landing requires thrust greater than local weight to decelerate and hover; electric thrusters produce milli-newtons to newtons — orders of magnitude too little to counter gravity during a powered descent (Ch. 20). Electric works only where you can spread a small thrust over months (cruise, orbit-raising, station-keeping), never for a minutes-long, high-g descent-and-hover. Hence chemical for the landing; electric can deliver the vehicle to lunar orbit (SMART-1), but not set it down.

40.27 (odd) The idealized coplanar heliocentric two-burn to 1.2 AU is only ~2.6 km/s (dep 1.32 + arr 1.27), and a rendezvous must cancel the arrival excess. The extra ~3.3 km/s in the ~4.6 km/s budget comes mostly from (1) inclination/eccentricity matching to a real target (even 3° at ~28 km/s heliocentric is ~1.5 km/s), and (2) low-thrust inefficiency — a spiral trajectory lacks the impulsive/Oberth advantage and effectively costs more delta-v than the ideal impulsive value. Both are named in §40.5.

Part G

40.28 † Two-stage lunar lander (qualitative): a TLI stage does the 3,100 m/s trans-lunar burn and is jettisoned; the descent stage then does only capture + descent (~2,400 m/s ideal) with a much smaller "landed" mass because it no longer hauls empty TLI tanks to the surface. Sizing the descent stage against 2,400 rather than 5,500 m/s slashes its mass ratio (from 6.72 toward ~2), and the TLI stage pushes the lighter descent stack — total LEO mass drops sharply. Principle: optimal staging / drop dead mass (Ch. 22, Ch. 3).

40.29 (odd) Model (Track C Mars orbiter): legs = launch to LEO (LV) → trans-Mars injection (LV upper stage, from C3) → cruise TCMs (propulsive) → Mars orbit insertion (propulsive) → aerobraking (aerodynamic, ~0 delta-v) → science-orbit maintenance (propulsive). Dominant leg for the spacecraft: the MOI capture burn. Aerodynamic leg: aerobraking. No gravity assist in the minimum-energy version.

40.30 † The dominant GEO station-keeping perturbation is the third-body (luni-solar) gravity that tugs the orbit plane, causing the inclination to drift ~0.75–0.95°/yr — corrected by north–south station-keeping, which is the large cost (~45–50 m/s/yr of the ~50). East–west keeping (against the J2/triaxiality longitude drift and solar radiation pressure) is far cheaper (~2–4 m/s/yr). So N–S dominates the 750 m/s (Ch. 12).

40.31 (odd) Around the loop (§29.5): +120 W → bigger array → +array mass, partly offsetting the −30 kg dry saving; net dry-mass change Δ, then Δ×(mass ratio) into wet mass, then launch/cost. To decide you need the array's specific mass (kg/W, ~0.02–0.05 kg/W for the extra 120 W ≈ 2.4–6 kg) plus battery/harness knock-ons: if the array+battery growth (~5–10 kg) is less than the 30 kg saved, it is a net win — the dry mass still drops, and the drop is amplified by the mass ratio into a larger wet-mass saving.

40.32 † The single trade running through all of propulsion is high thrust vs. high efficiency (specific impulse) — you cannot maximize both, because high $v_e$ means low mass flow for a given power. High thrust wins where you must fight gravity fast (Track B lunar descent — chemical; also launch). High efficiency wins where you have time and a big delta-v (Track D asteroid — ion; also GEO orbit-raising). It is the same tension as the rocket-equation-vs-clock trade the whole chapter turns on.

40.33 (odd) One-page synthesis (rubric): a single connected narrative for the reader's mission — objective → driving requirement → orbit → delta-v budget (margined) → mass ratio (the amplifier) → wet mass → launch vehicle (closure) → operations/timeline → dominant risk/SPOF → rough cost. Rewards: every link explicit, every number defensible and unit-carrying, uncertainties flagged, and the driving requirement traced through the whole chain. This is the reader's MDR in prose — the deliverable the whole book was built to produce.