Chapter 6: Energy in Space — The Vis-Viva Equation

"There is a fact, or if you wish, a law, governing all natural phenomena that are known to date. There is no known exception to this law — it is exact, so far as we know. The law is called the conservation of energy." — Richard P. Feynman, The Feynman Lectures on Physics (1963)

Overview

Ask someone how fast the International Space Station is moving and they will usually guess wrong in a revealing way: they picture it drifting lazily, a slow float against a black sky. It is in fact tearing around the Earth at about $7.7\ \text{kilometers every second}$ — fast enough to cross the United States in under seven minutes. Now ask the harder question, the one this chapter is built around: if the astronauts fired a thruster forward, along their direction of travel, to add speed, what would happen? The honest answer is one of the strangest facts in all of physics, and by the end of this chapter you will not only accept it but be able to prove it. They would climb to a higher orbit — and end up going slower. Speed up to slow down. Fire forward to fall behind. It sounds like nonsense, and it is exactly how every rendezvous in spaceflight is actually flown.

The key that unlocks this, and almost everything else about orbits, is energy. In Chapter 2 we treated gravity as a force and used it to find escape velocity. In Chapter 3 we learned to buy velocity change — delta-v — with propellant. This chapter fuses the two ideas by asking not "what force acts?" but "how much energy does an orbit carry, and where does it go?" That single shift in viewpoint pays off enormously. It gives us one beautifully compact equation — the vis-viva equation — that tells you the speed of a spacecraft at any point in any orbit, from a lazy circle to a comet's plunge to a probe leaving the solar system, all from two numbers: where you are and how big your orbit is. It is the single most useful equation in orbital mechanics, and mission designers reach for it a hundred times a day.

Energy is also the honest currency of spaceflight's biggest questions. Getting to orbit, transferring to another orbit, escaping a planet, flying to Mars — every one of these is, underneath, a transaction in energy. And coming home, as we will see at the chapter's end, is the same transaction run in reverse and paid in fire. This is where the theme that orbital mechanics is beautiful earns its keep: the same conservation of energy that governs a tossed ball, written down carefully for an inverse-square field, turns out to govern every trajectory in the cosmos.

In this chapter, you will learn to:

• Combine kinetic and potential energy into the specific orbital energy $\varepsilon$, and see that it is conserved around an entire orbit.
• Connect that energy to the size of the orbit through the elegant relation $\varepsilon = -\mu/(2a)$.
• Derive the vis-viva equation $v^2 = \mu\!\left(\tfrac{2}{r} - \tfrac{1}{a}\right)$ and use it on real orbits, from LEO to GEO to an interplanetary escape.
• Understand why a higher orbit is a slower orbit — the threshold idea that reorganizes everything.
• Read a trajectory's fate straight off its energy: bound, escaping, or gone forever.

Learning Paths

🚀 Space Enthusiast: Read 6.1 for the picture and then spend your time on 6.4 and 6.6 — the two sections that will permanently change how you watch a spacecraft maneuver. You can skim the algebra of the derivation in 6.3 and just take the vis-viva equation as a gift.

📐 Engineering Student: Read everything, and do the derivation in 6.3 yourself before you read ours. Vis-viva is a workhorse you will use in every chapter of Part II; the ⭐⭐/⭐⭐⭐ exercises here are foundational for Hohmann transfers (Chapter 10) and interplanetary trajectories (Chapter 11).

🎮 KSP Player: You have felt every idea in this chapter, especially the paradox in 6.6 — the maddening dance of trying to dock by burning the "wrong" way. Focus on 6.4 and 6.6 to finally understand why the game behaves as it does, then use the vis-viva calculator in the Checkpoint to plan burns.

🛰️ Industry Prep: The vis-viva equation and $\varepsilon = -\mu/(2a)$ are the two relations you will use most often in any real trajectory work. Section 6.5 (escape energy and $C_3$) is the vocabulary of launch-vehicle performance charts; your Mission Design Checkpoint records your target orbit's energy.

6.1 Kinetic and potential energy in orbit

A spacecraft in orbit carries two kinds of energy, and the whole chapter is the story of how they trade back and forth without the total ever changing. Let us name them carefully, because the entire subject is built from just these two terms.

The first is kinetic energy, the energy of motion, which you have met since school as $KE = \tfrac{1}{2}mv^2$ for a body of mass $m$ moving at speed $v$. A fast spacecraft carries a lot of it; the ISS at $7.7\ \text{km/s}$ carries a spectacular amount.

The second is gravitational potential energy, the energy of position in a gravity field. We derived its exact form in Chapter 2 by integrating the true inverse-square force from a distance $r$ out to infinity, and it is worth reprinting because the minus sign in it drives everything that follows:

$$ U(r) = -\frac{\mu m}{r}, $$

where $\mu = GM$ is the body's gravitational parameter (for Earth, $\mu_\oplus = 3.986\times10^{5}\ \text{km}^3/\text{s}^2$, introduced in Chapter 2). The convention is that $U = 0$ infinitely far away, so everywhere closer than infinity the potential energy is negative — a bound spacecraft sits down inside an energy "well," and climbing out (moving to larger $r$) raises $U$ toward zero.

The total mechanical energy of the spacecraft is simply the sum:

$$ E = \tfrac{1}{2}m v^2 - \frac{\mu m}{r}. $$

Strategy first. Everything in this chapter flows from one physical claim: as a spacecraft coasts along its orbit with the engine off, this total $E$ never changes. Kinetic energy and potential energy each rise and fall — the craft speeds up as it swoops close to the planet and slows as it climbs away — but they do so in perfect lockstep, trading one for the other so their sum holds fixed. If we can pin down that one conserved number for an orbit, we can find the speed at every point. That is the plan.

Why is $E$ conserved? Because the only force acting on a coasting spacecraft is gravity, and gravity is a conservative force — the work it does moving a body between two points depends only on the two points, not on the path taken. A conservative force stores its work as potential energy and hands it back perfectly; nothing is lost to friction, because in the vacuum there is no friction. So as the spacecraft falls inward, gravity does positive work, potential energy converts into kinetic, and the craft speeds up. As it climbs back out, it pays that kinetic energy back into potential and slows. The books balance to the last joule, forever. This is the same conservation of energy Feynman is describing in the epigraph, applied to the cleanest system in nature: a single body falling freely around another.

💡 Intuition: Think of an orbit as a frictionless skateboarder in a smooth valley. At the bottom of the valley she is moving fastest (maximum kinetic, minimum potential); as she coasts up the far slope she trades speed for height until, at her highest point, she is briefly slowest (minimum kinetic, maximum potential) — then she rolls back down and the trade reverses. An elliptical orbit is exactly this, wrapped into a closed loop around a planet: periapsis (closest approach) is the bottom of the valley where the craft is fastest, and apoapsis (farthest point) is the top of the slope where it is slowest. No energy is ever added or removed; it just sloshes between the two forms.

The circular orbit, and the speed to stay in one

Before we generalize, let us nail down the simplest orbit — the circle — because it gives us a speed we will use constantly. In a circular orbit the radius $r$ never changes, so the potential energy is constant and, therefore, so is the kinetic energy: a circular orbit is a constant-speed orbit. What speed?

For circular motion, gravity must supply exactly the centripetal force that bends the path into a circle. The gravitational acceleration is $\mu/r^2$ (Chapter 2), and the centripetal acceleration needed to hold a circle of radius $r$ at speed $v$ is $v^2/r$. Setting them equal:

$$ \frac{v^2}{r} = \frac{\mu}{r^2} \quad\Longrightarrow\quad v_{\text{circ}} = \sqrt{\frac{\mu}{r}}. $$

Definition (circular velocity). The circular velocity at radius $r$ from a body of gravitational parameter $\mu$ is $v_{\text{circ}} = \sqrt{\mu/r}$ — the exact speed at which gravity curves your path into a closed circle. Too slow and you fall inward; too fast and you swing outward. It depends only on where you are and what you orbit, never on your own mass.

Worked Example: how fast is low Earth orbit? Take a satellite in a circular orbit at $400\ \text{km}$ altitude — a typical LEO, near the ISS. Its orbital radius is $r = R_\oplus + h = 6{,}371 + 400 = 6{,}771\ \text{km}$. Then $$v_{\text{circ}} = \sqrt{\frac{\mu_\oplus}{r}} = \sqrt{\frac{3.986\times10^{5}\ \text{km}^3/\text{s}^2} > {6{,}771\ \text{km}}} = \sqrt{58.87\ \text{km}^2/\text{s}^2} = 7.67\ \text{km/s}.$$ That is about $27{,}600\ \text{km/h}$. The units check: $(\text{km}^3/\text{s}^2)/\text{km} = \text{km}^2/\text{s}^2$, whose square root is $\text{km/s}$. Sanity check against Chapter 2, which quoted $\approx 7.9\ \text{km/s}$ at the surface ($r = 6{,}371\ \text{km}$); our satellite is a little higher, so a little slower, and $7.67 < 7.9$ as it must be. This one number — $7.67\ \text{km/s}$ — is why reaching orbit is so hard: it is not the altitude that costs you, it is building up this ferocious sideways speed.

🔗 Connection: Compare this with GEO, the geostationary orbit at $r = 42{,}164\ \text{km}$ where a satellite circles once per day and so appears to hover over one spot on Earth. There, $v_{\text{circ}} = \sqrt{3.986\times10^{5}/42{,}164} = \sqrt{9.454} = 3.07\ \text{km/s}$ — less than half the LEO speed, even though GEO is the more "prestigious" orbit six times higher up. The higher orbit is the slower orbit. Hold onto that; by Section 6.6 it will stop being a curiosity and become a law you can prove. We meet the whole zoo of orbit types — LEO, GEO, and the rest — in Chapter 9.

6.2 Specific orbital energy

Notice a small annoyance in every equation so far: the spacecraft's mass $m$ keeps tagging along. $KE = \tfrac12 m v^2$, $U = -\mu m/r$, $E = \tfrac12 m v^2 - \mu m/r$ — each term carries an $m$. But we saw in Chapter 2 that the motion of a spacecraft never depends on its own mass: a bolt and a battleship in the same orbit fly identical paths. So the $m$ is dead weight in our bookkeeping. Let us divide it out once and for all by working with energy per kilogram.

Definition (specific orbital energy). The specific orbital energy $\varepsilon$ is the total mechanical energy of an orbiting body per unit of its mass: $$\varepsilon = \frac{E}{m} = \frac{v^2}{2} - \frac{\mu}{r}.$$ It has units of energy per mass — $\text{J/kg} = \text{m}^2/\text{s}^2$ (or, conveniently for orbits, $\text{km}^2/\text{s}^2$, which equals $\text{MJ/kg}$). Because it drops the mass, $\varepsilon$ is a property of the orbit itself, shared by everything flying that orbit, and — the crucial fact — it is conserved: as the body coasts, $v$ and $r$ both change, but the combination $\tfrac{v^2}{2} - \tfrac{\mu}{r}$ stays constant.

The sign of $\varepsilon$ is the single most informative number about a trajectory, and it is worth learning to read at a glance:

• $\varepsilon < 0$ — the body is bound. It has less kinetic energy than it would need to climb out of the well, so it can never reach infinity; it is trapped on a closed orbit (a circle or an ellipse) and will return again and again. Every satellite, every moon, every planet has $\varepsilon < 0$.
• $\varepsilon = 0$ — the body is exactly on the knife-edge. It has just enough energy to coast to infinity, arriving there with zero speed to spare. This is escape, and its path is a parabola (Section 6.5).
• $\varepsilon > 0$ — the body is unbound. It has more than enough energy to escape and will sail away forever with speed left over. Its path is a hyperbola (Section 6.5). Voyager 1 has $\varepsilon > 0$ with respect to the Sun; it is never coming back.

The energy of a circular orbit, and a beautiful pattern

Let us compute $\varepsilon$ for a circular orbit and see something lovely fall out. In a circle, $v^2 = v_{\text{circ}}^2 = \mu/r$, so

$$ \varepsilon_{\text{circ}} = \frac{v^2}{2} - \frac{\mu}{r} = \frac{\mu/r}{2} - \frac{\mu}{r} = \frac{\mu}{2r} - \frac{\mu}{r} = -\frac{\mu}{2r}. $$

Two things are worth pausing on. First, the total energy is negative, confirming that a circular orbit is bound. Second, look at how the three energies relate: the kinetic energy per kilogram is $\tfrac{v^2}{2} = \tfrac{\mu}{2r}$, the potential is $-\tfrac{\mu}{r}$, and the total is $-\tfrac{\mu}{2r}$. So $KE = -\varepsilon$ and $U = 2\varepsilon$: the kinetic energy is exactly the negative of the total energy, and the potential energy is exactly twice the total. This is a special case of the virial theorem, and it holds the key to the whole "higher is slower" paradox. It says that in a circular orbit, the faster you move (more $KE$), the more negative — the lower — your total energy, which means the smaller and lower your orbit. We will cash this in shortly.

Worked Example: the energy ledger of LEO and GEO. For our $400\ \text{km}$ LEO ($r = 6{,}771\ \text{km}$): $$\varepsilon_{\text{LEO}} = -\frac{\mu_\oplus}{2r} = -\frac{3.986\times10^{5}}{2 \times 6{,}771} > = -29.4\ \text{km}^2/\text{s}^2 = -29.4\ \text{MJ/kg}.$$ For GEO ($r = 42{,}164\ \text{km}$): $$\varepsilon_{\text{GEO}} = -\frac{3.986\times10^{5}}{2 \times 42{,}164} = -4.73\ \text{MJ/kg}.$$ The GEO orbit has the higher (less negative) energy — as it must, being farther out of the well — yet we computed in Section 6.1 that it has the lower speed ($3.07$ versus $7.67\ \text{km/s}$). How can more total energy mean less speed? Because in climbing from LEO to GEO the potential energy rockets upward (from $-58.9$ to $-9.45\ \text{MJ/kg}$, a gain of $49.4$) while the kinetic energy actually falls (from $29.4$ to $4.73\ \text{MJ/kg}$, a drop of $24.7$). You put in energy, all of it and more went into "height," and the motion slowed down. That is the paradox stated in the language of energy, and Section 6.6 turns it into an operating rule.

🔄 Check Your Understanding 1. A spacecraft's specific orbital energy is measured to be $+2\ \text{km}^2/\text{s}^2$. Is it bound to the planet, exactly escaping, or gone forever? 2. Why does the spacecraft's own mass not appear in $\varepsilon = v^2/2 - \mu/r$? 3. Two satellites share the same circular orbit, one twice as massive as the other. Which has the greater specific orbital energy $\varepsilon$?

Answers

1. Gone forever — $\varepsilon > 0$ means the trajectory is hyperbolic (unbound); it will coast to infinity with speed to spare. 2. Because $\varepsilon$ is energy per unit mass: both the kinetic term ($\tfrac12 m v^2$) and the potential term ($-\mu m/r$) are proportional to $m$, so dividing the total energy by $m$ removes it entirely. The orbit's energy is a property of the orbit, not the vehicle.
2. Neither — they are equal. $\varepsilon = -\mu/(2r)$ depends only on the radius and the planet, not on the satellite's mass. (Their total energies $E = m\varepsilon$ differ by a factor of two, but the specific energy is identical.)

6.3 The vis-viva equation

We now have everything we need to derive the workhorse of orbital mechanics. The name is Latin — vis viva, "living force," the term Gottfried Leibniz used in the 1690s for what we now call kinetic energy. The equation is nothing but energy conservation, rearranged into the form a navigator actually wants: given where you are, how fast are you going?

To get there we need one more fact, which we will connect to energy rather than derive from geometry (that geometric derivation is Chapter 8's job, from Kepler's laws). Every bound orbit — every ellipse — has a size, captured by its semi-major axis $a$, half the length of the ellipse's longest diameter. For a circle, $a$ is just the radius. The remarkable fact, which we will treat as the defining energy property of $a$, is that the specific orbital energy depends only on this size:

Definition (semi-major axis, energy form). The semi-major axis $a$ of an orbit is the size parameter that fixes its energy through $$\boxed{\ \varepsilon = -\frac{\mu}{2a}\ }$$ Equivalently $a = -\mu/(2\varepsilon)$. All orbits with the same $a$ have the same energy and the same period, regardless of their shape (how stretched-out they are). A big, lazy orbit has a large $a$ and a high (near-zero) energy; a small, tight orbit has a small $a$ and a deeply negative energy. For a bound orbit $a > 0$; we will see in Section 6.5 that escape corresponds to $a \to \infty$ and hyperbolic orbits to $a < 0$.

We verified this already for the circle, where $a = r$ gave $\varepsilon = -\mu/(2r) = -\mu/(2a)$. The claim is that the same relation holds for any ellipse if you use its semi-major axis — a fact we borrow now and Chapter 8 proves in full. With it, the derivation of vis-viva is three lines.

Strategy first. We have two expressions for the one conserved number $\varepsilon$: the definition in terms of the instantaneous speed and position, $\varepsilon = \tfrac{v^2}{2} - \tfrac{\mu}{r}$, and the size relation, $\varepsilon = -\tfrac{\mu}{2a}$. Both equal $\varepsilon$, so they equal each other. Set them equal and solve for $v^2$. That is the whole trick — vis-viva is just "energy here" equals "energy of the orbit."

Setting the two equal:

$$ \frac{v^2}{2} - \frac{\mu}{r} = -\frac{\mu}{2a}. $$

Add $\mu/r$ to both sides:

$$ \frac{v^2}{2} = \frac{\mu}{r} - \frac{\mu}{2a}. $$

Multiply through by $2$:

$$ \boxed{\ v^2 = \mu\!\left(\frac{2}{r} - \frac{1}{a}\right)\ } $$

That is the vis-viva equation, and it is worth reading aloud. $\blacksquare$

Definition (vis-viva equation). For any orbit around a body of gravitational parameter $\mu$, the speed $v$ at a point a distance $r$ from the center is $$v^2 = \mu\!\left(\frac{2}{r} - \frac{1}{a}\right),$$ where $a$ is the orbit's semi-major axis. It gives the speed anywhere on any conic orbit — circle, ellipse, parabola, or hyperbola — from just two inputs: your current distance $r$ and the orbit's size $a$. It is the most-used single equation in orbital mechanics.

Look at what it contains and what it does not. It does not care where on the orbit you are except through your distance $r$; it does not care about the orbit's orientation, or its eccentricity, or the date. Two numbers — $r$ and $a$ — and out comes the speed. And it gracefully includes the circular case we already know: set $a = r$ and you get $v^2 = \mu(2/r - 1/r) = \mu/r$, recovering $v_{\text{circ}} = \sqrt{\mu/r}$. The circle is just the special orbit whose size equals your distance at every point.

Worked Example: speed at the ends of a transfer orbit. Imagine an orbit that skims LEO at its low point and reaches all the way out to GEO at its high point — precisely the "transfer ellipse" a satellite rides from a parking orbit up to geostationary, and a preview of the Hohmann transfer we derive in Chapter 10. Its low point (periapsis) is at $r_p = 6{,}771\ \text{km}$ (LEO, $400\ \text{km}$ up) and its high point (apoapsis) at $r_a = 42{,}164\ \text{km}$ (GEO). The semi-major axis is half the long diameter: $$a = \frac{r_p + r_a}{2} = \frac{6{,}771 + 42{,}164}{2} = 24{,}468\ \text{km}.$$ Speed at periapsis (down low), from vis-viva: $$v_p^2 = \mu\!\left(\frac{2}{6{,}771} - \frac{1}{24{,}468}\right) > = 3.986\times10^{5}\,(2.954\times10^{-4} - 4.087\times10^{-5}) = 101.5\ \text{km}^2/\text{s}^2,$$ so $v_p = 10.07\ \text{km/s}$. Speed at apoapsis (up high): $$v_a^2 = \mu\!\left(\frac{2}{42{,}164} - \frac{1}{24{,}468}\right) > = 3.986\times10^{5}\,(4.743\times10^{-5} - 4.087\times10^{-5}) = 2.62\ \text{km}^2/\text{s}^2,$$ so $v_a = 1.62\ \text{km/s}$. On one and the same orbit, the spacecraft screams through its low point at $10.07\ \text{km/s}$ and crawls through its high point at $1.62\ \text{km/s}$ — a factor of six, purely from trading kinetic energy for potential and back. Sanity check: the low point is where it is closest and deepest in the well, so it should be fastest, and it is. This single orbit, computed with one equation, is the skeleton of every satellite delivery to GEO.

🐛 Find the Error. A student wants the speed of a satellite in a circular orbit at $r = 7{,}000\ \text{km}$ and writes vis-viva as $v^2 = \mu\left(\tfrac{2}{r} - \tfrac{1}{a}\right)$ with $a = 7{,}000\ \text{km}$, but then, "to be safe," also sets $r = 2a = 14{,}000\ \text{km}$ because "$a$ is the semi-major axis, which is half the diameter." They get a strange answer. What went wrong?

Answer

They confused two different distances. In vis-viva, $r$ is the spacecraft's actual current distance from the center of the planet — here $7{,}000\ \text{km}$ — while $a$ is the orbit's semi-major axis, which for a circle equals that same radius, $7{,}000\ \text{km}$. The "half the diameter" idea is right for $a$ but does not change $r$: the satellite really is $7{,}000\ \text{km}$ from the center. Using $r = a = 7{,}000\ \text{km}$ gives the correct $v^2 = \mu(2/r - 1/r) = \mu/r$, i.e. the circular velocity. The lesson: $r$ and $a$ are independent inputs — one is where you are, the other is how big the orbit is — and only for a circle do they happen to coincide.

6.4 Trading altitude for speed

The transfer-orbit example just gave us the physics of the whole chapter in a single picture: on one orbit, low means fast and high means slow. Let us make the principle explicit, because it is the hinge the counterintuitive results all turn on.

Look again at vis-viva, $v^2 = \mu(2/r - 1/a)$, and hold the orbit fixed so $a$ is a constant. The only thing that varies as the spacecraft travels is $r$. As $r$ gets smaller (the craft dives toward the planet), the term $2/r$ grows, so $v^2$ grows — the craft speeds up. As $r$ gets larger (the craft climbs away), $2/r$ shrinks and $v^2$ falls — the craft slows. Along any single orbit, you trade altitude for speed and speed for altitude, continuously, with the total energy $\varepsilon$ standing perfectly still. This is the frictionless skateboarder of Section 6.1, and it is exact.

There is a wonderfully useful special case. In an elliptical orbit the two extremes are periapsis (lowest, $r_p$) and apoapsis (highest, $r_a$). Because energy is conserved, the speed you gain diving down to periapsis is exactly the speed you lose climbing back to apoapsis. And there is a clean relation between the two, which comes from conserving angular momentum (a quantity we develop fully in Chapter 8): at the two ends, where velocity is purely sideways, $r_p v_p = r_a v_a$. The closer-in point moves proportionally faster. Our transfer orbit obeyed it: $6{,}771 \times 10.07 \approx 68{,}200$ and $42{,}164 \times 1.62 \approx 68{,}300\ \text{km}^2/\text{s}$ — equal to within rounding. Kepler saw this in the planets four centuries ago and called it his second law; we will prove it from gravity in Chapter 8.

⚠️ Common Misconception: "Higher orbits are faster because they have more energy." The word "energy" hides a trap. A higher orbit does have more total energy $\varepsilon$ (it is less negative). But total energy is kinetic plus potential, and climbing higher pours energy into the potential term so lavishly that the kinetic term — the speed — actually drops. More total energy, less speed. Confusing "more energy" with "more speed" is the single most common error in orbital thinking, and it is exactly what makes the next section feel like a paradox. Always separate the two: raising an orbit raises its total energy and lowers its speed.

🔧 Engineering Reality: this is how you get to GEO. A launch vehicle rarely delivers a communications satellite straight to its circular geostationary home. Instead it drops the satellite onto exactly the transfer ellipse we computed — low point in LEO, high point at GEO altitude — called a geostationary transfer orbit (GTO). The satellite coasts (for free, burning nothing) up the slope from $10.07\ \text{km/s}$ at the bottom to $1.62\ \text{km/s}$ at the top, and then fires its own engine at apoapsis to speed up from $1.62$ to the $3.07\ \text{km/s}$ of a circular GEO. Vis-viva sizes that final burn: $\Delta v = 3.07 - 1.62 = 1.45\ \text{km/s}$, which matches the "GTO → GEO" figure on the delta-v map of Chapter 3 almost exactly. The equation you just derived is not academic; it is the number the satellite's fuel load is sized against.

🔄 Check Your Understanding 1. In an elliptical orbit, at which point — periapsis or apoapsis — is the spacecraft moving fastest, and why, in terms of the two energies? 2. A spacecraft is at distance $r$ on an orbit of semi-major axis $a$. Without a calculator, how does its speed change if you keep $r$ the same but consider a larger orbit (bigger $a$)?

Answers

1. Fastest at periapsis, the lowest point. There it is deepest in the gravity well, so its potential energy is most negative; since the total energy is fixed, the kinetic energy — and thus the speed — is greatest. Climbing to apoapsis trades that kinetic energy for potential, so it is slowest at the top.
2. It speeds up. In $v^2 = \mu(2/r - 1/a)$, increasing $a$ makes $1/a$ smaller, so $-1/a$ is less negative and $v^2$ is larger. Physically: passing through the same point $r$ on a bigger (higher-energy) orbit, you must be moving faster there — for instance, a craft merely passing through LEO altitude on its way out to GEO is moving faster at that altitude than a craft that circularly orbits at LEO.

6.5 Escape energy: parabolic and hyperbolic trajectories

We have been living among bound orbits, all with $\varepsilon < 0$. Now we push to the boundary and beyond — to the trajectories that leave. This is where a mission stops orbiting a planet and starts going somewhere, and it is all governed by the sign of one number.

Recall the reading of $\varepsilon$ from Section 6.2. A bound orbit has $\varepsilon < 0$: not enough energy to reach infinity. The threshold is $\varepsilon = 0$: just enough energy to coast to infinity, arriving with exactly zero speed. Set $\varepsilon = 0$ in the definition and see what speed that demands at a distance $r$:

$$ \varepsilon = \frac{v^2}{2} - \frac{\mu}{r} = 0 \quad\Longrightarrow\quad v = \sqrt{\frac{2\mu}{r}} = v_{\text{esc}}. $$

This is precisely the escape velocity we derived in Chapter 2 — and now we see its true meaning. Escape velocity is simply the speed at which your specific orbital energy is exactly zero. It is not a magic barrier; it is the energy break-even point. Notice, too, that $v_{\text{esc}} = \sqrt{2\mu/r} = \sqrt{2}\, \sqrt{\mu/r} = \sqrt{2}\, v_{\text{circ}}$: escape speed is always $\sqrt{2} \approx 1.414$ times circular speed at the same radius, the clean factor Chapter 2 promised.

Definition (escape energy). The escape energy of a spacecraft in a given orbit is the additional specific energy it must gain to reach $\varepsilon = 0$ and become unbound — that is, $|\varepsilon|$ for a bound orbit. From a circular orbit of radius $r$, where $\varepsilon = -\mu/(2r)$, the escape energy is $\mu/(2r)$ per kilogram. Equivalently, it is the energy that raises your speed from $v_{\text{circ}}$ to $v_{\text{esc}} = \sqrt{2}\,v_{\text{circ}}$.

What does the path look like at each energy? This is where the trajectory earns a new name.

Definition (parabolic trajectory). A parabolic trajectory is the marginal escape path, with $\varepsilon = 0$ exactly. Its speed everywhere equals the local escape velocity, and it corresponds to an infinite semi-major axis: as $a \to \infty$, the term $1/a \to 0$ and vis-viva becomes $v^2 = 2\mu/r$, the escape-speed law. A parabolic orbit reaches infinity but arrives there with zero speed left. It is a mathematical idealization — the exact boundary between staying and leaving — that real missions approach but never sit on.

Definition (hyperbolic trajectory). A hyperbolic trajectory is an escape path with energy to spare, $\varepsilon > 0$. It reaches infinity still moving, with a leftover speed called the hyperbolic excess velocity $v_\infty$, given by $\varepsilon = v_\infty^2/2$. By the energy-form relation $\varepsilon = -\mu/(2a)$, a positive energy forces the semi-major axis to be negative, $a < 0$ — the algebra's way of telling you the orbit is an open hyperbola, not a closed ellipse. Every interplanetary probe leaves Earth on a hyperbola.

Worked Example: the energy cost of leaving. How much speed must you add to a $400\ \text{km}$ LEO to just barely escape Earth? At $r = 6{,}771\ \text{km}$, escape velocity is $$v_{\text{esc}} = \sqrt{\frac{2\mu_\oplus}{r}} = \sqrt{\frac{2 \times 3.986\times10^{5}}{6{,}771}} > = \sqrt{117.7\ \text{km}^2/\text{s}^2} = 10.85\ \text{km/s}.$$ The circular speed there is $7.67\ \text{km/s}$, so the delta-v to reach escape is $\Delta v = 10.85 - 7.67 = 3.18\ \text{km/s}.$ Compare this with the delta-v map of Chapter 3, which listed "LEO → escape Earth" at about $3.2\ \text{km/s}$ — a direct hit. And notice the beautiful economy hiding here: escaping from the surface takes $11.2\ \text{km/s}$, but escaping from LEO takes only $3.18\ \text{km/s}$ of extra speed, because in LEO you are already moving at $7.67\ \text{km/s}$ and already partway up the well. This is the exact, quantitative meaning of the old line that "once you are in orbit, you are halfway to anywhere."

🚪 Threshold Concept: escape is a budget, not a wall. The most freeing idea in this section is that leaving a planet is not about overpowering gravity — it is about accumulating enough energy. At $\varepsilon < 0$ you are bound and will always fall back, no matter how far you coast. At $\varepsilon = 0$ you can just reach infinity. At $\varepsilon > 0$ you leave with speed to spare. Because energy, not altitude or force, is the real currency, a spacecraft already in a fast orbit has most of the money: it needs only the small top-up from $\varepsilon_{\text{orbit}}$ up to zero (and then a little more for the hyperbolic excess it wants at the far end). This reframing — that every journey outward is a controlled purchase of positive energy — is what makes interplanetary travel tractable, and it is the direct foundation of the Mars trajectory we build across Chapters 10 and 11.

🔗 Connection: aiming at Mars. A probe headed for Mars does not want to just barely escape Earth (parabolic, $\varepsilon = 0$); it needs to escape and still be moving fast enough, relative to Earth, to coast outward to Mars's orbit around the Sun. Suppose that requires a hyperbolic excess of about $v_\infty \approx 2.94\ \text{km/s}$ (a representative value for a minimum-energy Mars transfer; we compute it properly in Chapter 11 — treat it here as illustrative). Then the speed needed at LEO is $v = \sqrt{v_\infty^2 + v_{\text{esc}}^2} = \sqrt{2.94^2 + 10.85^2} = \sqrt{8.6 + 117.7} = 11.24\ \text{km/s}$, so the departure burn from LEO is $\Delta v = 11.24 - 7.67 = 3.57\ \text{km/s}$ — again matching the delta-v map's "LEO → trans-Mars injection $\approx 3.6\ \text{km/s}$." The leftover energy that gets you to Mars is exactly $\varepsilon = v_\infty^2/2 \approx 4.3\ \text{km}^2/\text{s}^2 > 0$, a hyperbolic orbit. This positive-energy departure, and the quantity $C_3 = v_\infty^2$ that launch providers quote for it, are the heart of Chapter 11.

🔄 Check Your Understanding 1. A trajectory has semi-major axis $a = -18{,}000\ \text{km}$ (negative). Is it bound, parabolic, or hyperbolic? What is the sign of its energy? 2. Why is escaping from a high orbit "cheaper" in delta-v than escaping from a low orbit, even though the high orbit is farther from the planet?

Answers

1. Hyperbolic (unbound). A negative semi-major axis means $\varepsilon = -\mu/(2a) > 0$, i.e. positive energy — the signature of a hyperbola that escapes with speed to spare. 2. Escape speed falls off as $\sqrt{2\mu/r}$, so it is smaller at large $r$; and although a high circular orbit is slower, the gap between its circular speed and its escape speed (a factor of $\sqrt2$ of a smaller number) is smaller, so less delta-v is needed to bridge it. In energy terms, a high orbit is already near the top of the well ($\varepsilon$ close to zero), so little extra energy remains to buy.

6.6 The counterintuitive orbit: speed up to end up slower

We arrive at the idea this chapter has been circling from its first paragraph — the one that sounds like a riddle and turns out to be a law. Here it is, stated as plainly as it can be: to move to a higher orbit, you fire your engine forward (prograde) to speed up; but once you get there, you are moving slower than before. And its mirror image, which is even more useful and even more surprising: to catch up to something ahead of you in the same orbit, you must slow down, dropping to a lower, faster orbit. Every astronaut, every flight controller, every serious player of orbital games has had to unlearn their ground instincts to master this. Let us prove it, and then feel it.

The proof is two facts you already have

Everything follows from combining two results from earlier in this chapter:

1. Firing prograde raises your energy (Section 6.2). A prograde burn increases your speed $v$ at the burn point, which increases your specific orbital energy $\varepsilon = v^2/2 - \mu/r$. Higher energy means (by $\varepsilon = -\mu/2a$) a larger semi-major axis $a$ — a bigger orbit.
2. A bigger orbit is a slower orbit (Sections 6.1, 6.4). The circular speed $\sqrt{\mu/r}$ decreases as $r$ grows, and by vis-viva the same holds for the average speed of any larger orbit.

Put them together: the prograde burn that raises your orbit is the very thing that lowers your eventual speed. You spend energy to climb, and climbing costs you speed, because the energy you added — and then some — goes into potential energy, not kinetic. The virial pattern from Section 6.2 ($KE = -\varepsilon$ for circular orbits) is the fingerprint of this: raise the total energy $\varepsilon$ toward zero and the kinetic energy $-\varepsilon$ necessarily falls.

Worked Example: raising your orbit by 100 km, and slowing down. A spacecraft is in a circular orbit at $400\ \text{km}$ ($r = 6{,}771\ \text{km}$), moving at $v_{\text{circ}} = 7.673\ \text{km/s}$. We want to raise it to a circular orbit at $500\ \text{km}$ ($r = 6{,}871\ \text{km}$), where the circular speed is $v_{\text{circ}} = \sqrt{3.986\times10^{5}/6{,}871} = 7.617\ \text{km/s}$ — already we can see the destination is $56\ \text{m/s}$ slower than the start. To get there we ride a small transfer ellipse with $r_p = 6{,}771$, $r_a = 6{,}871$, so $a = 6{,}821\ \text{km}$.

• Burn 1 (prograde, at $400\ \text{km}$). We need the periapsis speed of the transfer ellipse: $v_p = \sqrt{\mu(2/6{,}771 - 1/6{,}821)} = \sqrt{59.30} = 7.701\ \text{km/s}$. The burn is $\Delta v_1 = 7.701 - 7.673 = +0.028\ \text{km/s} = +28\ \text{m/s}$ — a speed-up.
• Coast up to apoapsis at $500\ \text{km}$, arriving at $v_a = \sqrt{\mu(2/6{,}871 - 1/6{,}821)} = \sqrt{57.59} = 7.589\ \text{km/s}$.
• Burn 2 (prograde, at $500\ \text{km}$). Circularize by speeding up from $7.589$ to $7.617\ \text{km/s}$: $\Delta v_2 = +0.028\ \text{km/s} = +28\ \text{m/s}$ — another speed-up.

You fired prograde twice, spending about $56\ \text{m/s}$ of delta-v to go faster each time — and you ended up in an orbit that is $56\ \text{m/s}$ slower than where you began ($7.617$ versus $7.673\ \text{km/s}$). Every instantaneous burn added speed; the net result subtracted it. That is not a trick of arithmetic. It is the whole geometry of gravity: the energy you added went to lifting you higher, and higher is slower.

🚪 Threshold Concept: a higher orbit is a slower orbit — so to speed up, you must slow down. This is the idea that reorganizes your entire intuition about spaceflight, and it is worth stating in both directions. To reach a higher orbit (longer period, more energy), you burn prograde and end up moving slower. To catch something ahead of you in the same orbit, you cannot simply "hit the gas" — burning prograde raises your orbit, slows you down, and makes you fall behind. Instead you burn retrograde, dropping into a slightly lower orbit, where you move faster and circle in less time, so you pull ahead; then you raise back up to rendezvous. On the ground, "go faster" and "catch up" are the same command. In orbit they are opposites. Once this clicks, an enormous amount of orbital mechanics — phasing, rendezvous, docking, the counterrotating dance of two spacecraft — stops being mysterious. It is all just "higher is slower," applied with nerve. We build the full machinery of rendezvous on this foundation in Chapter 10.

📜 From History: the "orbital paradox" and Buzz Aldrin's dilemma. When NASA planned the first orbital rendezvous in the Gemini program (1965–66), the pilots' instinct — thrust toward the target to close the gap — kept making things worse: burning prograde raised their orbit and dropped them behind the target. Gemini 4 tried it in 1965 and failed to close on its target for exactly this reason. The fix was to fly the counterintuitive way — slow down to catch up — and it required pilots to trust the mathematics over their reflexes. Edwin "Buzz" Aldrin earned the nickname "Dr. Rendezvous" by working out the orbital mechanics of it in his 1963 MIT doctoral thesis, and that theory flew on Gemini and then Apollo. The paradox in this section is not a textbook curiosity; it is a problem real astronauts had to master before anyone could dock two spacecraft, let alone go to the Moon.

💡 Intuition: why the period stretches. There is a second way to feel "higher is slower" — through time. A larger orbit has a longer period: it takes longer to go around. A quick preview of Kepler's third law (which Chapter 8 derives) gives the period as $T = 2\pi\sqrt{a^3/\mu}$. Our $400\ \text{km}$ orbit has $T = 2\pi\sqrt{6{,}771^3/\mu} \approx 92.4\ \text{minutes}$; the $500\ \text{km}$ orbit has $T \approx 94.5\ \text{minutes}$. The higher satellite not only moves slower, it has farther to go, so it falls behind on both counts — which is exactly why a chaser must drop to a lower, shorter-period orbit to gain ground. Speed and period both say the same thing: down is fast, up is slow.

🔄 Check Your Understanding 1. Two astronauts are drifting in the same circular orbit, one $100\ \text{m}$ directly ahead of the other. The one behind wants to reach the one ahead. Which way should they burn — toward the target (prograde) or away from it (retrograde) — and why? 2. A satellite fires prograde and settles into a new, higher circular orbit. Is its new speed higher or lower than before? Is its new energy higher or lower?

Answers

1. Retrograde (away from the target, slowing down). Burning prograde toward the target would raise the chaser's orbit, slow it down, and make it drift farther behind. Burning retrograde drops the chaser into a slightly lower, faster, shorter-period orbit; it laps around more quickly, gains on the target, and then re-raises its orbit to close in. Slow down to catch up. 2. Its new speed is lower (a higher orbit is a slower orbit), but its new energy is higher (the prograde burn added energy, which went into raising the orbit). Higher energy, lower speed — the whole paradox in one sentence.

Mission Design Checkpoint: your orbit's energy and orbits.py

Across this book you are designing one real mission — a communications satellite to GEO (Track A), a lunar lander (Track B), a Mars orbiter (Track C), or an asteroid rendezvous (Track D) — and, if you like, building a small Python package, astrotools, to compute it. Chapter 3 started your delta-v budget and rocket.py; Chapter 2 recorded your primary body's $\mu$. This chapter gives you the energy of the orbit you are aiming for, and the module that will compute every orbital speed for the rest of the book.

The design. Open your Mission Design Review (MDR) and add an "Orbital Energy" line to your target orbit's data block. Using the primary-body $\mu$ you recorded in Chapter 2 and your target orbit's semi-major axis $a$, write down: the specific orbital energy $\varepsilon = -\mu/(2a)$; the circular speed $v_{\text{circ}} = \sqrt{\mu/r}$ at your operating radius; and, if your mission escapes its primary body (Tracks C and D leave Earth; Track B leaves the Moon's vicinity), the escape velocity $\sqrt{2\mu/r}$ and the delta-v from your parking orbit up to it. A Track-A comsat, for instance, records $\varepsilon_{\text{GEO}} = -4.73\ \text{MJ/kg}$ and $v_{\text{circ}} = 3.07\ \text{km/s}$, plus the GTO transfer speeds ($10.07\ \text{km/s}$ at perigee, $1.62\ \text{km/s}$ at apogee) that size its circularization burn. These numbers feed directly into the transfer delta-v you will compute in Chapter 10.

The code. Create astrotools/orbits.py and add the three energy functions of this chapter. (Chapter 8 will add period and Chapter 9 the orbital-element conversions, so keep these signatures stable.)

import math

def circular_velocity(mu, r):
    """Circular orbital speed (m/s) at radius r (m): sqrt(mu/r)."""
    return math.sqrt(mu / r)

def specific_energy(mu, a):
    """Specific orbital energy (J/kg = m^2/s^2) of an orbit with
    semi-major axis a (m): -mu/(2a).  Negative=bound, 0=escape, positive=hyperbolic."""
    return -mu / (2 * a)

def vis_viva(mu, r, a):
    """Speed (m/s) at radius r (m) on an orbit of semi-major axis a (m):
    sqrt(mu*(2/r - 1/a)).  The workhorse of orbital mechanics."""
    return math.sqrt(mu * (2 / r - 1 / a))

MU_EARTH = 3.986e14        # m^3/s^2
R_LEO = 6.771e6            # m  (400 km altitude)
R_GEO = 42.164e6           # m
print(round(circular_velocity(MU_EARTH, R_LEO)))       # LEO circular speed
print(round(circular_velocity(MU_EARTH, R_GEO)))       # GEO circular speed
a_gto = (R_LEO + R_GEO) / 2
print(round(vis_viva(MU_EARTH, R_LEO, a_gto)))         # speed at GTO perigee
# Expected output:
# 7673
# 3075
# 10072

Those three numbers — LEO circular speed, GEO circular speed, and GTO perigee speed — are the ones we hand-computed in Sections 6.1 and 6.3, now produced by the module you will point at your mission's orbit. By the capstone, orbits.py will compute the speeds at every waypoint of your trajectory, and vis-viva will turn each of those speeds into a delta-v your rocket must pay for.

Summary

The energy viewpoint compresses all of orbital speed into two relations. Carry these forward:

Key numbers worth memorizing: LEO ($400\ \text{km}$) speed $\approx 7.67\ \text{km/s}$; GEO speed $\approx 3.07\ \text{km/s}$; $\varepsilon_{\text{LEO}} \approx -29.4\ \text{MJ/kg}$, $\varepsilon_{\text{GEO}} \approx -4.73\ \text{MJ/kg}$; $v_{\text{esc}} = \sqrt{2}\,v_{\text{circ}}$; escape from LEO $\approx 3.2\ \text{km/s}$; trans-Mars injection from LEO $\approx 3.6\ \text{km/s}$.

Spaced Review

Retrieval strengthens memory. Answer from memory before checking, then look back at the cited section. This chapter revisits Chapter 2 and Chapter 3.

1. (§6.5, Ch. 2) Chapter 2 derived escape velocity from energy conservation as $v_{\text{esc}} = \sqrt{2\mu/R}$. In the energy language of this chapter, what is the specific orbital energy $\varepsilon$ of an object moving at exactly escape velocity?
2. (Ch. 2) Chapter 2 wrote gravitational potential energy as $U = -\mu m/r$, negative everywhere. Why must the total energy of any bound orbit also be negative?
3. (§6.4, Ch. 3) Chapter 3's delta-v map listed "GTO → GEO" at about $1.5\ \text{km/s}$. Using the transfer-orbit speeds in §6.3–6.4, reconstruct that number.
4. (Ch. 3) Chapter 3 called delta-v "the currency of spaceflight." This chapter added a second currency. How are the two related when you raise an orbit?

Answers

1. Exactly zero. Escape velocity is defined by $\varepsilon = 0$: $\tfrac{v^2}{2} - \tfrac{\mu}{r} = 0$ gives $v = \sqrt{2\mu/r}$. An object at escape speed sits on the boundary between bound ($\varepsilon<0$) and unbound ($\varepsilon>0$). 2. A bound orbit cannot reach infinity, so its kinetic energy ($\tfrac12 mv^2$, always positive) is never enough to cancel the negative potential energy; the sum $E = \tfrac12 mv^2 - \mu m/r$ stays below zero. Reaching $E \ge 0$ is escaping. 3. At GTO apogee (GEO altitude) the craft moves at $v_a = 1.62\ \text{km/s}$; a circular GEO needs $3.07\ \text{km/s}$; the circularization burn is $3.07 - 1.62 = 1.45 \approx 1.5\ \text{km/s}$. ✓ 4. Delta-v buys energy: a prograde burn spends delta-v (from the rocket-equation "account") to add specific orbital energy $\varepsilon$, which raises the orbit. Vis-viva is the exchange rate — it converts a desired change in orbit ($r$, $a$) into the speed change, and hence the delta-v, the rocket must supply.

What's Next

We have learned to count the energy an orbit carries and to spend it, prograde and retrograde, to move between orbits. But every joule we have so carefully added to a spacecraft is a joule that must, eventually, come back out — because a spacecraft returning from orbit arrives at the top of the atmosphere still carrying the full ferocity of orbital energy: $7.67\ \text{km/s}$ of speed, nearly $30\ \text{MJ}$ for every kilogram of vehicle. There is no gentle way to give that back. It cannot be un-burned; it must be converted, and the only place to put it is heat — a spacecraft glowing at thousands of degrees inside a sheath of plasma of its own making. In Chapter 7 we turn the energy accounting of this chapter around and ask the hardest homecoming question in spaceflight: how do you shed thirty megajoules per kilogram without burning up? Re-entry is vis-viva run in reverse, and it is paid in fire.