Case Study: Sizing a Lunar-Lander Descent Engine

"On the Moon there is no runway and no second attempt. The engine either throttles, or you crater."

Executive Summary

In the first case study we took a finished engine apart. Here we do the harder thing: we specify one from a mission requirement. Our vehicle is a Track B cargo lunar lander, and its most demanding propulsive job is the powered descent from low lunar orbit to a soft touchdown. Using nothing but this chapter's definitions and the rocket equation of Chapter 3, we will size the descent propulsion end to end: how much propellant it needs, how much thrust, what mass flow and burn time, how deeply the engine must throttle, and what total impulse it must deliver. Along the way we will meet a design constraint that never arises for a launch vehicle — the need to throttle deeply — and we will discover a reassuring contrast with Earth launch: for a fixed destination, a lander's propulsion scales linearly, not exponentially, with the cargo it sets down.

Skills applied

  • Inverting the rocket equation to size propellant for a required descent delta-v (§16.2; Ch. 3).
  • Sizing thrust from a thrust-to-weight requirement against lunar gravity (§16.4).
  • Deriving mass flow, burn time, and total impulse for the sized engine (§16.2, §16.6).
  • Discovering the deep-throttle requirement from the change in $T/W$ during descent (§16.4).

Background

The mission requirement

  • Vehicle: a robotic cargo lander (Track B), delivering payload to the lunar surface.
  • Critical maneuver: powered descent from low lunar orbit to a soft landing. From the delta-v map of Chapter 3, a soft lunar landing costs roughly $1.7\ \text{km/s}$; we add a margin for the gravity losses of a real, non-instantaneous descent and budget $\Delta v_{\text{descent}} = 1{,}900\ \text{m/s}$.
  • Target landed mass (dry lander structure + cargo, tanks empty): $m_f = 2{,}000\ \text{kg}$.
  • Engine: a storable-propellant engine with $I_{sp} = 311\ \text{s}$, so $c = 311 \times 9.80665 = 3{,}050\ \text{m/s}$. (Storables are chosen for a lander because they need no cryogenic boiloff management during the coast to the Moon — a Chapter 18 tradeoff.)
  • Lunar gravity: $g_{\text{Moon}} = 1.62\ \text{m/s}^2$.

Phase 1: Size the propellant

Invert the rocket equation. The mass ratio the descent demands is

$$ R = \frac{m_0}{m_f} = e^{\Delta v/c} = e^{1900/3050} = e^{0.623} = 1.86. $$

So the lander must begin powered descent at $1.86$ times its landed mass:

$$ m_0 = R\, m_f = 1.86 \times 2{,}000 = 3{,}729\ \text{kg}, \qquad m_p = m_0 - m_f = 1{,}729\ \text{kg}. $$

The lander ignites the descent with about 1,729 kg of propellant aboard a 3,729 kg vehicle — very nearly half its starting mass is propellant, just to lower two tonnes gently onto the Moon. (Sanity check: propellant fraction $m_p/m_0 = 1{,}729/3{,}729 = 46\%$, comfortably below the $\sim 90\%$ of an orbital vehicle, because $1.9\ \text{km/s}$ is far less than orbital delta-v — the rocket equation is being kind to us here.)

Phase 2: Size the thrust

Propellant tells us nothing about thrust; for that we impose a thrust-to-weight requirement. A lander must be able to arrest its descent and hover with margin, so we require an initial (start-of-descent) thrust-to-weight of $T/W = 2.0$ against lunar gravity. At the start of descent the mass is the full $m_0 = 3{,}729\ \text{kg}$, so the lunar weight is

$$ W_{\text{start}} = m_0\, g_{\text{Moon}} = 3{,}729 \times 1.62 = 6{,}041\ \text{N}, $$

and the required thrust is

$$ F = \frac{T}{W}\, W_{\text{start}} = 2.0 \times 6{,}041 \approx 12{,}082\ \text{N} \approx 12.1\ \text{kN}. $$

A 12-kilonewton engine — modest, roughly the thrust of a small aircraft engine, and entirely reasonable for a two-tonne lander. (For scale, the Apollo Lunar Module's descent engine topped out near $45\ \text{kN}$ for a far heavier crewed vehicle.)

Phase 3: Mass flow, burn time, and the throttle trap

At full thrust the mass flow is

$$ \dot m = \frac{F}{c} = \frac{12{,}082}{3{,}050} \approx 3.96\ \text{kg/s}. $$

If the engine ran wide open the whole way, the burn would last $t_b = m_p/\dot m = 1{,}729/3.96 \approx 437\ \text{s}$ — about seven minutes. But it cannot run wide open the whole way, and seeing why is the heart of lander design. Watch what happens to thrust-to-weight as propellant burns off. Just before touchdown the lander is nearly empty, mass $\approx m_f = 2{,}000\ \text{kg}$, so at full thrust

$$ \frac{T}{W}\bigg|_{\text{touchdown}} = \frac{12{,}082}{2{,}000 \times 1.62} = \frac{12{,}082}{3{,}240} \approx 3.7. $$

A thrust-to-weight of $3.7$ at touchdown is a disaster — the lander would rocket upward off the surface rather than settle. To land softly the engine must throttle down to about $T/W = 1$, i.e. to $\approx 3{,}240\ \text{N}$. So the descent engine must be **throttleable from $12{,}082\ \text{N}$ down to roughly $3{,}240\ \text{N}$ — a throttle ratio of nearly $3.7 : 1$** — and in practice more, to allow controlled hover and hazard avoidance.

🔧 Engineering Reality: Deep throttling is the signature demand of a landing engine, and a genuinely hard one — many rocket engines can only run near full thrust without combustion becoming unstable. The Apollo Lunar Module Descent Engine was throttleable roughly $10 : 1$, a landmark achievement of 1960s propulsion; modern lunar landers face the same requirement. A launch vehicle almost never needs this, because it fights a roughly constant Earth gravity while gaining speed, not a shrinking weight while trying to stop. This is why you cannot simply repurpose a launch engine as a landing engine.

Because the engine throttles down as the lander lightens, the average mass flow is below the full-thrust $3.96\ \text{kg/s}$; a representative burn time for the powered descent is therefore closer to $10$ minutes ($\sim 600\ \text{s}$), consistent with real lunar descents.

Phase 4: Total impulse and a consistency check

The descent's total impulse is the whole momentum budget of the landing:

$$ I_t = c\, m_p = 3{,}050 \times 1{,}729 \approx 5.27 \times 10^{6}\ \text{N·s} = 5.27\ \text{MN·s}. $$

Check it the other way, $I_t = I_{sp}\, g_0\, m_p = 311 \times 9.80665 \times 1{,}729 \approx 5.27\times 10^{6}\ \text{N·s}$ — identical, as it must be. And a final closure check on the whole design: does the sized vehicle actually deliver its required delta-v? With $m_0 = 3{,}729$ and $m_f = 2{,}000$,

$$ \Delta v = c\,\ln\!\left(\frac{m_0}{m_f}\right) = 3{,}050 \times \ln(1.86) = 3{,}050 \times 0.623 = 1{,}900\ \text{m/s}. \quad\blacksquare $$

The design closes exactly on the $1{,}900\ \text{m/s}$ we budgeted. Every number — propellant, thrust, flow, throttle range, total impulse — descends from that one requirement and the engine's $I_{sp}$.

💡 Intuition: Notice how cleanly the design decomposed. The rocket equation (Chapter 3) set the propellant; the thrust-to-weight requirement set the thrust; and the two never interfered. That separation is a gift of working in the impulse-and-delta-v language of this chapter: size the tank from the delta-v, size the engine from the gravity you must beat, then check that the engine can throttle across the range the descent demands. Three independent questions, three independent answers — which is why a sizing calculation that keeps them separate almost never goes wrong, and one that confuses them almost always does.

Phase 5: Iterate — what heavier cargo costs

Suppose the customer doubles the payload, pushing the landed mass to $m_f = 4{,}000\ \text{kg}$. Because the destination (and hence $\Delta v$ and the mass ratio $R = 1.86$) is unchanged, everything scales linearly:

$$ m_0 = 1.86 \times 4{,}000 = 7{,}458\ \text{kg}, \quad m_p = 3{,}458\ \text{kg}, \quad F = 2.0 \times 7{,}458 \times 1.62 \approx 24.2\ \text{kN}, \quad I_t \approx 10.5\ \text{MN·s}. $$

Every figure simply doubled. This is a genuinely different feeling from Earth launch, and worth savoring: for a fixed destination, a lander's propulsion scales linearly with the mass it sets down, because the mass ratio is fixed and only $m_f$ changes. The exponential tyranny of the rocket equation lives in the delta-v, not in the payload — so doubling cargo to a fixed orbit or surface doubles the propellant, while doubling the delta-v would have squared the mass ratio. Knowing where the exponential hides is half of mission design.

💡 Intuition: Contrast this with Chapter 3's Earth-launch sizing, where the payload fraction was a brutal $1$–$4\%$. There, $\Delta v \approx 9.4\ \text{km/s}$ sat far up the exponential curve; here, $1.9\ \text{km/s}$ sits low on it ($R$ only $1.86$), so propellant is a gentle $46\%$ and cargo scales cheaply. Same equation, opposite feeling — because the delta-v is what the exponential magnifies.

🔧 Engineering Reality: Would a higher-$I_{sp}$ propellant help? Swap the storable engine for liquid hydrogen at $c = 4{,}400\ \text{m/s}$ and the descent mass ratio drops from $1.86$ to $R = e^{1900/4400} = e^{0.432} = 1.54$. For the original $2{,}000\ \text{kg}$ landed mass, propellant falls from $1{,}729\ \text{kg}$ to $m_p = 2{,}000\,(1.54 - 1) = 1{,}080\ \text{kg}$ — a $37\%$ saving. And yet real landers usually decline it: hydrogen must be held at $-253\,^\circ\text{C}$ and boils off steadily during the days-long coast to the Moon, so a storable propellant that sits quietly in its tanks can win on the whole system even while losing on paper $I_{sp}$. It is a Chapter 18 tradeoff between propellant performance and propellant manageability — and a reminder that the highest $I_{sp}$ is not always the right one.

Discussion Questions

  1. Why does a landing engine need deep throttling while a launch engine generally does not? Frame your answer in terms of how $T/W$ changes during each maneuver.
  2. We required $T/W = 2$ at the start of descent. How would the sized thrust change if we required $T/W = 1.5$ instead? Would the propellant change? Explain why thrust and propellant are sized by different requirements.
  3. The propellant fraction came out at $46\%$, versus $\sim 90\%$ for an orbital launch. Trace this difference to a single quantity in the rocket equation.
  4. Storable propellant ($I_{sp} = 311\ \text{s}$) was chosen over higher-$I_{sp}$ hydrogen ($\sim 450\ \text{s}$). Estimate the propellant you would save with hydrogen, and name one reason a lander might still avoid it. (Hint: recompute $R$ with $c = 4{,}400\ \text{m/s}$.)

Your Turn: Extensions

  • Option A (design). Add an ascent stage: after landing, a $600\ \text{kg}$ vehicle must return to low lunar orbit ($\Delta v \approx 1.9\ \text{km/s}$). Size its propellant and thrust (require $T/W = 3$ for a brisk liftoff from the surface). Do not run any code.
  • Option B (computation). Write size_lander(mf, dv, isp, g_body, twr) returning propellant mass, start mass, required thrust, mass flow, and total impulse. Reproduce this study's numbers, then run it (by hand) for the $4{,}000\ \text{kg}$ case of Phase 5. Add # Expected output:.
  • Option C (your mission). If you are on Track B, adopt these numbers into your MDR. If you are on another track, size the single most thrust-critical burn of your mission the same way (Track A: the apogee-raising burn into GEO; Track C/D: orbit insertion at the target body), and record thrust, propellant, and total impulse.

Key Takeaways

  1. Thrust and propellant are sized by different requirements. Propellant comes from the delta-v via the rocket equation; thrust comes from a thrust-to-weight requirement against the local gravity. Confusing the two is the most common sizing error.
  2. Landing engines must throttle deeply, because $T/W$ climbs as propellant burns off — a demand launch engines escape. Our lander needed nearly $4 : 1$; real crewed landers reach $10 : 1$.
  3. Always close the loop. After sizing masses, recompute the delta-v from them to confirm the design delivers what the mission asked ($1{,}900\ \text{m/s}$, exactly).
  4. For a fixed destination, lander propulsion scales linearly with payload — the exponential of the rocket equation lives in the delta-v, not the cargo. Knowing where the exponential hides is central to mission design.