Optimal staging — equal mass ratios (for same Isp): Exercises

This page is a graded workout for Optimal staging — equal mass ratios (for same Isp). Every problem states its difficulty level and hides a full solution you should only open after a genuine attempt. If a symbol here feels unfamiliar, that's a signal to revisit the parent note or its prerequisites: Tsiolkovsky rocket equation, Staging basics, Structural fraction, Payload fraction.

Throughout, the tools we lean on are:

Recall The three formulas you will reuse

Tsiolkovsky (one burn): $Δ v = v_{e} ln R$ , where $R = \frac{m _{before}}{m _{after}}$ is the mass ratio and $v_{e}$ is exhaust speed. The true staging constraint (chain each stage's leftover mass down to the payload): $\frac{m _{p}}{M _{0}} = \prod_{i = 1}^{n} \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} . (⋆)$ Optimal single ratio for $n$ identical stages ( $ϵ$ = structural fraction): $R^{*} = \frac{1}{ϵ + ( 1 - ϵ ) k}, k \equiv (\frac{m _{p}}{M _{0}})^{1/ n} .$

Level 1 — Recognition

These test whether you can read the formulas and plug numbers in. No cleverness required.

Exercise 1.1 — Read a mass ratio

A stage ignites at $M_{before} = 8000$ kg and burns out at $M_{after} = 2000$ kg. Find its mass ratio $R$ and the $Δ v$ it delivers if $v_{e} = 2500$ m/s.

Recall Solution

What we do: mass ratio is just "before over after". $R = \frac{8000}{2000} = 4.$ Why $ln$ ? The rocket equation says each equal fractional loss of mass buys the same additive velocity — that repeated multiplication of mass fractions becomes addition under the logarithm. So: $Δ v = v_{e} ln R = 2500 \cdot ln 4 = 2500 \cdot 1.3863 = 3466 m/s .$

Exercise 1.2 — Structural fraction from masses

A stage holds $m_{f} = 4500$ kg of fuel and $m_{s} = 500$ kg of structure. Compute the structural fraction $ϵ$ .

Recall Solution

Definition: $ϵ$ is structure divided by (structure + fuel) — the "dead weight fraction" of the stage. $ϵ = \frac{m _{s}}{m _{s} + m _{f}} = \frac{500}{500 + 4500} = \frac{500}{5000} = 0.1.$

Exercise 1.3 — State the optimum in words

Fill in the blanks (reveal to check):

For

n

stages sharing the same

ϵ

and same

v_{e}

, maximum

Δ v

occurs when all mass ratios are

equal,

R_{1} = R_{2} = \dots = R_{n} = R^{*}

The tool that proves this (because we maximize one quantity under a constraint) is

Lagrange multipliers.

Level 2 — Application

Now you assemble two or three formulas in sequence.

Exercise 2.1 — Optimal $R^{*}$ for a two-stage rocket

$m_{p} = 800$ kg, $M_{0} = 12800$ kg, $ϵ = 0.08$ , two stages. Find $R^{*}$ .

Recall Solution

Step 1 — get $k$ . $k$ is the $n$ -th root of the payload fraction: $k = (\frac{m _{p}}{M _{0}})^{1/2} = (\frac{800}{12800})^{1/2} = (0.0625)^{1/2} = 0.25.$ Step 2 — plug into the optimal-ratio formula. $R^{*} = \frac{1}{ϵ + ( 1 - ϵ ) k} = \frac{1}{0.08 + 0.92 \cdot 0.25} = \frac{1}{0.08 + 0.23} = \frac{1}{0.31} = 3.226.$ Why this formula and not $(M_{0} / m_{p})^{1/ n}$ ? That simpler form only holds when $ϵ = 0$ . Here structure is real, so we must use the version that includes $ϵ$ .

Exercise 2.2 — Total $Δ v$ from optimal staging

Continue Exercise 2.1 with $v_{e} = 3200$ m/s. Find $Δ v_{total}$ .

Recall Solution

Both stages share $R^{*} = 3.226$ , so $Δ v_{total} = n v_{e} ln R^{*} = 2 \cdot 3200 \cdot ln 3.226 = 6400 \cdot 1.1714 = 7497 m/s .$

Exercise 2.3 — Reconstruct the stage-1 fuel

For the rocket of Exercise 2.1, how much fuel does stage 1 burn?

Recall Solution

What we use: fuel burned in a stage is the initial mass times the fraction lost, $m_{f} = M_{before} (1 - 1/ R)$ . Stage 1 ignites at $M_{0} = 12800$ kg: $m_{f 1} = 12800 (1 - \frac{1}{3.226}) = 12800 (1 - 0.31) = 12800 \cdot 0.69 = 8832 kg .$

Level 3 — Analysis

Here you compare scenarios, or run a formula "backwards".

Exercise 3.1 — Equal vs unequal with real structure

Two stages, $ϵ = 0.1$ , $v_{e} = 3000$ m/s, $M_{0} = 10000$ kg, $m_{p} = 1000$ kg. The parent note found the equal solution $R^{*} = 2.60$ giving $Δ v = 5734$ m/s. Now try the unequal split $R_{1} = 3.5$ . Using constraint $(⋆)$ , solve for the forced $R_{2}$ , and compute $Δ v$ . Which wins?

Recall Solution

Step 1 — write $(⋆)$ for two stages. $\frac{m _{p}}{M _{0}} = \frac{1 - ϵ R _{1}}{( 1 - ϵ ) R _{1}} \cdot \frac{1 - ϵ R _{2}}{( 1 - ϵ ) R _{2}} = 0.1.$ Step 2 — plug $R_{1} = 3.5$ . $\frac{1 - 0.1 \cdot 3.5}{0.9 \cdot 3.5} = \frac{0.65}{3.15} = 0.20635.$ Step 3 — solve for the second factor. $\frac{1 - 0.1 R _{2}}{0.9 R _{2}} = \frac{0.1}{0.20635} = 0.48462.$ Cross-multiply: $1 - 0.1 R_{2} = 0.43616 R_{2} \Rightarrow 1 = 0.53616 R_{2} \Rightarrow R_{2} = 1.8651$ . Step 4 — compute $Δ v$ . $Δ v = 3000 (ln 3.5 + ln 1.8651) = 3000 (1.2528 + 0.6234) = 3000 \cdot 1.8762 = 5629 m/s .$ Conclusion: $5629 < 5734$ . Equal ratios win by ~105 m/s. This is the strict advantage that only appears when $ϵ > 0$ — exactly what the Lagrange argument predicts.

Exercise 3.2 — Single-stage ceiling

With $ϵ = 0.1$ , what is the absolute maximum mass ratio a single stage can reach (payload $\to 0$ )? Can a single stage reach $M_{0} / m_{p} = 30$ ?

Recall Solution

Take $n = 1$ , so $k = m_{p} / M_{0}$ . As $m_{p} \to 0$ , $k \to 0$ : $R_{m a x} = \frac{1}{ϵ + ( 1 - ϵ ) \cdot 0} = \frac{1}{ϵ} = \frac{1}{0.1} = 10.$ A single stage can never exceed $R = 10$ with this $ϵ$ . To reach $M_{0} / m_{p} = 30$ you would need $R = 30 > 10$ : impossible in one stage. This is the whole reason staging exists — you shed structure to break past the $1/ ϵ$ ceiling.

Exercise 3.3 — Compounding beats the ceiling

Two identical stages, each with $R = 5.5$ . What effective mass ratio $M_{0} / m_{p}$ does the pair achieve in the ideal ( $ϵ \to 0$ ) limit? Compare to the single-stage ceiling of 3.2.

Recall Solution

In the $ϵ \to 0$ limit the constraint is $\prod R_{i} = M_{0} / m_{p}$ , so ratios multiply: $\frac{M _{0}}{m _{p}} = R_{1} R_{2} = 5.5 \times 5.5 = 30.25.$ Each individual stage's $R = 5.5$ is comfortably below the $R_{m a x} = 10$ ceiling, yet together they exceed $30$ — impossible for one stage. Staging compounds mass ratios.

Level 4 — Synthesis

Multi-part problems that stitch the whole chapter together.

Exercise 4.1 — Full three-stage design

Design a three-stage rocket: $M_{0} = 27000$ kg, $m_{p} = 1000$ kg, $ϵ = 0.1$ per stage, $v_{e} = 3000$ m/s. (a) Find $R^{*}$ . (b) Find $Δ v_{total}$ . (c) Compute the mass entering stage 2, $M_{1}$ .

Recall Solution

(a) $k = (m_{p} / M_{0})^{1/3} = (1000/27000)^{1/3} = (0.037037)^{1/3} = 0.33333$ . $R^{*} = \frac{1}{0.1 + 0.9 \cdot 0.33333} = \frac{1}{0.1 + 0.3} = \frac{1}{0.4} = 2.5.$ (b) $Δ v = n v_{e} ln R^{*} = 3 \cdot 3000 \cdot ln 2.5 = 9000 \cdot 0.91629 = 8247 m/s .$ (c) Each stage multiplies the mass by the factor $\frac{1 - ϵ R ^{*}}{( 1 - ϵ ) R ^{*}} = k = 0.33333$ : $M_{1} = M_{0} \cdot k = 27000 \cdot 0.33333 = 9000 kg .$ (Check: $M_{2} = 9000 \cdot 0.3333 = 3000$ , $M_{3} = 3000 \cdot 0.3333 = 1000 = m_{p}$ . ✓)

Exercise 4.2 — Is a 4th stage worth it?

Same rocket as 4.1 but now split into $n = 4$ stages ( $ϵ = 0.1$ each). Compute $Δ v$ and compare to the 3-stage answer. Interpret.

Recall Solution

$k = (1000/27000)^{1/4} = (0.037037)^{0.25} = 0.43869$ . $R^{*} = \frac{1}{0.1 + 0.9 \cdot 0.43869} = \frac{1}{0.1 + 0.39482} = \frac{1}{0.49482} = 2.0209.$ $Δ v = 4 \cdot 3000 \cdot ln 2.0209 = 12000 \cdot 0.70359 = 8443 m/s .$ Compare: 3 stages gave 8247 m/s, 4 stages gives 8443 m/s — a gain of only 196 m/s (~2.4%). Each extra stage helps less (diminishing returns) while adding real hardware, interstages and complexity the model ignores. Real rockets stop at 2–3 stages.

Exercise 4.3 — Back-solve required $M_{0}$

You need $Δ v_{total} = 9000$ m/s with $n = 2$ , $ϵ = 0.05$ , $v_{e} = 3400$ m/s, $m_{p} = 500$ kg. What launch mass $M_{0}$ is required?

Recall Solution

Step 1 — needed $R^{*}$ per stage. $Δ v = 2 v_{e} ln R^{*}$ : $ln R^{*} = \frac{9000}{2 \cdot 3400} = 1.32353 \Rightarrow R^{*} = e^{1.32353} = 3.7566.$ Step 2 — the per-stage mass factor $k$ . $k = \frac{1 - ϵ R ^{*}}{( 1 - ϵ ) R ^{*}} = \frac{1 - 0.05 \cdot 3.7566}{0.95 \cdot 3.7566} = \frac{0.81217}{3.5688} = 0.22758.$ Step 3 — invert the constraint. $m_{p} / M_{0} = k^{2}$ , so $M_{0} = \frac{m _{p}}{k ^{2}} = \frac{500}{0.2275 8 ^{2}} = \frac{500}{0.051795} = 9653 kg .$

Level 5 — Mastery

These demand you derive, not just plug.

Exercise 5.1 — Prove equal ratios from scratch (2 stages, $ϵ \to 0$ )

With $ϵ \to 0$ the constraint is $R_{1} R_{2} = C$ (fixed). Maximize $f = ln R_{1} + ln R_{2}$ and prove the optimum is $R_{1} = R_{2} = C$ , using calculus.

Recall Solution

Eliminate the constraint. $R_{2} = C / R_{1}$ , so $f (R_{1}) = ln R_{1} + ln \frac{C}{R _{1}} = ln R_{1} + ln C - ln R_{1} = ln C .$ Surprise: $f$ is constant in $R_{1}$ ! In the $ϵ \to 0$ ideal case every split gives the same $Δ v$ (this matches the parent note's "equal beats unequal is not strict here"). The optimum is degenerate — any split, including $R_{1} = R_{2} = C$ , is optimal. Why the theorem still names equal ratios: the strict preference for equal ratios only appears once $ϵ > 0$ makes the constraint nonlinear (see Ex 3.1). Equal ratios is the split that survives both regimes, so it is the safe design rule.

Exercise 5.2 — Lagrange derivation shows all $R_{i}$ satisfy one equation

For general $ϵ$ , we maximize $f = \sum_{i} ln R_{i}$ subject to $g = \sum_{i} ln \frac{1 - ϵ R _{i}}{( 1 - ϵ ) R _{i}} = const$ . Show $\partial g / \partial R_{i}$ depends on $R_{i}$ alone, and conclude all $R_{i}$ are equal.

Recall Solution

Differentiate one term of $g$ . Write $g_{i} = ln (1 - ϵ R_{i}) - ln (1 - ϵ) - ln R_{i}$ . Then $\frac{\partial g}{\partial R _{i}} = \frac{- ϵ}{1 - ϵ R _{i}} - \frac{1}{R _{i}} .$ And $\partial f / \partial R_{i} = 1/ R_{i}$ . The Lagrange condition $\nabla f = λ \nabla g$ reads, for each $i$ : $\frac{1}{R _{i}} = λ (\frac{- ϵ}{1 - ϵ R _{i}} - \frac{1}{R _{i}}) .$ The key observation: every quantity in this equation is a function of $R_{i}$ only (because $ϵ$ and $λ$ are shared constants). So all $n$ stages satisfy the identical equation in one unknown. A well-behaved equation like this has one physical root ⇒ $R_{1} = R_{2} = \dots = R^{*}$ . Equal mass ratios are optimal. $■$

Exercise 5.3 — Sensitivity: how a wrong $ϵ$ costs you

You designed the Ex 4.1 rocket ( $n = 3$ , planned $ϵ = 0.1$ , $R^{*} = 2.5$ , $Δ v = 8247$ m/s). Manufacturing came in heavier: actual $ϵ = 0.12$ . Recompute $Δ v$ (keep the same physical masses, i.e. same $R_{i} = 2.5$ ? No — the extra structure changes burnout). Treat it cleanly: with the same $M_{0}, m_{p}, n$ but $ϵ = 0.12$ , find the new optimal $Δ v$ and the loss.

Recall Solution

Same $k = (1000/27000)^{1/3} = 0.33333$ (unchanged — it depends only on $m_{p} / M_{0}$ and $n$ ). $R_{new}^{*} = \frac{1}{0.12 + 0.88 \cdot 0.33333} = \frac{1}{0.12 + 0.29333} = \frac{1}{0.41333} = 2.4194.$ $Δ v_{new} = 3 \cdot 3000 \cdot ln 2.4194 = 9000 \cdot 0.88354 = 7952 m/s .$ Loss: $8247 - 7952 = 295$ m/s. A 2-percentage-point rise in structural fraction costs nearly 300 m/s — showing why lightweight structures (Structural fraction) are obsessed over in real design.

Recall Master check — reveal one at a time

Optimal rule for $n$ equal- $I s p$ stages ::: all mass ratios equal, $R_{1} = \dots = R_{n} = R^{*}$ . Formula for $R^{*}$ ::: $R^{*} = 1/ (ϵ + (1 - ϵ) k)$ with $k = (m_{p} / M_{0})^{1/ n}$ . Single-stage ceiling ::: $R_{m a x} = 1/ ϵ$ . Why we need Lagrange not just AM–GM ::: for $ϵ > 0$ the product $\prod R_{i}$ is not fixed; the true constraint $(⋆)$ is nonlinear.