Case Study: Designing a Two-Stage Rocket from Scratch

"Design is where the equation stops being a fact and starts being a decision."

Executive Summary

In the first case study we took a finished rocket apart. Here we do the harder thing: we build one on paper. Given only a payload and a destination, we will size a two-stage launcher — how much propellant, how much structure, how big overall — using nothing but the rocket equation and one honest engineering constraint (structures cannot be made arbitrarily light). Along the way we will derive a reusable sizing formula, discover why the payload fraction of an orbital rocket is so painfully small, and see how the choice of propellant reshapes the whole vehicle. This is exactly the calculation that begins every real launch-vehicle design, and the one you will run on your own mission in later chapters.

Skills applied

  • Inverting the rocket equation to size a stage, not just score it (§3.1–3.3).
  • Modeling structure with a structural coefficient and deriving a stage-mass formula (§3.5).
  • Splitting a delta-v budget between stages and building from the payload down (§3.5–3.6).
  • Judging a design by its payload fraction and iterating on propellant choice (§3.4).

Background

The requirement

  • Payload: $m_L = 500\ \text{kg}$ (a small satellite).
  • Destination: low Earth orbit, so the vehicle must supply $\Delta v = 9.4\ \text{km/s}$ (including the losses of Chapter 4).
  • Architecture: two stages, because Section 3.5 showed a single chemical stage cannot do this with anything left to carry.

The one hard constraint

Real stages are not massless shells. We capture that with a structural coefficient $$\varepsilon = \frac{m_s}{m_s + m_p},$$ the fraction of a loaded stage (structure + propellant) that is dry structure. Good kerosene stages reach $\varepsilon \approx 0.06$–$0.10$; we will use $\varepsilon = 0.08$. This single number is where all the difficulty lives: if $\varepsilon$ could be zero, orbit would be easy.

The design tool: a stage-sizing formula

To size a stage we invert the rocket equation. A stage that must deliver $\Delta v$ with exhaust velocity $v_e$ needs mass ratio $R = e^{\Delta v/v_e}$. Let $m_{\text{above}}$ be everything the stage carries (its payload — for the upper stage that's the satellite; for the lower stage that's the entire fueled upper stage plus satellite), and let $m_{\text{stage}} = m_s + m_p$ be the stage's own loaded mass. Then $$m_0 = m_{\text{above}} + m_{\text{stage}}, \qquad m_f = m_{\text{above}} + \varepsilon\, m_{\text{stage}},$$ because at burnout only the structural fraction $\varepsilon\,m_{\text{stage}}$ remains. Setting $R = m_0/m_f$ and solving for the stage mass:

Strategy first. We want $m_{\text{stage}}$ in terms of the things we know — $R$, $\varepsilon$, and $m_{\text{above}}$. Substitute the two mass expressions into $R = m_0/m_f$, clear the fraction, and collect the $m_{\text{stage}}$ terms.

$$ R\,(m_{\text{above}} + \varepsilon\, m_{\text{stage}}) = m_{\text{above}} + m_{\text{stage}} \;\;\Longrightarrow\;\; \boxed{\ m_{\text{stage}} = m_{\text{above}}\,\frac{R - 1}{1 - R\varepsilon}\ } $$

Two things to notice before we use it. The formula blows up as $R\varepsilon \to 1$, i.e. as $R \to 1/\varepsilon = 12.5$: no stage with $\varepsilon = 0.08$ can deliver a mass ratio of 12.5 or more, no matter how much propellant you add, because the structure to hold that propellant would eat the whole gain. That wall is exactly why we must stage. Each of our two stages must therefore keep its own $R$ comfortably below 12.5.

Phase 1: Split the delta-v

We divide the $9.4\ \text{km/s}$ between the two stages. A simple, near-optimal choice when the two stages use similar propellants is to split it evenly: $\Delta v_1 = \Delta v_2 = 4.7\ \text{km/s}$. We'll use a sea-level-to-vacuum kerosene first stage ($v_e = 2.9\ \text{km/s}$) and a vacuum kerosene upper stage ($v_e = 3.4\ \text{km/s}$).

Their mass ratios: $$R_2 = e^{4700/3400} = e^{1.382} = 3.98, \qquad R_1 = e^{4700/2900} = e^{1.621} = 5.06.$$ Both are well under 12.5 — good, the design is feasible. (Check: $R_2\varepsilon = 0.32$, $R_1\varepsilon = 0.40$.)

Phase 2: Size the upper stage

The upper stage carries only the satellite, so $m_{\text{above}} = 500\ \text{kg}$: $$m_{\text{stage},2} = 500 \times \frac{3.98 - 1}{1 - 3.98\times0.08} = 500 \times \frac{2.98}{0.682} = 2{,}185\ \text{kg}.$$ Of that, structure is $\varepsilon\,m_{\text{stage},2} = 0.08 \times 2{,}185 = 175\ \text{kg}$ and propellant is $2{,}010\ \text{kg}$. The loaded upper stage plus payload — what the first stage must carry — is $$m_{\text{above},1} = 500 + 2{,}185 = 2{,}685\ \text{kg}.$$

Phase 3: Size the first stage

Now feed that up-mass into the same formula with the first stage's numbers: $$m_{\text{stage},1} = 2{,}685 \times \frac{5.06 - 1}{1 - 5.06\times0.08} = 2{,}685 \times \frac{4.06}{0.595} = 18{,}320\ \text{kg}.$$ Structure $= 0.08 \times 18{,}320 = 1{,}466\ \text{kg}$; propellant $= 16{,}854\ \text{kg}$.

Phase 4: Total the vehicle and judge it

The lift-off mass is everything: $$m_0 = 500 + 2{,}185 + 18{,}320 = 21{,}005\ \text{kg} \approx 21\ \text{t}.$$

So a 21-tonne rocket delivers a half-tonne satellite to orbit. The payload fraction is $$\frac{m_L}{m_0} = \frac{500}{21{,}005} \approx 2.4\%.$$

That is not a mistake, and it is not bad — it is typical. Real orbital launchers deliver 1–4% of their lift-off mass as payload. The other ~97% is propellant and the structure to hold it. You are looking, in one number, at the entire reason spaceflight is expensive: to send anything to orbit, you must build, fuel, and throw away something roughly forty times heavier.

💡 Intuition: Feel where the mass went. The upper stage more than quadrupled the payload it carries (500 kg became a 2,685 kg up-mass). The first stage then had to lift that, and multiplied again to 21 tonnes. Staging is a chain of multiplications, each one the mass ratio of a stage — and because the ratios multiply, the lift-off mass grows fast as the delta-v climbs. This compounding is the rocket equation's exponential wearing an engineering costume.

Phase 5: Iterate — what if the upper stage burned hydrogen?

Design is iteration. Suppose we upgrade the upper stage to liquid hydrogen, $v_e = 4.4\ \text{km/s}$ (at a cost in tank size and complexity we will study in Chapter 18). Its mass ratio drops to $R_2 = e^{4700/4400} = e^{1.068} = 2.91$, so $$m_{\text{stage},2} = 500 \times \frac{2.91 - 1}{1 - 2.91\times0.08} = 500 \times \frac{1.91}{0.767} = 1{,}245\ \text{kg},$$ and the up-mass the first stage carries falls to $500 + 1{,}245 = 1{,}745\ \text{kg}$ — down from 2,685. The first stage shrinks in proportion, and the whole vehicle gets dramatically lighter. Higher exhaust velocity, applied to the upper stage where its benefit is carried the whole way, is one of the most powerful levers a designer has. (We put hydrogen upstairs, not downstairs, for reasons of thrust and tank size that Chapters 17–18 will make clear.)

🔧 Engineering Reality: Our even 4.7/4.7 split was a guess. The truly optimal split — which minimizes lift-off mass — depends on each stage's $v_e$ and $\varepsilon$, and generally gives the higher-$v_e$ stage a bit more of the delta-v. Finding it is an optimization problem we set up properly in Chapter 22. For a first design, an even split gets you within a few percent, which is close enough to size tanks and move on.

Discussion Questions

  1. Why does the stage-sizing formula have $1 - R\varepsilon$ in the denominator, and what does it mean physically when that quantity approaches zero?
  2. The first stage ended up nearly ten times heavier than the upper stage. Trace why, in terms of what each stage carries.
  3. Payload fraction came out around 2.4%. If you halved the structural coefficient to $\varepsilon = 0.04$ (much lighter structure), would payload fraction roughly double, more than double, or less than double? Reason about the formula before computing.
  4. We split the delta-v evenly. Argue physically why giving the higher-$v_e$ stage more of the delta-v might reduce total mass.

Your Turn: Extensions

  • Option A (design). Redo the whole design for a $2{,}000\ \text{kg}$ payload. Does the lift-off mass scale linearly with payload? Explain why or why not.
  • Option B (computation). Turn the boxed formula into a Python function size_stage(m_above, dv, ve, eps) and chain it to size a two-stage rocket from the top down. Reproduce the 21-tonne result. (Do not run it; hand-trace and add # Expected output:.)
  • Option C (your mission). Apply this method to your mission's launch requirement (or to the spacecraft's own propulsion stage), and record the sized propellant mass in your MDR.

Key Takeaways

  1. Sizing inverts the rocket equation. Given a required delta-v, a $v_e$, and a structural coefficient $\varepsilon$, the stage mass is $m_{\text{stage}} = m_{\text{above}}(R-1)/(1-R\varepsilon)$.
  2. Structure sets a hard ceiling. With $\varepsilon = 0.08$, no single stage can exceed a mass ratio of $1/\varepsilon = 12.5$ — the mathematical reason staging is mandatory.
  3. Build from the payload down, because each stage carries everything above it; the masses compound.
  4. Payload fraction is small on purpose (~1–4%): the price the rocket equation charges to reach orbit. Raising upper-stage $v_e$ is among the most effective ways to lower the whole vehicle's mass.