Why rockets are built in stages

Every rocket that has reached orbit has thrown most of itself away on the way. A Falcon 9 drops its first stage about two and a half minutes after lift-off. The Saturn V dropped two whole stages before its third reached orbit. It looks wasteful, and for anyone who wants to reuse a rocket it is expensive to undo. This course explains why chemical rockets leaving the Earth have no way round it, and how to decide where to make the cut.

Dead mass

Course 01 ended with the mass ratio, and it is worth looking at what the final mass contains. The payload, and everything that held the propellant and burned it: tanks, engines, pipes, pumps, wiring, and the structure that carries the loads between them. Call all of that except the payload the structure. It is dead mass. It is needed at the start, but by the end of the burn it does nothing except sit in the bottom of the mass ratio, making it smaller.

Picture a hiker crossing a desert with her water in heavy steel bottles. By the end of the first day half the bottles are empty, and she is still carrying them. If she leaves each bottle behind once it is empty, she walks the second day lighter and further. Staging is leaving the bottles behind.

The structural ceiling

Describe a stage by its structural fraction:

ε = \frac{m _{s}}{m _{s} + m _{p}}

where $m_{s}$ is the stage's structure and $m_{p}$ its propellant. Good modern stages manage roughly 4 to 10 per cent. In the simulator, Aster's first stage is 5.7 per cent structure; its upper stage, which carries landing legs, grid fins and a heat shield, is 7.8 per cent.

Now fly a single stage carrying a payload $m_{pl}$ . Its mass ratio is

R = \frac{m _{s} + m _{p} + m _{pl}}{m _{s} + m _{pl}}

The payload appears above and below the line. Take it away altogether and the mass ratio is as large as it can ever be:

R_{ma x} = \frac{m _{s} + m _{p}}{m _{s}} = \frac{1}{ε}

This is a ceiling, and adding propellant does not raise it, because the tanks grow with the propellant they hold. A stage that is 6 per cent structure can never have a mass ratio above $1/0.06 = 16.7$ . With a kerosene engine's vacuum exhaust velocity of 3.05 km/s, the most it could reach carrying nothing at all is

Δ v_{ma x} = 3.05 ln 16.7 = 8.58 km/s

Low Earth orbit needs about 9.4. A single stage of this kind cannot reach orbit even empty. Hydrogen engines exhaust faster, at about 4.4 km/s, but liquid hydrogen is so light that its tanks are enormous, and hydrogen stages carry proportionally more structure. At 10 per cent the ceiling is $4.43 ln 10 = 10.2$ km/s with nothing on board, which leaves almost nothing for a payload once the climb has taken its share. No launcher has yet reached orbit on a single stage.

Adding stages

Now stack two stages. The first burns, then drops its empty structure, and the second starts with a fresh mass ratio of its own. Their Δv adds:

Δ v = v_{e} ln R_{1} + v_{e} ln R_{2} = v_{e} ln (R_{1} R_{2})

using the property of the logarithm from course 01. The Δv adds; the mass ratios multiply. Each stage is still held under its own ceiling of $1/ ε$ , but their product is not. Two stages with a mass ratio of 5 each give the Δv of a single stage with a mass ratio of 25 — a ratio no stage of 6 per cent structure could ever reach.

The two mass ratios are worth writing out, because each stage has to carry everything above it:

R_{1} = \frac{M _{0}}{M _{0} - m _{p 1}} R_{2} = \frac{m _{2} + m _{pl}}{m _{2} + m _{pl} - m _{p 2}}

$M_{0}$ is the whole vehicle at lift-off, $m_{p 1}$ and $m_{p 2}$ are the two stages' propellant, and $m_{2}$ is the upper stage's total mass. The first stage's ratio is poor, because it pushes the whole fuelled upper stage along with its own tanks. The upper stage's ratio is good, because its load is only the payload. That imbalance is where the design choices live.

Where to cut

Fix the lift-off mass, the payload and the structural fraction, and ask where the boundary between the stages should go.

Put almost everything in the first stage, and the second is a tiny stage that barely adds anything. Put almost everything in the second, and the first is a thin booster that barely moves the heavy stack above it. Somewhere between, the total peaks.

For two stages with the same engines and the same structural fraction, the peak is where the two mass ratios are equal. Equivalently, each stage carries the same multiple of whatever sits above it. With a lift-off mass $M_{0}$ and a payload $m_{pl}$ , the upper stage and payload together should weigh

m_{2} + m_{pl} = M_{0} m_{pl}

the geometric mean of the two. Here is why. Call $r$ the ratio of the mass a stage lifts, itself included, to the mass sitting on top of it: for the first stage $r_{1} = M_{0} / (m_{2} + m_{pl})$ , for the second $r_{2} = (m_{2} + m_{pl}) / m_{pl}$ . Their product is fixed at $M_{0} / m_{pl}$ , wherever the cut goes. A stage's mass ratio depends only on its own $r$ :

R = \frac{r}{1 + ε ( r - 1 )}

which comes from writing its propellant as $(1 - ε)$ of its own mass. Moving the cut raises one $r$ and lowers the other by the same factor. Because $ln R$ grows ever more slowly as $r$ grows — the logarithm again, and the structure dragging it down further — the stage that gains always gains less than the other loses, unless they started equal. So equal is best.

A worked case shows the size of the effect. Take a 500 t vehicle with a 10 t payload, structure at 8 per cent in each stage, and an exhaust velocity of 3.05 km/s.

As one stage, its mass ratio is $500/ (500 - 0.92 \times 490) = 10.16$ , worth $3.05 ln 10.16 = 7.07$ km/s. Not an orbital rocket.
As two stages, the upper stage and payload should weigh $500 \times 10 = 70.7$ t, so the upper stage is 60.7 t and the first stage 429.3 t. Each stage then lifts 7.07 times the mass above it, each mass ratio is $7.07/ (1 + 0.08 \times 6.07) = 4.76$ , and the total is $2 \times 3.05 ln 4.76 = 9.52$ km/s.

The same tanks, the same engines and the same propellant: 2.44 km/s more, bought by dropping 34 t of empty first stage at the right moment.

More stages help less each time. Split optimally, the same vehicle reaches 10.15 km/s on three stages, 10.41 on four and 10.55 on five. On infinitely many it would approach $v_{e} (1 - ε) ln (M_{0} / m_{pl}) = 10.98$ km/s, as if its structure fell away continuously as it burned. Every separation also costs hardware — an interstage, a separation system, another set of engines — and another chance for something to go wrong. After two or three stages those costs outweigh the gain, which is why nearly every launcher stops there.

Figure · where to put the mass

STAGE 1 SHARE OF STAGE MASS70.0 %

STRUCTURAL FRACTION8.0 %

PAYLOAD10 t

TOTAL Δv: 9.07 km/s
STAGE 1: 3.04 km/s
STAGE 2: 6.03 km/s
MASS RATIOS: 2.71 · 7.22
GAIN OVER ONE STAGE: +2.00 km/s

500 t at lift-off, both stages with the same engines (exhaust velocity 3.05 km/s) and the same structural fraction. Vacuum Δv only. The dashed line is the same tanks, engines and propellant flown as one stage, nothing dropped. With identical stages the peak falls where the two mass ratios are equal; the curve is flat near it, which is the freedom designers spend on everything else.

Drag the split and watch the two burns trade places in the bar. At the peak the two mass ratios in the readout agree. Notice how flat the top of the curve is: anywhere from 80 to 90 per cent of the mass in the first stage, the total is within 130 m/s of its best.

Then change the other two. A heavier structure lowers both curves and widens the gap between one stage and two, so it makes staging matter more — but with identical stages it does not move the peak at all. A heavier payload does move it, towards a bigger upper stage, and lowers everything.

Why real rockets sit off the peak

The equal-ratio rule assumes identical stages in vacuum. Real stages differ in ways that move the peak, and the flatness of the curve gives designers room to follow them.

Different engines. A first stage fires through the atmosphere on sea-level nozzles; an upper stage can carry a vacuum engine with a higher specific impulse (course 02). Δv is cheaper where the exhaust is faster, which pushes more of the job onto the upper stage.
Thrust. The first stage has to lift the whole vehicle off the pad with room to spare, and an upper stage with too little thrust spends its Δv holding itself up against gravity (course 04).
Coming home. A booster that lands keeps propellant back for the return and carries legs and fins, so it gives the upper stage less than its tanks could.
Cost. Engines are expensive, and one kind is cheaper to build than two. Falcon 9 flies Merlins on both stages, the upper one with a vacuum bell.

Vivapse's own design model shows the flatness directly. Keep Aster's 500 t of propellant and move it between the stages:

Upper-stage propellant	First-stage propellant	Total Δv in the Vehicle panel
60 t	440 t	11.12 km/s
80 t	420 t	11.15 km/s
100 t, the preset	400 t	11.11 km/s
120 t	380 t	11.05 km/s
140 t	360 t	10.96 km/s

Eighty tonnes of change moves the total by less than 200 m/s, and the preset sits 34 m/s below the model's best. Something else decides where the cut goes. Between 80 and 100 t, for instance, the upper stage's thrust-to-weight ratio at ignition falls from about 1.05 to 0.86.

A worked example: Aster in one piece

At lift-off Aster weighs 535,477 kg, and it carries 500,000 kg of propellant in its two stages. Everything left once both are empty — both stages' structure, 900 kg of attitude-thruster gas and the payload in its fairing — weighs 35,477 kg.

Fly it as one piece, keeping all that structure to the end, and its mass ratio is $535 477/35 477 = 15.09$ . With the Merlin's vacuum exhaust velocity of 3,050 m/s,

Δ v = 3050 \times ln 15.09 = 3050 \times 2.714 = 8278 m/s

Staged, the same propellant gives 4,192 m/s from the first stage and 7,116 m/s from the second (course 01 worked both): 11,308 m/s. The difference, 3.0 km/s, comes from a single act — dropping 24.7 t of empty booster, its 24,137 kg of structure and 600 kg of thruster gas, once it has nothing left to give.

In Vivapse

buildDesign() in src/sim/parts.ts estimates each stage's structure from its size and hardware, then walks the stack from the top down, so that each stage's mass ratio counts everything it carries:

let above = payload.totalMass;
for (let i = nSt - 1; i >= 0; i--) {
  const s = stages[i];
  s.massAtIgnition = s.wetMass + above;
  s.massAtBurnout = s.dryMass + s.rcsGas + above;
  // ... the rocket equation, as in course 01 ...
  above += s.wetMass;
}

The structure comes from a model rather than a fixed fraction. Tank walls scale with the tanks' surface area, the material's density and the stage's diameter; the thrust structure scales with the engines' thrust; legs, grid fins and a heat shield add their own mass. That is why the same Aster first stage is 2.5 t heavier with its landing hardware than without it, and why its structural fraction improves as its tanks grow longer. The real-scale presets use structureScale to bring an upper stage's walls down to its published dry mass.

For missions beyond Earth orbit, spareDeltaV() walks the stages the same way to find the Δv a stack still has once 9.4 km/s has been spent reaching orbit — the number a trip to the Moon has to fit inside.

In flight, staging is your program's decision. fc.separate() drops the spent stage. The reference program full-mission.js shuts the engines down, separates a second later, fires the upper stage's thrusters to settle its propellant, and lights its engine a few seconds after cut-off. The vehicle and flight program pages cover the two halves.

Try it

In the Vehicle panel, start from Aster and note its Total Δv: 11.11 km/s. Now remove stage 2, and raise stage 1's propellant mass from 400 t to 500 t, so that the rocket carries the same propellant in one piece. The total falls to 8.38 km/s, and the design review warns that low Earth orbit needs 9.3 to 9.6 km/s.

Look at the dry mass as well. The single stage is lighter than the two stages were — one set of avionics, no interstage, no second engine — and it is still 2.7 km/s short of the rocket it replaced. Carrying less structure did not help, because none of it could be left behind.

What carries forward

Every Δv so far has been ideal: engines firing in empty space, with nothing to fight. The real climb starts on the ground, in thick air, pulling against gravity, and a good part of the 9.4 km/s budget goes on those fights rather than on speed. The next course follows where it goes, and why the answer is for the rocket to turn gently rather than steer.