The rocket equation

A rocket in space has nothing to push against. There is no road under it and no air for a propeller to grip. Whatever it does to its own motion, it does by throwing part of itself away. So the question that decides every mission is this: how much can a rocket change its own speed before it runs out of things to throw?

The answer is a single number, the delta-v, written Δv. It is the velocity budget from the overview. Every burn you command spends some of it, and when it is gone the flight is over, however well it has been flown. This course derives the equation that sets the budget, from one law of physics and some patient arithmetic. By the end you should be able to look at a rocket's masses and say how far it can go — and why making it bigger would not help.

Throwing bricks

Picture yourself on a trolley on perfectly smooth ice, with a pile of bricks. Nothing touches you but the ice, and the ice cannot push you along. You pick up a brick and throw it backwards as hard as you can. The brick goes one way; you, the trolley and the remaining bricks drift the other. Throw another and you speed up again.

Two things are worth noticing before any equation.

The first brick is the least useful. When you throw it you are pushing against yourself, the trolley and every brick still on it. That heavy load barely moves. The last brick is the most useful, because by then it pushes against nothing but you and an empty trolley.

And your speed does not matter. A brick thrown at the same speed relative to you gives the same shove whether you are at rest or already rolling fast. The ice does not know how fast you are going, and neither does the brick. This is what lets a rocket keep accelerating long after a propeller or a wheel would have given up.

A rocket engine is a brick-thrower that throws very small bricks very fast — about three kilometres per second for the engines in this course — and a great many of them every second. Everything else follows from those two observations.

One throw

Momentum is mass times velocity. With no outside force on a system — no gravity, no air, no ground — its total momentum stays the same. That is the only law this course needs.

Take a rocket of mass $m$ , moving at velocity $v$ , which includes a chunk of propellant of mass $Δ m$ that it is about to throw. It throws the chunk backwards. Afterwards the rocket, now of mass $m - Δ m$ , moves at $v + δ v$ , where $δ v$ is the small change in speed we are after. The chunk leaves at speed $v_{e}$ relative to the rocket, which in our frame puts it at $v + δ v - v_{e}$ .

Before the throw the total momentum is $m v$ . After it, adding the rocket and the chunk, it is

(m - Δ m) (v + δ v) + Δ m (v + δ v - v_{e})

Multiply out and the two $Δ m (v + δ v)$ terms cancel, leaving $m (v + δ v) - Δ m v_{e}$ . Set that equal to $m v$ and

m δ v = Δ m v_{e} so δ v = v_{e} \frac{Δ m}{m}

Read it slowly. The speed a throw buys is the exhaust speed times the chunk's share of the mass at that moment. The starting velocity $v$ has cancelled out: the trolley's speed really does not matter. Three assumptions went in, and they are worth keeping in view. Nothing outside pushes on the rocket. The exhaust always leaves at the same speed $v_{e}$ relative to it. And that speed is measured after the throw.

Many throws

A real engine throws its propellant in a continuous stream, but we can get there one chunk at a time. Take a rocket of 10 tonnes, 9 of them propellant, so that it ends at 1 tonne. How much speed can it gain?

Throw all 9 tonnes at once and the formula gives $v_{e} \times 9/10 = 0.90 v_{e}$ .

Throw it in two chunks of 4.5 tonnes. The first is 4.5 of 10 tonnes, worth $0.45 v_{e}$ . The second is 4.5 of the 5.5 tonnes left, worth $0.818 v_{e}$ — nearly twice as much, for the same propellant. Together, $1.268 v_{e}$ .

Keep cutting the propellant finer and the total keeps growing, but more and more slowly.

Chunks	Speed gained, in units of the exhaust velocity
1	0.900
2	1.268
3	1.479
5	1.715
10	1.960
100	2.263
1,000	2.299
100,000	2.3025

It settles on a limit, 2.3026, and no amount of cutting takes it past. That limit is what a real engine achieves, because a stream is what you get from infinitely many infinitely small throws. The limit also has a name. It is the natural logarithm of 10, written $ln 10$ — and 10 is the ratio of the rocket's starting mass to its final mass.

The logarithm

Repeat the experiment with any masses and the limit is always the natural logarithm of the ratio of the starting mass to the final one. That is the rocket equation, which Konstantin Tsiolkovsky published in 1903:

Δ v = v_{e} ln \frac{m _{0}}{m _{f}}

Here $Δ v$ is the total change in speed, $v_{e}$ is the exhaust velocity relative to the rocket, $m_{0}$ is the mass at the start of the burn, $m_{f}$ the mass at the end, and $ln$ is the natural logarithm. The ratio $m_{0} / m_{f}$ is the mass ratio, and we will call it $R$ .

If you know a little calculus, the table is a sum turning into an integral. Shrink $Δ m$ to a vanishingly small $d m$ , note that the rocket's own mass falls by exactly that amount, and each throw gives $d v = - v_{e} d m / m$ . Adding these up from $m_{0}$ down to $m_{f}$ is the integral of $1/ m$ , which is the logarithm. If you do not, the table is the whole argument: the logarithm is what "each small piece of propellant, divided by the mass carrying it" adds up to.

One property of the logarithm does most of the work from here on. The logarithm of a product is the sum of the logarithms:

ln (a b) = ln a + ln b

So doubling the mass ratio always adds the same Δv, $v_{e} ln 2 \approx 0.693 v_{e}$ , whether you double it from 2 to 4 or from 10 to 20. The second doubling costs far more propellant than the first and buys exactly the same.

Size does not appear

Look at what is missing from the equation. There is no length, no volume, no thrust and no total mass — only the ratio of two masses and the speed of the exhaust. Multiply every mass in a rocket by ten and its mass ratio does not change, so neither does its Δv.

A rocket the size of a pencil and a rocket the size of a building reach the same Δv if they have the same mass ratio and the same exhaust velocity. Size buys payload, not speed. A bigger rocket carries more to the same destination; it does not reach a further one. The only two ways to go further are a faster exhaust or a better mass ratio.

Size does matter elsewhere, in ways the equation leaves out. Tanks and engines do not shrink in proportion (course 03), and air slows small rockets more than large ones (course 04). But the equation itself has no place for size.

Why the ratio is so punishing

Turn the equation round to ask what mass ratio a given Δv needs:

\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}}

where $e \approx 2.718$ is the base of the natural logarithm. The mass ratio grows exponentially with the Δv you ask for: every additional $v_{e}$ of speed multiplies it by 2.718.

Reaching low Earth orbit costs about 9.4 km/s once the losses of the climb are counted. A kerosene engine like the Merlin exhausts at about 3.05 km/s in vacuum. The mass ratio needed is $e^{9.4/3.05} = 21.8$ , so the rocket must be 95.4 per cent propellant at lift-off, and everything else — tanks, engines, wiring, structure and the payload — must fit into the remaining 4.6 per cent. That is about the proportion of drink to aluminium in a full drinks can. Even hydrogen and oxygen, the fastest exhaust in practical use at about 4.4 km/s, need a mass ratio of 8.3: 88 per cent propellant.

The returns diminish in the same way. At an exhaust velocity of 3.05 km/s, raising the mass ratio from 5 to 6 adds 556 m/s. From 10 to 11 it adds 291 m/s, and from 20 to 21 only 149. Each extra tonne of propellant has to carry itself for most of the burn.

Drag the mass ratio in the figure and watch the Δv readout crawl as it climbs.

Figure · the rocket equation

EXHAUST VELOCITY3.05 km/s

MASS RATIO4.0

Δv: 4.23 km/s
PROPELLANT SHARE OF LIFT-OFF MASS: 75.0 %
THE NEXT +1 OF MASS RATIO ADDS: 681 m/s
MASS RATIO FOR 9.4 km/s: 21.8

Δv = v_e ln(m₀/m_f), with no gravity and no air. Each tread of the grey staircase doubles the mass ratio, and each riser is the same height, v_e ln 2, however far along the curve it sits. The dotted line is roughly what low Earth orbit costs once the losses of the climb are counted (course 04).

The staircase is the logarithm made visible. Each riser is the same height; each tread is twice as long as the one before. Everything to the right of a mass ratio of about 15 is expensive country, and it is where orbital rockets have to live.

A worked example: Aster's upper stage

Aster is Vivapse's default vehicle, a two-stage kerosene rocket in the class of a Falcon 9. Its upper stage carries 100,000 kg of propellant in 8,447 kg of dry structure, with 300 kg of gas for its attitude thrusters and a 1,993 kg payload and fairing on top. At ignition it weighs

m_{0} = 100000 + 8447 + 300 + 1993 = 110740 kg

and when its propellant is gone, $m_{f} = 10740$ kg. The mass ratio is $110 740/10 740 = 10.311$ , and $ln 10.311 = 2.3332$ . Its Merlin engine exhausts at 3,050 m/s in vacuum (course 02 explains where that figure comes from), so

Δ v = 3050 \times 2.3332 = 7116 m/s

That is the figure the simulator's Vehicle panel shows for stage 2.

Now the first stage. It burns four times as much propellant, 400,000 kg, but it has to push everything, the fully fuelled upper stage included. At lift-off Aster weighs 535,477 kg; after the first stage's burn, 135,477 kg. That mass ratio is only 3.953, so in vacuum the first stage gives $3050 \times ln 3.953 = 3050 \times 1.3744 = 4192$ m/s. Four times the propellant buys less than 60 per cent of the upper stage's Δv. It is the first brick, on an industrial scale.

Together the two stages come to 11.31 km/s. The Vehicle panel shows 11.11, because it counts the first stage at the average of its sea-level and vacuum performance: that stage starts on the pad and finishes almost in vacuum, and course 02 explains why the two differ by about ten per cent.

In Vivapse

The simulator computes every stage's Δv with this equation when it builds a design. buildDesign() in src/sim/parts.ts walks the stack from the top down, so that each stage's starting mass includes everything it carries:

s.massAtIgnition = s.wetMass + above;
s.massAtBurnout = s.dryMass + s.rcsGas + above;
const lr = Math.log(s.massAtIgnition / s.massAtBurnout);
s.deltaVVac = s.ispVac * C.G0 * lr;
s.deltaVSL = s.ispSL * C.G0 * lr;

ispVac * C.G0 is the exhaust velocity, in the form engine datasheets quote it. The Vehicle panel adds the stages into its Total Δv, marks low Earth orbit at 9.4 km/s on its Δv budget bar, and warns in its design review when a vehicle falls short of 9.3 km/s.

In flight, fc.deltaV gives your program the Δv left in the current stage, from the same equation with the current mass, the propellant still in the tanks and the engines' present performance. On a real launch site it counts only the propellant the engines can use: one per cent of every load stays in the tanks as residuals. The flight computer reference lists it with everything else your program can read.

The physics does not use the equation to fly the rocket. It integrates thrust and mass flow fifty times a second, and the rocket equation falls out of the motion; the simulator's physics audit found a vacuum burn matching Tsiolkovsky to within 0.015 per cent. The physics model describes how.

Try it

Open the playground, choose the Hopper preset in the Vehicle panel, and note its Total Δv: 4.67 km/s from 24 t of propellant on one Merlin.

Double the propellant mass to 48 t. The Δv rises to 6.07 km/s. Double it again, to 96 t: 7.35 km/s. Once more, to 192 t: 8.37 km/s. Each doubling buys less than the one before — 1.40, then 1.28, then 1.02 km/s — because doubling the propellant does not double the mass ratio, and the tanks that hold it grow as well. Eight times the propellant, and the Hopper still could not reach orbit.

Watch the thrust-to-weight ratio as you go. At 96 t it falls to 0.83, and the design review says the vehicle cannot leave the pad. One engine cannot lift that much propellant, and every engine you add is mass the rocket equation has to carry. The equation sets the budget; it is not the only bill.

What carries forward

Everything here hangs on one engine number, the exhaust velocity, and this course has treated it as given. The next one takes it apart: where thrust comes from, why the same engine is about ten per cent better in vacuum than on the pad, and why the specific impulse printed on every engine datasheet is this velocity divided by a constant that has nothing to do with gravity. Keep the mass ratio in mind too. Tanks and engines put a ceiling on it, and course 03 shows how rockets get past that ceiling by throwing whole pieces of themselves away.