Engines · Rocket10 min read

The rocket equation

A rocket is the only vehicle that carries everything it pushes against, and one short equation — written by a deaf Russian schoolteacher in 1903 — prices that decision. It says speed is bought with exponentially more propellant, and every rocket ever flown is a negotiation with it.

Nothing to push against

Every other engine pushes on something outside the vehicle. A car engine twists tyres against the road; a jet engine grabs air at the front and hurls it out of the back; a propeller screws itself through water. In space there is nothing to grab — so a rocket brings its own stuff to throw. It carries tonnes of propellant for exactly one purpose: to fling it backwards as fast as physically possible, so that momentum conservation shoves the vehicle forwards.

That is all thrust is. If the engine throws out $\dot m$ kilograms of exhaust every second at a speed $v_e$ , the momentum leaving per second — which is a force — is:

F = \dot m \, v_e

The F-1 engines under the Saturn V each pumped about 2.6 tonnes of kerosene and oxygen per second out of the nozzle at roughly 2.6 km/s: multiply those and you get its ~6.8 MN of thrust. No aerodynamic mystery, just mass × velocity, per second. And notice what the equation does not contain: the surrounding air. Rockets do not push on the atmosphere, which is why they work better in vacuum, not worse.

The equation

Here is the catch: the propellant a rocket will throw later, it must carry — and accelerate — now. Early in the burn the engine is pushing a vehicle that is mostly its own fuel. As the tank drains the rocket lightens, so the same thrust produces more and more acceleration. Add up that whole draining, lightening burn and you get the speed gain, and the answer turns out to depend on just two numbers: how fast the exhaust leaves, and the ratio of the full mass $m_0$ to the empty mass $m_1$ :

\Delta v = v_e \ln\!\left(\frac{m_0}{m_1}\right)

This is the Tsiolkovsky rocket equation. Not thrust, not burn time, not the size of the rocket — for the ideal case, only exhaust velocity and mass ratio decide how much speed a stage can buy. Everything in rocketry is downstream of those two numbers.

The full derivation, from momentum conservationfor engineers

Take a rocket of mass $m$ coasting at speed $v$ (no gravity, no drag). In a short moment it expels a small mass $\mathrm{d}\mu$ backwards at speed $v_e$ relative to itself. Afterwards the rocket has mass $m - \mathrm{d}\mu$ and speed $v + \mathrm{d}v$ ; the exhaust moves at $v - v_e$ . No external forces act, so momentum before equals momentum after:

m v = (m - \mathrm{d}\mu)(v + \mathrm{d}v) + \mathrm{d}\mu\,(v - v_e)

Multiply out and drop the doubly-small term $\mathrm{d}\mu\,\mathrm{d}v$ :

m\,\mathrm{d}v = v_e\,\mathrm{d}\mu

The expelled mass is exactly what the rocket loses, $\mathrm{d}\mu = -\mathrm{d}m$ , so $\mathrm{d}v = -v_e\,\mathrm{d}m/m$ . Integrate from $m_0$ down to $m_1$ :

\Delta v = -v_e \int_{m_0}^{m_1} \frac{\mathrm{d}m}{m} = v_e \ln\!\left(\frac{m_0}{m_1}\right)

The logarithm appears because each kilogram of propellant is less effective the heavier the rocket currently is — the integrand is $1/m$ . Equivalently, inverting: the mass ratio needed grows exponentially with the Δv you want, $m_0/m_1 = e^{\Delta v / v_e}$ . That inverse form is where the tyranny lives.

The tyranny of the logarithm

Because the mass ratio sits inside a logarithm, propellant obeys a brutal law of diminishing returns. A kerosene rocket that is half propellant gains about 2.1 km/s. To gain the next2.1 km/s it must be three-quarters propellant; the next, seven-eighths. Every equal step of speed doubles the required mass ratio — and doubling the propellant never doubles the Δv, because the extra tonnes must themselves be hauled up to speed before they are burned.

Equal speed steps cost exponentially more propellant fraction. Each +2.1 km/s tick doubles the mass ratio a kerosene stage needs — and 100% propellant is unreachable, because tanks, engines and the payload itself all weigh something.

Now put a number on the mission. Orbital speed is about 7.8 km/s, but a real ascent also pays gravity and drag on the way up, so the practical bill for low Earth orbit is roughly 9.4 km/s of Δv. Feed that into the inverted equation and ask: what share of the rocket on the pad must be propellant, in one stage?

The whole launch industry in one chart. Even hydrogen — the best chemistry we fly — leaves only ~12% of liftoff mass for tanks, engines, structure and payload combined. Every bar here assumes vacuum Isp for the whole flight, so reality is slightly crueller still.

Why rockets stage

You cannot cheat the logarithm with more fuel — but you can cheat the mass ratio. Once a tank is empty it is pure dead weight, yet it still sits inside $m_1$ , spoiling the ratio for every second of the rest of the flight. So: throw it away. Cut loose the empty tank and its heavy engines, light a fresh, smaller rocket that was riding on top, and the equation starts over with a clean $m_0/m_1$ . The Δv of the stages simply adds.

The price is complexity — separation events, extra engines that must light in flight — and everything jettisoned is money spent per launch. That trade is also why reusability changed the economics: a booster that flies home gives back the hardware that staging throws away.

Isp vs thrust

Look back at the equation: thrust is not in it. A feeble engine burning for an hour and a monster burning for a minute deliver the same Δv if they expel the same propellant at the same $v_e$ . For reaching speed, exhaust velocity is everything — it is the only term outside the logarithm, so 10% more Isp buys 10% more Δv everywhere, while 10% more propellant buys ever less.

But a launch pad adds a second, blunter test: before any of that matters, thrust must exceed weight. A rocket with a thrust-to-weight ratio below 1 does not fly slowly — it does not fly. And every second spent climbing costs about 9.8 m/s of Δv to gravity, so limping off the pad wastes the very Δv the good Isp was buying. Hence the classic division of labour: high-thrust, medium-Isp engines (kerosene, solids) to get moving, high-Isp engines (hydrogen) upstairs where thrust barely matters. The Shuttle flew exactly this compromise in one stack: solid boosters for the shove, hydrogen RS-25s for the speed.

Gravity and drag losses — where the extra 1.6 km/s goesfor engineers

The ideal equation assumes all thrust becomes speed. On a real ascent, thrust also fights gravity and the atmosphere, so the delivered Δv is:

\Delta v_{\text{real}} = v_e \ln\!\frac{m_0}{m_1} \;-\; \underbrace{\int g \sin\gamma \; \mathrm{d}t}_{\text{gravity loss}} \;-\; \underbrace{\int \frac{D}{m} \, \mathrm{d}t}_{\text{drag loss}}

where $\gamma$ is the climb angle and $D$ the drag force. Gravity loss dominates: typically 1.1–1.5 km/s for an orbital launcher, versus only 0.05–0.15 km/s of drag. Because the gravity integral runs over time spent climbing steeply, higher liftoff T/W shortens it — but bigger engines add dry mass and hit the atmosphere harder, so real boosters settle around T/W ≈ 1.2–1.5 and pitch over toward the horizontal as early as the air allows. This loss budget is also why low-thrust ion engines, with Isp in the thousands of seconds, are superb in orbit and useless for launch: on the pad their gravity loss would be total.

Choosing a propellant

Since $v_e$ is the strongest lever, why doesn't everyone burn the fastest exhaust available? Because propellant must also be stored, and the tank is part of $m_1$ . The real choice is a three-way trade between exhaust velocity, density, and handling — and each major combination picks a different corner:

Combination	Vacuum Isp	Bulk density	The trade	Flown on
Kerolox — RP-1 / LOX	300–340 s	~1,030 kg/m³	Dense and room-temperature-friendly: compact tanks, big thrust, honest Isp. Kerosene cokes engines, which complicates reuse.	F-1 (Saturn V), Merlin (Falcon 9), RD-180 (Atlas V)
Hydrolox — LH₂ / LOX	420–465 s	~360 kg/m³	The best exhaust velocity chemistry allows — paid for with enormous, heavily insulated tanks at 20 K and modest thrust.	RS-25 (Shuttle/SLS), RL10 (Centaur), Vulcain (Ariane 5)
Methalox — CH₄ / LOX	340–370 s	~830 kg/m³	The deliberate middle: most of kerosene's density, most of hydrogen's Isp, burns clean for rapid reuse — and can in principle be made on Mars.	Raptor (Starship), BE-4 (Vulcan, New Glenn)
Solid — APCP	250–290 s	~1,800 kg/m³	Propellant that is its own tank wall: cheap, storable for years, colossal thrust. Once lit it cannot be throttled, stopped or restarted.	Shuttle & SLS boosters, Ariane 5 EAP

Why hydrogen's Isp comes with a tankage penaltyfor engineers

Liquid hydrogen's density is a farcical 71 kg/m³ — kerosene is eleven times denser. At a typical 6:1 oxygen-to-fuel mass ratio, a hydrolox stage needs nearly three times the tank volume of a kerolox stage holding the same propellant mass, all of it insulated against boil-off at 20 K and stiff enough to fly. Volume is skin, skin is structure, and structure lands in $m_1$ — exactly where the rocket equation punishes it. The flown record shows the bill: the hydrolox S-IVB stage of the Saturn V was about 11% dry mass, while a Falcon 9 kerolox first stage is about 6%. Fold that heavier dry fraction back into $\ln(m_0/m_1)$ and hydrogen's ~45% Isp advantage over kerosene shrinks appreciably — still a win, which is why upper stages love it, but far from the free lunch the Isp column suggests. Add ground handling (liquefaction energy, leaks through everything, embrittled metals) and you see why several modern launchers skipped hydrogen for methane despite the smaller number in the table.