Most of a rocket is fuel, and better engineering will not change that

By the 42Math team · 13 May 20267 min read

A rocket that reaches low orbit is about 93% propellant by mass. That is not a sign the engineers gave up; it is what the arithmetic permits.

Every rocket obeys one equation, written down by Konstantin Tsiolkovsky in 1903:

Δv = v_e · ln(m₀ / m_f)

The speed you gain is the exhaust velocity multiplied by the logarithm of the ratio between your starting mass and your finishing mass. That logarithm is the whole story, and it is bad news.

What the logarithm costs

A good chemical engine has a specific impulse around 350 seconds, which is an exhaust velocity of 3.43 km/s. Reaching low Earth orbit takes roughly 9.4 km/s once you include the losses to gravity and air.

Divide and exponentiate: the mass ratio you need is e^(9.4/3.43) = 15.5. Your rocket must start 15.5 times heavier than it finishes, which means 93.5% of it is propellant before you have added a single kilogram of payload, tank, engine or structure.

That is the number people find shocking. It is not the one that decides how a mission is designed.

Orbit is the expensive part

Once you are in low orbit, the transfer to Mars costs about 3.6 km/s. Same engine, same equation: a mass ratio of 2.85, or 65% propellant.

So the journey divides into a brutal first stage and a comparatively cheap everything-else. Getting 300 km off the ground demands 15.5 against 2.85 — five and a half times the mass ratio of crossing 200 million kilometres of empty space. Robert Heinlein's line — that once you are in orbit you are halfway to anywhere — is an understatement about the energy and roughly right about the difficulty.

The reason is that orbit is not about altitude. A rocket that went straight up 300 km and stopped would fall back. Orbit is about going sideways fast enough that your fall keeps missing, and that horizontal speed is around 7.8 km/s. The altitude is nearly free; the speed is what you are buying.

Why staging exists

The logarithm also explains the least elegant thing about rockets: they throw away most of themselves on the way up.

Empty tanks are still mass, and mass in the denominator of that ratio is mass you are paying to accelerate for the rest of the flight. Discarding a spent stage does not merely lighten the vehicle; it resets the ratio, so the next stage's own e^(Δv/v_e) applies to a much smaller number. Staging is not a workaround for weak engines. It is the only way to beat an exponential without a better exhaust velocity.

And better exhaust velocity is exactly what chemistry will not give you. The energy per kilogram of a chemical bond sets v_e, and hydrogen-oxygen at about 4.4 km/s is close to the ceiling. Every gain since Apollo has come from structures, materials and reuse — not from the term the equation actually cares about.

Falcon 9, weighed

That estimate is worth testing against a vehicle that exists. SpaceX publishes the stage masses: the first stage carries 410,600 kg of propellant and weighs 22,200 kg empty, the second carries 107,500 kg and weighs 4,000 kg.

So 518,100 kg of propellant against 26,200 kg of empty rocket. The vehicle by itself is 95.2% propellant. Load a full expendable payload of 22,800 kg and the stack is 91.4%.

The prediction from two numbers and a logarithm was 93.5%. The real vehicle, with all its plumbing and margins and landing legs, sits two points away. You can size a rocket you have never seen with an exhaust velocity and a target speed, which is the practical reason the equation is worth knowing.

One caution if you go checking. SpaceX also quotes a liftoff mass of 549,000 kg, and that is smaller than the stages plus a full payload. The two figures describe different configurations, so pick one basis and stay on it rather than dividing a number from one by a number from the other.

What landing costs, in payload

Falcon 9 lifts 22,800 kg to low orbit when the booster is thrown away, and 18,500 kg when it flies home. The difference is 4,300 kg, or 18.9% of the payload.

That is the price of reuse, and it is charged in exactly the currency the equation deals in. Coming back means a boostback burn, an entry burn and a landing burn, and the propellant for all three has to be aboard at liftoff, sitting in the same tanks and counted in the same mass ratio. The rocket equation does not know or care that you intend to fly the stage again.

So reuse is not a way around the arithmetic. It is a decision that a recovered booster is worth more than the 4.3 tonnes of payload it displaces, which for a vehicle that has flown the same first stage more than twenty times is not a close call. The economics changed; the physics did not move at all.

Starship, and the one exit

Starship is far larger and lands in the same place. Its booster holds 3,400 t of propellant and the ship another 1,500 t, which is 4,900 t of a stack around 5,300 t: 92.5% propellant. Being the biggest rocket ever built does not move the fraction, because the fraction is set by an exponential and not by scale.

Its engines do move it. Raptor runs at 380 seconds of specific impulse in vacuum against the 350 assumed above, an 8.6% better exhaust velocity. That drops the required mass ratio from 15.5 to 12.5 and the propellant fraction from 93.5% to 92.0%, which sounds like almost nothing.

Look at the other side of the same subtraction. Everything that is not propellant — structure, engines, tanks, payload — goes from 6.47% of the vehicle to 8.03%. An 8.6% gain in exhaust velocity bought a 24% larger budget for everything you actually wanted to launch. This is the one place in the whole subject where the logarithm works in your favour, and it is why engine efficiency is fought over by the second.

The vehicle is not there yet. The published payload for the Block 1 ship is 15 t to low orbit, which is less than Falcon 9 carries. The largest rocket ever flown lifts less than the rocket it is meant to replace, because early articles are heavy and the mass comes out over successive versions.

The genuinely new idea is not the size. You cannot beat the logarithm, but you can reset the number it acts on: refuel in orbit. The ship's tanks hold 1,500 t, and Musk's estimate is eight tanker flights to fill one in low orbit, which implies roughly 190 t delivered per trip.

Read that as arithmetic rather than ambition. A single stack going from the ground to Mars needs one enormous mass ratio covering the whole trip. Refuelling splits it into nine ordinary ones: eight climbs to orbit and then a departure that begins with full tanks, above the atmosphere, already moving at 7.8 km/s. Each is an exponential you can afford. Their sum is not.

Every architecture for going further than the Moon eventually arrives at that same move, because it is the only known way to make m0 larger than the vehicle that carried it.

Where the numbers go next

Ion engines break the ceiling by not being chemical at all: exhaust velocities of 30 km/s and up, which turns a mass ratio of 15 into one of about 1.4. The catch is thrust measured in millinewtons, so they cannot lift anything off a planet and take months to do what a chemical burn does in minutes. They are the right answer for the cheap part of the trip and useless for the expensive part.

This is why the shape of spaceflight has not changed much in sixty years, and why proposals that assume it will are usually assuming a v_e nobody has. The equation does not forbid going to Mars. It forbids doing it in a small vehicle.

You can run the ratios yourself in the rocket equation tool. Try holding the payload fixed and dragging the exhaust velocity, and watch how little the mass ratio moves compared with dragging the Δv.

Published 13 May 2026 · corrections welcome via the corrections page.