Tsiolkovsky Rocket Equation

Δv = Isp · g₀ · ln(m₀/mf) — the equation that rules spaceflight

Starting values — Tsiolkovsky Rocket Equation

Specific impulse (Isp, s)	311

Change any control above and the results recompute from these.

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the tyranny of the rocket equation 🖖

The logarithm is the enemy of rocketry. To double your Δv you must square your mass ratio — a rocket that needs ratio 4 for one stage needs ratio 16 for two stages of the same Δv. This is why reaching LEO from the ground (~9.4 km/s total) demands that roughly 85–90% of liftoff mass is propellant, and why staging exists: each discarded empty tank resets the mass ratio for the next burn. The animation below shows how Δv accumulates as propellant burns off. Watch how quickly the tank empties even for a modest Δv.

why rockets throw mass backward 🖖

A rocket has nothing to push against, so it moves by hurling propellant out the back: every kilogram flung rearward at exhaust speed v_e nudges the ship forward (Newton's third law). Add up those nudges as the ship keeps getting lighter and the logarithm appears. The practical lesson is that how fast you throw the exhaust — what specific impulse measures — matters more than how much fuel you pack: doubling v_e doubles your Δv at the very same mass ratio.

the equation survives relativity 🖖

Replace Newtonian Δv with rapidity — the relativistic quantity that still adds linearly — and the exact same log-of-mass-ratio form holds at any speed. So a rocket's Δv budget has no ceiling even though its speed can never reach c. The catch is brutal: a matter–antimatter photon rocket (v_e = c) reaching 0.9c needs a mass ratio near 4.4, and a round trip that also brakes at each end raises that to the fourth power — over 350.

Problems solved in full

Reaching orbit with a mass ratio of 4.5 and 311 s specific impulse 5 steps
A stage carries 3.5 tonnes of propellant for every 1 tonne of everything else — tanks, engine and payload combined — behind an engine of 311 s specific impulse. Find its Δv, then find whether a single stage built like that could ever reach orbit. That loaded-to-empty ratio is a mass ratio of 4.5, and 311 s is the figure the tool's Saturn V first-stage preset carries — near the top of the kerosene–oxygen range, which runs from about 260 s at sea level to 310 s in vacuum.
1. Specific impulse quoted in seconds is exhaust speed in disguise: it is the exhaust speed divided by standard gravity, so multiplying it back recovers the number the rocket equation actually wants. Seconds are used because the figure then comes out the same in imperial units, not because anything lasts 311 s.
  
  $v_{e} = I_{s p} g_{0} = 311 \times 9.80665 = 3050 m/s$
2. The logarithm takes a ratio, so only the proportions matter. A 3-tonne stage and a 3000-tonne stage built in these proportions deliver the same 4.587 km/s.
  
  $Δ v = v_{e} ln (\frac{m _{0}}{m _{f}}) = 3.050 \times ln 4.5 = 3.050 \times 1.5041 = 4.587 km/s$
3. Read the same mass ratio as a fraction of the launch mass instead. Nearly four fifths of the stage is propellant, and it has bought less than half of what a launch to orbit costs.
  
  $f_{p} = 1 - \frac{m _{f}}{m _{0}} = 1 - \frac{1}{4.5} = 77.78%$
4. So run the equation backwards from the target. Reaching low Earth orbit from the ground takes about 9.4 km/s once drag and gravity losses are counted, and at this exhaust speed that fixes the mass ratio required — leaving 4.6% of the liftoff mass for tanks, engine, structure and satellite together.
  
  $\frac{m _{0}}{m _{f}} = e^{Δ v / v_{e}} = e^{9.4/3.050} = e^{3.082} = 21.80 ⟹ f_{p} = 1 - \frac{1}{21.80} = 95.4%$
5. Nobody builds tankage that light. A well-made kerosene stage is around 8% structure by loaded mass, and that alone caps the mass ratio at 12.5 before a single kilogram of payload is added. What is left of 9.4 km/s is the shortfall.
  
  $ε = 0.08 ⟹ (\frac{m _{0}}{m _{f}})_{m a x} = \frac{1}{0.08} = 12.5, Δ v_{m a x} = 3.050 ln 12.5 = 7.703 km/s (9.4 - 7.703 = 1.70)$
Answer

The tool prints an exhaust velocity of 3050 m/s, 4.587 km/s of Δv and a propellant fraction of 77.78%. The last step is what the equation is for, because it does not say a single stage would be hard — it says the Δv is not there, with a payload of zero. Even at 6% structure, around the best a large kerosene stage achieves, the total only reaches 8.58 km/s. Closing 9.4 km/s in one stage at 8% structure would need a specific impulse of 380 s, a fifth more than this engine and beyond any kerosene engine ever flown; hydrogen-oxygen engines reach about 450 s, which is why every serious single-stage-to-orbit study has been a hydrogen one. The other escape is to stop carrying the empty tank: drop it and the equation restarts with a far smaller m₀, which makes staging not an optimisation but the workaround for an impossibility.

$I_{s p} \geq \frac{Δ v}{g _{0} ln ( 1/ ε )} = \frac{9400}{9.807 \times 2.526} = 380 s$
Propellant cost for a rocket managing 4.587 km/s to reach orbit 6 steps
The panel says this rocket manages 4.587 km/s with 77.78% of its lift-off mass as propellant. Orbit needs about 9.4. Twice the speed does not mean twice the fuel — work out what it actually costs, and then how anybody got around it.
1. Start where the panel is. Exhaust speed is just the specific impulse times g₀, and the Δv follows from the mass ratio and nothing else.
  
  $v_{e} = I_{s p} g_{0} = 311 \times 9.807 = 3050 m s^{- 1}, Δ v = v_{e} ln 4.5 = 4.587 km s^{- 1}$
2. Read the propellant fraction as what it literally is: with a mass ratio of 4.5, everything except one part in 4.5 is fuel that gets thrown away.
  
  $\frac{m _{p}}{m _{0}} = 1 - \frac{1}{4.5} = 0.7778$
3. Now ask the same engine for 9.4 km/s. Invert the rocket equation and the mass ratio does not double — it goes exponential, to 21.80.
  
  $Δ v = 9.4 km s^{- 1} : \frac{m _{0}}{m _{f}} = e^{9400/3050} = e^{3.082} = 21.80 ⟹ \frac{m _{p}}{m _{0}} = 0.9541$
4. Look at what 95.41% means before moving on. Tanks, engines, structure, guidance and payload together get 4.59% of the mass on the pad. That is not a hard engineering target; it is roughly a soap bubble that has to survive launch.
5. So split the job in two. Each stage does 4.7 km/s and needs a mass ratio of 4.669 — and multiplied together those give back exactly 21.80.
  
  $Δ v_{1} = Δ v_{2} = 4.7 : R_{1} = e^{4700/3050} = 4.669, R_{1} R_{2} = 4.66 9^{2} = 21.80$
6. Why multiplied: Δv adds, and Δv is a logarithm of the ratio. A sum of logarithms is the logarithm of a product.
  
  $i \sum Δ v_{i} = v_{e} i \sum ln R_{i} = v_{e} ln i \prod R_{i}$
Answer

Staging does not reduce the mass ratio. It factorises it. 21.80 is unreachable in one piece and trivial as two copies of 4.669, and the reason the trick is honest is that the first stage's empty tanks stop being cargo the moment they are let go. Each stage is then 78.58% propellant — and look again at the panel, which has been showing 77.78% all along. The preset is not a rocket that reaches orbit. It is one stage of a rocket that does.

$i \prod R_{i} = 21.80 : 95.41% ⟷ 78.58% + 78.58%$

Learning path

Orbits from two numbers

References (1)

The equation itself: K. E. Tsiolkovsky, "Исследование мировых пространств реактивными приборами" ("Exploration of Outer Space by Means of Rocket Devices"). Nauchnoye Obozreniye, 1903.

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Saturn V first stage - Isp=311s, mass ratio=4.5 → Δv≈4.6 km/s, propellant fraction 78%
Falcon 9 (sea level) - Isp=282s, mass ratio=6 → Δv≈5.0 km/s, propellant fraction 83%
Ion thruster - Isp=3000s, mass ratio=12 → Δv≈73 km/s (ion thruster: tiny thrust, huge Δv)
Hobby rocket - Isp=130s, mass ratio=2.5 → Δv≈1.2 km/s, propellant fraction 60%