Hail Mary Science
Spaceflight and relativity

Rocket equation calculator

Enter the velocity you want to gain, the exhaust speed and the ship's dry mass. The calculator applies the Tsiolkovsky equation and tells you how much fuel you need to load.

Key formula

Δv = vₑ · ln(m₀ / m_f) · m_fuel = m_f · (e^(Δv/vₑ) − 1)

Try it

How to use the simulator

The Tsiolkovsky equation links a rocket's change in velocity (Δv) to its exhaust speed (vₑ) and the ratio of its starting and final mass: Δv = vₑ · ln(m₀/mf). The calculator tells you how much fuel a ship of a given mass needs. By default the exhaust speed is the speed of light, as in an engine that emits photons.

Controls

  • Desired velocity (km/s): 1,000 to 280,000 km/s.
  • Exhaust velocity (km/s): from 3 km/s, typical of a good chemical rocket, up to the speed of light.
  • Ship mass (kg): dry mass, without fuel.

Things to try

  1. Set the exhaust speed to 3 km/s and ask for just 30 km/s: the ship would need about 22,000 times its own mass in fuel. That's why chemical interstellar travel is impossible.
  2. Go back to vₑ = c and raise the desired velocity. Fuel grows exponentially, not linearly.
  3. Note that this is the classical form of the equation. Near light speed you use the relativistic version, Δv = c·tanh((vₑ/c)·ln(m₀/m_f)), which never reaches c. With photon exhaust (vₑ = c) and a mass ratio of 21 you could hit ~0.995c one way; if you have to speed up and then brake, each leg gets √21 ≈ 4.6 and the top speed drops to ~0.91c.

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Where the equation comes from

A rocket moves by throwing mass backwards. Each kilogram of gas leaving at speed vₑ pushes the ship the other way through conservation of momentum. The catch is that unburned fuel has to be accelerated too, so each extra kilometer per second costs more than the last.

Konstantin Tsiolkovsky solved the problem in 1903. The result is that the change in velocity (Δv) depends on the logarithm of the ratio between the initial mass (m₀, with fuel) and the final mass (m_f, empty). Put the other way round: the fuel you need grows exponentially with the speed you want.

Exhaust speed is everything

Because Δv is divided by vₑ inside an exponential, doubling the exhaust speed cuts the fuel you need dramatically. The best chemical engines (hydrogen and oxygen) exhaust at about 4.4 km/s. Ion engines reach 30–50 km/s, but with tiny thrust.

The theoretical limit is to exhaust light, which leaves at 300,000 km/s. Such an engine would need very little fuel mass to reach interstellar speeds, as long as it could turn that mass into light efficiently. No current technology comes close.

Limits of the classical form

The classical equation stops being exact near light speed. The relativistic version is Δv = c · tanh((vₑ/c) · ln(m₀/m_f)), which never reaches c however much fuel you carry. Below about 10–20% of c, both versions give very similar results.

Worked example

Example: putting a 1,000 kg craft into orbit

  1. Reaching low Earth orbit costs about 9 km/s of Δv once gravity and air drag losses are included.
  2. With a hydrogen-oxygen engine, vₑ ≈ 4.4 km/s. The mass ratio is e^(9/4.4) = e^2.05 ≈ 7.7.
  3. The fueled craft must weigh 7.7 times its empty mass: about 6,700 kg of fuel for every 1,000 kg of craft.
  4. With vₑ = 3 km/s (a more modest chemical engine), the ratio rises to e³ ≈ 20. That is why real rockets are mostly tank and use several stages.

In Project Hail Mary

The novel's ship runs on a fictional fuel that stores energy very densely and releases it as light. The Tsiolkovsky equation is exactly the tool for checking how much of that fuel an interstellar trip would take.

This tool appears in:Chapter 4The Hail Mary Ship and the Suicide Mission

Frequently asked questions

Why do rockets have several stages?

Dropping empty tanks and spent engines lowers each stage's final mass and improves the m₀/m_f ratio. With a single stage, reaching orbit on chemical fuel is barely possible.

What is delta-v?

It is the total change in velocity a craft can produce with the fuel it carries. Missions are planned by adding up the Δv of every maneuver, like a budget.

Could chemical rockets reach another star?

Not in any reasonable time. Reaching 30 km/s with vₑ = 3 km/s would take about 22,000 times the craft's mass in fuel, and at that speed the trip to the nearest star would last around 40,000 years.

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