Rocket Equation
Connect consumed fuel, remaining rocket mass and ideal velocity gain through the logarithmic rocket equation.
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Velocity gain from ejecting mass
A rocket accelerates by ejecting fuel in the opposite direction. In an inertial, one-dimensional model without external forces, momentum balance gives dv = −u dm/m. The positive constant u is the exhaust speed relative to the rocket, and m is the rocket's current total mass, including fuel that has not yet been burned.
Integrating from initial mass m0 to current mass m gives Δv = u ln(m0/m). If the dry mass is md and a fraction b of available fuel has been burned, m = m0 − b(m0 − md). Only a complete burn makes the current mass equal to the dry mass, unless the rocket had no fuel to begin with.
Worked example: a partial burn
Take u = 2000 m/s, m0 = 1000 kg, md = 250 kg and b = 0.5. There are initially 750 kg of fuel. Burning half ejects 375 kg, leaving another 375 kg of fuel aboard. The current mass is 250 + 375 = 625 kg. Thus Δv = 2000 ln(1000/625) = 2000 ln(1.6), approximately 940.00725849 m/s.
| u (m/s) | Initial / dry mass (kg) | b | Current mass (kg) | Δv (m/s, approximate) |
|---|---|---|---|---|
| 1000 | 1000 / 500 | 1 | 500 | 693.14718056 |
| 2000 | 1000 / 250 | 0.5 | 625 | 940.00725849 |
| 3000 | 600 / 200 | 1 | 200 | 3295.8368660 |
Experiment with mass ratio and fuel consumption
Open the Rocket Equation experiment. Each edit selects an exact burn state; the curve is not a simulated clock. Compare the current total mass and velocity cards with the separate dry, remaining-fuel and burned-fuel ledger entries.
- Set Exhaust speed u to 1000, Initial total mass m0 to 1000, Dry mass md to 500 and Burned fuel fraction b to 1. Check current mass 500 kg and velocity gain approximately 693.14718056 m/s.
- Change Burned fuel fraction b to 0. The current mass returns to 1000 kg, the remaining fuel is 500 kg and velocity gain is 0.
- Set Exhaust speed u to 2000, Dry mass md to 250 and Burned fuel fraction b to 0.5. Check 375 kg each of remaining and burned fuel, current mass 625 kg and velocity gain approximately 940.00725849 m/s.
- Set Exhaust speed u to 3000, Initial total mass m0 to 600, Dry mass md to 200 and Burned fuel fraction b to 1. Check 3000 ln(3), approximately 3295.8368660 m/s.
- Set Initial total mass m0 and Dry mass md both to 500. Change Burned fuel fraction b between 0 and 1: there is no fuel, so current mass stays 500 kg and velocity gain stays 0.
- Set Initial total mass m0 to 100 and Dry mass md to 200. The mass-order error replaces the dependent results; the model does not silently swap the masses.
- Select Reset inputs. Pan the graph with a drag or arrow key, then select Reset view: the original inputs and numerical answers stay unchanged.
- Clear Burned fuel fraction b and inspect the invalid raw input. Enter 0.5 to recover the valid half-burn calculation, with current mass 750 kg and velocity gain approximately 287.68207245 m/s.
Ideal-model limitations
The model assumes constant exhaust speed relative to the rocket and no gravity, drag or other external forces. It describes a single stage and a velocity change, not an absolute speed, trajectory, burn duration or thrust history. Burn fraction measures consumed available fuel. The dry mass remains aboard throughout; unburned fuel also contributes to current mass. Inputs must obey m0 ≥ md > 0. Displayed decimal values are rounded evaluations of the logarithm, not exact terminating values.
Reference and derivation
OpenStax University Physics, Rocket Propulsion derives the variable-mass momentum equation and logarithmic rocket equation. The mass ledger and burn-fraction examples here apply that ideal derivation to explicitly bounded, reproducible inputs; they do not include atmospheric or gravitational losses.
Experience it interactively
Adjust parameters, observe in real time, and build deep intuition with Riano’s interactive Rocket Equation module.
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