Ideal Rocket Equation Calculator

The ideal rocket equation, published by Konstantin Tsiolkovsky in 1903, is Δv = Isp · g₀ · ln(m₀/mf). It says a rocket's change in velocity (Δv) equals its specific impulse times standard gravity times the natural logarithm of the mass ratio (initial/wet mass over final/dry mass). Example: Isp = 450 s and mass ratio 10 gives vₑ = 450 × 9.80665 = 4,413.0 m/s, so Δv = 4,413.0 × ln(10) = 10,161.3 m/s (10.16 km/s).

Because it describes a perfectized burn: propellant is expelled at constant exhaust velocity, the burn is effectively instantaneous, and there are no gravity losses, no atmospheric drag, no steering losses and no leftover residuals. Real launches need extra Δv on top of the ideal figure (about 1.5-2 km/s of gravity and drag losses just to reach low Earth orbit), so treat the result as the theoretical budget your vehicle must at least provide.

The logarithm appears when you integrate the momentum-conservation differential equation dm/m, and that integral is the natural logarithm by definition. Using log₁₀ is a classic mistake: for the worked example above (vₑ = 4,413.0 m/s, mass ratio 10), log₁₀(10) = 1 gives 4,413.0 m/s, less than half of the correct 10,161.3 m/s. Always use ln.

Specific impulse measures how efficiently an engine turns propellant into thrust, expressed in seconds. It is related to exhaust velocity by vₑ = Isp · g₀, so an Isp of 450 s means an exhaust velocity of about 4,413 m/s. Higher Isp means less propellant for the same Δv: chemical rockets manage roughly 280-450 s, while ion thrusters reach about 3,000 s at tiny thrust levels.

Because Δv grows only logarithmically with the mass ratio, each extra km/s of Δv demands an exponentially larger propellant load: doubling the required Δv squares the mass ratio. That is why a rocket that delivers 1 tonne to orbit can weigh 20-50 tonnes on the pad, and why staging (dropping empty tanks) is used to reset the mass ratio mid-flight instead of dragging dead mass all the way.

It does not apply to air-breathing engines (jets, scramjets) that take oxidizer from the atmosphere, since their reaction mass is not all carried aboard. It also breaks down if exhaust velocity varies strongly during the burn, at relativistic speeds where Newtonian momentum no longer holds, and whenever you forget that gravity and drag losses must be budgeted separately on top of the ideal Δv.

Apply the equation to each stage separately and add the results: Δv_total = Δv₁ + Δv₂ + …. For stage i, use its own Isp, its wet mass at ignition and its dry mass at burnout (including the upper stages as payload). Staging helps because jettisoning empty tanks raises the effective mass ratio of what remains. The multi-stage section of this calculator does exactly this summation.

Rearrange the equation: m₀ = mf · e^(Δv/vₑ), then propellant = m₀ − mf. Example: for Δv = 9.4 km/s (low Earth orbit) with Isp = 450 s (vₑ = 4,413 m/s) and 1,000 kg of dry mass, m₀ ≈ 8,415.5 kg, so you need about 7,415.5 kg of propellant, over 7 times the dry mass. Use the "Propellant mass" solve mode above for any combination.

Approximate from-surface budgets including losses: low Earth orbit ≈ 9.4 km/s, geostationary orbit ≈ 13.7 km/s, Mars transfer ≈ 13.7 km/s (typical), lunar landing ≈ 15.3 km/s. These are reference values (source: Wikipedia "Delta-v budget"); real mission planning adds margins, plane-change costs and reserves. The calculator checks your Δv against these references automatically.

No. The g₀ in Δv = Isp · g₀ · ln(m₀/mf) is always the standard 9.80665 m/s² by definition: it is the conversion factor between specific impulse in seconds and exhaust velocity in m/s, not the local gravity where the rocket flies. A Moon or Mars "g₀" option in a calculator only explores a what-if rescaling of that constant; it does not model gravity losses, which must be budgeted as extra Δv instead.