Rocket equation calculator

What is the rocket equation?

Δv = u ln(m₀/m), where u is the exhaust speed relative to the rocket, m₀ the mass before the burn and m the mass after it. OpenStax University Physics credits it to Konstantin Tsiolkovsky (1897). The change in speed depends only on the exhaust speed and the mass ratio m₀/m — not on how fast the fuel is burned, as long as there is no gravity or drag.

How do I convert specific impulse to exhaust velocity?

Multiply by standard gravity: u = Isp × g₀, with g₀ = 9.80665 m/s² (32.174 ft/s²). NASA Glenn defines specific impulse as the equivalent exhaust velocity divided by g₀, so it is measured in seconds and is the same in US and metric units. An Isp of 300 s means u = 300 × 9.80665 ≈ 2,942 m/s, or about 9,652 ft/s.

Example: the OpenStax spacecraft

In OpenStax Example 9.20 a 2.0 × 10⁴ kg spacecraft ejects 200 kg of fuel per second at 250 m/s for 30 s. That burns 6,000 kg, so the mass drops to 14,000 kg and Δv = 250 × ln(20,000 ÷ 14,000) ≈ 89 m/s. The book also finds the thrust: 250 m/s × 200 kg/s = 5 × 10⁴ N.

How much propellant do I need for a given delta-v?

Turn the equation around: the mass ratio is m₀/m = e^(Δv/u), and the propellant is m₀ − m. When the Δv you want equals the exhaust speed, the ratio is e ≈ 2.718, so about 63% of the launch mass must be propellant. Every further Δv of one exhaust speed multiplies the ratio by e again — that is why big Δv needs staging.

What about gravity during launch?

Going straight up near Earth, OpenStax subtracts gΔt, where Δt is the burn time: Δv = u ln(m₀/m) − gΔt (Equation 9.39). The longer the burn, the more of the Δv is lost, which is why launches burn fuel as fast as possible. Enter a burn time to include this loss; leave it at 0 for the ideal Δv in space. Air drag and steering losses are not included.

Why is a high exhaust speed so valuable?

Propellant mass grows exponentially with the Δv you need divided by the exhaust speed. For a 3,000 m/s change in speed and a 1,000 kg dry mass, an engine with an Isp of 300 s needs 1,772 kg of propellant, while an Isp of 250 s needs 2,400 kg.

Reading the equation

The change in speed is Δv = u ln(m₀ ÷ m), with u the exhaust speed and m₀ ÷ m the ratio of the mass before the burn to the mass after it. The equation assumes that the rocket moves in empty space with no gravity or drag, and the burned propellant leaves at a constant speed u relative to the rocket. Specific impulse in seconds is converted with u = Isp × g₀, where g₀ = 9.80665 m/s². An engine with an Isp of 300 s therefore has an exhaust speed of 2,942 m/s.

A worked example

A stage that weighs 5,000 kg at ignition and 1,000 kg after burning out, with an Isp of 300 s (u = 2,942 m/s), has a mass ratio of 5.0 and a Δv of 2,942 × ln(5.0) = 4,735 m/s. Working backward, a Δv of 3,000 m/s with a final mass of 1,000 kg needs a mass ratio of e^(3000 ÷ 2,942) = 2.77, so 2,772 kg at ignition and 1,772 kg of propellant, 64% of the starting mass. If the burn lasts 120 s while climbing straight up from Earth, gravity subtracts about g₀ × 120 = 1,177 m/s, leaving 3,558 m/s.

Common mistakes

Using the mass of the propellant alone as m instead of the mass remaining (structure, engines and payload) is the most frequent error. Mixing units is the second: Isp is always in seconds, but exhaust speeds are in m/s or ft/s, and the masses must be in the same unit as each other. A third is expecting the idealized Δv to equal the speed achieved on a launch; gravity and drag take a large share, and the equation says nothing about the time it takes. Finally, stages are added up one by one, each with its own mass ratio, because dropping empty tanks improves the ratio of the rest.

Sources: OpenStax University Physics Volume 1, section 9.7 (Rocket Propulsion), for the equation and the gravity term (Eq. 9.39); NASA Glenn Research Center, “Specific Impulse”, for the definition of Isp; g₀ = 9.80665 m/s² from NIST. Related: escape velocity calculator, orbital period calculator, momentum calculator and kinematics calculator.