Last updated: 2026-06-15

Escape Velocity

Escape velocity is the speed at which an object, given one push and no further power, can coast away from a planet or star forever instead of falling back. It is the headline number behind every rocket launch: about 11.2 km/s to leave Earth, but only 2.4 km/s to leave the Moon. This calculator finds the escape velocity from a body's mass and radius, and also reports the circular orbital velocity at the same distance.

Where the formula comes from

An object escapes when its kinetic energy is just enough to climb out of the gravitational well. Setting the kinetic energy $\tfrac{1}{2}mv^2$ equal to the gravitational potential energy $GMm/r$ and solving for $v$ gives the escape velocity. The object's own mass $m$ appears on both sides and cancels, which is why escape velocity depends only on the body you are leaving and your distance from its centre — never on what you are launching.

Formula

The two speeds differ by a constant factor: $v_{esc} = \sqrt{2}\,v_{orb}$. So escaping takes only about 41% more speed than circling.

Worked example

For Earth, $M = 5.972\times10^{24}\ \text{kg}$ and $r = 6.371\times10^{6}\ \text{m}$:

The circular orbital velocity at the same radius is this divided by $\sqrt{2}$, about 7.9 km/s — the speed of a satellite skimming low Earth orbit.

Escape velocities across the Solar System

Because escape velocity grows with the square root of mass and shrinks with the square root of radius, it varies enormously between bodies. From the surface: the Moon needs about 2.4 km/s, Mars 5.0 km/s, Earth 11.2 km/s, Jupiter 59.5 km/s, and the Sun a colossal 617.5 km/s. This is why missions from the Moon need far less rocket power than launches from Earth, and why a gas giant's gravity is so hard to leave.

Limitations

This is the ideal escape velocity: the speed an unpowered projectile would need, ignoring air resistance and assuming a single spherical body. Real rockets accelerate continuously, so they never have to reach the full escape velocity in one instant; the number instead measures the energy required to break free. The formula also ignores atmospheric drag (significant near the surface) and the gravity of other bodies, and it assumes the launch point is at distance $r$ from a spherically symmetric mass.