Escape Velocity Calculator
Calculate the speed needed to escape the gravity of Earth, the Moon, Mars, Venus or Jupiter from the surface or from any altitude, and the extra Δv required from a circular orbit.
What escape velocity means
Escape velocity is the speed at which an object’s kinetic energy exactly equals the energy needed to climb out of a body’s gravity well to infinite distance. Below it, a coasting object eventually falls back or stays in orbit; at or above it, the object never returns. The formula is v = √(2μ/r), where μ is the gravitational parameter and r is the distance from the centre. It applies to a coasting object with no thrust and no air. A rocket burning its engine continuously can leave Earth slowly; escape velocity is a statement about energy, not a minimum speed that must be reached at the launch pad.
With μ = 398,600.4418 km³/s² and the equatorial radius of 6,378.137 km, Earth’s surface escape velocity is 11.180 km/s, about 40,250 km/h. Textbooks often quote 11.186 km/s because they use the mean radius of 6,371 km; both are correct for their radius. The Moon’s is only 2.376 km/s, Mars needs 5.022 km/s, Venus 10.361 km/s and Jupiter 59.53 km/s at its cloud tops.
The √2 rule
Compare the escape formula with the circular orbit speed √(μ/r) at the same radius and the ratio is always √2, about 1.414. That gives a quick way to plan a departure. From a circular orbit at 400 km above Earth, where the spacecraft already moves at 7.669 km/s, escape requires 10.845 km/s, so the burn needs only 3.176 km/s more. That is far less than the 11.18 km/s figure suggests, because most of the work was done getting into orbit. Departure burns are made as low as possible for the same reason: the extra speed gained deep in a gravity well buys more energy than the same burn made higher up, an effect named after Hermann Oberth.
Common misconceptions
Escape velocity does not depend on direction, provided the path does not hit the planet, and it does not depend on the mass of the spacecraft. It also falls with altitude: at the distance of the Moon, about 384,400 km from the Earth’s centre, it is only about 1.44 km/s. Escaping Earth is not the same as leaving the Solar System. A probe at exactly Earth’s escape velocity ends up in an orbit around the Sun almost identical to Earth’s, moving at about 29.8 km/s with the planet; reaching another planet or leaving the Sun entirely requires extra speed beyond escape, called the hyperbolic excess speed. Finally, the atmosphere does not appear in the formula. A projectile fired at 11.2 km/s from the ground would lose a large part of its speed to drag in the first few seconds.
How to use it
- Choose the body. Each uses its own gravitational parameter and radius, listed with the result.
- Enter the starting altitude above the surface in kilometres. The default 0 gives the surface value of 11.180 km/s for Earth.
- Read the escape velocity, the circular orbit speed at the same radius and the difference between them.
- To plan a departure from orbit, enter the parking orbit altitude, for example 400 km, and read the extra Δv: 3.176 km/s.
- Remember that arriving at another planet needs speed beyond escape; use the launch window and Hohmann tools for that part of the budget.
Questions from the flight desk
What is the escape velocity of Earth?
About 11.18 km/s from the surface at the equatorial radius of 6,378 km, or 11.19 km/s using the mean radius of 6,371 km, roughly 40,000 km/h or 25,000 mph. It ignores air drag and assumes a coasting object. From a 400 km orbit the figure is 10.85 km/s, and the spacecraft already has 7.67 km/s of it.
Do rockets have to reach escape velocity to leave Earth?
Not in the sense of reaching it at the ground. A rocket that keeps thrusting can climb at any speed. Escape velocity is the speed a coasting object needs at a given distance. In practice spacecraft first reach orbit, then make a short departure burn of a little over 3 km/s to exceed the local escape velocity.
Why is escape velocity √2 times orbital velocity?
Escape requires kinetic energy equal to the full gravitational binding energy, μ/r per kilogram, while a circular orbit has kinetic energy of exactly half of that, μ/2r. Kinetic energy grows with speed squared, so doubling the energy multiplies the speed by √2. The relation holds at every radius and around every body.
Does escape velocity depend on the mass of the object?
No. Both the kinetic energy and the gravitational energy of the object scale with its own mass, so it cancels out. A grain of dust and a spacecraft need the same escape speed. The mass of the planet matters, through μ, together with the distance from its centre. A heavier spacecraft simply needs more propellant to reach the speed.
Disclaimer: Orbital mechanics results are ideal two-body figures with instantaneous burns, computed from the values you enter and the constants stated on each page. They are not mission analysis, flight dynamics or launch safety advice. Real orbits are perturbed by Earth’s oblateness, the Moon, the Sun and atmospheric drag, and real burns carry gravity and steering losses. Never use these numbers to operate a spacecraft or to fly a rocket; follow the safety code of your rocketry association and the advice of qualified engineers.