Escape Velocity Calculator
Calculate the escape velocity of a planet, moon or star, v = √(2GM/r), from its mass in kilograms and the distance from its center in meters.
Last updated: October 2026
Escape Velocity
11,186.17 m/s
Formula below · 2 sources (nssdc.gsfc.nasa.gov, arxiv.org) · Updated Oct 2026
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About this calculator
Escape velocity is the speed an unpowered object needs to coast away from a body's gravity and never fall back, ignoring air resistance. Setting the kinetic energy equal to the gravitational potential energy gives:
v = √(2GM ÷ r)
G is the gravitational constant, 6.6743 × 10⁻¹¹ N·m²/kg² (CODATA 2022), M the body's mass in kilograms and r the distance from its center in meters: its radius for a launch from the surface, or radius plus altitude from higher up. For Earth (M = 5.9722 × 10²⁴ kg, r = 6,371 km) this is 11,186 m/s, about 11.2 km/s.
The mass of the escaping object does not appear: a pebble and a spacecraft need the same speed, though the spacecraft needs far more energy. Orbiting just above the surface takes less, the escape velocity divided by √2 (about 7.9 km/s for Earth).
The formula assumes a non-rotating sphere and no atmosphere. Real launches need extra speed to overcome drag and gravity losses, and gain a little from Earth's rotation when launched eastward.
| Body | Escape velocity (km/s) |
|---|---|
| Mercury | 4.3 |
| Venus | 10.4 |
| Earth | 11.2 |
| Mars | 5 |
| Jupiter | 59.5 |
| Saturn | 35.5 |
| Uranus | 21.3 |
| Neptune | 23.5 |
How to use
Enter the body's mass in kilograms and the distance from its center in meters; scientific notation such as 5.9722e24 works. The default is Earth (5.9722e24 kg, 6.371e6 m): √(2 × 6.6743e-11 × 5.9722e24 ÷ 6.371e6) = 11,186.17 m/s. For the Moon (7.346e22 kg, 1.7374e6 m) the answer is about 2,376 m/s; for Mars (6.4169e23 kg, 3.3895e6 m) about 5,027 m/s. From 400 km above Earth, enter r = 6.771e6 m: 10,850.72 m/s. Divide by 1,000 for km/s.
Frequently asked questions
Does escape velocity depend on the mass of the rocket?
No. Only the mass and radius of the body being escaped appear in v = √(2GM/r). A heavier object needs more energy (½mv²) to reach that speed, but the speed is the same.
What is the difference between escape velocity and orbital velocity?
Circular orbital velocity at a given distance is √(GM/r), which is escape velocity divided by √2. Near Earth's surface that is about 7.9 km/s, against 11.2 km/s to escape.
What units should I use?
Mass in kilograms and distance in meters give m/s. Kilometers entered as meters make the answer about 32 times too large (√1,000), the most common mistake.
Why does escape velocity fall with altitude?
Gravity weakens with distance from the center, so less speed is needed to escape from higher up. The speed scales with 1 ÷ √r.
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