Case 1
- Mass of the body
- 5.97210²⁴ kg
- Radius from the centre
- 6371000m
11,185.978m/s
Open with these valuesEscape Velocity Calculator
11,185.978m/s
Result: 11,185.978 m/sEarth's escape velocity is 11186 m/s, about 11.19 km/s. It depends only on the body's mass and the radius you leave from — never on the mass of the object escaping. Enter the mass in units of 10²⁴ kg (Earth is 5.972, the Moon 0.07342) and the radius in metres.
Worked examples
Case 2
- Mass of the body
- 0.0734210²⁴ kg
- Radius from the centre
- 1737400m
2,375.063m/s
Open with these valuesCase 3
- Mass of the body
- 189810²⁴ kg
- Radius from the centre
- 69911000m
60,199.545m/s
Open with these valuesHow it's calculated
v = √(2 × G × M ÷ r)
- StepEnter the mass in units of 10²⁴ kg — Earth is 5.972, Jupiter 1898.
- StepEnter the radius in metres, measured from the centre of the body.
- ResultRead the escape velocity in m/s; divide by 1000 for km/s.
What this number means
The escaping object's mass cancels out
A pebble and a spacecraft both need 11186 m/s to leave Earth, because the object's mass drops out when kinetic and gravitational potential energy are set equal. The heavier one needs more energy to reach that speed, not a higher speed.
It assumes nothing pushes after launch
This is the speed an unpowered projectile would need at the surface never to fall back. A vehicle that keeps its engines running is a different problem; the number here is the one-shot case.
Leaving from farther out costs less
The radius sits in the denominator, so the speed falls the farther from the centre you start. Earth asks 11186 m/s at the surface and less at the top of the atmosphere, while the much larger Jupiter still asks 60200 m/s because its mass outweighs its size.
G is the worst-known constant in physics
CODATA 2022 gives 6.674 30(15)·10⁻¹¹ m³·kg⁻¹·s⁻², a relative uncertainty of 2.2·10⁻⁵, and it is measured rather than fixed by agreement. This calculator uses 6.6743e-11, so the trailing decimals of a result are arithmetic, not accuracy.
Commonly misread
A heavy spacecraft needs a higher escape velocity than a light one.
Both need the same speed, because the object's mass cancels out of √(2GM/r). The heavier one needs more energy to get there, which is not the same thing.
Entering Earth's mass as 5972000000000000000000000 kg.
The mass box already counts in units of 10²⁴ kg, exactly as planetary tables do, so Earth is 5.972 and the Moon 0.07342. The calculation multiplies by 10²⁴ for you.
Four times the mass should mean four times the escape velocity.
The square root halves the effect, so four times the mass gives twice the speed. That is why the Sun, some 333000 Earth masses, comes out at 55 times Earth's escape velocity rather than more.
Reference table
| Mass, radius | Body | Escape velocity (m/s) |
|---|---|---|
| 0.07342, 1737400 | The Moon | 2375.063039 |
| 0.6417, 3389500 | Mars | 5027.083051 |
| 5.972, 6371000 | Earth | 11185.977892 |
| 1898, 69911000 | Jupiter | 60199.544692 |
| 1989000, 695700000 | The Sun | 617766.351625 |
Questions
How do I calculate escape velocity?
Take the square root of twice the gravitational constant times the body's mass, divided by the radius: v = √(2 × G × M / r), with G = 6.6743e-11 m³·kg⁻¹·s⁻². Here the mass box counts in units of 10²⁴ kg, so Earth is 5.972 and the radius 6371000 m. That gives 11186 m/s.
What is escape velocity?
Escape velocity is the minimum speed an object needs to break free of a body's gravity and never fall back, without further propulsion. It depends only on the body's mass and the distance from its centre, not on the mass of the escaping object. It is measured in metres per second.
What is Earth's escape velocity?
About 11186 m/s, roughly 11.19 km/s or 40270 km/h. That is the speed an unpowered projectile would need at the surface to leave Earth's gravity entirely — one reason orbital rockets are so large.
Does the escaping object's own mass matter?
No. Escape velocity is the same for a pebble and a spacecraft, because the object's mass cancels out when kinetic and gravitational potential energy are set equal. A heavier object needs more energy to reach that speed, not a higher speed.
Why does a larger radius lower the escape velocity?
Because the radius sits in the denominator of v = √(2GM/r). Gravity weakens with distance from the centre, so leaving from farther out means there is less of it to overcome. Escape velocity at the top of an atmosphere is therefore lower than at the surface.
Why is the mass entered in units of 10²⁴ kg?
Because that is how planetary tables list it, and because it keeps every planet inside the input range: Earth is 5.972, the Moon 0.07342, the Sun 1989000. Multiply by 10²⁴ to get kilograms again. The calculation itself works in kilograms.
Sources and last check
Information, not professional advice.