Library / Foundations: Physics / FPH-43
Foundations: Physics
Escape speed
Minimum launch speed to leave a body’s gravity without further thrust.
- Ref
- FPH-43
- Rev
- v1
- Field
- Foundations: Physics
- Result
- ve [m/s]
M is the mass of the body and r the distance from its centre at the point of launch. The result is the speed at which kinetic energy exactly cancels the gravitational potential energy, so the object just reaches infinity with no speed left. It is independent of the object’s mass and of its direction (ignoring the atmosphere and the ground).
Minimum launch speed to leave a body’s gravity without further thrust. It is √2 times the circular orbital speed at the same radius.
Calculator
- Circular orbital speed at the same radius7,910 m/s
- Kinetic energy per kilogram needed, ½v²62.56 MJ/kg
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Escape speed from the Earth’s surface is 11.2 km/s, which is √2 times the 7.9 km/s circular speed at the same radius: a satellite in low orbit already has half the energy needed to leave.
- The Moon’s escape speed is only 2.4 km/s, so its surface lost its atmosphere long ago.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (m/s) | Local step | Change in result |
|---|---|---|---|---|---|---|
| M Mass of the body | 1e20 to 1e31 kg | 45.77 | 1.447e7 | 1.447e7 | +10% | +4.88 % |
| r Starting radius | 100,000 to 1e9 m | 892.8 | 89,280 | 88,390 | +10% | −4.65 % |
Sweep of Starting radius
| Starting radius (m) | Escape speed (m/s) |
|---|---|
| 100,000 | 89,280 |
| 316,200 | 50,210 |
| 1,000,000 | 28,230 |
| 3,162,000 | 15,880 |
| 1e7 | 8,928 |
| 3.162e7 | 5,021 |
| 1e8 | 2,823 |
| 3.162e8 | 1,588 |
| 1e9 | 892.8 |
Sweep one variable
Variables and how each one contributes
Escape speed
What the equation gives you. Everything else feeds it.
Mass of the body
Mass of the body being escaped from. Default is Earth. Square root: four times the mass is needed to double the escape speed.
Starting radius
Distance from the centre of the body where the object starts. Inverse square root: starting from a higher point lowers the speed needed, but only slowly. A 10 % rise in r lowers it by about 5 %.
Worked example with the values above
- 2GM/r = 2 × 6.674×10−11 × 5.972×1024 / 6,371,000 = 1.251×108 m²/s².
- Take the square root.
- ve = 11,190 m/s, about 11.19 km/s.
Assumptions and limits
- A spherical, non-rotating body with no atmosphere.
- Only the gravity of this body acts; the pull of the Sun and Moon is ignored.
Common mistakes
- Treating it as the speed a rocket must reach at lift-off. Rockets climb under power, so they never need 11 km/s at the ground.
- Using the radius of the planet when the launch point is at altitude.
Statistics and uncertainty
The result goes as √(M/r), so errors are halved: a 4 % error in M gives 2 % in ve. The absolute figure is a useful bound rather than a launch requirement, because rockets reach space by thrusting for a long time rather than by one launch speed.
Where it is used
- Judging the energy needed for an interplanetary departure.
- Deciding whether a body can hold an atmosphere.
Maths you need
- Gravitational potential energy: −GMm/r
- Kinetic energy: ½mv²