Escape Velocity Formula
Learn the escape velocity formula v = sqrt(2GM/r) to calculate the speed needed to leave a gravitational field.
The Formula
Escape velocity is the minimum speed an object must reach to break free from the gravitational pull of a celestial body without any further propulsion. At this speed, the object's kinetic energy exactly equals the gravitational potential energy binding it to the body, allowing it to coast to infinity and arrive with zero residual velocity.
The formula is derived by setting the total mechanical energy (kinetic plus gravitational potential) equal to zero. Starting from the energy equation ½mv2 - GMm/r = 0, solving for v gives the escape velocity expression. Notice that the mass of the escaping object (m) cancels out, meaning escape velocity depends only on the mass and radius of the body you are escaping from, not on the mass of the object itself. A marble and a spacecraft need the same escape velocity from Earth's surface.
For Earth, using M = 5.972 times 1024 kg, r = 6.371 times 106 m, and G = 6.674 times 10-11 N m2/kg2, the escape velocity works out to approximately 11.2 km/s (about 40,320 km/h or 25,020 mph). This is roughly 33 times the speed of sound at sea level.
Different celestial bodies have vastly different escape velocities. The Moon's escape velocity is only 2.38 km/s due to its smaller mass and radius. Jupiter, being massive, requires 59.5 km/s. The Sun's escape velocity from its surface is a staggering 617.5 km/s. For a black hole, the escape velocity at the event horizon equals the speed of light, which is precisely why nothing can escape from within it.
In practical rocketry, spacecraft do not actually need to reach escape velocity instantaneously. Rockets provide continuous thrust, gradually building speed. The escape velocity concept is most useful as a theoretical benchmark and for understanding ballistic trajectories where no further thrust is applied after launch.
Variables
| Symbol | Meaning |
|---|---|
| ve | Escape velocity (m/s) |
| G | Gravitational constant (6.674 × 10-11 N m2/kg2) |
| M | Mass of the celestial body (kg) |
| r | Distance from the center of the body (m) |
Example 1
Problem: Calculate Earth's escape velocity. M = 5.972 × 1024 kg, r = 6.371 × 106 m.
ve = √(2 × 6.674 × 10-11 × 5.972 × 1024 / 6.371 × 106)
ve = √(2 × 3.986 × 1014 / 6.371 × 106) = √(1.251 × 108)
ve ≈ 11,186 m/s ≈ 11.2 km/s
Example 2
Problem: Calculate the Moon's escape velocity. M = 7.342 × 1022 kg, r = 1.737 × 106 m.
ve = √(2 × 6.674 × 10-11 × 7.342 × 1022 / 1.737 × 106)
ve = √(2 × 4.899 × 1012 / 1.737 × 106) = √(5.641 × 106)
ve ≈ 2,375 m/s ≈ 2.38 km/s
When to Use It
The escape velocity formula is essential in astrophysics and space engineering.
- Determining the minimum launch speed for spacecraft and probes
- Understanding why some celestial bodies retain atmospheres and others do not
- Calculating whether a gas molecule can escape a planet's gravity (atmospheric retention)
- Comparing gravitational binding of different planets, moons, and stars
- Understanding the physics of black holes and event horizons