免费逃逸速度计算器
计算脱离天体引力所需的最小速度。
逃逸速度
11,185.98 m/s
Escape Velocity vs Mass of Celestial Body
公式
## Escape Velocity Escape velocity is the minimum speed an object must have to permanently leave a gravitational field without further propulsion. ### Derivation from Energy Conservation Setting kinetic energy equal to gravitational potential energy: **(1/2)mv² = GMm/r** Solving for v: **v = sqrt(2GM/r)** Note that escape velocity is independent of the escaping object's mass.
计算示例
Escape velocity from Earth's surface.
- 01v = sqrt(2 * G * M / r)
- 02v = sqrt(2 * 6.6743e-11 * 5.972e24 / 6.371e6)
- 03v = sqrt(1.251e8)
- 04v ≈ 11,186 m/s ≈ 40,270 km/h
常见问题
Does escape velocity depend on the object's mass?
No. Escape velocity depends only on the mass and radius of the body being escaped from, not on the mass of the escaping object.
What is Earth's escape velocity?
Approximately 11,186 m/s or about 40,270 km/h from Earth's surface.
Does direction matter for escape velocity?
The magnitude is the same regardless of direction (as long as the path doesn't intersect the body). However, launching parallel to the surface requires slightly more energy due to the need to also gain altitude.
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