Escape Velocity⚠ unverified
Physics / Mechanics · Compute the escape velocity from a spherical body
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| mass | mass | kg | 1.0 | Mass of the gravitating body |
| radius | radius | m | 1.0 | Distance from the centre of the body at which the velocity is evaluated (e.g. its surface radius) |
| G | G | N*m^2/kg^2 | 1.0 | Universal gravitational constant |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | vesc | m/s | Escape velocity, in metres per second (m/s) |
The science & history
Understanding the Parameters
- $M$, $r$ — larger $M$ or smaller $r$ raises escape speed.
-
$G$ — use $\approx 6.674\times 10^{-11}$ N·m²/kg² in SI; do not leave the default
1.0for physical answers. -
$v_{\mathrm{esc}}$ — speed only; direction must be outward enough that the trajectory is not recaptured (energy condition is isotropic in the ideal point-mass field).
Derivation (Approaching a Proof)
Gravitational potential energy of $m$ at distance $r$ (zero at infinity) is $U = -G M m / r$. Kinetic energy $\tfrac12 m v^{2}$. Total mechanical energy for a parabolic escape trajectory is zero:
$$\frac{1}{2} m v_{\mathrm{esc}}^{2} - \frac{G M m}{r} = 0 \quad\Rightarrow\quad v_{\mathrm{esc}} = \sqrt{\frac{2 G M}{r}}.$$
Compare Circular Orbital Velocity $v_c = \sqrt{G M / r}$: $v_{\mathrm{esc}} = \sqrt{2}\,v_c$.
History
Escape speed follows directly from Newtonian gravity and energy conservation; it is a standard result in orbital mechanics and planetary science (Earth surface escape $\approx 11.2$ km/s).
Related Concepts: Circular Orbital Velocity, Gravitational Force, Kinetic Energy, Potential Energy
Notes: Registry calculator mechanics-escape-velocity (unverified). Non-rotating, drag-free,
point-mass gravity. Set $G$ correctly.