Escape Velocity⚠ unverified
Aerospace / Orbital · Escape velocity from a radius
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| r | r | m | 6771000.0 | Radius |
| mu | μ | m^3/s^2 | 398600000000000.0 | Gravitational parameter |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| v | ve | m/s | Escape velocity |
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.