Celestial Escape Velocity Formula
Calculates the minimum initial speed a ballistic projectile requires to break free from the gravitational field of a celestial body without further propulsion.
\[ v_{esc} = \sqrt{\frac{2 G M}{R}} = \sqrt{2 g R} \]
Plain Text Formula:
v_esc = sqrt((2 * G * M) / R) = sqrt(2 * g * R)
📐 Variable Definitions & Parameters
| Symbol | Variable Name | Standard Unit | Physical Definition |
|---|---|---|---|
| \[ v_{esc} \] | Escape Velocity | m/s or km/s | Speed required where kinetic energy equals gravitational potential energy |
| \[ G \] | Gravitational Constant | 6.67430×10⁻¹¹ m³/(kg·s²) | Newton's gravitational constant |
| \[ M \] | Planetary Mass | kg | Mass of the celestial planet, moon, or star |
| \[ R \] | Equatorial Radius | meters (m) | Distance from planet center to surface |
| \[ g \] | Surface Gravity | m/s² | Surface acceleration: G*M / R² |
💡 Worked Engineering Example
Problem Statement:
Calculate the escape velocity of Earth (M = 5.972e24 kg, R = 6,371,000 m).
Step-by-Step Solution:
- 2 * G * M = 2 * 6.67430e-11 * 5.972e24 = 7.9718e14
- Divide by R: 7.9718e14 / 6,371,000 = 125,126,353 m²/s²
- Square root: sqrt(125,126,353) = 11,186 m/s = 11.19 km/s
🎯 Result: Earth's surface escape velocity is 11.19 km/s (approx. 25,000 mph).