Escape Velocity Formula
Escape velocity is the ideal speed a coasting object needs so its total mechanical energy reaches zero instead of remaining gravitationally bound. With mass input, the calculator uses:
vesc = √(2GM / r) with r = R + h
With the advanced μ input, it uses the equivalent form vesc = √(2μ / r). Circular orbital speed at the same radius is vcirc = √(μ / r), so escape speed is always √2 times circular speed in this ideal model.
Worked example: Earth surface
- Use G = 6.67430 × 10-11 m³/(kg·s²), M = 5.9722 × 1024 kg, and R = 6,371,000 m.
- Set altitude h = 0, so r = 6,371,000 m.
- Compute v = √(2GM/r) = √(2 × 6.67430e-11 × 5.9722e24 / 6,371,000).
- The result is about 11,186 m/s, or 11.186 km/s.
- Convert to miles per hour: 11.186 km/s × 2,236.936 ≈ 25,020 mph.
Worked example: 400 km above Earth
- Use the same Earth mass and G, but add altitude: r = 6,371,000 m + 400,000 m = 6,771,000 m.
- Compute v = √(2GM/r) with the larger radius from Earth's center.
- The ideal escape speed is about 10.85 km/s, or 24,270 mph.
- The circular speed there is about 7.67 km/s, so the extra speed from circular orbit to escape is about 3.18 km/s.
