Understanding the escape and orbital velocity calculator — concept and principle
Escape velocity is the minimum launch speed an unpowered object needs to leave a planet's gravity for good, while orbital velocity is the speed needed to stay in a circular orbit at a given altitude. This calculator gives both, along with the orbital period and the surface gravity, from a planet's mass and radius. The defaults use the Earth and a 400 km orbit, close to the altitude of the International Space Station.
How it is calculated
With GM = G × M (G = 6.6743 × 10⁻¹¹ N·m²/kg²), radius R and orbit radius r = R + h (converted to metres):
- Escape velocity at the surface: v_esc = √(2GM ÷ R)
- Circular orbit speed: v = √(GM ÷ r)
- Orbital period: T = 2π √(r³ ÷ GM), shown in minutes
- Surface gravity: g = GM ÷ R²
Speeds are shown in km/s.
The principle
Escape velocity comes from energy: kinetic energy ½mv² must equal the work needed to climb out of the gravity well, GMm ÷ R. Solving gives v = √(2GM/R); the object's own mass cancels.
Orbital speed comes from force: for a circular orbit, gravity supplies exactly the centripetal force, GMm ÷ r² = mv² ÷ r, so v = √(GM/r). The period is the circumference 2πr divided by that speed. At the same radius, escape speed is always √2 ≈ 1.414 times the circular speed.
Worked example
- Inputs: M = 5.972 × 10²⁴ kg, R = 6,371 km, h = 400 km.
- GM = 6.6743 × 10⁻¹¹ × 5.972 × 10²⁴ = 3.98589 × 10¹⁴ m³/s².
- v_esc = √(2 × 3.98589 × 10¹⁴ ÷ 6,371,000) ≈ 11,186 m/s = 11.18597789 km/s.
- r = 6,771,000 m, so v = √(3.98589 × 10¹⁴ ÷ 6,771,000) ≈ 7.672490413 km/s.
- T = 2π √(6,771,000³ ÷ 3.98589 × 10¹⁴) ≈ 5,545 s ≈ 92.41555528 min.
- g = 3.98589 × 10¹⁴ ÷ 6,371,000² ≈ 9.819973426 m/s².
| Radius (km) | Escape velocity at surface (km/s) |
|---|---|
| 1,593 | 22.37195578 |
| 3,186 | 15.81936164 |
| 4,778 | 12.91645469 |
| 6,371 | 11.18597789 |
| 7,964 | 10.00504278 |
| 9,556 | 9.133312703 |
| 12,742 | 7.909680822 |
How to read the result
Higher orbits are slower and take longer: at the surface radius the circular speed would be about 7.91 km/s, at 400 km it is 7.67 km/s, and at geostationary distance about 3.07 km/s. The surface gravity here (9.82 m/s²) is a little above the standard 9.80665 m/s² because the calculation ignores Earth's rotation and flattening. Escape velocity is a speed limit for a coasting object only; a rocket that keeps thrusting can leave slowly.
Common mistakes
- Using altitude instead of distance from the centre in the orbit formula.
- Mixing km and m.
- Thinking escape velocity must point straight up; any direction works if nothing blocks the path, ignoring air drag.
Frequently asked questions
Why do rockets launch eastward?
Earth's rotation adds up to about 0.46 km/s at the equator, reducing the speed the rocket must supply.
What is the escape velocity from the Moon?
About 2.38 km/s, much lower than Earth's because the Moon has far less mass.
Is escape velocity the same at the ISS altitude?
No. At r = 6,771 km it is √2 × 7.67 ≈ 10.85 km/s, lower than at the surface.
Sources: Halliday, Resnick and Walker, Fundamentals of Physics, chapter on gravitation · H. D. Curtis, Orbital Mechanics for Engineering Students, Elsevier