Escape velocity calculator

This escape velocity calculator works out the speed needed to leave a planet or moon from its mass and radius: 11.18 km/s for Earth. Pick a body and an altitude, or enter your own values.

Example: Celestial body: Earth · Altitude above the surface 0 km → Result: Escape velocity: 11.18 km/s (40,248 km/h). Source: Wikipedia – Kosmische Geschwindigkeit (Fluchtgeschwindigkeit, Kreisbahngeschwindigkeit). Updated: .

km
0 for a launch from the surface
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Result

Result
Escape velocity: 11.18 km/s (40,248 km/h)
Orbit data
MeasureValueFormula
Escape velocity11.18 km/s (40,248 km/h)v = √(2·G·M/r)
Circular orbit speed7.91 km/s (28,459 km/h)v = √(G·M/r)
Orbital period (circular)84.5 minT = 2π·√(r³/(G·M))
Gravitational acceleration9.8 m/s²g = G·M/r²
Distance from the centre6,378.1 kmr = R + h
Schwarzschild radius8.87 mmr_s = 2·G·M/c²
Altitude above the surface: 0 km. Values from the NASA fact sheet (equatorial radius, ignoring atmosphere and rotation). Escape velocity applies to an unpowered object that never falls back.

How it is calculated

Escape velocity v = √(2·G·M/r). For Earth it is 11.18 km/s (about 40,250 km/h), for the Moon 2.38 km/s and for Mars 5.02 km/s.

What escape velocity means

It is the speed at which an unpowered object leaves a gravitational field without ever falling back. It depends only on the mass M of the body and the distance r from its centre, not on the mass of the object. It drops with altitude because r grows.

Circular orbit and orbital period

The speed of a circular orbit is √(G·M/r), which is the escape velocity divided by √2. At 400 km above Earth that is 7.67 km/s and one lap takes about 92.6 minutes. The calculator also shows the gravitational acceleration at that altitude and the Schwarzschild radius, the radius to which a body would have to shrink for not even light to escape (about 9 mm for Earth).

Frequently asked questions

What is the escape velocity of Earth?

About 11.18 km/s, or roughly 40,250 km/h. You get it from v = √(2·G·M/r) with M = 5.972·10²⁴ kg and r = 6,378 km.

How do you calculate escape velocity?

With v = √(2·G·M/r). G is the gravitational constant 6.6743·10⁻¹¹ m³/(kg·s²), M the mass of the body and r the distance from its centre.

Why is the orbital speed lower?

A circular orbit needs only 1/√2 (about 71 %) of the escape velocity. Exceeding it by 41 % lets you leave the gravitational field.

Sources and legal basis

As of: · Editorial policy

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