Escape Velocity Calculator
Work out the minimum surface speed needed to leave a body’s gravity for ever, given the body’s mass and radius. Pick a preset (Earth, Moon, Mars, the Sun…) or enter a custom mass and radius.
Escape velocity from Earth
11.186 km/s
- Velocity (m/s)
- 11186.2 m/s
- Velocity (km/h)
- 40270 km/h
- Velocity (mph)
- 25023 mph
- Mach number (sea-level air)
- 32.9 × Mach
- Surface gravity
- 9.8203 m/s²
- Mass used
- 5.9722e+24 kg
- Radius used
- 6e+6 m
Newtonian escape velocity from the surface of a non-rotating spherical body, v = √(2GM/r), with G = 6.674 30 × 10⁻¹¹ m³ kg⁻¹ s⁻² (CODATA 2018). For Earth, that works out to about 11.19 km/s.
How to use this calculator
Pick the body you want the escape velocity for from the dropdown — Earth is the default and gives the familiar 11.2 km/s answer. The presets cover the Moon, every planet in the Solar System, Pluto and the Sun, with mass and radius values taken from the NASA Planetary Fact Sheet. To compute escape velocity from something not on the list — a star, an exoplanet, an asteroid, a black hole — choose "Custom body…" and type the mass in kilograms and the surface radius in metres into the two number fields. Scientific notation is fine (e.g. 5.972e24 for Earth’s mass). The result is shown in km/s as the headline number, with metres per second, km/h, mph, the Mach number relative to sea-level air and the body’s surface gravity in the breakdown.
How the calculation works
Escape velocity is the speed at which the kinetic energy of an object exactly equals the magnitude of its gravitational potential energy, so that the total energy is zero and the object can recede to infinity without ever turning back. Equating ½mv² with GMm/r and cancelling the test mass m gives v = √(2GM/r), where G is Newton’s gravitational constant (6.674 30 × 10⁻¹¹ m³ kg⁻¹ s⁻², CODATA 2018), M is the mass of the body, and r is the distance from its centre — usually the surface radius. Three things to notice: the answer does not depend on the mass of the thing trying to escape (a feather and a rocket need the same speed); the formula assumes a non-rotating spherical body and ignores other gravity sources (it gives the local escape speed, not what is needed to leave the Solar System); and the speed is the *initial* surface speed in any direction other than straight into the body — gravity does the rest of the work of slowing it as it climbs out.
Worked example
Earth has M = 5.9722 × 10²⁴ kg and a volumetric mean radius r = 6 371 000 m (NASA Earth Fact Sheet). v = √(2 × 6.6743 × 10⁻¹¹ × 5.9722 × 10²⁴ / 6 371 000) = √(1.2516 × 10⁸) ≈ 11 187 m/s, or 11.19 km/s — about 25 020 mph, Mach 32.9 relative to sea-level air, which is roughly the speed Apollo reached at trans-lunar injection. Doing the Moon the same way (M = 7.342 × 10²², r = 1 737 400 m) gives 2.38 km/s, and the Sun (M = 1.989 × 10³⁰, r = 6.957 × 10⁸) gives 617.7 km/s — the latter is why probes use Jupiter gravity assists to leave the Solar System rather than burning straight outward from low Earth orbit.
Frequently asked questions
What is escape velocity in plain English?
It is the slowest speed at which something can be flung straight up from the surface of a planet (or moon, or star) and never fall back, ignoring air resistance and any other gravity sources. Above that speed, the object’s kinetic energy is greater than the well of gravitational potential energy it sits in, so it can climb to infinity and still have some speed left over. Exactly at that speed, it just barely escapes, ending up at rest infinitely far away. Below it, gravity wins and the object falls back.
Why does escape velocity not depend on the mass of the rocket?
In the equation ½mv² = GMm/r, the mass m of the escaping object appears on both sides and cancels out, leaving v = √(2GM/r). Physically, this is the same reason a heavy rock and a feather fall at the same rate in a vacuum — gravity accelerates everything equally, so a heavier object has more kinetic energy to gain but also a deeper potential well to climb out of, and the two effects exactly balance. What does depend on rocket mass is the *fuel* required to reach that speed: a heavier rocket needs proportionally more propellant for the same delta-v.
What is Earth’s escape velocity?
About 11.186 km/s from the equator at sea level — call it 11.2 km/s or 25 020 mph. NASA and most physics textbooks quote 11.186 km/s as the standard value. It varies very slightly with latitude (Earth is an oblate spheroid, so radius is smaller at the poles) and altitude (escape velocity falls with √(1/r), so it is about 10.9 km/s from low Earth orbit at 400 km up). Rotation also helps: launching east from near the equator gives you ~0.46 km/s for free from Earth’s spin.
Is escape velocity the same as orbital velocity?
No, but they are closely related. Orbital velocity at radius r for a circular orbit is v_orb = √(GM/r), while escape velocity at the same point is v_esc = √(2GM/r). So escape velocity is exactly √2 ≈ 1.414 times the local circular orbital velocity. At the surface of the Earth, that gives orbital velocity ≈ 7.9 km/s and escape velocity ≈ 11.2 km/s — the well-known factor-of-1.4 gap that separates "in orbit" from "on its way out".
What is the escape velocity of a black hole?
A black hole is defined by the surface (the event horizon) where the escape velocity equals the speed of light, c ≈ 299 792 458 m/s. Setting v = c in v = √(2GM/r) and solving for r gives the Schwarzschild radius r_s = 2GM/c², which is about 3 km for one solar mass and roughly 12 million km for the supermassive black hole at the centre of the Milky Way. The Newtonian formula in this calculator gives the right Schwarzschild radius even though it is a coincidence — a fully relativistic derivation (general relativity) gives the same numerical answer for a non-rotating black hole.
Where do the planetary masses and radii in the presets come from?
From the NASA Planetary Fact Sheet (nssdc.gsfc.nasa.gov/planetary/factsheet/), which is the standard reference for Solar System parameters and is maintained by NASA Goddard. Masses are given in kg, radii are equatorial in metres. Pluto is included for nostalgia even though the IAU re-classified it as a dwarf planet in 2006 — the escape-velocity formula does not care about taxonomy.