Escape Velocity Calculator
Calculates the escape velocity from a planet, moon, or star — the minimum speed an object needs to break free of its gravity without further propulsion. It uses √(2GM/r) with the universal gravitational constant.
Last updated: May 2026
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About this calculator
Escape velocity is the minimum speed an unpowered object must have to escape a celestial body’s gravitational pull and never fall back, ignoring atmospheric drag and other forces. It comes from equating kinetic energy to gravitational potential energy and solving for speed, giving v = √(2GM ÷ r), where G is the universal gravitational constant (6.674 × 10⁻¹¹ N·m²/kg²), M is the mass of the body in kilograms, and r is the distance from its centre in metres (usually the body’s radius if launching from the surface). For Earth, with mass 5.972 × 10²⁴ kg and radius 6.371 × 10⁶ m, the result is about 11,186 m/s — roughly 11.2 km/s, the famous figure rockets must approach to leave Earth. A striking feature of the formula is that escape velocity does not depend on the mass of the escaping object: a pebble and a spacecraft need the same speed (though the spacecraft needs far more energy to reach it). Escape velocity scales with the square root of the body’s mass and inversely with the square root of its radius, so massive, compact bodies have very high escape velocities — the Moon’s is only about 2.4 km/s, the Sun’s is 618 km/s, and a black hole’s exceeds the speed of light at its event horizon. Note the formula assumes a non-rotating, spherically symmetric body and ignores atmospheric drag, so real rockets must supply extra energy. It also gives the speed needed to escape to infinity; reaching orbit requires less (orbital velocity is escape velocity divided by √2).
How to use
Example 1 — Earth. Enter mass 5.972e24 and radius 6.371e6. Result: about 11,186 m/s (11.2 km/s). Verify: 2 × 6.674e-11 × 5.972e24 ≈ 7.97e14; ÷ 6.371e6 ≈ 1.251e8; √ ≈ 11,186 m/s. ✓ This matches the textbook value for Earth’s escape velocity. Example 2 — The Moon. Enter mass 7.342e22 and radius 1.737e6. Result: about 2,376 m/s (2.4 km/s). Verify: 2 × 6.674e-11 × 7.342e22 ≈ 9.80e12; ÷ 1.737e6 ≈ 5.64e6; √ ≈ 2,375 m/s. ✓ The Moon’s far lower escape velocity is why the lunar lander needed only a small ascent engine.
Frequently asked questions
Does escape velocity depend on the mass of the object escaping?
No — and this surprises many people. The formula v = √(2GM/r) contains the mass of the body being escaped (M) but not the mass of the escaping object. A feather, a person, and a spaceship all need the same escape velocity from Earth, about 11.2 km/s. What does differ is the energy required: kinetic energy is ½mv², so a heavier object needs proportionally more energy (and fuel) to reach that same speed. This is why launching a massive rocket is so demanding even though the target speed is identical for any object. The independence from the escaping mass falls directly out of the energy balance the formula is derived from.
What is the difference between escape velocity and orbital velocity?
Escape velocity is the speed needed to leave a body’s gravity entirely and coast to infinity, while orbital velocity is the speed needed to maintain a stable circular orbit at a given altitude. For a circular orbit just above the surface, orbital velocity equals escape velocity divided by the square root of 2 — about 7.9 km/s for low Earth orbit versus 11.2 km/s to escape. In other words, reaching orbit takes substantially less speed than escaping. Rockets going to orbit therefore target the lower figure; missions leaving Earth for the Moon or beyond must reach or exceed escape velocity (or use orbital maneuvers to build up to it).
What units should I enter, and where do I get mass and radius?
Enter mass in kilograms and radius in metres, and the result comes out in metres per second. Astronomical values are large, so use scientific notation: Earth’s mass is 5.972e24 kg and radius 6.371e6 m; the Moon is 7.342e22 kg and 1.737e6 m; the Sun is 1.989e30 kg and 6.957e8 m. You can find these figures in NASA fact sheets or most physics references. The radius should be the distance from the body’s centre to your launch point — usually the surface radius, but if you are escaping from altitude, add that height to the radius, which lowers the required escape speed.
What mistakes do people make calculating escape velocity?
The most common is unit error — mixing kilometres with metres for the radius, or using grams instead of kilograms for mass — which skews the result dramatically. Another is forgetting scientific notation and entering, say, 5972 instead of 5.972e24 for Earth’s mass. People sometimes confuse escape velocity with orbital velocity, expecting the orbital figure (about 7.9 km/s for Earth) from this formula, which actually gives the higher escape value. A conceptual mistake is assuming heavier rockets need a higher escape velocity; they need more energy, not more speed. Finally, the formula ignores atmospheric drag and rotation, so it underestimates the real velocity a surface-launched rocket must achieve.
When does this simple formula not apply?
The formula assumes a non-rotating, spherically symmetric body and ignores air resistance, so it is an idealisation. For a real rocket launched through an atmosphere, drag means more energy is needed than the bare escape velocity implies, while a planet’s rotation can give a small free boost if you launch eastward. It also does not apply once relativistic speeds are involved — near a black hole, where escape velocity approaches the speed of light, you need general relativity, not this Newtonian equation. And it gives escape to infinity from a single body; in multi-body systems (escaping the Sun’s gravity from Earth’s orbit, for example) you must account for several gravitational sources.