Escape velocity

Escape velocity is the minimum speed an object needs, starting from the surface of a planet or star, to escape its gravity completely without any further push. The formula is v = √(2GM/r), where G is the gravitational constant, M is the mass of the planet and r is its radius. For Earth, the escape velocity is about 11.2 km/s, roughly 40 000 km/h.

The escape velocity formula

Two things stand out. The mass of the escaping object is not in the formula, so a pebble and a spacecraft need the same speed. And r is measured from the centre, so starting higher up, from an orbit, needs less speed.

The Escape Velocity Calculator works this out for any mass and radius, including the planets.

Escape velocity derivation

The derivation uses energy. Give an object of mass m a speed v at the surface. To escape, its kinetic energy must be enough to carry it all the way out to infinity, where it would just come to rest.

At distance r, its gravitational potential energy is −GMm/r. At infinity, both its kinetic energy and potential energy are zero. So, to escape, its kinetic energy plus its potential energy at the start needs to add up to zero or more. At the minimum speed they add to exactly zero:

The m cancels, which is why the object's own mass does not matter. Multiply both sides by 2 and take the square root:

The same formula with g

At the surface, g = GM/r², so GM = gr². Substituting gives a shortcut when you know the surface gravity instead of the mass:

Worked examples

Example 1: the escape velocity of Earth

Earth has a mass of 5.97 × 10²⁴ kg and a mean radius of 6371 km. Find its escape velocity. Use G = 6.67 × 10⁻¹¹ N m² kg⁻².

Convert the radius to metres first: 6371 km = 6.371 × 10⁶ m.

Earth's escape velocity is 11.2 km/s, the value NASA lists. Using v = √(2gr) with g = 9.81 m/s² gives the same 11.2 km/s.

Example 2: the escape velocity of the Moon

The Moon has a mass of 7.35 × 10²² kg and a radius of 1737 km.

The Moon's escape velocity is about 2.4 km/s, under a quarter of Earth's. That is why the Apollo lunar modules could leave the Moon with a small rocket, while leaving Earth took a Saturn V.

Example 3: finding a mass

Mars has a radius of 3390 km and an escape velocity of 5.0 km/s. Estimate its mass.

Rearrange the formula: square both sides to get v² = 2GM/r, so M = v²r ÷ 2G.

That matches NASA's figure of 0.642 × 10²⁴ kg to 2 significant figures.

How escape velocity depends on mass and radius

Because v = √(2GM/r), escape velocity goes with the square root of mass divided by radius. Four times the mass, with the same radius, doubles it. A quarter of the radius, with the same mass, also doubles it. The graph shows what happens if you keep Earth's mass and squeeze it smaller.

Here x is the radius in thousands of kilometres and y is the escape velocity in km/s. Shrinking the radius makes the escape velocity climb faster and faster. Try a scaling question like this one, then press Mark to check your answer in the course.

Escape velocity of the planets and the Sun

These values come from NASA's planetary and Sun fact sheets.

Escape velocity and black holes

If you squeeze a mass into a small enough radius, its escape velocity reaches the speed of light, c. Nothing can travel faster, so not even light can escape: it is a black hole. Set v = c in the escape velocity formula and rearrange for r:

This is the Schwarzschild radius, the size of the event horizon. For the Sun, with M = 1.99 × 10³⁰ kg and c = 3.00 × 10⁸ m/s, r = 2950 m, about 3 km. For Earth it is just under 9 mm. This simple escape velocity argument happens to give the right answer, though the full explanation needs general relativity. The Black Holes course starts from escape speed and builds up to event horizons.

Escape velocity and orbital velocity

For a circular orbit at distance r from the centre, the speed needed is v = √(GM/r). Compare it with the escape velocity at the same radius, √(2GM/r): escape velocity is always √2, about 1.41, times the orbital speed. Just above Earth's surface, a circular orbit would need 7.9 km/s, and 7.9 × √2 = 11.2 km/s.

Common mistakes

• Radius in km. G is in SI units, so r must be in metres. 6371 km is 6.371 × 10⁶ m.
• Using height above the surface for r. r is measured from the centre. From an orbit 400 km up, r = 6371 + 400 = 6771 km.
• Forgetting the 2 or the square root. √(GM/r) is the orbital speed, not the escape velocity.
• Thinking the object's mass matters. It cancels in the derivation.
• Thinking rockets must reach escape velocity. The formula is for something given one push at the start and then left to coast. A rocket that keeps firing its engines can leave Earth at a lower speed, because it keeps adding energy as it goes.

Is the escape velocity formula in the data booklet?

Not for AQA. Escape velocity is in the A-level only part of AQA's gravitational fields topic, and it also comes up in the astrophysics option for black holes. The AQA data and formulae booklet gives g = GM/r², V = −GM/r and F = GMm/r², but not v = √(2GM/r), so you need to be able to derive it from the potential. This was checked in November 2026.

Practise gravitational fields in a short timed paper

A-level Physics: Year 2 gives you a short timed paper every day, with 8 questions worth 25 to 35 marks across the second year of the course, including gravitational fields and orbits. In the US, the same practice opens as AP Physics 2.

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Physics · SpaceEscape velocity calculator: v = √(2GM/r) with stepsFind the escape velocity from any planet, moon or star from its mass and radius, in m/s, km/s, mph or as a fraction of light speed, with worked steps.