Physics · Astrophysics · A2 Level

Escape Velocity

Throw a ball upward and it always comes back down. Throw it hard enough, though, and something changes: gravity never wins. There's a threshold speed above which an object leaves a massive body's gravitational pull for good — and that threshold has a strikingly simple formula.

7 min read

What "escaping" means

Fire a projectile straight up from the Earth's surface at increasing speeds, and its maximum height increases too — but for any finite launch speed, gravity always eventually wins, decelerating the projectile to a stop and pulling it back down. Escape velocity is the minimum launch speed at which this stops being true: the speed at which the object can travel to an arbitrarily large distance from the central body, with, in the limiting case, exactly zero speed remaining once it gets there.

It's easy to misread "escape velocity" as some kind of permanent cruising speed needed to keep moving away forever. It isn't. An object launched at exactly escape velocity keeps decelerating the whole way — gravity never stops pulling on it — but it decelerates more and more slowly, approaching (but never quite reaching) zero speed as its distance from the central body grows without bound. Launch faster than escape velocity, and the object still has some speed left over even at effectively infinite distance. Launch slower, and it never escapes at all: it either falls back down or, if given some sideways motion, settles into a bound orbit or falls back on a curved path.

Deriving escape velocity from energy conservation

The cleanest route to escape velocity is energy conservation, using gravitational potential energy. The usual convention is to define gravitational potential energy as zero at infinite separation, which makes it negative at any finite distance (since gravity is attractive, energy must be added to increase separation). An object "just escapes" if its total mechanical energy — kinetic plus gravitational potential — is exactly zero: it arrives at infinite distance with exactly zero kinetic energy remaining, having exactly used up its initial kinetic energy fighting against the (negative) potential energy along the way.

DerivationEscape velocity from a body of mass M

Require total mechanical energy to be exactly zero for the limiting "just escapes" case:

$$\underbrace{\tfrac{1}{2}mv_{esc}^2}_{\text{kinetic energy}} + \underbrace{\left(-\dfrac{GMm}{r}\right)}_{\text{grav. PE at radius } r} = 0$$

Zero total energy means the object reaches infinite distance with zero kinetic energy left — the threshold case of escaping.

Rearrange so kinetic energy exactly balances the magnitude of the potential energy:

$$\tfrac{1}{2}mv_{esc}^2 = \dfrac{GMm}{r}$$

The initial kinetic energy must equal the gravitational potential energy that needs to be "overcome."

Cancel the mass $m$ of the escaping object from both sides:

$$\tfrac{1}{2}v_{esc}^2 = \dfrac{GM}{r}$$

Just as with orbital speed, the escaping object's own mass cancels out entirely.

Multiply both sides by 2 and take the square root:

$$v_{esc} = \sqrt{\dfrac{2GM}{r}}$$

This is the escape velocity from radius $r$ (measured from the centre) of a body of mass $M$.

Result

The key insight

Escape velocity depends only on the mass $M$ of the body being escaped and the starting distance $r$ from its centre — never on the mass of the escaping object, and (for a spherically symmetric body, ignoring atmospheric drag) not on the direction it's launched in either. A pebble and a spacecraft launched from the same point need exactly the same speed to escape. This mirrors the same mass-cancellation seen in the orbital speed formula from the previous lesson — both are consequences of gravitational force and gravitational potential energy each being directly proportional to the mass of the object experiencing them.

Worked example: escaping the Earth

Find the escape velocity from the surface of the Earth, given the Earth's mass $M = 5.97\times10^{24}\,\text{kg}$, radius $r = 6.37\times10^{6}\,\text{m}$, and $G = 6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$.

Worked exampleEarth's escape velocity

Substitute directly into $v_{esc} = \sqrt{2GM/r}$:

$$v_{esc} = \sqrt{\dfrac{2(6.67\times10^{-11})(5.97\times10^{24})}{6.37\times10^{6}}}$$

Using Earth's mass and mean radius.

Evaluate:

$$v_{esc} \approx 1.12\times10^{4}\,\text{m s}^{-1} \approx 11.2\,\text{km s}^{-1}$$

This is the well-known figure for Earth's surface escape velocity — about 40,300 km/h, roughly 33 times the speed of sound in air.

Checks out

Escape velocity taken to its extreme: black holes

The escape velocity formula becomes especially dramatic when a huge mass is compressed into a small enough radius. As $r$ shrinks for a fixed mass $M$, escape velocity $v_{esc} = \sqrt{2GM/r}$ grows without bound. Nothing in principle stops this growth — until it runs into the ultimate speed limit in physics, the speed of light $c$.

This is precisely the condition that defines a black hole, covered earlier in this module. A black hole is a region of space compressed so densely that its escape velocity, calculated from exactly this formula, equals or exceeds $c$ — meaning not even light can escape once inside that radius. Setting $v_{esc} = c$ in the formula above and solving for $r$ is exactly how the Schwarzschild radius was derived in the earlier Black Holes lesson: the boundary at which escape velocity reaches the speed of light. Everything covered in that derivation is just this same escape velocity relationship pushed to its most extreme possible limit.

Key takeaways

Escape velocity is the minimum launch speed needed to reach an arbitrarily large distance from a massive body, arriving there with (in the limiting case) zero speed remaining.
It is derived by setting total mechanical energy to zero: $\tfrac{1}{2}mv_{esc}^2 = \dfrac{GMm}{r}$, using the convention that gravitational potential energy is zero at infinite separation.
Cancelling the escaping object's mass and rearranging gives $v_{esc} = \sqrt{2GM/r}$.
Escape velocity depends only on the central mass $M$ and starting radius $r$ — not on the escaping object's own mass or (ignoring drag) its launch direction.
Earth's surface escape velocity works out to approximately 11.2 km/s.
As $r$ shrinks for fixed mass $M$, escape velocity increases without bound.
A black hole is defined by exactly this formula: the region within which escape velocity reaches or exceeds the speed of light $c$, giving the Schwarzschild radius derived in the earlier Black Holes lesson.

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