Physics · Ch 7 — Gravitation

Escape Speed

7.8

Escape Speed

The Meaning of Escape Speed

When you throw a ball upward, it rises, stops, and falls back. The harder you throw it, the higher it goes. If you could throw it fast enough, it would keep going away from the Earth forever, never falling back. That minimum speed — the speed needed to break free from a planet's gravitational pull without any further propulsion — is called the escape speed.

It is a common mistake to think escape speed means you need to reach a certain height. It does not. Escape speed is about kinetic energy being large enough to overcome the gravitational potential energy that binds an object to the planet. If you give an object that much kinetic energy at the surface, it will coast to infinity, slowing down but never stopping until it is infinitely far away.

Watch out

Escape speed does not depend on the mass of the escaping object. A feather and a rocket both need the same speed to escape Earth's gravity (ignoring air resistance). The energy required is different, but the speed is the same.

Deriving the Escape Speed

Consider an object of mass $m$ on the surface of a planet of mass $M$ and radius $R$ . The gravitational potential energy of the object at the surface is:

U = -\frac{G M m}{R}

We want to give the object just enough kinetic energy so that its total mechanical energy becomes zero. Why zero? Because at infinity, the gravitational potential energy is zero (by convention), and if the object has zero total energy, it will have zero kinetic energy at infinity — meaning it just barely escapes, reaching infinity with zero speed.

The kinetic energy we give it at the surface is:

K = \frac{1}{2} m v_e^2

where $v_e$ is the escape speed. The total mechanical energy at the surface is:

E = K + U = \frac{1}{2} m v_e^2 - \frac{G M m}{R}

For the object to escape, we set $E = 0$ :

\frac{1}{2} m v_e^2 - \frac{G M m}{R} = 0

The mass $m$ cancels out. Solving for $v_e$ :

\frac{1}{2} v_e^2 = \frac{G M}{R}

v_e^2 = \frac{2 G M}{R}

v_e = \sqrt{\frac{2 G M}{R}}

This is the escape speed from the surface of a spherical body of mass $M$ and radius $R$ .

Escape Speed from Earth

For Earth, $M = 5.98 \times 10^{24} \text{ kg}$ , $R = 6.37 \times 10^6 \text{ m}$ , and $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$ . Plugging in:

v_e = \sqrt{\frac{2 \times (6.67 \times 10^{-11}) \times (5.98 \times 10^{24})}{6.37 \times 10^6}}

v_e = \sqrt{\frac{7.98 \times 10^{14}}{6.37 \times 10^6}} \approx \sqrt{1.25 \times 10^8}

v_e \approx 1.12 \times 10^4 \text{ m/s} = 11.2 \text{ km/s}

So the escape speed from Earth is about 11.2 km/s. For comparison, a typical satellite in low Earth orbit moves at about 7.8 km/s — noticeably slower.

Note

The escape speed from Earth is independent of the direction of launch, as long as you are not aiming into the ground. Going straight up is the most direct, but any direction that avoids the planet works — the required speed is the same.

Escape Speed in Terms of g

Since the acceleration due to gravity at the surface is $g = G M / R^2$ , we can rewrite the escape speed in a convenient form. Substitute $G M = g R^2$ into the formula:

v_e = \sqrt{\frac{2 (g R^2)}{R}} = \sqrt{2 g R}

For Earth, using $g = 9.8 \text{ m/s}^2$ and $R = 6.37 \times 10^6 \text{ m}$ :

v_e = \sqrt{2 \times 9.8 \times 6.37 \times 10^6} \approx \sqrt{1.25 \times 10^8} \approx 11.2 \text{ km/s}

This gives the same result, confirming the consistency.

Escape Speed from Other Bodies

The formula $v_e = \sqrt{2 G M / R}$ applies to any spherical body. Here are some notable values:

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