Skip to content

Escape velocity

The escape speed of an object within a planet's gravitation field is the minimum speed it requires to reach a location where it is free from the planet's gravitational pull. To achieve this, the object must have enough kinetic energy to move infinitely far away from the planet.

Suppose an object of mass \(m\) is projected from the Earth's surface with an initial speed \(v\).

\[U_\infty=0\]
\[U_{planet}=-\frac{GMm}{R}\]

Applying conservation of energy:

\[\frac{1}{2} m v^2 -\frac{GMm}{R}=0\]

where \(R\) is the radius of the planeta dn \(M\) is the mass of the planet.

Hence, escape speed is:

\[v=\sqrt{\frac{2GM}{R}}\]

Question

What is the escape speed of the Earth?

a) \(35 km s^-1\)

b) \(11 km s^-1\)

c) \(23 km s^-1\)

d) Not possible to determine without knowing mass of the satellite.

B.

\[v=\sqrt{\frac{2GM}{R}}\approx11 km s^-1\]

Question

What is the escape speed of the Earth, but now its a black hole? The radius of a black hole is the Schwarzschild radius, given by \(\frac{2GM}{c^2}\).

a) \(2c \space ms^-1\)

b) \(\sqrt{c} \space ms^-1\)

c) \(c \space ms^-1\)

d) \(\sqrt{2}c \space ms^-1\)

C.

\(v=\sqrt{\frac{2GM}{R}}\)

The Schwarzschild radius is given by:

\[r=\frac{2GM}{c^2}\]

Hey that looks oddly similar. Rearranging:

\[v=\sqrt{c^2}=c\]

And so the escape speed of a black hole is famously the speed of light.

Another tangent:

The Schwarzschild radius was obtained by well, Schwarzschild as a solution to the Einstein field equations. However, Laplace found this quantity just by playing around with the escape speed formula and saw what the radius of an object would have to be such that light would be the minimum speed required to escape.