## The Schwarzschild Radius

For a body of mass\[m\]

on the Earth's surface, the gravitational potential energy is \[GPE=- \frac{GM_{EARTH}m}{R_{EARTH}}\]

.This means that in order for that body to just escape the Earth's gravitational influence it must be given an amount of kinetic energy

\[\frac{1}{2}mv^2\]

sufficient to cancel out the negative gravitational potential; energy.\[\frac{1}{2}mv^2 = \frac{GM_{EARTH}m}{R_{EARTH}} \rightarrow v^2=\frac{2GM_{EARTH}}{R_{EARTH}} \]

.The Schwarzschild radius is the answer to this question: Suppose a mass is on the surface of a spherically uniform non rotating body of mass

\[M\]

. What would the radius of the mass \[M\]

have to be for the escape velocity to equal the speed of light \[c\]

.This is really a question in General Relativity, but surprising the result is exactly (1), that is

\[c^2=\frac{2GM}{R_{SCWARZSCHILD}} \]

.We can rearrange this to give

\[R_{SCWARZSCHILD}=\frac{2GM}{c^2}\]

.Of course, nothing can travel faster than light, not even light, so anything on the surface of such a body will stay there forever. The body is called a black hole.