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.

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