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. 