Let us consider the work done in moving an object of mass [tex]m[/tex] from the surface of an object of mass [tex]M[/tex] and radius [tex]R[/tex] out to infinity. Recall that work [tex]W[/tex] is given by the product of the force [tex]F [/tex] and distance [tex]r[/tex]. If the force is constant we can use the formula

[tex]W = F r. [/tex]

However in this case the force is a function of distance and we need to integrate to get the correct result,

[tex]W = \int_a^b F(r) dr. [/tex]

Newton’s law of gravitation, gives the force between two bodies separated by a distance [tex]r[/tex] as

[tex] F(r) = - \frac{G m M} {r^2} ,[/tex] where [tex]G[/tex] is the gravitational constant.

Substituting this into our formula and integrating gives,

[tex]W =\int_{R}^{\infty} - \frac{G m M} {r^2} dr [/tex]

[tex] = \left[ \frac{G m M}{r} \right]_R^{\infty}[/tex]

[tex] = \lim_{r \rightarrow \infty} \frac{G m M} { r} - \frac{G m M} {R} [/tex]

[tex] = - \frac{G m M} R [/tex]

The work done is given by the difference between the final kinetic energy [tex]KE_f[/tex] and the initial kinetic energy [tex]KE_i[/tex]. At infinity the object has zero velocity and consequently [tex]KE_f =0[/tex]. The initial kinetic energy of the object is given by

[tex] KE_i = \frac{1}{2} m v_e^2,[/tex]
where [tex]v_e[/tex] is the initial velocity (escape velocity).

[tex]W = KE_{f}-KE_{i} = 0 - \frac{1}{2} m v_e^2 [/tex]
so that

[tex] - \frac{1}{2} m v_e^2 = - \frac{G m M}{R} [/tex]
after some rearrangement, we find

[tex]v_e = \sqrt{\frac{2 G M} {R}}.[/tex]

Notice that the escape velocity [tex]v_e[/tex] is dependant only on the mass of the body [tex]M[/tex].

We can use this classical (i.e. non-relativistic) result to find the Schwarzschild radius of a black hole. To do this we set [tex]v_e = c [/tex] in the previous formula, where [tex]c[/tex] is the speed of light.

Some algebraic rearrangement yields,

[tex] R = \frac{2 G M} {c^2} [/tex]