Two masses m₁ and m₂ are initially at rest and are separated by a very large distance. If the masses approach each other subsequently, due to gravitational attraction between them, their relative velocity of approach at a separatation of d is

As there is no external force acting on the system, linear momentum of the system is conserved.

$$\begin{gathered} 0={\mathrm{m}}_1\overrightarrow{{\mathrm{V}}_1}+{\mathrm{m}}_2\overrightarrow{{\mathrm{V}}_2} \\ \mathrm{magnitude} \\ {\mathrm{V}}_2=\frac{{\mathrm{m}}_1{\mathrm{V}}_1}{{\mathrm{m}}_2}---\left(1\right) \end{gathered}$$

the relative velocity of approach is $\left|\overrightarrow{{\mathrm{V}}_2}-\overrightarrow{{\mathrm{V}}_1}\right|={\mathrm{V}}_2+{\mathrm{V}}_1=\left(\frac{{\mathrm{m}}_1+{\mathrm{m}}_2}{{\mathrm{m}}_2}\right){\mathrm{V}}_1----\left(2\right)$

Applying conservation of mechanical energy for the system

$$\begin{gathered} 0=-\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{\mathrm{d}}+\frac{1}{2}{\mathrm{m}}_1{{\mathrm{V}}_1}^2+\frac{1}{2}{\mathrm{m}}_2{{\mathrm{V}}_2}^2 \\ \mathrm{substituting} \left(1\right) \mathrm{in} \mathrm{the} \mathrm{above} \mathrm{we} \mathrm{get} \\ {\mathrm{V}}_1=\sqrt{\frac{2{{\mathrm{Gm}}_2}^2}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\mathrm{d}}} \end{gathered}$$

Substitute V₁ in (2) we get

relative velocity of approach $=\sqrt{\frac{2\mathrm{G}\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}{\mathrm{d}}}$