Two bodies of masses ${\mathrm{m}}_1 \mathrm{and} {\mathrm{m}}_2$ are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is 

Let velocities of these masses at r distance from each other be ${\mathrm{v}}_1 \mathrm{and} {\mathrm{v}}_2$ respectively.

By conservation of momentum

${\mathrm{m}}_1{\mathrm{v}}_1 - {\mathrm{m}}_2{\mathrm{v}}_2 = 0$

$\Rightarrow {\mathrm{m}}_1{\mathrm{v}}_1 = {\mathrm{m}}_2{\mathrm{v}}_2--------\left(\mathrm{i}\right)$

By conservation of energy

Change in PE = change in KE

       $\frac{{\mathrm{Gm}}_1{\mathrm{m}}_2}{\mathrm{r}} = \frac{1}{2}{\mathrm{m}}_1{{\mathrm{v}}_1}^2+\frac{1}{2}{\mathrm{m}}_2{{\mathrm{v}}_2}^2$

$\Rightarrow \frac{{{\mathrm{m}}_1}^2{{\mathrm{v}}_1}^2}{{\mathrm{m}}_1}+\frac{{{\mathrm{m}}_2}^2{{\mathrm{v}}_2}^2}{{\mathrm{m}}_2}= \frac{2{\mathrm{Gm}}_1{\mathrm{m}}_2}{\mathrm{r}}-------\left(\mathrm{ii}\right)$

On solving Eqs. (i) and (ii)

${\mathrm{v}}_1 = \sqrt{\frac{2{{\mathrm{Gm}}_2}^2}{\mathrm{r}({\mathrm{m}}_1+{\mathrm{m}}_2)}} \mathrm{and}$

${\mathrm{v}}_2 = \sqrt{\frac{2{{\mathrm{Gm}}_1}^2}{\mathrm{r}({\mathrm{m}}_1+{\mathrm{m}}_2)}}$

$\therefore {\mathrm{v}}_{\mathrm{app}} = \left|{\mathrm{v}}_1\right|+\left|{\mathrm{v}}_2\right| = \sqrt{\frac{2\mathrm{G}}{\mathrm{r}}({\mathrm{m}}_1+{\mathrm{m}}_2)}$