Two point masses m and M are held at rest at a large distance from each other. When released, they begin moving under their mutual gravitational pull. Then at a relative seperation x
 

 Force between the two masses $\mathrm{F}=\frac{\mathrm{GmM}}{{\mathrm{x}}^2}$

$$\begin{gathered} \therefore \text{ Acceleration of }\mathrm{m}, {\mathrm{a}}_1=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{GM}}{{\mathrm{x}}^2} \\ \text{ Acceleration of }\mathrm{M}, {\mathrm{a}}_2=\frac{\mathrm{F}}{\mathrm{M}}=\frac{\mathrm{Gm}}{{\mathrm{x}}^2} \\ \text{ Relative acceleration }\mathrm{a}={\mathrm{a}}_1+{\mathrm{a}}_2=\frac{\mathrm{G}(\mathrm{M}+\mathrm{m})}{{\mathrm{x}}^2} \end{gathered}$$
$\therefore \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{dv}}{\mathrm{dx}}\cdot \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dx}}$ [v = relative velocity]
$-\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{G}(\mathrm{M}+\mathrm{m})}{{\mathrm{x}}^2}$
[- sign has been placed because v is increasing with decreasing x]
$\begin{array}{l}\therefore \mathrm{vdv}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-\mathrm{G}(\mathrm{M}+\mathrm{m})\frac{\mathrm{dx}}{{\mathrm{x}}^2}\\ \therefore {\int }_0^{\mathrm{v}} \mathrm{vdv}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}=-\mathrm{G}(\mathrm{M}+\mathrm{m}){\int }_{\mathrm{\infty }}^{\mathrm{x}} \frac{\mathrm{dx}}{{\mathrm{x}}^2}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{{\mathrm{v}}^2}{2}=-\mathrm{G}(\mathrm{M}+\mathrm{m}){\left[-\frac{1}{\mathrm{x}}\right]}_{-\mathrm{\infty }}^{\mathrm{x}}=\frac{\mathrm{G}(\mathrm{M}+\mathrm{m})}{\mathrm{x}}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{v}=\sqrt{\frac{2\mathrm{G}(\mathrm{M}+\mathrm{m})}{\mathrm{x}}}\end{array}$
 The centre of mass has no acceleration as there is no external force.
Hence, velocity of COM is zero throughout.