 Gravitaional potential energy
Question

# Two bodies of masses m1 and m2 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

Difficult
Solution

## Let velocities of these masses at r distance from each other be v1 and v2 respectively.By conservation of momentumBy conservation of energy change in P.E.=change in K.E.$\begin{array}{l}\frac{{\mathrm{Gm}}_{1}{\mathrm{m}}_{2}}{\mathrm{r}}=\frac{1}{2}{\mathrm{m}}_{1}{\mathrm{v}}_{1}^{2}+\frac{1}{2}{\mathrm{m}}_{2}{\mathrm{v}}_{2}^{2}\\ ⇒\text{\hspace{0.17em}}\frac{{\mathrm{m}}_{1}^{2}{\mathrm{v}}_{1}^{2}}{{\mathrm{m}}_{1}}+\frac{{\mathrm{m}}_{2}^{2}{\mathrm{v}}_{2}^{2}}{{\mathrm{m}}_{2}}=\frac{2\text{\hspace{0.17em}}{\mathrm{Gm}}_{1}{\mathrm{m}}_{2}}{\mathrm{r}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\end{array}$On solving equation (i) and (ii)${\mathrm{v}}_{1}=\sqrt{\frac{2\text{\hspace{0.17em}}{\mathrm{Gm}}_{2}^{2}}{\mathrm{r}\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)}}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{v}}_{2}=\sqrt{\frac{2\text{\hspace{0.17em}}{\mathrm{Gm}}_{1}^{2}}{\mathrm{r}\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)}}$$\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{v}}_{\text{app}}=\text{\hspace{0.17em}}|\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{v}}_{1}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}}|\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{v}}_{2}\text{\hspace{0.17em}}|\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\sqrt{\frac{2\mathrm{G}}{\mathrm{r}}\left({\mathrm{m}}_{1}+{\mathrm{m}}_{2}\right)}$

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