First slide

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

A

${[2 G \frac{(m_{1} - m_{2})}{r}]}^{1 / 2}$
B

${[\frac{2 G}{r} (m_{1} + m_{2}]}^{1 / 2}$
C

${[\frac{r}{2 G (m_{1} m_{2})}]}^{1 / 2}$
D

${[\frac{2 G}{r} m_{1} m_{2}]}^{1 / 2}$

Solution

Let velocities of these masses at r distance from each other be v₁ and v₂ respectively.
By conservation of momentum
$\begin{array}{l} m_{1} v_{1} - m_{2} v_{2} = 0 \\ \Rightarrow m_{1} v_{1} = m_{2} v_{2} . . . (i) \end{array}$
By conservation of energy
change in P.E.=change in K.E.
$\begin{array}{l} \frac{{Gm}_{1} m_{2}}{r} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} \\ \Rightarrow \frac{m_{1}^{2} v_{1}^{2}}{m_{1}} + \frac{m_{2}^{2} v_{2}^{2}}{m_{2}} = \frac{2 {Gm}_{1} m_{2}}{r} . .. (ii) \end{array}$
On solving equation (i) and (ii)
$v_{1} = \sqrt{\frac{2 {Gm}_{2}^{2}}{r (m_{1} + m_{2})}} and v_{2} = \sqrt{\frac{2 {Gm}_{1}^{2}}{r (m_{1} + m_{2})}}$
$∴ v_{app} = | v_{1} | + | v_{2} | = \sqrt{\frac{2 G}{r} (m_{1} + m_{2})}$

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