First slide

Gravitational potential and 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.
$m_{1} v_{1} - m_{2} v_{2} = 0$
$\Rightarrow m_{1} v_{1} = m_{2} v_{2}$ ……..(i)
By conservation of energy
change in PE = change in KE
$\frac{{Gm}_{1} m_{2}}{r} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$
$\Rightarrow \frac{m_{1}^{2} v_{1}^{2}}{m_{1}} + \frac{m_{2}^{2} v_{2}^{2}}{m_{2}} = \frac{2 {Gm}_{1} m_{2}}{r}$ ……(ii)
On solving equations (i) and (ii),
$\begin{array}{l} v_{1} = \sqrt{\frac{2 {Gm}_{2}^{2}}{r (m_{1} + m_{2})}} and v_{2} = \sqrt{\frac{2 {Gm}_{1}^{2}}{r (m_{1} + m_{2})}} \\ ∴ v_{app} = |v_{1}| + |v_{2}| = \sqrt{\frac{2 G}{r} (m_{1} + m_{2})} \end{array}$

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