First slide

Gravitational force and laws

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

Moderate

A

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

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

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

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

Solution

If v be the relative velocity of approach , then we can write , $μ (v \frac{d v}{d r}) = - \frac{G m_{1} m_{2}}{r} w h e r e μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}} = R e d u c e d m a s s . \Rightarrow \int_{0}^{v} v d v = - G (m_{1} + m_{2}) \int_{\infty}^{r} \frac{1}{r} d r \Rightarrow v = \sqrt{\frac{2 G (m_{1} + m_{2})}{r}}$

Hence the correct choice is (1).

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