Question

Two bodies of masses $m_1$ and $m_2$ 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

• A. $\left[2 G \frac{\left(m_1-m_2\right)}{r}\right]^{1 / 2}$.
• B. $\left[\frac{2 G}{r}\left(m_1+m_2\right)\right]^{1 / 2}$
• C. $\left[\frac{r}{2 G\left(m_1 m_2\right)}\right]^{1 / 2}$
• D. $\left[\frac{2 G}{r} m_1 m_2\right]^{1 / 2}$

Answer 1

Answer: (B)

Apply the principle of conservation of momentum and conservation of energy.
Let velocities of these masses at $r$ distance from each other be $v_1$ and $v_2$ 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{G m_1 m_2}{r}=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2$
$\Rightarrow \frac{m_1^2 v_1^2}{m_1}+\frac{m_2^2 v_2^2}{m_2}=\frac{2 G m_1 m_2}{r}$(ii)
On solving Eqs. (i) and (ii)
$v_1 =\sqrt{\frac{2 G m_2^2}{r\left(m_1+m_2\right)}} \text { and } v_2=\sqrt{\frac{2 G m_1^2}{r\left(m_1+m_2\right)}}$
$\therefore v_{ app } =\left|v_1\right|+\left|v_2\right|=\sqrt{\frac{2 G}{r}\left(m_1+m_2\right)}$