Two bodies of mass `m_(1) and m_(2)` are initially at rest placed infinite distance apart. They are then allowed to move towards each other under mutual gravitational attaction. Show that their relative velocity of approach at separation r betweeen them is
`v=sqrt(2G(m_(1)+m_(2)))/(r)`

Options

A. `[2G((m_1 - m_2))/(r)]^(1//2)`
B. `[(2G)/(r)(m_1 - m_2)]^(1//2)`
C. `[(r)/(2G(m_1 m_2))]^(1//2)`
D. `[(2G)/(r) m_1 m_2]^(1//2)`

Answer

B

Explanation

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`
`rArr m_1 v_1 = m_2 v_2` ...(i)
By conservation of energy
Change in `PE` = change in `KE`
`(G m_1 m_2)/( r) = (1)/(2) m_1 v_1^2 + (1)/(2) m_2 v_2^2`
`rArr (m_1^2 v_1^2)/(m_1) + (m_2^2 v_2^2)/(m_2) = (2 Gm_1 m_2)/( r)` ...(ii)
On solving Eqs. (i) and (ii)
`v_1 = sqrt((2 Gm_2^2)/(r(m_1 + m_2))) and v_2 = sqrt((2 Gm_1^2)/(r(m_1 + m_2)))`
`:. v_(app) = |v_1|+|v_2|= sqrt((2 G)/(r)(m_1 + m_2))`.