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_1m_2))]^(1//2)`
D. `[(2G)/(r)m_1m_2]^(1//2)`

Answer

B

Explanation

Low of conservation of momentum
`m_1v_1=m_2v_2`
Conservation of energy –
Change in P.E. = change in K.E.
`rArr(Gm_1m_2)/r=1/2m_1v_1^2+1/2m_2v_2^2`
`rArr(m_1^2v_1^2)/(m_1)+(m_2^2v_2^2)/(m_2)=(2Gm_1m_2)/r`
(i) and (ii)
`v_1=sqrt((2Gm_2^2)/(r(m_1+m_2)))` and `v_2=sqrt((2Gm_1^2)/(r(m_1+m_2)))`
`v_r=v_1+v_2=sqrt((2G)/r(m_1+m_2))`