Two bodies of masses `m_1` and `m_2` initially at rest at infinite distance apart move towards each other under gravitational force of attraction. Their relative velocity of approach when they are separated by a distance r is (G= universal gravitational constant.)

Options

A. `[(2G(m_1-m_2))/r]^(1//2)`
B. `[(2G(m_1+m_2))/r]^(1//2)`
C. `[r/(2G(m_1m_2))]^(1//2)`
D. `[r/(2G)m_1m_2]^(1//2)`

Explanation

Initially when the two masses are at an infinite distance from each other, their gravitational potential energy is zero. When they are at a distance r from each,the gravitational PE is
`PE=(-Gm_m_2)/r^2`
The minus sign-indicates that there is a distance in PE. This gives rise to an increase in kinetic energy. IF `v_1` and `v_2` are their distance r apart then, from the law of conservation of energy, we have
`1/2m_1v_1^2=(Gm_1m_2)/r or v_1=sqrt((2Gm_2)/r)`
and `1/2 m_2v_2^2=(Gm_1m_2)/r or v_2=sqrt((2Gm_1)/r)`
Therefore, their relative velocity if approach is
`v_1+v_2=sqrt((2Gm_2)/r)+sqrt((2Gm_1)/r)=sqrt((2G)/r(m_1+m_2))`
`=((2G (m_1 timesm_2))/r)^(1/2)`