InterviewSolution
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InterviewSolution
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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)`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)` |
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Answer» Correct 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` `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))`. |
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