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.

Two bodies of masses `m_1` and `m_2` are initially at rest at infinite

1.	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.
Answer» Let `upsilon_(r)` be the relative velocity of approach of two bodies at distance `r` apart. The reduced mass of the system of two particles is, `mu = (m_(1) m_(2))/(m_(1) + m_(2))`. According to law of conservation of mechincal energy. Decrease in potential energy = increase in K.E. `0 - (-(Gm_(1) m_(2))/(r)) = (1)/(2) mu upsilon_(r)^(2)` or `(Gm_(1) m_(2))/(r) = (1)/(2) ((m_(1) m_(2))/(m_(1) + m_(2))) upsilon_(r)^(2)` or `upsilon_(r) = sqrt((2G (m_(1) + m_(2)))/(r))`

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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.

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