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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.

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)/rm_(1)m_(2)]^(1//2)`

Text Solution

Verified by Experts

The correct Answer is:
B

`1/2muv^(2)=(Gm_(1)m_(2))/r`,
`mu=((m_(1)+m_(2))/(m_(1)+m))`: reduced mass
`1/2(m_(1)m_(2))/(m_(1)+m_(2))v^(2)=(Gm_(1)m_(2))/r`
`v=sqrt((2G(m_(1)+m_(2)))/r)`
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