Text Solution
Solution : Let `upsilon_(r)` be the relative velocity of approach of two bodies at distance `r` apart. The reduced mass of <br> the system of two particles is, `mu = (m_(1) m_(2))/(m_(1) + m_(2))`. <br> According to law of conservation of mechincal energy. <br> Decrease in potential energy = increase in K.E. <br> `0 - (-(Gm_(1) m_(2))/(r)) = (1)/(2) mu upsilon_(r)^(2)` <br> or `(Gm_(1) m_(2))/(r) = (1)/(2) ((m_(1) m_(2))/(m_(1) + m_(2))) upsilon_(r)^(2)` <br> or `upsilon_(r) = sqrt((2G (m_(1) + m_(2)))/(r))`
Text Solution
`(2G)/(r(m_(1)+m_(2)))` `[(2G(m_(1)+m_(2)))/(r )]^((1)/(2))` `((2Gr)/(m_(1)+m_(2)))^((1)/(2))` `((2Gr)/(m_(1)+m_(2)))`.
Answer : B
Solution : When masses are infinite distance apart then their K.E, P.E and hence total energy is zero. <br> If v is the relative velocity of approach then by law of conservation of energy. <br> `(1)/(2) ((m_(1)m_(2))/(m_(1)+m_(2)))v^(2)-(Gm_(1)m_(2))/(r )=0` <br> or `v=[(2G(m_(1)+m_(2)))/(r )]^((1)/(2))` <br> Thus correct choice is (b).
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