A body is released from a point of distance `R'` from the centre of earth. Its velocity at the time of striking the earth will be `(R’gtR_(e))`

Explanation

Initial velocity of the body is zero.
Total energy of the body `= K.E. + P.E.`
`= + ((- Gm)/(r )) = - (mg R^(2))/(r )`
Let `upsilon` be the velocity acquired by body on reaching the surface of the Earth.
Total energy on the surface of Earth
`= (1)/(2) m upsilon^(2) +(-(mgR^(2))/(R )) = (1)/(2) m upsilon^(2) - mg R`
According to law of conservation of energy, we
have `(1)/(2) m upsilon^(2) - mg R = - (mg R^(2))/(r )`
or `upsilon^(2) = 2 g R-(2 gR^(2))/(r ) = 2 g R^(2) ((1)/(R ) - (1)/(r ))`
or `upsilon = R [2g ((1)/(R ) - (1)/(r ))]^(1//2)`