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A body is released from a point of dista...

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))`

A

`sqrt(2 gr)`

B

`sqrt(2g(R + r))`

C

`R[2g((1)/(R ) - (1)/(r ))]^(1//2)`

D

`[2g((1)/(r ) - (1)/(R))]^(1//2)`

Text Solution

Verified by Experts

The correct Answer is:
C
Initial velocity of body is zero. Total energy of the
body `= -(GM m)/(r ) = - (mg R^(2))/(r )`
Let `upsilon` be the velocity acquired by body on reaching the surface of earth than total energy of the body on the surface of earth
`=(1)/(2) m upsilon^(2) -(mg R^(2))/(R ) = (1)/(2) m upsilon^(2) - mg R`
According to law of conservation of energy
`(1)/(2)m upsilon^(2) -mg R = - (mg R^(2))/(r )`
or `upsilon^(2) = 2gR -(2g R^(2))/(r ) = 2g R^(2) [(1)/(R ) -(1)/(r )]`
or `upsilon = R [2g ((1)/(R ) -(1)/(r ))]^(1//2)`
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