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))`

Options

A. `sqrt(2gR_(e))`
B. `sqrt(R_(e)g)`
C. `sqrt(2g(R'-R_(e)))`
D. `sqrt(2gR_(e)(1-(R_(e))/(R^(r))))`

Answer

D

Explanation

Let the mass of the particle be `m`
`PE` at a distance of `R'=(GMm)//R'`
`PE` at a distance of `R_(e)=-(GMm)//(R_(e))`
Decrease in `PE=` Increase in `KE` in
`implies-(Gmm)/R'+(GMm)/(R_(e))=1/2mv^(2)`
`v^(2)=2GM[1/R_(e)-1/R']impliesv^(2)=(2GM)/(R_(e))[1-(R_(e))/R']`
`implies v^(2)=(2GM)/(R_(e))(1-(R_(e))/(R^(')))=v=sqrt((2GMR_(e)))/R_(r)^(2) (1-(R_(e))/(R^(')))`
`:. v=sqrt(2R_(e)(1-(R_(e))/(R^(')))`