A body which is initially at rest at a height R above the surface of the earth of radius R, falls freely towards the earth. Find out its velocity on reaching the surface of earth. Take `g=` acceleration due to gravity on the surface of the Earth.

Options

A. `sqrt(2gR)`
B. `sqrt(3/2 gR)`
C. `sqrt(gR)`
D. `sqrt(1/2 gR)`

Answer

C

Explanation

Potential energy at ground surface potential energy=`(-GMm)/R`
Potential energy at a height of R is
potential energy `=(-GMm)/(2R)`
When a body comes to ground
Loss in potential energy =Gain in kinetic energy
`rArr (-GMm)/(2R)-((-GMm)/R)=1/2 mv^(2) rArr (GMm)/(2R)=1/2 mv^(2)`
`rArr gR =v^(2) ( :' (GM)/(R^(2))=g)`
`rArr v=sqrt(gR)`