A body of mass m falls from earth's surface at a height equal to twice the radius (R) each. Then the change in P.E. of body will be

Options

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

Answer

C

Explanation

Gravitational potential energy at any point at a distance r from the centre of the Earth is
`U=-(GMm)/(r)`
Where M and m are the masses of the Earth and the body respectively.
At the surface of the Earth,
r = R
`therefore" "U_(i)=-(GMm)/(R)`
At a height h from the surface, `r=R+h=R+2R=3R[becauseh=2R(given)]`
`thereforeU_(f)=-(GMm)/(3R)`
Change in potential energy,
`DeltaU=U_(f)-U_(i)=-(GMm)/(3R)-(-(GMm)/(R))`
`=(GMm)/(R)(1-(1)/(3))`
`=(2)/(3)(GMm)/(R)`
`=(2)/(3)mgR(becauseg=(GM)/(R^2))`