A body of mass `m` taken form the earth's surface to the height is equal to twice the radius `(R)` of the earth. The change in potential energy of body will be

Options

A. 3mgR
B. `1/3 mgR`
C. mg2R
D. `2/3 mgR`

Answer

D

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 be 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
` therefore U_f = - (GMm)/(3R)`
Change in potential energy, `Delta U = U_f - U_i`
`= - (GMm)/(3R) - (- (GMm)/( R)) = (GMm)/( R) (1 - 1/3) = 2/3 (GMm)/(R )`
` = 2/3 mg R`