The change in the gravitational potential energy when a body of a mass `m` is raised to a height `nR` above the surface of the earth is (here `R` is the radius of the earth)

Options

A. `mgR_(E) n/((n-1))`
B. `mgR_(E)`
C. `(mgR_(E)) n/((n+1))`
D. `(mgR_(E))/n`

Answer

C

Explanation

Gravitational potential energy at any point at a distance `r` from the centre of the earth is
`U=-(Gm_(E)m)/r`
Where `M_(E)` and `m` be masses of earth and body respectively.
At the surface of the earth, `r=R_(E)`
`U_(1)=-(GM_(E)m)/(R_(E))`
At a height `h` form the surface,
`r=R_(E)+h=R_(E)+nR_(E)=(1+n)R_(E)`
`:. U_(2)=-(GM_(E)m)/((n+1)R_(E))`
Change in potential energy is
`DeltaU =U_(2)-U_(1)=-(GM_(E)m)/(R_(E))-(-(GM_(E)m)/((n+1)R_(E)))`
`=(GM_(E)m)/(R_(E))(1-1/((n+1)))=(GM_(E)mn)/((n+1)R_(E))`
`=mgR_(E) n/((n+1)) ( :' g=(GM_(E))/(R_(E)^(2)))`