Question

The change in potential energy when body of mass m is raised to height $nR$ from earth's surface is ($R$=radius of the earth)

• A. $mgR$
• B. $mgR\frac{n}{(n-1)}$
• C. $mgR\frac{n}{(n+1)}$
• D. $mgR\frac{n^2}{(n^2+1)}$

Answer 1

Answer: (C)

The correct option is C $mgR\frac{n}{(n+1)}$
Gravitational potential energy at any point at a distance r from the centre of the earth is $U=-\frac{G{M}_{E}m}{r}$
where $M_E$ and m be masses of earth and body respectively. At the surface of the earth, $r=R_E$
$\therefore U_1=-\frac{G{M}_{E}m}{R_E}$
At a height h from the surface, $r=R_E+h=R_E+n{R}_E=(1+n)R_E$
$\therefore U_2=-\frac{G{M}_{E}m}{(n+1)R_E}$
change in potential energy is $\Delta U=U_2-U1$

$=-\frac{G{M}_{E}m}{(n+1)R_E}-(-\frac{G{M}_{E}m}{R_E})$

$=\frac{G{M}_{E}m}{R_E}(1-\frac{1}{(n+1)})=\frac{G{M}_{E}mn}{(n+1)R_E}$

$=mg{R}_E\frac{n}{(n+1)}(\because g=\frac{G{M}_E}{R_E^2})$