Question

A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $\left(R_e\right)$ the inerease in potential energy will be:( $g$ = acceleration due to gravity on the surface of Earth)

• A. $\frac{1}{2} m g R_e$
• B. $3 m g R_e$
• C. $\frac{2}{3} m g R_e$
• D. $\frac{1}{3} m g R_e$

Answer 1

The Correct Option is C

The correct answer is (C) : $\frac{2}{3} m g R_e$
$U=\frac{-G{M}_{e}m}{r}$
$U_i=\frac{-G{M}_{e}m}{R_e}$
$U_f=\frac{-G{M}_{e}m}{\left(R_e+h\right)}=\frac{-G{M}_{e}m}{R_e+2R_e}$
$\frac{-G{M}_{e}m}{3R_e}$
Increase in internal energy $\Delta U=U_f-U_i$
$=\frac{2}{3}\frac{G{M}_{e}m}{R_e}$
$\frac{2}{3}\frac{G{M}_e}{R_e^2}m{R}_e$
$=\frac{2}{3}mg{R}_e$

Answer 2

The potential energy at a distance \( R \) from the center of the Earth is given by: \[ U = -\frac{GM_em}{R} \] At the surface (\( R = R_e \)): \[ U_i = -\frac{GM_em}{R_e} \] At height \( h = 2R_e \) above the surface (\( R = 3R_e \)): \[ U_f = -\frac{GM_em}{3R_e} \] The change in potential energy (\( \Delta U \)) is: \[ \Delta U = U_f - U_i = \frac{2 G M_e m R_e}{3 R_e^2} = \frac{2}{3} m g R_e \]