Gravitation Question 7
Question 7 - 25 January - Shift 2
A body of mass is taken from earth surface to the height $h$ equal to twice the radius of earth $(R_e)$, the increase in potential energy will be : ( $g=$ acceleration due to gravity on the surface of Earth)
(1) $3 mgR_e$
(2) $\frac{1}{3} mgR_e$
(3) $\frac{2}{3} mgR_e$
(4) $\frac{1}{2} mgR_e$
Show Answer
Answer: (3)
Solution:
Formula: Gravitational Potential
$U=\frac{-GM_e m}{r}$
$U_i=\frac{-GM_e m}{R_e}$
$U_f=\frac{-GM_e m}{(R_e+h)}=\frac{-GM_e m}{R_e+2 R_e}$
$-GM_e m$
$3 R_e$
Increase in internal energy $\Delta U=U_f-U_i$
$=\frac{2}{3} \frac{GM_e m}{R_e}$
$\frac{2}{3} \frac{GM_e}{R_e^{2}} mR_e$
$=\frac{2}{3} mgR_e$
