Gravitation Ques 32
32. The potential energy of a satellite, having mass $m$ and rotating at a height of $6.4 \times 10^{6} $ $m$ above the earth surface, is
[2001]
(a) $-m g R_e$
(b) $-0.67 , m g R_e$
(c) $-0.5$ $ m g R_e$
(d) $-0.33$ $ m g R_e$
Show Answer
Answer:
Correct Answer: 32.(c)
Solution:
- (c) Mass of the satellite $=m$ and height of satellite from earth $(h)=6.4 \times 10^{6} m$.
We know that gravitational potential energy of the satellite at height $h$ is given by $U = -\frac{GMm}{R+h}$, where $G$ is the gravitational constant, $M$ is the mass of the Earth, $m$ is the mass of the satellite, $R$ is the radius of the Earth, and $h$ is the height above the Earth’s surface.
$GPE=-\frac{G M_e m}{R_e+h}=-\frac{g R_e^{2} m}{2(R_e+h)}$
$=-\frac{g R_e m}{2}=-0.5$ $ m g R_e$
(where, $G M_e=g R_e{ }^{2}$ and $h=R_e$ )
BETA
