Gravitation - Result Question 24
25. Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass $m$, when taken to a height $h$ from the surface of earth (of radius $R$ ), is given by,
[NEET Odisha 2019]
(a) $\frac{G M m}{R+h}$
(b) $-\frac{G M m}{R+h}$
(c) $\frac{G M m h}{R(R+h)}$
(d) $m g h$
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Answer:
Correct Answer: 25. (c)
Solution:
- (c) As $U=-\frac{G M m}{r}$
$\therefore \quad(P . E) _{A}=-\frac{G M m}{R} ;(P . E) _{B}=-\frac{G M m}{R+h}$
$\therefore \Delta U=(P . E) _{B}-(P . E) _{A}$
$=-\frac{G M m}{R+h}+\frac{G M m}{R}=\frac{G M m h}{(R)(R+h)}$
