Gravitation - Result Question 27
28. A body of mass ’ $m$ ’ is taken from the earth’s surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be
[2013]
(a) $\frac{2}{3} mgR$
(b) $3 mgR$
(c) $\frac{1}{3} mgR$
(d) $mg 2 R$
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Answer:
Correct Answer: 28. (a)
Solution:
- (a) Initial P. E., $U_i=\frac{-GMm}{R}$,
Final P.E., $U_f=\frac{-GMm}{3 R}[\because R^{\prime}=R+2 R=3 R]$
$\therefore$ Change in potential energy,
$\Delta U=\frac{-GMm}{3 R}+\frac{GMm}{R}$
$=\frac{GMm}{R}(1-\frac{1}{3})=\frac{2}{3} \frac{GMm}{R}=\frac{2}{3} mgR$
$(\because \frac{GMm}{R}=mgR)$
$\Delta U=\frac{m g h}{1+\frac{h}{R}}$
By placing the value of $h=2 R$ we get
$\Delta U=\frac{2}{3} mgR$.
Work done against gravitational forces in taking a body of mass $m$ from the surface of earth to a height $h$ is the change in potential energy of the body and is given by
$ \Delta u=\frac{G M m h}{R(R+h)} $
(a) if $h \ll R$, then $W=\frac{G M m h}{R(R+h)}$
(b) if $h=R$, then $W=\frac{G M m}{2 R}=\frac{1}{2} m g h$
