Gravitation Ques 14
14. Imagine a new planet having the same density as that of earth but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is $g$ and that on the surface of the new planet is $g^{\prime}$, then
[2005]
(a) $g^{\prime}=g / 9$
(b) $g^{\prime}=27 $ $g$
(c) $g^{\prime}=9 $ $g$
(d) $g^{\prime}=3 $ $g$
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Answer:
Correct Answer: 14.(d)
Solution:
- (d) We know that
$g=\frac{G M}{R^{2}}=\frac{G(\frac{4}{3} \pi R^{3}) \rho}{R^{2}}=\frac{4}{3}$ $\pi G R \rho$
$g$ depends on radius of the planet
$\frac{g^{\prime}}{g}=\frac{R^{\prime}}{R}=\frac{3 R}{R}=3 \quad [$ As G & $\rho$ are constants]
$\therefore g^{\prime}=3 $ $g$
If the radius of a planet decreases by $n \%$, keeping its mass unchanged, the acceleration due to gravity on its surface increases by $2 n \%$
$ \begin{gathered} \frac{\Delta g}{g}=-\frac{2 \Delta R}{R} \\ \quad=-2(-n) \\ \quad=2 n \% \end{gathered} $
BETA
