Gravitation - Result Question 18
18. 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
(a) $g^{\prime}=g / 9$
(b) $g^{\prime}=27 g$
(c) $g^{\prime}=9 g$
(d) $g^{\prime}=3 g$
[2005]
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Answer:
Correct Answer: 18. (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
$\frac{g^{\prime}}{g}=\frac{R^{\prime}}{R}=\frac{3 R}{R}=3[$ 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} $
