Gravitation Ques 15
15. The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$, the radius of the planet would be
[2004]
(a) $1 / 2 $ $R$
(b) $2 $ $R$
(c) $4 $ $R$
(d) $1 / 4 $ $R$
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Answer:
Correct Answer: 15.(a)
Solution:
- (a) $g=\frac{G M}{R^{2}}$. Also, $M=\rho \times \frac{4}{3} $ $\pi R^{3}$
$\therefore \quad g=G \frac{4}{3} \rho \pi R$.
At the surface of planet, $g_p=\frac{4}{3} G(2 \rho) \pi R^{\prime}$,
At the surface of the earth $g_e=\frac{4}{3} G \rho \pi R$
$g_e=g_p \Rightarrow \rho R=2 \rho R^{\prime} \Rightarrow R^{\prime}=R / 2$
BETA
