Question 341: Gravitation

Question

Consider a planet whose density is same as that of the earth but whose radius is three times the radius '$$R$$' of the earth. The acceleration due to gravity '$$\mathrm{g}_{\mathrm{n}}$$' on the surface of planet is $$\mathrm{g}_{\mathrm{n}}=\mathrm{x}$$. $$\mathrm{g}$$ where $$\mathrm{g}$$ is acceleration due to gravity on surface of earth. The value of '$$\mathrm{x}$$' is

Accepted Answer

$$\begin{array}{ll} 
& 	ext { As, } g=\frac{G M}{R^2} 	ext { and } M=ho V \
	herefore \quad & g=\frac{G ho V}{R^2}=\frac{G ho \frac{4}{3} \pi R^3}{R^2} \
	herefore \quad & g \propto R \
& 	ext { For the planet: Radius } R=3 R \
	herefore \quad & g_{	ext {planet }}=\frac{G ho V_{	ext {planet }}}{(3 R)^2} \
& 	ext { where } V_{	ext {planet }}=\frac{4}{3} \pi(3 R)^3 \
	herefore \quad & g_{	ext {planet }}=\frac{G ho \frac{4}{3} \pi(3 R)^3}{(3 R)^2} \
	herefore \quad & g_{	ext {planet }} \propto 3 R \
	herefore \quad & \frac{g}{g_{	ext {planet }}}=\frac{R}{3 R} \quad 	ext{....(since $ho$ is constant)}\
	herefore \quad & g_{	ext {planet }}=3 g \
	herefore & x=3
\end{array}$$

Answer

9

Answer

3

Answer

$$\frac{1}{3}$$

Answer

$$\frac{1}{9}$$

Gravitation

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