Question 340: Gravitation

Question

The radius of a planet is twice the radius of the earth. Both have almost equal average mass densities. If '$$V_P$$' and '$$V_E$$' are escape velocities of the planet and the earth respectively, then

Accepted Answer

Escape velocity $$\mathrm{v}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$$
$$M=\frac{4}{3} \pi R^3 \cdot ho$$
$$\begin{aligned}
& 	herefore v=\sqrt{\frac{2 G 	imes \frac{4}{3} R^3 \cdot ho}{R}}=\sqrt{\frac{8 G}{3} R^2 ho}=R \sqrt{\frac{8 G}{3} ho} \
& 	herefore v \propto R \sqrt{ho}
\end{aligned}$$
$$	herefore$$ If $$ho$$ is constant, the $$\mathrm{v} \propto \mathrm{R}$$
$$\begin{aligned}
& 	herefore \frac{v_P}{v_E}=\frac{R_P}{R_E}=2 \
& 	herefore v_p=2 v_E
\end{aligned}$$

Answer

$$\mathrm{V}_{\mathrm{E}}=1.5 \mathrm{~V}_{\mathrm{P}}$$

Answer

$$\mathrm{V}_{\mathrm{P}}=1.5 \mathrm{~V}_{\mathrm{E}}$$

Answer

$$\mathrm{V_P=2 V_E}$$

Answer

$$\mathrm{V_E=3 V_P}$$

Gravitation

The radius of a planet is twice the radius of the earth. Both have almost equal average mass densities. If '$V_P' and 'V_E$' are escape velocities of the planet and the earth respectively, then

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