Question 291: Gravitation

Question

The escape velocity from earth surface is $11 \mathrm{~km} / \mathrm{s}$. The escape velocity from a planet having twice the radius and same mean density as earth is

Accepted Answer

Escape velocity is given by,
$$\begin{aligned}
v_e & =\sqrt{\frac{2 G M}{R}} \
& =\sqrt{\frac{2 G}{R} 	imes \frac{4}{3} \pi R^3 ho}=\sqrt{\frac{8 G}{3} \pi R^2 ho}=\sqrt{\frac{8 G \pi ho}{3}} 	imes R
\end{aligned}$$
As the planets have the same density,
$$\begin{aligned}
& \mathrm{v}_{\mathrm{e}} \propto \mathrm{R} \
& \frac{\mathrm{v}_{\mathrm{e}}^{\prime}}{\mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\frac{2 \mathrm{R}}{\mathrm{R}}=2 \
	herefore \quad & \mathrm{v}_{\mathrm{e}}^{\prime}=2 \mathrm{v}_{\mathrm{e}}=2 	imes 11=22 \mathrm{~km} / \mathrm{s}
\end{aligned}$$

Answer

22 km/s

Answer

11 km/s

Answer

5.5 km/s

Answer

15.5 km/s

Gravitation

The escape velocity from earth surface is 11 \mathrm{~km} / \mathrm{s}. The escape velocity from a planet having twice the radius and same mean density as earth is

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