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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Aicharya · Accepted Answer

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

The escape velocity from earth surface is $11 \mathrm{~km} / \mathrm{s}$. The escape velocity from a planet…

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