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Gravitational acceleration on the surface of a planet is $\frac{\sqrt 6}{11}g$ , where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $\frac{2}{3}\, times$ that of the earth. If the escape speed on the surface of the earth is taken to be $11\, kms^{-1}$, the escape speed on the surface of the planet in $kms^{-1}$ will be
$2$
$3$
$4$
$6$
Solution
$\mathrm{g}_{\mathrm{p}}=\frac{\mathrm{GM}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}^{2}}=\frac{4}{3} \mathrm{G} \pi \mathrm{R}_{\mathrm{p}} \rho_{\mathrm{p}}$
$\Rightarrow \frac{g_{p}}{g_{e}}=\frac{R_{p} p_{p}}{R_{e} p_{e}}$
Also, $\mathrm{v}_{\mathrm{e}}=\sqrt{2 \mathrm{g} \mathrm{R}}$
$\Rightarrow \frac{\mathrm{v}_{\mathrm{p}}}{\mathrm{v}_{\mathrm{e}}}=\sqrt{\frac{\mathrm{g}_{\mathrm{p}} \mathrm{R}_{\mathrm{p}}}{\mathrm{g}_{\mathrm{e}} \mathrm{R}_{\mathrm{e}}}}=\left(\frac{\mathrm{g}_{\mathrm{p}}}{\mathrm{g}_{\mathrm{e}}}\right) \sqrt{\frac{\rho_{\mathrm{e}}}{\rho_{\mathrm{p}}}}=\frac{\sqrt{6}}{11} \times \sqrt{\frac{3}{2}}$
$\Rightarrow \mathrm{v}_{\mathrm{p}}=3 \mathrm{km} / \mathrm{s}$
