The radius and mean density of the planet are four times as that of the earth. The ratio of escape velocity…

The radius and mean density of the planet are four times as that of the earth. The ratio of escape velocity at the earth to the escape velocity at a planet is
  1. $1: \sqrt{8}$
  2. $1: 8$
  3. $1: \sqrt{3}$
  4. $1: 3$

Solution

$\begin{aligned} \mathrm{v}_{\mathrm{e}} & =\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ \therefore \quad \mathrm{v}_{\mathrm{e}} & =\mathrm{R} \sqrt{\frac{8}{3} \pi \mathrm{G} \rho} \quad \ldots .\left(\because \mathrm{M}=\frac{4}{3} \pi \mathrm{R}^3 \rho\right)\end{aligned}$ $\begin{aligned} & \text { Also, } \rho_{\mathrm{p}}=4 \rho_{\mathrm{E}} \text { and } \mathrm{R}_{\mathrm{p}}=4 \mathrm{R}_{\mathrm{E}} \\ & \frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{p}}}=\frac{\mathrm{R}_{\mathrm{E}} \sqrt{\rho_{\mathrm{E}}}}{4 \mathrm{R}_{\mathrm{E}} \sqrt{4 \rho_{\mathrm{E}}}}=\frac{1}{8}=1: 8\end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 1)

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