The radius of planet is twice the radius of the earth. Both have almost equal average mass densities. If '…

The radius of 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
  1. $\mathrm{V}_{\mathrm{E}}=1.5 \mathrm{~V}_{\mathrm{P}}$
  2. $\mathrm{V}_{\mathrm{P}}=1.5 \mathrm{~V}_{\mathrm{E}}$
  3. $\mathrm{V}_{\mathrm{P}}=2 \mathrm{~V}_{\mathrm{E}}$
  4. $\mathrm{V}_{\mathrm{E}}=3 \mathrm{~V}_{\mathrm{P}}$

Solution

$\begin{aligned} & \text { Escape velocity } v=\sqrt{\frac{2 G M}{R}} \\ & M=\frac{4}{3} \pi R^3 \cdot \rho \\ & \therefore v=\sqrt{\frac{2 G \times \frac{4}{3} \pi R^3 \cdot \rho}{R}}=\sqrt{\frac{8 G}{3} \pi R^2 \rho}=R \sqrt{\frac{8 \pi G}{3} \rho} \\ & \therefore v \propto R \sqrt{\rho} \\ & \therefore \text { If } \rho \text { is constant, the } v \propto R \\ & \therefore \frac{v_P}{v_E}=\frac{R_P}{R_E}=2 \\ & \therefore v_P=2 v_E\end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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