Q103 / 150 · MHT CET 2021 (23 Sep Shift 2) · Physics

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

  • A
    $\mathrm{V}_{\mathrm{E}}=1.5 \mathrm{~V}_{\mathrm{P}}$
  • B
    $\mathrm{V}_{\mathrm{P}}=1.5 \mathrm{~V}_{\mathrm{E}}$
  • Correct answer
    C
    $\mathrm{V}_{\mathrm{P}}=2 \mathrm{~V}_{\mathrm{E}}$
  • D
    $\mathrm{V}_{\mathrm{E}}=3 \mathrm{~V}_{\mathrm{P}}$

Solution

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

Hence, the correct answer is option: 'C'.

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