Q133 / 150 · MHT CET 2022 (07 Aug Shift 1) · Physics

The escape velocity for a planet whose mass is six times the mass of earth and the radius is twice the radius of earth will be $\left[\mathrm{V}_{\mathrm{e}}=\right.$ escape velocity from the earth $]$

  • A
    $\sqrt{2} \mathrm{~V}_{\mathrm{e}}$
  • B
    $\frac{1}{2} \mathrm{~V}_{\mathrm{e}}$
  • Correct answer
    C
    $\sqrt{3} \mathrm{~V}_{\mathrm{e}}$
  • D
    $2 \sqrt{2} \mathrm{~V}_{\mathrm{e}}$

Solution

Escape velocity is defined using relation: $\begin{aligned} & \frac{1}{2} \mathrm{mv}_{\mathrm{e}}^2=\frac{\mathrm{GMm}}{\mathrm{R}} \\ & \Rightarrow \mathrm{V}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\end{aligned}$ If $M$ is replaced $6 M \& R$ is replaced by $2 R$ then $\mathrm{V}_{\mathrm{e}}^{\prime}=\sqrt{3} \mathrm{~V}_{\mathrm{e}}$

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

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