Q90 / 160 · JIPMER 2013 · Physics

The escape velocity for the earth is $11.2 \mathrm{~km} / \mathrm{s}$. The mass of another planet 100 times mass of earth and its radius is 4 times radius of the earth. The escape velocity for the planet is

  • A
    $280 \mathrm{~km} / \mathrm{s}$
  • Correct answer
    B
    $56 \mathrm{~km} / \mathrm{s}$
  • C
    $112 \mathrm{~km} / \mathrm{s}$
  • D
    $28 \mathrm{~km} / \mathrm{s}$

Solution

The escape velocity of the the planet $\begin{aligned} v_{\text {escape }} & =\sqrt{\frac{2 G m}{R}} \propto \sqrt{\frac{M}{R}} \\ \frac{v_p}{v_e} & =\sqrt{\frac{M_p}{R_p} \times \frac{R_e}{M_e}} \\ \frac{v_p}{v_e} & =\sqrt{100 \times \frac{1}{4}}=5 \\ v_p & =5 \times 11.2=56 \mathrm{~km} / \mathrm{s}\end{aligned}$

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

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