Q131 / 155 · WBJEE 2018 · Physics

The ratio of accelerations due to gravity $g_{1}:g_{2}$ on the surfaces of two planets is 5: 2 and the ratio of their respective average densities $\rho_{1}: \rho_{2}$ is 2: 1 . What is the ratio of respective escape velocities $v_{1}: v_{2}$ from the surface of the planets?

  • A
    5: 2
  • B
    $\sqrt{5}: \sqrt{2}$
  • Correct answer
    C
    $5: 2 \sqrt{2}$
  • D
    25.4

Solution

Escape velocity from a planet. $v_{0}=\sqrt{\frac{2 G M}{R}}=\sqrt{\frac{2 G M}{R^{2}}} R$ $$ =\sqrt{2 g R} $$ According due to gravity $g=\frac{G M}{R^{2}}=\frac{G\frac{4}{3}\pi R^{3} \cdot \rho}{R^{2}}=\frac{4}{3} G\pi R_{p}$ $\therefore \quad$ Radius $R=\frac{3 g}{4 \pi G \rho}$ From Eqs. (i) and (ii), we get $v_{e}=\sqrt{2 g \cdot \frac{3 g}{4 \pi G \rho}}=\sqrt{\frac{3}{2} \frac{g^{2}}{\pi G \rho}}$ Thus, $\quad v_{e} \propto \frac{g}{\sqrt{\rho}}$ $\therefore \quad \frac{v_{n}}{v_{r_{2}}}=\frac{g_{1}}{\sqrt{\rho_{1}}} \times \frac{\sqrt{\rho_{2}}}{g_{2}}=\frac{5}{2} \times \frac{1}{\sqrt{2}}$ (given, $\frac{g_{1}}{g_{2}}=\frac{5}{2}$ and $\frac{Q}{\rho_{2}}=2: 1$ ) $=\frac{5}{2 \sqrt{2}}$

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

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