Question

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 plants ?

• A. $5 : 2$
• B. $\sqrt{5}:\sqrt{2}$
• C. $5:2\sqrt{2}$
• D. $25 : 4$

Answer 1

Answer: (C)

$\because$ Escape velocity from a planet,
$v_{e} =\sqrt{\frac{2 G M}{R}}=\sqrt{\frac{2 G M}{R^{2}}} R $
$=\sqrt{2 g R} \ldots$(i)
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 $ Radius $R=\frac{3 g}{4 \pi G \rho} \ldots$(ii)
From Bqs. (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, $v_{e} \propto \frac{g}{\sqrt{\rho}}$
$\therefore \frac{v_{a_{1}}}{v_{\epsilon_{2}}}= \frac{g_{1}}{\sqrt{\rho_{1}}} \times \frac{\sqrt{\rho_{2}}}{g_{2}}=\frac{5}{2} \times \frac{1}{\sqrt{2}} $
(given, $ \left.\frac{g_{1}}{g_{2}}=\frac{5}{2} \text{ and} \frac{\rho_{1}}{\rho_{2}}=2: 1\right) $
$= \frac{5}{2 \sqrt{2}}$