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,
$v_e=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2GM}{R^2}}R$
$=\sqrt{2gR}\dots$(i)
According due to gravity
$g=\frac{GM}{R^2}=\frac{G\frac{4}{3}\pi R^3\cdot \rho }{R^2}=\frac{4}{3}G\pi Rp$
$\therefore$ Radius $R=\frac{3g}{4\pi G\rho }\dots$(ii)
From Bqs. (i) and (ii), we get
$v_e=\sqrt{2g\cdot \frac{3g}{4\pi G\rho }}=\sqrt{\frac{3}{2}\frac{g^2}{\pi G\rho }}$
Thus, $v_e\propto \frac{g}{\sqrt{\rho }}$
$\therefore \frac{v_{a_1}}{v_{{ϵ}_2}}=\frac{g_1}{\sqrt{{\rho }_1}}\times \frac{\sqrt{{\rho }_2}}{g_2}=\frac{5}{2}\times \frac{1}{\sqrt{2}}$
(given, $\frac{g_1}{g_2}=\frac{5}{2}\text{ and}\frac{{\rho }_1}{{\rho }_2}=2:1)$
$=\frac{5}{2\sqrt{2}}$