Question

Pertaining to two planets, the ratio of escape velocities from respective surfaces is $1:2$ , the ratio of the time period of the same simple pendulum at their respective surfaces is $2:1$ (in same order). Then the ratio of their average densities is

• A. $1:1$
• B. $1:2$
• C. $1:4$
• D. $8:1$

Answer 1

The correct option is C $1:4$

The escape velocity from a surface of a planet of mass $M$ , radius $R$ and mean density $\rho$ is given by,$$v_e=\sqrt{\frac{8G\pi R^2\rho }{3}}$$$\therefore v_e\propto R\sqrt{\rho }...\left(1\right)$

Time period of a simple pendulum is given by,
$$T=2\pi \sqrt{\frac{l}{g}}$$ As, $g=\frac{GM}{R^2}$ and $M=\frac{4}{3}\pi R^3\rho$

$\Rightarrow T=2\pi \sqrt{\frac{l}{\frac{G\left(\frac{4}{3}\pi R^3\rho \right)}{R^2}}}$

$\therefore T\propto \sqrt{\frac{1}{R\rho }}....\left(2\right)$

From equation $\left(1\right)$,

$\sqrt{\rho }\propto \frac{v_e}{R}........\left(3\right)$ and

from equation $\left(2\right)$,

$\rho \propto \frac{1}{R{T}^2}........\left(4\right)$

So, combining equation $\left(3\right)$ and $\left(4\right)$ we get,

$\sqrt{\rho }\propto \frac{1}{v_e{T}^2}$

$\Rightarrow \rho \propto \frac{1}{v_e^2{T}^4}$

$\Rightarrow \frac{{\rho }_1}{{\rho }_2}={\left[\frac{(v_e{)}_2}{(v_e{)}_1}\right]}^2\times {\left[\frac{T_2}{T_1}\right]}^4$

Given,

$\frac{(v_e{)}_1}{(v_e{)}_2}=\frac{1}{2}$ and $\frac{T_1}{T_2}=\frac{2}{1}$

$\Rightarrow \frac{{\rho }_1}{{\rho }_2}={\left[\frac{2}{1}\right]}^2\times {\left[\frac{1}{2}\right]}^4$

$\Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{1}{4}$

Hence, option (c) is the correct answer.

Why this question: To make students figure out dependence of escape velocity on various quantities like mass, radius, average density, time period, acceleration due to gravity etc.