Q. Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the ratio of densities of earth ${(\rho}_{e})$ and moon ${(\rho}_{m})$ is $\left( \frac{\rho_{e}}{\rho_{m}} \right) = \frac{5}{3}$ then radius of moon $(R_{m})$ in terms of $R_{e}$ will be
A) $\frac{5}{18}R_{e}$
B) $\frac{1}{6}R_{e}$
C) $\frac{3}{18}R_{e}$
D) $\frac{1}{2\sqrt{3}}R_{e}$
Explanation
(a) $g = \frac{4}{3}\pi G\rho R \Rightarrow g \propto \rho R \Rightarrow \frac{g_{e}}{g_{m}} = \frac{\rho_{e}}{\rho_{m}} \times \frac{R_{e}}{R_{m}}$
$\Rightarrow \frac{6}{1} = \frac{5}{3} \times \frac{R_{e}}{R_{m}} \Rightarrow R_{m} = \frac{5}{18}R_{e}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.