Question

Acceleration due to gravity on moon is $\frac{1}{6}$ of the acceleration due to gravity on earth. If the ratio of densities of earth $\left({\rho }_m\right)$ and moon $\left({\rho }_e\right)$ 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. $\frac{1}{2\sqrt{3}}{R}_e$

Answer 1

The correct option is A

$\frac{5}{18}{R}_e$

Acceleration due to gravity $g=\frac{4}{3}\pi \rho GR\therefore g\propto \rho Ror\frac{g_m}{g_e}=\frac{{\rho }_m}{{\rho }_e}.\frac{R_m}{R_e}[As\frac{g_m}{g_e}=\frac{1}{6}and\frac{{\rho }_e}{{\rho }_e}=\frac{5}{3}(given\left)\right]$

$\therefore \frac{R_m}{R_e}=\left(\frac{g_m}{g_e}\right)\left(\frac{{\rho }_e}{{\rho }_m}\right)=\frac{1}{6}\times \frac{5}{3}\therefore R_m=\frac{5}{18}{R}_e$