Question

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 $(\frac{\rho_e}{\rho_m}) = \frac{5}{3}$ 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. $\frac{7}{18}R_e$
• D. $- \frac{1}{2\sqrt{3}} R_e$

Answer 1

Answer: (A)

Acceleration due to gravity $g = \frac{4}{3} \pi \rho GR$
$\therefore g \propto \rho R$
or $\frac{g_m}{g_e} = \frac{\rho_m}{\rho_e} .\frac{R_m}{R_e}$
[As $\frac{g_m}{g_e} = \frac{1}{6} $ and $\frac{\rho_e}{\rho_m} = \frac{5}{3}$ (given)]
$\therefore \frac{R_m}{R_e} = (\frac{g_m}{g_e})(\frac{\rho_e}{\rho_m})$
$ = \frac{1}{6} \times \frac{5}{3} $
$\therefore R_m = \frac{5}{18} R_e$