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)]

$\therefore \frac{R_m}{R_e}=(\frac{g_m}{g_e})(\frac{{\rho }_e}{{\rho }_m})$

$=\frac{1}{6}\times \frac{5}{3}$

$\therefore R_m=\frac{5}{18}{R}_e$