Physics - Gravitation Question with Solution | TestHub

The acceleration due to gravity on the surface of a planet is given by $g=\frac{GM}{R^2}$.

Substituting mass $M=\frac{4}{3}\pi R^3\rho$, we get:

$g=\frac{G}{R^2}\left(\frac{4}{3}\pi R^3\rho \right)=\frac{4}{3}\pi GR\rho$

This implies $g\propto R\rho$.

Given that the acceleration due to gravity on the planet is the same as that on the earth ($g_p=g_e$), we have:

$R_p{\rho }_p=R_e{\rho }_e$

It is given that the density of the planet is thrice that of the earth, so ${\rho }_p=3{\rho }_e$.

$R_p(3{\rho }_e)=R{\rho }_e$

$R_p=\frac{R}{3}$

Answer: $\frac{\mathrm{R}}{3}$