Question

Suppose that the acceleration of a free fall at the surface of a distant planet was found to be equal to that at the surface of the earth. If the diameter of the planet were twice the diameter of the earth, then the ratio of mean density of the planet to that of the earth would be:

• A. $4:1$
• B. $2:1$
• C. $1:1$
• D. $1:2$

Answer 1

The correct option is D $1:2$

Acceleration of a free fall=$g=\frac{GM}{R^2}=\frac{G\left(\rho \frac{4}{3}\pi R^3\right)}{R^2}\propto \rho R$

Acceleration at both earth's surface and planet's surface is same.

$⟹{\rho }_e{R}_e={\rho }_p{R}_p$

$⟹\frac{{\rho }_p}{{\rho }_e}=\frac{R_e}{R_p}=\frac{d_e}{d_p}=1:2$