Question

The density of newly discovered planet is twice that of the earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is It the radius of the plane would be

• A. $2R$
• B. $4R$
• C. $\frac{1}{4}R$
• D. $\frac{1}{2}R$

Answer 1

The correct option is D $\frac{1}{2}R$

Acceleration due to gravity $g=\frac{GM}{R^2}$
According to the question, acceleration due to gravity is same on both planets.
$\therefore$ $\frac{G{M}_p}{R_p^2}=\frac{G{M}_e}{R_e^2}$
$\Rightarrow \frac{G\times \frac{4}{3}\pi R_p^3{\rho }_p}{R_p^2}=\frac{G\times \frac{4}{3}\pi R_e^3{\rho }_e}{R_e^2}$
We get $R_p{\rho }_p=R_e{\rho }_e$
Given : ${\rho }_p=2{\rho }_e$
$\therefore$ $R_p\times 2{\rho }_e=R_e{\rho }_e$
$⟹R_p=\frac{R_e}{2}=\frac{R}{2}$