Question

The density of a newly discovered planet is twice that of 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 $R$, the radius of the planet would be

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

Answer 1

Answer: (D)

From equation of acceleration due to gravity.
$g_e=\frac{G{M}_e}{R_e^2}=\frac{G(4/3)\pi R_e^3}{R_e^2}{\rho }_e$
$g_e\propto R_e{\rho }_e$
acceleration due to gravity of planet $g_p\propto R_p{\rho }_p$
$R_e{\rho }_e=R_p{\rho }_p$
$\Rightarrow R_e{\rho }_e=R_{p}2{\rho }_e$
$\Rightarrow R_p=\frac{1}{2}R$
$\left(\because R_e=R\right)$