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 $
$\left(\because R_{e}=R\right)$