The acceleration due to gravity at the surface of the planet is same as that at the surface of the earth, but the density of planet is thrice that of the earth. If 'R' is the radius of the earth, the radius of the planet will be

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Aicharya · Accepted Answer

$\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}=\frac{4}{3} \pi \mathrm{R} \rho \mathrm{G}$
As, $g$ is same on earth and the planet, $\begin{array}{ll} & \mathrm{R} \propto \frac{1}{\rho} \\ \therefore \quad & \frac{R_2}{R}=\frac{\rho}{3 \rho} \\ \therefore \quad & R_2=\frac{R}{3} \end{array}$

The acceleration due to gravity at the surface of the planet is same as that at the surface of the earth,…

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