Physics - Gravitation Question with Solution | TestHub

$\because$ acceleration due to gravity on surface is given by $\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}_{\mathrm{e}}^2}$ Now since diameter is reduced to $1/3^{\text{rd }}$, radius also reduces to $1/3^{\text{rd }}$, keeping mass constant New value of acceleration due to gravity on Earth's surface is ${\mathrm{g}}^{\prime }=\frac{\mathrm{GM}}{{\left(\frac{{\mathrm{R}}_{\mathrm{e}}}{3}\right)}^2}=9\frac{\mathrm{GMe}}{{\mathrm{R}}_{\mathrm{e}}^2}=9\mathrm{g}$