If the diameter of earth becomes half keeping mass to be constant, then the acceleration due to gravity at surface of earth becomes

Answer :

Solution :

$g=\frac{GM}{R^2}$

$g\propto <!--\; \propto \; -->\frac{1}{R^2}$

$\frac{g^{\prime }}{g}={\left(\frac{R}{\frac{R}{2}}\right)}^2=4$