Question

$g$ is the acceleration due to gravity and $K$ is the rotational kinetic energy of the earth. If the earth's radius increases by $2\%$ keeping mass constant, then the value of decrement in $g$ and $K$ is $\ldots \ldots \%$

Answer 1

Acceleration due to gravity is $g=\frac{G M}{R^{2}}$
and if $L$ be the angular momentum of the earth, then rotational $KE=\frac{L^{2}}{2 I}$
where, $I$ be the moment of inertia.
So, $I=\frac{2}{5}MR^{2}$ (for sphere)
$\therefore $ Rotational $KE=\frac{5 L^{2}}{4 M R^{2}}$
since, angular momentum remains conserved.
So, rotational $KE\left(K\right) \propto \frac{1}{R^{2}}$
$\therefore $ Both $g$ and $K$ are $ \propto R^{- 2}$
$\therefore \frac{\Delta g}{g}=\frac{\Delta K}{K}=-2\times \frac{\Delta R}{R}$
$\therefore $ Both $g$ and $K$ would decrease by $2\times 2\%=4\%$