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 $\dots \dots \%$

Answer 1

Answer: 4

Acceleration due to gravity is $g=\frac{GM}{R^2}$
and if $L$ be the angular momentum of the earth, then rotational $KE=\frac{L^2}{2I}$
where, $I$ be the moment of inertia.
So, $I=\frac{2}{5}M{R}^2$ (for sphere)
$\therefore$ Rotational $KE=\frac{5L^2}{4M{R}^2}$
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 $2\times 2\%=4\%$