Let g be the acceleration due to gravity at t | vedclass

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(C) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$. Since $G$ and $M$ are constant,$g \propto R^{-2}$. Taking the derivative,$\frac{

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$g$ increases by $4 \%$ and $K$ increases by $4 \%$

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(C) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$. Since $G$ and $M$ are constant,$g \propto R^{-2}$. Taking the derivative,$\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$. Given $\frac{\Delta R}{R} = -2 \%$,we get $\frac{\Delta g}{g} = -2(-2 \%) = +4 \%$. Thus,$g$ increases by $4 \%$.The rotational kinetic energy is $K = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$,where $L$ is the angular momentum. Assuming $L$ remains constant,$K \propto \frac{1}{I}$. Since $I = \frac{2}{5} MR^2$,we have $I \propto R^2$. Therefore,$K \propto R^{-2}$. Taking the derivative,$\frac{\Delta K}{K} = -2 \frac{\Delta R}{R}$. Given $\frac{\Delta R}{R} = -2 \%$,we get $\frac{\Delta K}{K} = -2(-2 \%) = +4 \%$. Thus,$K$ increases by $4 \%$. Therefore,both $g$ and $K$ increase by $4 \%$.

Let $g$ be the acceleration due to gravity at the Earth's surface and $K$ be the rotational kinetic energy of the Earth. Suppose the Earth's radius decreases by $2 \%$ keeping its mass the same,then:

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