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

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(C) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$. If the mass $M$ is constant,then $g \propto R^{-2}$.Using the error formula,$\fr

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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 $g = \frac{GM}{R^2}$. If the mass $M$ is constant,then $g \propto R^{-2}$.Using the error formula,$\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$.Given $\frac{\Delta R}{R} = -2\% = -0.02$,we get $\frac{\Delta g}{g} = -2(-0.02) = 0.04 = 4\%$. Thus,$g$ increases by $4\%$.The rotational kinetic energy is $K = \frac{L^2}{2I}$,where $I = \frac{2}{5}MR^2$ is the moment of inertia. Since angular momentum $L$ is conserved,$K = \frac{L^2}{2(\frac{2}{5}MR^2)} \propto R^{-2}$.Using the error formula,$\frac{\Delta K}{K} = -2 \frac{\Delta R}{R}$.Given $\frac{\Delta R}{R} = -0.02$,we get $\frac{\Delta K}{K} = -2(-0.02) = 0.04 = 4\%$. 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 $2\%$ while keeping all other quantities the same,then:

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