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(C) The acceleration due to gravity $g$ on the surface of the earth is given by $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is th

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Increase by $2\%$

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(C) The acceleration due to gravity $g$ on the surface of the earth is given by $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the earth,and $R$ is the radius of the earth.Taking the natural logarithm on both sides,we get $\ln g = \ln G + \ln M - 2 \ln R$.Differentiating both sides,we get the relative error formula: $\frac{\Delta g}{g} = -2 \left( \frac{\Delta R}{R} ight)$.Given that the radius shrinks by $1\%$,we have $\frac{\Delta R}{R} = -0.01$.Substituting this value,we get $\frac{\Delta g}{g} = -2 	imes (-0.01) = 0.02$.This represents an increase of $2\%$ in the acceleration due to gravity.

If the radius of the earth were to shrink by $1\%$ and its mass remained the same,the acceleration due to gravity on the earth's surface would

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