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(D) The acceleration due to gravity $g$ on the surface of the earth is given by the formula: $g = \frac{GM}{R^2}$,where $G$ is the gravitational const

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increase by $4 \%$

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(D) The acceleration due to gravity $g$ on the surface of the earth is given by the formula: $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the earth,and $R$ is the radius of the earth.Since the mass $M$ remains constant,we have $g \propto \frac{1}{R^2}$.Taking the natural logarithm and differentiating,we get the relative error formula: $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$.Given that the radius shrinks by $2 \%$,we have $\frac{\Delta R}{R} = -0.02$.Substituting this into the error formula: $\frac{\Delta g}{g} = -2 	imes (-0.02) = 0.04$.To express this as a percentage,multiply by $100$: $\frac{\Delta g}{g} 	imes 100 = 4 \%$.Since the result is positive,the acceleration due to gravity will increase by $4 \%$.

If the radius of earth shrinks by $2 \%$ while its mass remains the same,the acceleration due to gravity on the earth's surface will approximately:

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