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(B) The acceleration due to gravity on the surface of the earth is given by $g = \frac{GM}{R^2}$.Substituting mass $M = \rho V = \rho \left( \frac{4}{

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$1/5$

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(B) The acceleration due to gravity on the surface of the earth is given by $g = \frac{GM}{R^2}$.Substituting mass $M = \rho V = \rho \left( \frac{4}{3} \pi R^3 \right)$,we get:$g = \frac{G \rho \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} G \pi \rho R$.Since $G$,$\pi$,and $\frac{4}{3}$ are constants,we have $g \propto \rho R$.To keep $g$ constant,the product $\rho R$ must remain constant.Let the initial density be $\rho_1$ and radius be $R_1$. Let the final density be $\rho_2$ and radius be $R_2 = 5R_1$.Then,$\rho_1 R_1 = \rho_2 R_2$.$\rho_1 R_1 = \rho_2 (5R_1)$.$\rho_2 = \frac{\rho_1}{5}$.Thus,the density must be changed by a factor of $1/5$.

If the radius of the earth is increased by a factor of $5$,by what factor must its density be changed to keep the value of $g$ the same?

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