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(D) The escape velocity is given by $V_e = \sqrt{\frac{2GM}{R}}$.Since $M = \rho \cdot \frac{4}{3}\pi R^3$,we have $V_e = \sqrt{\frac{2G \rho \frac{4}

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$\frac{3}{4}$

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(D) The escape velocity is given by $V_e = \sqrt{\frac{2GM}{R}}$.Since $M = \rho \cdot \frac{4}{3}\pi R^3$,we have $V_e = \sqrt{\frac{2G \rho \frac{4}{3}\pi R^3}{R}} = \sqrt{\frac{8}{3}G\pi\rho} \cdot R$.Thus,$V_e \propto \rho^{1/2} R$.Given $\frac{V_{eA}}{V_{eB}} = \frac{1}{2}$ and $\frac{R_A}{R_B} = \frac{1}{3}$.From $\frac{V_{eA}}{V_{eB}} = \sqrt{\frac{\rho_A}{\rho_B}} \cdot \frac{R_A}{R_B}$,we get $\frac{1}{2} = \sqrt{\frac{\rho_A}{\rho_B}} \cdot \frac{1}{3}$,which implies $\sqrt{\frac{\rho_A}{\rho_B}} = \frac{3}{2}$,so $\frac{\rho_A}{\rho_B} = \frac{9}{4}$.The acceleration due to gravity is $g = \frac{GM}{R^2} = \frac{G(\rho \cdot \frac{4}{3}\pi R^3)}{R^2} = \frac{4}{3}G\pi\rho R$.Therefore,$\frac{g_A}{g_B} = \frac{\rho_A R_A}{\rho_B R_B} = \left(\frac{\rho_A}{\rho_B}\right) \left(\frac{R_A}{R_B}\right) = \left(\frac{9}{4}\right) \left(\frac{1}{3}\right) = \frac{3}{4}$.

The escape velocities of two planets $A$ and $B$ are in the ratio $1:2$. If the ratio of their radii respectively is $1:3$,then the ratio of acceleration due to gravity of planet $A$ to the acceleration due to gravity of planet $B$ will be:

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