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(D) The acceleration due to gravity at the surface of a planet is given by $g = \frac{GM}{R^2}$.Since $M = V \rho = \frac{4}{3} \pi R^3 \rho$,we can w

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$2:1$

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(D) The acceleration due to gravity at the surface of a planet is given by $g = \frac{GM}{R^2}$.Since $M = V \rho = \frac{4}{3} \pi R^3 \rho$,we can write $g = \frac{G (\frac{4}{3} \pi R^3 \rho)}{R^2} = \frac{4}{3} \pi G \rho R$.Assuming the radius of the planet is the same as the Earth $(R_p = R_e)$ unless otherwise specified,or evaluating the ratio based on the given constants:Given $\rho_p = \rho_e$ and $G_p = 2G_e$.The ratio of acceleration due to gravity is $\frac{g_p}{g_e} = \frac{\frac{4}{3} \pi G_p \rho_p R_p}{\frac{4}{3} \pi G_e \rho_e R_e}$.Assuming $R_p = R_e$,we get $\frac{g_p}{g_e} = \frac{G_p}{G_e} = \frac{2G_e}{G_e} = 2$.Thus,the ratio is $2:1$.

$A$ planet has the same density as that of the Earth,and the universal gravitational constant $G$ is twice that of the Earth. The ratio of the acceleration due to gravity on the planet to that on the Earth is:

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