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(C) The escape velocity $v_e$ of a planet is given by the formula: $v_e = \sqrt{\frac{2GM}{R}}$.Since the mass $M$ of a planet can be expressed in ter

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$2$

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(C) The escape velocity $v_e$ of a planet is given by the formula: $v_e = \sqrt{\frac{2GM}{R}}$.Since the mass $M$ of a planet can be expressed in terms of its density $\rho$ and radius $R$ as $M = \frac{4}{3}\pi R^3 \rho$,we substitute this into the escape velocity formula:$v_e = \sqrt{\frac{2G}{R} \cdot \frac{4}{3}\pi R^3 \rho} = \sqrt{\frac{8}{3}G\pi\rho R^2} = R \sqrt{\frac{8}{3}G\pi\rho}$.Given that the densities $\rho$ are the same for both the planet and the earth,the escape velocity is directly proportional to the radius: $v_e \propto R$.Therefore,the ratio of the escape velocities is: $\frac{V_p}{V_E} = \frac{R_p}{R_E}$.Given that the radius of the planet is double that of the earth $(R_p = 2R_E)$,we have:$\frac{V_p}{V_E} = \frac{2R_E}{R_E} = 2$.Thus,the value of $x$ is $2$.

The radius of the planet is double that of the earth,but their average densities are same. $V_{p}$ and $V_E$ are the escape velocities of planet and earth respectively. If $\frac{V_p}{V_E}=x$,the value of '$x$' is

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