The ratio of accelerations due to gravity g_{ | vedclass

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(C) The escape velocity $v_{e}$ from the surface of a planet is given by $v_{e} = \sqrt{\frac{2GM}{R}}$.Since $g = \frac{GM}{R^{2}}$, we have $v_{e} =

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$5:2\sqrt{2}$

Vedclass · Answer

(C) The escape velocity $v_{e}$ from the surface of a planet is given by $v_{e} = \sqrt{\frac{2GM}{R}}$.Since $g = \frac{GM}{R^{2}}$, we have $v_{e} = \sqrt{2gR}$.The acceleration due to gravity is $g = \frac{GM}{R^{2}} = \frac{G}{R^{2}} \cdot \frac{4}{3}\pi R^{3}\rho = \frac{4}{3}G\pi R\rho$.Thus, the radius $R$ is proportional to $\frac{g}{\rho}$, i.e.,$R \propto \frac{g}{\rho}$.Substituting this into the expression for $v_{e}$:$v_{e} = \sqrt{2g \cdot \left(\frac{3g}{4\pi G\rho}\right)} = \sqrt{\frac{3g^{2}}{2\pi G\rho}} \propto \frac{g}{\sqrt{\rho}}$.Given $\frac{g_{1}}{g_{2}} = \frac{5}{2}$ and $\frac{\rho_{1}}{\rho_{2}} = \frac{2}{1}$, the ratio of escape velocities is:$\frac{v_{1}}{v_{2}} = \frac{g_{1}}{g_{2}} \cdot \sqrt{\frac{\rho_{2}}{\rho_{1}}} = \frac{5}{2} \cdot \sqrt{\frac{1}{2}} = \frac{5}{2\sqrt{2}}$.Therefore, the ratio is $5:2\sqrt{2}$.

The ratio of accelerations due to gravity $g_{1}:g_{2}$ on the surfaces of two planets is $5:2$ and the ratio of their respective average densities $\rho_{1}:\rho_{2}$ is $2:1$. What is the ratio of respective escape velocities $v_{1}:v_{2}$ from the surface of the planets?

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