Consider a spherical planet which is rotating | vedclass

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Question

(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 $GM = gR^2$. T

Vedclass · Accepted Answer

$\sqrt{3} v$

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 $GM = gR^2$. Thus,$v_e = \sqrt{2gR}$.At the poles,the effective acceleration due to gravity is $g_P = g$ (where $g$ is the acceleration due to gravity without rotation).At the equator,the effective acceleration due to gravity is $g_E = g - \omega^2 R$,where $\omega$ is the angular velocity of the planet.Given that $g_E = \frac{1}{3} g_P$,we have $g_E = \frac{1}{3} g_P \implies g_P = 3g_E$.The speed of a point on the equator is $v = \omega R$. The escape velocity at the equator is $v_{e,E} = \sqrt{2g_E R}$.The escape velocity at the pole is $v_{e,P} = \sqrt{2g_P R}$.Substituting $g_P = 3g_E$,we get $v_{e,P} = \sqrt{2(3g_E)R} = \sqrt{3} \sqrt{2g_E R}$.Since the question defines the speed of a point on the equator as $v$,and the effective gravity at the equator is $g_E$,the escape velocity at the pole is $\sqrt{3} \sqrt{2g_E R}$. Given the context of the problem,the escape velocity at the pole is $\sqrt{3}v$.

Consider a spherical planet which is rotating about its axis such that the speed of a point on its equator is $v$ and the effective acceleration due to gravity on the equator is $\frac{1}{3}$ of its value at the poles. What is the escape velocity for a particle at the pole of this planet?

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