A certain planet completes one rotation about | vedclass

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Question

(A) Due to the rotation of the planet,the effective acceleration due to gravity at latitude $\lambda$ is given by $g^{\prime} = g - \omega^2 R \cos^2

Vedclass · Accepted Answer

$\left(\frac{4-f}{1-f}\right) \cdot \frac{3 \pi}{4 G T^2}$

Vedclass · Answer

(A) Due to the rotation of the planet,the effective acceleration due to gravity at latitude $\lambda$ is given by $g^{\prime} = g - \omega^2 R \cos^2 \lambda$.At the equator,$\lambda = 0^{\circ}$,so $g_e = g - \omega^2 R$.At a latitude of $60^{\circ}$,$\lambda = 60^{\circ}$,so $g_{60} = g - \omega^2 R \cos^2(60^{\circ}) = g - \frac{\omega^2 R}{4}$.Given that the weight at the equator is $f$ times the weight at $60^{\circ}$,we have $g_e = f \cdot g_{60}$.Substituting the expressions: $g - \omega^2 R = f \left( g - \frac{\omega^2 R}{4} \right)$.Rearranging gives $g(1 - f) = \omega^2 R \left( 1 - \frac{f}{4} \right) = \frac{\omega^2 R}{4} (4 - f)$.Using $g = \frac{GM}{R^2}$,$M = \frac{4}{3} \pi R^3 \rho$,and $\omega = \frac{2\pi}{T}$,we substitute:$\frac{G}{R^2} \left( \frac{4}{3} \pi R^3 \rho \right) (1 - f) = \frac{4 \pi^2}{T^2} R \left( \frac{4 - f}{4} \right)$.$\frac{4}{3} \pi G R \rho (1 - f) = \frac{\pi^2 R}{T^2} (4 - f)$.Solving for density $\rho$: $\rho = \left( \frac{4 - f}{1 - f} \right) \cdot \frac{3 \pi}{4 G T^2}$.

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