A certain planet completes one rotation about | vedclass

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Question

(a)

Due to rotation of earth, acceleration due to gravity at latitude $\lambda$ is

$\quad g^{\prime}=g-\omega^2 R \cos ^2 \lambda$

$	ext { A

Vedclass · Accepted Answer

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

Vedclass · Answer

(a)

Due to rotation of earth, acceleration due to gravity at latitude $\lambda$ is

$\quad g^{\prime}=g-\omega^2 R \cos ^2 \lambda$

$	ext { At equator } \lambda=0^{\circ}$

$\Rightarrow \quad g_e=g-\omega^2 R$

$	ext { At latitude of } 60^{\circ}, \lambda=60^{\circ}$

$\Rightarrow \quad g_\lambda=g-\omega^2 R\left(\frac{1}{2}ight)^2$

$	ext { Given, }$

$	ext { Weight at equator }=f 	imes 	ext { Weight at }$

$\Rightarrow \quad m g_e=f 	imes m g_\lambda \Rightarrow g_e=f g_\lambda$

$\Rightarrow \quad g-\omega^2 R=f\left(g-\frac{\omega^2 R}{4}ight)$

$\Rightarrow \quad g(1-f)=\frac{\omega^2 R}{4}(4-f) \quad \dots(i)$

$	ext { Now, as } g=\frac{G M}{R^2} 	ext { and } \omega=\frac{2 \pi}{T}$

Also, $\quad M=\frac{4}{3} \pi R^3 \cdot ho$

Substituting these in Eq. $(i)$, we get

$\frac{4}{3} \pi ho G R(1-f)=\frac{4 \pi^2 R}{4 T^2}(4-f)$

$\Rightarrow \quad ho=\left(\frac{4-f}{1-f}ight) \cdot \frac{3 \pi}{4 G T^2}$

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