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Consider earth to be a sphere of radius $R_e$ rotating about its own axis with angular speed $\omega$. If $g_{\mathrm{E}}$ and $g_{\mathrm{P}}$ are the accelerations due to gravity at the equator and the poles respectively, then $\left(g_{\mathrm{P}}-g_{\mathrm{E}}\right)$ is given by
$\left[\cos \left(0^{\circ}\right)=\sin \left(\frac{\pi}{2}\right)=1, \sin \left(0^{\circ}\right)=\cos \left(\frac{\pi}{2}\right)=0\right]$
Options:
$\left[\cos \left(0^{\circ}\right)=\sin \left(\frac{\pi}{2}\right)=1, \sin \left(0^{\circ}\right)=\cos \left(\frac{\pi}{2}\right)=0\right]$
Solution:
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Verified Answer
The correct answer is:
$R_{\mathrm{E}} \omega^2$
Acceleration due to gravity at a place of latitude $\lambda$ due to the rotation of earth is
$g^{\prime}=g-\mathrm{R}_{\mathrm{E}} \omega^2\left(\cos ^2 \lambda\right)$
At equator, $\lambda=0^{\circ}, \cos 0^{\circ}=1$
$\Rightarrow g_E=g-\mathrm{R}_{\mathrm{E}} \omega^2$
At poles, $\lambda=90^{\circ}, \cos (\lambda)=0$
$\begin{aligned} & \therefore g^{\prime}=g_{\mathrm{P}}=g \\ & \therefore g_{\mathrm{P}}-g_{\mathrm{E}}=g-\left(g-R_E \omega^2\right)=R_E \omega^2\end{aligned}$
$g^{\prime}=g-\mathrm{R}_{\mathrm{E}} \omega^2\left(\cos ^2 \lambda\right)$
At equator, $\lambda=0^{\circ}, \cos 0^{\circ}=1$
$\Rightarrow g_E=g-\mathrm{R}_{\mathrm{E}} \omega^2$
At poles, $\lambda=90^{\circ}, \cos (\lambda)=0$
$\begin{aligned} & \therefore g^{\prime}=g_{\mathrm{P}}=g \\ & \therefore g_{\mathrm{P}}-g_{\mathrm{E}}=g-\left(g-R_E \omega^2\right)=R_E \omega^2\end{aligned}$
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