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R is the radius of Earth and $\omega$ is its angular velocity and $g_p$ is the value of $g$ at the poles. The effective value of $g$ at a latitude $\lambda=60^{\circ}$.
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Verified Answer
The correct answer is:
$g_p-\frac{1}{4} R \omega^2$
The acceleration due to gravity at a latitude $\lambda$ is
$\mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R} \omega^2 \cos ^2 \lambda$
At the poles, $\lambda=90^{\circ}$
$\therefore \quad \mathrm{g}^{\prime}=\mathrm{g}=\mathrm{g}_{\mathrm{p}}$
For $\quad \lambda=60^{\circ}$,
$\mathrm{g}^{\prime}=\mathrm{g}_{\mathrm{p}}-\mathrm{R} \omega^2 \cos ^2 \lambda$
or $\quad g^{\prime}=g_p-R \omega^2 \cos ^2 60^{\circ}$
$=g_p-R \omega^2\left(\frac{1}{2}\right)^2=g_p-\frac{1}{4} R \omega^2$
$\mathrm{g}^{\prime}=\mathrm{g}-\mathrm{R} \omega^2 \cos ^2 \lambda$
At the poles, $\lambda=90^{\circ}$
$\therefore \quad \mathrm{g}^{\prime}=\mathrm{g}=\mathrm{g}_{\mathrm{p}}$
For $\quad \lambda=60^{\circ}$,
$\mathrm{g}^{\prime}=\mathrm{g}_{\mathrm{p}}-\mathrm{R} \omega^2 \cos ^2 \lambda$
or $\quad g^{\prime}=g_p-R \omega^2 \cos ^2 60^{\circ}$
$=g_p-R \omega^2\left(\frac{1}{2}\right)^2=g_p-\frac{1}{4} R \omega^2$
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