Earth is assumed to be a sphere of radius R. | vedclass

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Question

(A) The effective acceleration due to gravity at a latitude $\phi$ is given by the formula: $g_{\phi} = g_{0} - R \omega^2 \cos^2 \phi$,where $g_{0}$

Vedclass · Accepted Answer

$\frac{1}{4} \omega^2 R$

Vedclass · Answer

(A) The effective acceleration due to gravity at a latitude $\phi$ is given by the formula: $g_{\phi} = g_{0} - R \omega^2 \cos^2 \phi$,where $g_{0}$ is the acceleration due to gravity at the poles (ignoring rotation). At the equator,$\phi = 0^{\circ}$,so $g = g_{0} - R \omega^2 \cos^2 0^{\circ} = g_{0} - R \omega^2$. At latitude $\phi = 30^{\circ}$,$g_{\phi} = g_{0} - R \omega^2 \cos^2 30^{\circ} = g_{0} - R \omega^2 \left(\frac{\sqrt{3}}{2}\right)^2 = g_{0} - \frac{3}{4} R \omega^2$. Now,calculating the difference $|g - g_{\phi}|$: $|g - g_{\phi}| = |(g_{0} - R \omega^2) - (g_{0} - \frac{3}{4} R \omega^2)|$ $|g - g_{\phi}| = |-\frac{1}{4} R \omega^2| = \frac{1}{4} R \omega^2$.

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