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

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increases.

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(D) The acceleration due to gravity $g$ at a latitude $\phi$ is given by the formula: $g_{\phi} = g - \omega^2 R \cos^2 \phi$,where $g$ is the acceleration due to gravity at the poles,$\omega$ is the angular velocity of the earth,and $R$ is the radius of the earth.At the equator,$\phi = 0^{\circ}$,so $\cos 0^{\circ} = 1$,which gives $g_{eq} = g - \omega^2 R$ (minimum value).At the poles,$\phi = 90^{\circ}$,so $\cos 90^{\circ} = 0$,which gives $g_{pole} = g$ (maximum value).Additionally,the earth is an oblate spheroid,meaning the radius at the equator is greater than the radius at the poles $(R_{eq} > R_{pole})$.Since $g = \frac{GM}{R^2}$,a smaller radius at the poles results in a higher value of $g$.Therefore,as we move from the equator to the poles,the value of acceleration due to gravity increases.

As we go from the equator of the earth to the pole of the earth,the value of acceleration due to gravity

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