The value of acceleration due to gravity g is | vedclass

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(A) The acceleration due to gravity $g^{\prime}$ at a latitude $\lambda$ on the surface of the Earth is given by the formula:$g^{\prime} = g - \omega^

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poles

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(A) The acceleration due to gravity $g^{\prime}$ at a latitude $\lambda$ on the surface of the Earth is given by the formula:$g^{\prime} = g - \omega^2 R_e \cos^2 \lambda$where $g$ is the acceleration due to gravity at the poles,$\omega$ is the angular velocity of the Earth,and $R_e$ is the radius of the Earth.At the poles,the latitude $\lambda = 90^{\circ}$.Since $\cos 90^{\circ} = 0$,the expression becomes:$g^{\prime} = g - \omega^2 R_e (0)^2 = g$At the equator,$\lambda = 0^{\circ}$,so $\cos 0^{\circ} = 1$,which gives $g^{\prime} = g - \omega^2 R_e$,which is the minimum value.Therefore,the value of acceleration due to gravity is maximum at the poles.

The value of acceleration due to gravity $g$ is maximum at

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