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(C) The effective acceleration due to gravity at the equator (where $	heta = 0^\circ$) is given by $g' = g - \omega^2 R$.Given that the weight become

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$0.63 	imes 10^{-3} \ rad/s$

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(C) The effective acceleration due to gravity at the equator (where $	heta = 0^\circ$) is given by $g' = g - \omega^2 R$.Given that the weight becomes $\frac{3}{4} W$,the effective gravity must be $g' = \frac{3}{4} g$.Substituting this into the equation: $\frac{3}{4} g = g - \omega^2 R$.Rearranging for $\omega^2 R$: $\omega^2 R = g - \frac{3}{4} g = \frac{1}{4} g$.Solving for $\omega$: $\omega = \sqrt{\frac{g}{4R}}$.Given $g = 10 \ m/s^2$ and $R = 6400 \ km = 6.4 	imes 10^6 \ m$.$\omega = \sqrt{\frac{10}{4 	imes 6.4 	imes 10^6}} = \sqrt{\frac{10}{25.6 	imes 10^6}} = \sqrt{\frac{1}{2.56 	imes 10^6}} = \frac{1}{1.6 	imes 10^3} \ rad/s$.$\omega = 0.625 	imes 10^{-3} \ rad/s \approx 0.63 	imes 10^{-3} \ rad/s$.

If the Earth has no rotational motion,the weight of a person on the equator is $W$. Determine the speed with which the Earth would have to rotate about its axis so that the person at the equator will weigh $\frac{3}{4} W$. The radius of the Earth is $6400 \ km$ and $g = 10 \ m/s^2$.

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