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(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g - \omega^2 R \cos^2 \lambda$,where $\omega

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Decrease at the equator but remain unchanged at the poles

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(B) The effective acceleration due to gravity $g'$ at a latitude $\lambda$ is given by the formula: $g' = g - \omega^2 R \cos^2 \lambda$,where $\omega$ is the angular velocity of the Earth and $R$ is the radius of the Earth.At the equator,the latitude $\lambda = 0^o$,so $\cos 0^o = 1$. Thus,$g' = g - \omega^2 R$. If the Earth rotates faster,$\omega$ increases,which leads to a decrease in the effective gravity $g'$,and consequently,the weight of an object decreases at the equator.At the poles,the latitude $\lambda = 90^o$,so $\cos 90^o = 0$. Thus,$g' = g$. Since the term involving $\omega$ becomes zero,the weight of an object remains unchanged at the poles regardless of the speed of rotation.

If the earth rotates faster than its present speed,the weight of an object will

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