Physics - Gravitation Question with Solution | TestHub

At the poles, the effect of Earth's rotation is negligible (latitude $\lambda =90^{\circ }$, ${\cos }^2\lambda =0$). So, the weight at the poles is $W=mg$.

At the equator, the apparent acceleration due to gravity ($g^{\prime }$) is affected by the Earth's rotation. The formula for apparent gravity at latitude $\lambda$ is $g^{\prime }=g-{\omega }^{2}R{\cos }^2\lambda$. At the equator, $\lambda =0^{\circ }$, so ${\cos }^2\lambda =1$. Thus, at the equator, $g^{\prime }=g-{\omega }^{2}R$. The apparent weight ($W^{\prime }$) will be $m{g}^{\prime }$.$$W^{\prime }=m(g-{\omega }^{2}R)=mg-m{\omega }^{2}R$$ Since $W=mg$, substituting this gives:$$W^{\prime }=W-m{\omega }^{2}R$$