Variation Of G With Rotation

An object resting at the equator moves in a circle of radius $R$ as the Earth rotates, so part of the true gravitational pull must supply the required centripetal acceleration ${\omega }^{2}R$. The remaining force determines the apparent weight, giving $g_{eff}=g-{\omega }^{2}R{\cos }^2\lambda$, which is largest at the poles ($\lambda =90^{\circ }$, no rotation effect) and smallest at the equator ($\lambda =0$). The equatorial reduction is therefore $\Delta g={\omega }^{2}R=(7.27\times 10^{-5}{)}^2\times 6.37\times 10^6=5.29\times 10^{-9}\times 6.37\times 10^6=0.034$ m/s^2. The option $0.34$ m/s^2 is ten times too large from a power-of-ten slip. The option $0.0034$ m/s^2 is ten times too small. The option $0.068$ m/s^2 wrongly doubles the centripetal term. The cosine-squared dependence on latitude $\lambda$ means the effect is strongest at the equator and disappears entirely at the poles, where the rotation radius shrinks to zero and a point on the axis simply spins in place. This rotational reduction is separate from, and adds to, the larger difference caused by the Earth's slightly flattened, oblate shape, which also makes the polar radius smaller. Both effects together make true poles-to-equator weight variation larger than rotation alone. This reproduces the NCERT explanation of why $g$ varies with latitude. A magnitude check confirms the change is about 0.3% of $g$, consistent with the small measured equator-to-pole difference.