Variation Of G With Latitude And Rotation

NCERT Class 11, Chapter 8 (Gravitation) explains that on a rotating Earth a portion of the true gravitational pull must be diverted to provide the centripetal force for a stationary observer who is carried around in a daily circle of latitude. The apparent acceleration due to gravity is therefore reduced to $g^{\prime }=g-{\omega }^{2}R{\cos }^2\lambda$, where $\lambda$ is the latitude, $\omega$ is the Earth's angular speed of rotation, and $R$ is its radius. The reduction term is largest when ${\cos }^2\lambda$ is maximum, that is at the equator where $\lambda =0^{\circ }$ and the observer moves in the widest circle at the greatest linear speed. At the poles, where $\lambda =90^{\circ }$, $\cos \lambda =0$, so rotation has no effect on apparent gravity there at all. At $45^{\circ }$ latitude the effect is only intermediate and not the greatest. It is therefore certainly not uniform everywhere on the surface. As a plausibility check, points on the equator trace the largest daily circle and so require the most centripetal force, making the rotational reduction in measured gravity greatest there, exactly matching the ${\cos }^2\lambda$ dependence in the formula and explaining why the measured value of $g$ is slightly smaller at the equator than at the poles.