Physics - Gravitation Question with Solution | TestHub

Due to rotation of earth the effective acceleration due to gravity$g^{\prime }=g-R{\omega }^2{\mathrm{c}\mathrm{o}\mathrm{s}}^2\lambda .$

For a given point on the surface of earth$g$decreases as$\omega$increases. The angular speed of earth is maximum at equator hence, the value of$g$on the surface of the earth is smallest.