Variation of g Due to Rotation of the Earth

The Earth spins on its own axis once approximately every $24$ hours, and this rotation provides a third reason -- entirely separate from altitude or depth -- why the acceleration due to gravity actually measured (or felt as weight) at different points on the Earth's surface is not perfectly uniform.

Consider an object of mass $m$ resting on the Earth's surface at latitude $\lambda$ (measured from the equator, so $\lambda = 0$ at the equator and $\lambda = 90^\circ$ at either pole). Because the Earth is rotating, this object is not really at rest at all -- it is moving in a horizontal circle of radius $R\cos\lambda$ about the Earth's rotation axis (this radius shrinks steadily from a maximum of $R$ at the equator down to zero exactly at the poles, where a point on the axis itself does not move in a circle at all). Moving in this circle at the Earth's own angular speed $\omega = 2\pi/T$ (with $T \approx 24\ \text{h}$) requires a net inward (centripetal) force of magnitude $m\omega^2(R\cos\lambda)$, directed horizontally toward the rotation axis.

This required centripetal force is supplied by a small part of the true gravitational pull $GMm/R^2 = mg$ -- so the force the ground actually needs to supply upward on the object (its measured, apparent weight, $mg'$) is slightly LESS than the full gravitational pull $mg$, by exactly the component of that required centripetal force that acts along the local vertical (the radial direction at that latitude). Resolving the horizontal centripetal requirement $m\omega^2 R\cos\lambda$ along the local vertical introduces one further factor of $\cos\lambda$, giving

Two limiting cases make the physical meaning of this formula clear:

• At the equator ($\lambda = 0^\circ$, so $\cos\lambda = 1$): $g' = g - \omega^2 R$, the MAXIMUM possible reduction, since the full radius $R$ is doing the swinging and the entire centripetal requirement acts straight along the local vertical.
• At either pole ($\lambda = 90^\circ$, so $\cos\lambda = 0$): $g' = g$, i.e. NO reduction at all -- a point exactly on the Earth's rotation axis does not move in a circle as the Earth spins, and so needs no centripetal force whatsoever. …