(C) Variation in g with Latitude and Rotation of the Earth

So far g has been treated as though the Earth were a perfect, non-rotating sphere. In reality the Earth rotates about its polar axis (west to east) with angular velocity $\omega$, and every point on its surface except the two poles themselves is carried along a circular path (parallel to the equator) by this rotation, which requires a centripetal force -- and that force is SUPPLIED by part of the gravitational attraction, reducing the portion of gravity actually felt as weight.

Latitude $\theta$ is defined as the angle between the radius vector to a surface point P and the equatorial plane, ranging from $0^\circ$ at the equator to $90^\circ$ at the poles. As the Earth spins, point P traces a horizontal circle of radius $r=R\cos\theta$ about a centre $O'$ on the polar axis (Fig. 5.9), with centripetal acceleration $a=r\omega^2=R\omega^2\cos\theta$ directed along $PO'$. Only the COMPONENT of this centripetal acceleration directed along PO (i.e. towards the Earth's true centre) actually subtracts from gravity; resolving along PO gives $a_r=a\cos\theta=R\omega^2\cos^2\theta$. The effective (measured) acceleration due to gravity at latitude $\theta$ is therefore $g'=g-R\omega^2\cos^2\theta.$

At the EQUATOR ($\theta=0^\circ$, $\cos\theta=1$), the reduction is maximum: $g'=g-R\omega^2$. Using $R=6.4\times10^6$ m and $\omega=2\pi/T$ with $T=24$ hours (the Earth's rotation period), $R\omega^2\approx0.0339$ m/s^2 -- a small but measurable reduction. At the POLES ($\theta=90^\circ$, $\cos\theta=0$), there is NO reduction at all ($g'=g$), because points exactly on the rotation axis do not move in a circle and feel no centripetal effect whatsoever. Between these extremes, $g'$ increases smoothly and monotonically from equator to pole, exactly as tabulated in Table 5.2. …