Centrifugal Force due to Rotation of the Earth

Although the Earth is usually treated as (approximately) an inertial frame, it is strictly non-inertial — it spins about its own axis with angular velocity $\omega=\dfrac{2\pi}{T}$ ($T\approx24$ hours $\approx86{,}400$ s, so $\omega\approx7.27\times10^{-5}\text{ rad s}^{-1}$). Any object resting on the Earth's surface therefore experiences a small outward centrifugal force, directed perpendicular to (and away from) the Earth's rotation axis.

For a person of mass $m$ standing at latitude $\varphi$, the perpendicular distance from the rotation axis is $r=R\cos\varphi$ ($R$ = Earth's radius), so the centrifugal force is

This is largest at the equator ($\varphi=0°$, $r=R$, maximum) and zero at the poles ($\varphi=90°$, $r=0$). For a $60\text{ kg}$ person at Chennai (latitude $\approx13°$):

$F_c\approx60\times(7.27\times10^{-5})^2\times6.4\times10^6\times\cos13°\approx1.97\text{ N}$ …