Magnitude of force

Relevant wiki: Centrifugal Force

The person standing on the Earth is acted on by three forces in the non–inertial reference frame rotating with the Earth: the person is attracted with the force of gravity towards the Earth center, the person is lifted up by the centrifugal force (these two forces constitute the person’s weight) and the person is pushed up by the ground upon which he or she stands as a reaction to his or her weight.

The person is at rest with respect to the rotating Earth. The resultant of these forces is thus zero.

$\vec{F}_g + \vec{N} +\vec{F}_{cf} = \vec{o}$

The magnitudes of the forces are:

$F_g-N-F_{cf} =0$

So,

$F_g=N+F_{cf}$

If the person “took off” it would mean that the person would lose contact with the Earth and it would stop pushing him or her up. The magnitude of the force of gravity would be equal to the magnitude of the centrifugal force in that case. Let us calculate the magnitudes of both these forces.

Magnitude of the centrifugal force:

$F_{cf} =m\dfrac{v^2}{R} = mω^2 R= 80 \times(7.29 \cdot 10^{-5})^2 \times 6378 \times103N \approx 2.7N$

Magnitude of the force of gravity:

$F_g=G\dfrac{mM_E}{R^2} =6.67 \times10^{-11} \times 80 \times 5.97 \times 1024(6378 \times 103)2N \approx 783.1N$

Comparing the magnitudes of these two forces, we see that the person cannot fly away by the effect of the centrifugal force.