Consider an object sitting on a scale at the surface of the Earth. The scale reading is the magnitude of the normal force which the scale exerts on the object. To a first approximation, there is force balance, and the magnitude of the scale's force is the magnitude of the gravitational force:
where G is Newton's constant, MEarth is the mass of the Earth, and REarth is the radius of the Earth. The simple result is that the force of gravity, and the reading of the scale, is proportional to the mass:
Fgrav = mg (2)
where g has the value GMEarth/REarth2= 9.8 m/s2. We have made several idealizations, however, and if we want to calculate the scale reading, we need to be more careful.
For example, we have ignored the rotation of the Earth. Consider a man standing on a scale at the equator. Because he is moving in a circle, there is a centripetal acceleration. The result is that the scale will not give a reading equal to the force of gravity (equation ).
We have also assumed that the Earth is a perfect sphere. Because it is rotating, the distance from the center of the Earth to the equator is greater than the distance from center to pole by about 0.1%.
A third effect we have ignored is that the Earth has local irregularities which make it necessary to measure g in the local laboratory, if we need an exact value of the effective acceleration due to gravity.
For a man standing at the equator of a rotating Earth, which expression gives the best expression of his velocity? (Let Tday be the time of one rotation, 1 day.)