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*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:*

* (1)*

*where G is Newton's constant, M_{Earth} is the mass of the Earth, and R_{Earth} 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:*

*F*_{grav} = *mg* (2)

*where g has the value GM_{Earth}/R_{Earth}^{2}= 9.8 m/s^{2}. We have made several idealizations, however, and if we want to calculate the scale reading, we need to be more careful.*

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*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 [1]).*

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*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%.*

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*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.*

If the man at the equator stood on a scale, how would the scale read compared to the scale reading for an identical man standing at the equator of a nonrotating Earth?