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, 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,
where g has the value GM earth /R 2 earth = 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.
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)]
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.
(i) For a man standing at the equator of a rotating Earth, which expression gives the best expression of his velocity?
(Let T day be the time of one rotation, 1 day)
Text Solution
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Ans.
(i)
Sol. Each day the man travels a distance given by the circumference of the Earth C = 2 π R Earth
(ii)
Sol. In order to calculate the centripetal force F cent = ma cent . W e need the mass of the man and the centripetal acceleration. We can find the centripetal acceleration a cent = v 2 /R Earth , if we know his velocity and the radius of the Earth. It seems that D is the correct answer. The problem with choice D is that we can calculate the velocity once we know the period and radius (as in the previous problem).
(iii)
Sol. Since the man is traveling in a circle at constant speed, his acceleration vector points toward the center of rotation, and so does the net force vector. For this to be so, the magnitude of the gravitational force must be greater than the magnitude of the force of the ground.
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