# Apparent weight and the earth’s rotation

Because the earth rotates on its axis, it is not precisely an inertial frame of reference. For this reason the apparent weight of a body on earth is not precisely equal to the earth’s gravitational attraction, which we will call the true weight w ^{→} 0 of the body. Figure 13.26 is a cutaway view of the earth, showing three observers. Each one holds a spring scale with a body of mass m hanging from it. Each scale applies a tension force F ^{→} to the body hanging from it, and the reading on each

scale is the magnitude F of this force. If the observers are unaware of the earth’s rotation, each one thinks that the scale reading equals the weight of the body because he thinks the body on his spring scale is in equilibrium. So each observer thinks that the tension F^{ →} must be opposed by an equal and opposite force w S , which we call the apparent weight. But if the bodies are rotating with the earth, they are not precisely in equilibrium. Our problem is to find the relationship between the apparent weight w S and the true weight w^{→}_{0}.

If we assume that the earth is spherically symmetric, then the true weight w^{→}_{0} has magnitude Gm_{E}m/R_{E}^{ 2} , where m_{E} and R_{E} are the mass and radius of the earth. This value is the same for all points on the earth’s surface. If the center of the earth can be taken as the origin of an inertial coordinate system, then the body at the north pole really is in equilibrium in an inertial system, and the reading on that observer’s spring scale is equal to w0. But the body at the equator is moving in a circle of radius R_{E} with speed v, and there must be a net inward force equal to the mass times the centripetal acceleration:

So the magnitude of the apparent weight (equal to the magnitude of F) is

If the earth were not rotating, the body when released would have a free-fall acceleration g_{0} = w_{0}/m. Since the earth is rotating, the falling body’s actual acceleration relative to the observer at the equator is g = w/m. Dividing Eq. (13.27) by m and using these relationships, we find

To evaluate v^{2}/R_{E}, we note that in 86,164 s a point on the equator moves a distance equal to the earth’s circumference, 2pR_{E} = 2p16.37 * 106 m^{2}. (The solar day, 86,400 s, is 1 365 longer than this because in one day the earth also completes 1 365 of its orbit around the sun.) Thus we find

So for a spherically symmetric earth the acceleration due to gravity should be about 0.03 m/s^{ 2} less at the equator than at the poles. At locations intermediate between the equator and the poles, the true weight w^{→} and the centripetal acceleration are not along the same line, and we need to write a vector equation corresponding to Eq. (13.27). From Fig. 13.26 we see that the appropriate equation is

The difference in the magnitudes of g and g_{0 }lies between zero and 0.0339 m/s^{2} . As Fig. 13.26 shows, the direction of the apparent weight differs from the direction toward the center of the earth by a small angle ß, which is 0.1° or less.

Table 13.1 gives the values of g at several locations. In addition to moderate variations with latitude, there are small variations due to elevation, differences in local density, and the earth’s deviation from perfect spherical symmetry.

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