# 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 w0.

If we assume that the earth is spherically symmetric, then the true weight w0 has magnitude GmEm/RE 2 , where mE and RE 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 RE 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 g0 = w0/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 v2/RE, we note that in 86,164 s a point on the equator moves a distance equal to the earth’s circumference, 2pRE = 2p16.37 * 106 m2. (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 g0 lies between zero and 0.0339 m/s2 . 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.

+
Ans: We assumed at the beginning that the point mass m was outside the spherical shell, so our proof is valid only when m is outside a spherically symmetric mass distribution. view more..
+
Ans: Any spherically symmetric mass distribution can be thought of as a combination of concentric spherical shells. Because of the principle of superposition of forces, what is true of one shell is also true of the combination. So we have proved half of what we set out to prove: that the gravitational interaction between any spherically symmetric mass distribution and a point mass is the same as though all the mass of the spherically symmetric distribution were concentrated at its center. view more..
+
Ans: We start by considering a ring on the surface of a shell , centered on the line from the center of the shell to m. We do this because all of the particles that make up the ring are the same distance s from the point mass m. view more..
+
Ans: 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. view more..
+
Ans: In 1916 Albert Einstein presented his general theory of relativity, which included a new concept of the nature of gravitation. In his theory, a massive object actually changes the geometry of the space around it view more..
+
Ans: Think first about the properties of our own sun. Its mass M = 1.99 * 1030 kg and radius R = 6.96 * 108 m are much larger than those of any planet, but compared to other stars, our sun is not exceptionally massive view more..
+
Ans: The first expression for escape speed in Eq. (13.29) suggests that a body of mass M will act as a black hole if its radius R is less than or equal to a certain critical radius. view more..
+
Ans: At points far from a black hole, its gravitational effects are the same as those of any normal body with the same mass. If the sun collapsed to form a black hole, the orbits of the planets would be unaffected. But things get dramatically different close to the black hole. view more..
+
Ans: If light cannot escape from a black hole and if black holes are small . how can we know that such things exist? The answer is that any gas or dust near the black hole tends to be pulled into an accretion disk that swirls around and into the black hole, rather like a whirlpool view more..
+
Ans: HERE ISA SUMMARY OF GRAVITATION , FOR QUICK REVISION view more..
+
Ans: Many kinds of motion repeat themselves over and over: the vibration of a quartz crystal in a watch, the swinging pendulum of a grandfather clock, the sound vibrations produced by a clarinet or an organ pipe, and the back-and-forth motion of the pistons in a car engine. This kind of motion, called periodic motion or oscillation view more..
+
Ans: n. A body with mass m rests on a frictionless horizontal guide system, such as a linear air track, so it can move along the x-axis only. The body is attached to a spring of negligible mass that can be either stretched or compressed. The left end of the spring is held fixed, and the right end is attached to the body. The spring force is the only horizontal force acting on the body; the vertical normal and gravitational forces always add to zero view more..
+
Ans: Here are some terms that we’ll use in discussing periodic motions of all kinds: view more..
+
Ans: The simplest kind of oscillation occurs when the restoring force Fx is directly proportional to the displacement from equilibrium x. This happens if the spring in Figs. 14.1 and 14.2 is an ideal one that obeys Hooke’s law view more..
+
Ans: To explore the properties of simple harmonic motion, we must express the displacement x of the oscillating body as a function of time, x1t2. view more..
+
Ans: the period and frequency of simple harmonic motion are completely determined by the mass m and the force constant k. In simple harmonic motion the period and frequency do not depend on the amplitude A. view more..
+
Ans: We still need to find the displacement x as a function of time for a harmonic oscillator. Equation (14.4) for a body in SHM along the x-axis is identical to Eq. (14.8) for the x-coordinate of the reference point in uniform circular motion with constant angular speed v = 2k/m view more..
+
Ans: We can learn even more about simple harmonic motion by using energy considerations. The only horizontal force on the body in SHM in Figs. 14.2 and 14.13 is the conservative force exerted by an ideal spring. The vertical forces do no work, so the total mechanical energy of the system is conserved. We also assume that the mass of the spring itself is negligible. view more..

Rating - 3/5
466 views