# GRAVITATIONAL POTENTIAL ENERGY

When we assumed that the earth’s gravitational force on a body of mass m doesn’t depend on the body’s height. This led to the expression U = mgy. But Eq. (13.2), Fg = GmEm/r2 , shows that the gravitational force exerted by the earth (mass mE) does in general depend on the distance r from the body to the earth’s center. For problems in which a body can be far from the earth’s surface, we need a more general expression for gravitational potential energy

To find this expression, we follow the same steps as in Section 7.1. We consider a body of mass m outside the earth, and first compute the work Wgrav done by the gravitational force when the body moves directly away from or toward the center of the earth from r = r1 to r = r2, as in Fig. 13.10. This work is given by

where Fr is the radial component of the gravitational force F S —that is, the component in the direction outward from the center of the earth. Because F S points directly inward toward the center of the earth, Fr is negative. It differs from Eq. (13.2), the magnitude of the gravitational force, by a minus sign:

Substituting Eq. (13.7) into Eq. (13.6), we see that Wgrav is given by

The path doesn’t have to be a straight line; it could also be a curve like the one in Fig. 13.10. By an argument similar to that in Section 7.1, this work depends on only the initial and final values of r, not on the path taken. This also proves that the gravitational force is always conservative.

We now define the corresponding potential energy U so that Wgrav = U1 - U2, as in Eq. (7.3). Comparing this with Eq. (13.8), we see that the appropriate definition for gravitational potential energy is

Figure 13.11 shows how the gravitational potential energy depends on the distance r between the body of mass m and the center of the earth. When the body moves away from the earth, r increases, the gravitational force does negative work, and U increases (i.e., becomes less negative). When the body “falls” toward earth, r decreases, the gravitational work is positive, and the potential energy decreases (i.e., becomes more negative)

You may be troubled by Eq. (13.9) because it states that gravitational potential energy is always negative. But in fact you’ve seen negative values of U before. In using the formula U = mgy in Section 7.1, we found that U was negative whenever the body of mass m was at a value of y below the arbitrary height we chose to be y = 0—that is, whenever the body and the earth were closer together than some arbitrary distance. (See, for instance, Example 7.2 in Section 7.1.) In defining U by Eq. (13.9), we have chosen U to be zero when the body of mass m is infinitely far from the earth 1r = ∞2. As the body moves toward the earth, gravitational potential energy decreases and so becomes negative.

If we wanted, we could make U = 0 at the earth’s surface, where r = RE, by adding the quantity GmEm>RE to Eq. (13.9). This would make U positive when r 7 RE. We won’t do this for two reasons: One, it would complicate the expression for U; two, the added term would not affect the difference in potential energy between any two points, which is the only physically significant quantity. If the earth’s gravitational force on a body is the only force that does work, then the total mechanical energy of the system of the earth and body is constant, or conserved. In the following example we’ll use this principle to calculate escape speed, the speed required for a body to escape completely from a planet.

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Ans: We defined the weight of a body in Section 4.4 as the attractive gravitational force exerted on it by the earth. We can now broaden our definition and say that the weight of a body is the total gravitational force exerted on the body by all other bodies in the universe view more..
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Ans: gravitational forces are negligible between ordinary household-sized objects but very substantial between objects that are the size of stars. Indeed, gravitation is the most important force on the scale of planets, stars, and galaxies view more..
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Ans: HERE ARE SOME SOLVED EXAMPLES TO CLEAR YOUR CONCEPTS view more..
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Ans: When we first introduced gravitational potential energy in Section 7.1, we assumed that the earth’s gravitational force on a body of mass m doesn’t depend on the body’s height. This led to the expression U = mgy view more..
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Ans: As a final note, let’s show that when we are close to the earth’s surface, Eq. (13.9) reduces to the familiar U = mgy view more..
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Ans: Artificial satellites orbiting the earth are a familiar part of technology But how do they stay in orbit, and what determines the properties of their orbits? We can use Newton’s laws and the law of gravitation to provide the answers. In the next section we’ll analyze the motion of planets in the same way. view more..
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Ans: A circular orbit, like trajectory 4 in Fig. 13.14, is the simplest case. It is also an important case, since many artificial satellites have nearly circular orbits and the orbits of the planets around the sun are also fairly circular view more..
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Ans: The name planet comes from a Greek word meaning “wanderer,” and indeed the planets continuously change their positions in the sky relative to the background of stars. One of the great intellectual accomplishments of the 16th and 17th centuries was the threefold realization that the earth is also a planet, that all planets orbit the sun, and that the apparent motions of the planets as seen from the earth can be used to determine their orbits precisely view more..
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Ans: First consider the elliptical orbits described in Kepler’s first law. Figure 13.18 shows the geometry of an ellipse. The longest dimension is the major axis, with half-length a; this half-length is called the semi-major axis. view more..
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Ans: In a small time interval dt, the line from the sun S to the planet P turns through an angle du. The area swept out is the colored triangle with height r, base length r du, and area dA = 1 2 r2 du in . The rate at which area is swept out, view more..
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Ans: We have already derived Kepler’s third law for the particular case of circular orbits. Equation (13.12) shows that the period of a satellite or planet in a circular orbit is proportional to the 3 2 power of the orbit radius. view more..
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Ans: We have assumed that as a planet or comet orbits the sun, the sun remains absolutely stationary. This can’t be correct; because the sun exerts a gravitational force on the planet, the planet exerts a gravitational force on the sun of the same magnitude but opposite direction. In fact, both the sun and the planet orbit around their common center of mass view more..
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Ans: We have stated without proof that the gravitational interaction between two spherically symmetric mass distributions is the same as though all the mass of each were concentrated at its center. Now we’re ready to prove this statement. Newton searched for a proof for several years, and he delayed publication of the law of gravitation until he found one view more..
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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..
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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..
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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..
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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..
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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..

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