# satellites: Circular orbits

A circular orbit 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. The only force acting on a satellite in circular orbit around the earth is the earth’s gravitational attraction, which is directed toward the center of the earth and hence toward the center of the orbit (Fig. 13.15). As we discussed in Section 5.4, this means that the satellite is in uniform circular motion and its speed is constant. The satellite isn’t falling toward the earth; rather, it’s constantly falling around the earth. In a circular orbit the speed is just right to keep the distance from the satellite to the center of the earth constant.

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Let’s see how to find the constant speed v of a satellite in a circular orbit. The radius of the orbit is r, measured from the center of the earth; the acceleration of the satellite has magnitude arad = v2/r and is always directed toward the center of the circle. By the law of gravitation, the net force (gravitational force) on the satellite of mass m has magnitude Fg = GmEm/r2 and is in the same direction as the acceleration. Newton’s second law then tells us that

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Solving this for v, we find

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This relationship shows that we can’t choose the orbit radius r and the speed v independently; for a given radius r, the speed v for a circular orbit is determined

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The satellite’s mass m doesn’t appear in Eq. (13.10), which shows that the motion of a satellite does not depend on its mass. An astronaut on board an orbiting space station is herself a satellite of the earth, held by the earth’s gravity in the same orbit as the station. The astronaut has the same velocity and acceleration as the station, so nothing is pushing her against the station’s floor or walls. She is in a state of apparent weightlessness, as in a freely falling elevator; see the discussion following Example 5.9 in Section 5.2. (True weightlessness would occur only if the astronaut were infinitely far from any other masses, so that the gravitational force on her would be zero.) Indeed, every part of her body is apparently weightless; she feels nothing pushing her stomach against her intestines or her head against her shoulders (Fig. 13.16).

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Apparent weightlessness is not just a feature of circular orbits; it occurs whenever gravity is the only force acting on a spacecraft. Hence it occurs for orbits of any shape, including open orbits such as trajectories 6 and 7 in Fig. 13.14. We can derive a relationship between the radius r of a circular orbit and the period T, the time for one revolution. The speed v is the distance 2pr traveled in one revolution, divided by the period:

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We solve Eq. (13.11) for T and substitute v from Eq. (13.10)

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Equations (13.10) and (13.12) show that larger orbits correspond to slower speeds and longer periods. As an example, the International Space Station (Fig. 13.13) orbits 6800 km from the center of the earth (400 km above the earth’s surface) with an orbital speed of 7.7 km/s and an orbital period of 93 min. The moon orbits the earth in a much larger orbit of radius 384,000 km, and so has a much slower orbital speed (1.0 km/s) and a much longer orbital period (27.3 days).

It’s interesting to compare Eq. (13.10) to the calculation of escape speed in Example 13.5. We see that the escape speed from a spherical body with radius R is (2)1/2 times greater than the speed of a satellite in a circular orbit at that radius. If our spacecraft is in circular orbit around any planet, we have to multiply our speed by a factor of 21/2 to escape to infinity, regardless of the planet’s mass

Since the speed v in a circular orbit is determined by Eq. (13.10) for a given orbit radius r, the total mechanical energy E = K + U is determined as well. Using Eqs. (13.9) and (13.10), we have

The total mechanical energy in a circular orbit is negative and equal to one-half the potential energy. Increasing the orbit radius r means increasing the mechanical energy (that is, making E less negative). If the satellite is in a relatively low orbit that encounters the outer fringes of earth’s atmosphere, mechanical energy decreases due to negative work done by the force of air resistance; as a result, the orbit radius decreases until the satellite hits the ground or burns up in the atmosphere.

We have talked mostly about earth satellites, but we can apply the same analysis to the circular motion of any body under its gravitational attraction to a stationary body. Figure 13.17 shows an example.

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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: 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: 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: 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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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..
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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..
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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..

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