# Kepler's second Law

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Figure 13.19 shows Kepler’s second law. 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 Fig. 13.19b. The rate at which area is swept out, dA/dt, is called the sector velocity:

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The essence of Kepler’s second law is that the sector velocity has the same value at all points in the orbit. When the planet is close to the sun, r is small and du/dt is large; when the planet is far from the sun, r is large and du/dt is small.

To see how Kepler’s second law follows from Newton’s laws, we express dA/dt in terms of the velocity vector v S of the planet P. The component of v S perpendicular to the radial line is v. = vsinØ. From Fig. 13.19b the displacement along the direction of v. during time dt is r du, so we also have v# = r du/dt. Using this relationship in Eq. (13.14), we find

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Now rvsinØ is the magnitude of the vector product r X S : v X S , which in turn is 1/m times the angular momentum L X S = r X S : mv S of the planet with respect to the sun. So we have

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Thus Kepler’s second law—that sector velocity is constant—means that angular momentum is constant!

It is easy to see why the angular momentum of the planet must be constant. According to Eq. (10.26), the rate of change of L S equals the torque of the gravitational force F S acting on the planet:

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In our situation, r S is the vector from the sun to the planet, and the force F → is directed from the planet to the sun (Fig. 13.20). So these vectors always lie along the same line, and their vector product r S : F S is zero. Hence dL → /dt = 0. This conclusion does not depend on the 1/r2 behavior of the force; angular momentum is conserved for any force that acts always along the line joining the particle to a fixed point. Such a force is called a central force. (Kepler’s first and third laws are valid for a 1/r2 force only.)

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Conservation of angular momentum also explains why the orbit lies in a plane. The vector L S = r S : mv S is always perpendicular to the plane of the vectors r S and v S ; since L S is constant in magnitude and direction, r → and v → always lie in the same plane, which is just the plane of the planet’s orbit.

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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: 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: 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: 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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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..
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Ans: HERE ISA SUMMARY OF GRAVITATION , FOR QUICK REVISION view more..
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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..

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