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Unformatted text preview: These lecture notes were prepared for Rutgers Physics 341/342: Principles of Astrophysics by Prof. Chuck Keeton, and modified by Profs. Saurabh Jha and Eric Gawiser. All rights reserved. c 2011 Lecture 11: Tidal Forces (This discussion draws on § 19.2 of Carroll & Ostlie.) I. General Picture In the onebody and twobody problems we used point masses. What about the sizes of the bodies involved in gravitational motion? • source of gravity: we argued before that its size does not matter, if it is spherically symmetric • target of gravity: size does matter, because gravity pulls harder on one side than on the other The situation is depicted below. For definiteness, let me refer to the source of gravity as a star and the target as a planet, although we will consider a variety of situations. Since the force of gravity scales as 1 /r 2 , the side of the planet that faces the star (the “front”) feels a stronger force than the side of the planet away from the star (the “back”). This is shown in panel (a). What is more interesting is the force relative to the center of the planet. To find this, we subtract the force at the center, and obtain the differential force shown in panel (b). Figure 1: General picture of the tidal force. (a) The force of gravity from an object off to the right. (b) The differential force relative to the center. (From Carroll & Ostlie Fig. 19.4.) 1 Figure 2: Geometry of the tidal force. (From Carroll & Ostlie Fig. 19.3.) The differential force pulls “up” (relative to the planet’s surface) near the equator, and “down” near the poles. So we can think of it as squeezing the planet. If the planet is a little bit squishy, it will respond by flattening out a little bit. Now consider the gravity felt by Earth from our Moon. There is plenty of squishy stuff on the surface of Earth: the water in the oceans. The differential force pulls the oceans up on the “front” and “back” of Earth — thereby creating the ocean tides. Hence, the effect we are considering is referred to as the tidal force . This was another of Newton’s preoccupations. Now we can analyze the effect quantitatively. II. Derivation of the Tidal Force The geometry needed to analyze the tidal force is shown above. (In this analysis I work in the plane containing the center of Earth and the North and South poles. You can obtain everything else by symmetry, by rotating around the line from the Earth to the Moon.) Consider a test mass m located at radius R and latitude θ . Let r be the distance from the center of Earth to the Moon. It is helpful to complete the triangle by defining s to be the length of the third side, and φ to be the other angle. These are important because s is the distance from the test mass to the Moon, so it will enter the gravitational force; while φ indicates the direction of the force....
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This note was uploaded on 10/20/2011 for the course PH 341 taught by Professor Gawiser during the Fall '11 term at Rutgers.
 Fall '11
 Gawiser
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