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Location of Lecture 28 Material in the 7th Edition of your Textbook = Torque: Pages 248-250; l F (right hand rule for vector product) I = Moment of Inertia: pages 261-262; I is a scalar L = Angular Momentum: pages 271-274; L I Vector Nature of: , L, , . page 237, Fig. 8.17 for and . For see page 276, Eq. 9.1 and problem 9.9 on page 279. Newton’s 2nd law for rotational motion; Eq. 9.7 on page 262. Updated Kepler’s Laws 1. The Law of Orbits: All planets move in elliptical orbits, with the Sun at one of the foci. 2. The Law of Areas: A line that connects a planet to the Sun sweeps out equal areas in the plane of the planet’s orbit in equal times: that is the rate dA / dt constant L / 2mp L = angular momentum of the planet mp= mass of the planet L=I1ω1 = I2ω2 constant of the eliptic orbit, where v r This is the consequence of the fact that no net torque is exerted on the planet ( net 0) 3. The Law of Periods: The square of the period of any planet is proportional to the cube of the semi major axis of its orbit T 2 /a 3 = constant For a circle where e = 0, a = r 4 2 T /r GM s 2 3 1 The shape of an ellipse is determined by its eccentricity “e”. e ab ab Where a and b are the semi major and minor axis of the ellipse. This is tabulated for all the planets. It is largest for Pluto (eP=0.248) and Mercury (eM = 0.24). This results in a dramatic variation in velocity as Mercury executes a revolution along the ellipse. X Perigee Apogee X 2 Material from the last Lecture for the Final Exam No questions on impulse and eclipse Chapter 9: 1, 9, 57, 58 Conceptual questions 1 and Example 15 Understand Equations: 9.7 and 9.10 Understand the figures: 9.28 and 9.29 Understand that 0 on a planet in elliptical orbit. Understand that from Newton’s 2nd law for rotation. t I I If =0 , = (Eq. 9.7 page 262) 0 constant, t thus the angular momentum of a planet in elliptical orbit. L I (Eq. 9.10 page 272) is also constant. The final exam will cover homework set 13 to 15. 3