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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 7: TwoBody Problem I. Theory (This discussion draws from § 2.3 of Carroll & Ostlie.) Previously we considered the onebody problem, in which the source of the gravity (e.g., a star) is stationary and only one object (e.g., a planet) moves. This is a reasonable approxi mation when the planet is much less massive than the star. But by Newton’s third law, if the star pulls on the planet, the planet also pulls on the star, so both should actually move. Now we treat the general twobody problem . Consider the setup: Figure 1: Setup for the twobody problem. Objects of mass m 1 and m 2 are found at positions ~ r 1 and ~ r 2 . (From Carroll & Ostlie Fig. 2.10.) Let’s define the separation vector , ~ r = ~ r 2 ~ r 1 and the center of mass position , ~ R = m 1 ~ r 1 + m 2 ~ r 2 m 1 + m 2 1 and finally the total mass , M = m 1 + m 2 Now we can identify the force on each object (assuming there are no external forces, just the gravity between the two objects): force on #1: ~ F 1 = + Gm 1 m 2 r 2 ˆ r force on #2: ~ F 2 = Gm 1 m 2 r 2 ˆ r Notice that ~ F 1 = ~ F 2 , which is Newton’s third law. Now consider: d 2 ~ R dt 2 = 1 m 1 + m 2 m 1 d 2 ~ r 1 dt 2 + m 2 d 2 ~ r 2 dt 2 = 1 m 1 + m 2 ~ F 1 + ~ F 2 = 0 In other words, the center of mass does not accelerate . It might move with a constant velocity, but it does not accelerate. Now let’s rewrite the object positions: ~ R = m 1 ~ r 1 + m 2 ~ r 2 m 1 + m 2 ( m 1 + m 2 ) ~ R = m 1 ~ r 1 + m 2 ( ~ r 1 + ~ r ) ( m 1 + m 2 ) ~ R = ( m 1 + m 2 ) ~ r 1 + m 2 ~ r ⇒ ~ r 1 = ~ R μ m 1 ~ r where the reduced mass is μ = m 1 m 2 m 1 + m 2 A similar analysis applied to object #2 yields...
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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
 Physics

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