lecture-07

lecture-07 - These lecture notes were prepared for Rutgers...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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: Two-Body Problem I. Theory (This discussion draws from § 2.3 of Carroll & Ostlie.) Previously we considered the one-body 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 two-body problem . Consider the set-up: Figure 1: Set-up for the two-body 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...
View Full Document

This note was uploaded on 10/20/2011 for the course PH 341 taught by Professor Gawiser during the Fall '11 term at Rutgers.

Page1 / 8

lecture-07 - These lecture notes were prepared for Rutgers...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online