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lecture-05

# lecture-05 - These lecture notes were prepared for Rutgers...

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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 5: The Black Hole at the Center of the Galaxy I. Using Kepler III: Motion → Mass It’s all well and good that Newton gave a physical explanation of Kepler’s empirical laws. But what good does it do us? Let’s reconsider Kepler III. Rearranging, we can write M = 4 π 2 a 3 GP 2 This form is useful because the left-hand side is something we may want to know — the mass of an astronomical object — while the right-hand side involves stuff we can measure — the size and period of an orbit. In other words, we know what measurements we need to make, and calculations we need to do, in order to measure mass. A couple of things to keep in mind for later in the course are that M represents the total mass of the system, and that a is the average separation of the two bodies in three dimensions, whereas we typically observe a system projected onto the two-dimensional sky. The key concept is that we can observe motion, interpret it using Newton’s laws of motion and gravity, and infer mass. Or, as I will often summarize the principle: motion → mass . Example: : Jupiter. We can use the orbits of Jupiter’s moons (from Appendix C of Carroll & Ostlie) to compute the planet’s mass. Moon Orbital Period Semimajor Axis M J (d) (10 3 km) (10 30 g) Io 1.769 421.6 1.90 Europa 3.551 670.9 1.90 Ganymede 7.155 1070.4 1.90 Callisto 16.689 1882.7 1.90 1 II. Equations of Motion Redux Let’s look again at the complete set of equations of motion for the one-body problem: dr dt = v r dθ dt = ω = ‘ r 2 dv r dt = d 2 r dt 2 = a r + rω 2 =- GM r 2 + rω 2 Let’s make the finite difference approximation: Δ r ≈ dr dt Δ t = v r Δ t Δ θ ≈ dθ dt Δ t = ‘ r 2 Δ t Δ v r ≈ dv r dt Δ t =- GM r 2 + rω 2 Δ t Now suppose you specify r , θ , and v r at some time t , along with the angular momentum ‘ (which is constant). Then you can compute the changes Δ r , Δ θ , and Δ v r and take a step: ( t, r, θ, v r ) → ( t + Δ t, r + Δ r, θ + Δ θ v r + Δ v r ) Then you can repeat the process to take another step ... and then another ... and so on. This repetitive process sounds tedious, but it is perfect for computers. I have created a spreadsheet to do the calculation, and you will work with it on problem set #3. A word of warning: the finite difference approximation is valid only when the time step Δ t is small. If you want your calculation to be accurate, you can’t make Δ t too big. On the other hand, when Δ t is small it can take many, many steps to compute an orbit. Computers do make very small errors in calculations (because they have to round numbers to a large but finite number of digits), and if you take many, many steps those errors can build up....
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