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# a11f04 - 2 2 ~v 2 2-U r r = | ~ r | ~ r = ~ r 1-~ r 2 1 Is...

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ADVANCED DYNAMICS — PHY 4934 HOME AND CLASS WORK – SET 4 (October 1, 2011) Masses Initial Positions Initial Velocities # i m i x 1 i, 0 x 2 i, 0 x 3 i, 0 ˙ x 1 i, 0 ˙ x 2 i, 0 ˙ x 3 i, 0 1 1 0.651 0.585 -0.238 -0.755 -0.828 -0.865 -0.726 2 0.931 -0.096 0.000 0.357 -0.209 0.107 -0.660 2 1 1.510 0.460 -0.359 -0.234 -0.918 -0.941 -0.323 2 0.126 -0.066 -0.090 -0.809 0.789 0.788 0.620 3 1 1.328 -0.125 0.898 0.194 -0.452 0.172 0.125 2 1.999 -0.449 -0.085 -0.454 -0.976 -0.990 -0.968 4 1 0.180 0.204 -0.968 -0.753 -0.811 -0.632 0.784 2 1.560 -0.889 -0.979 0.854 -0.323 -0.774 -0.533 Table 1: Initial conditions for the Kepler problem (arbitrary units and set G = 1). (16) Calculate the angular momentum vectors in the CM frame for the initial condi- tions given in the table above (you can download the data as text ﬁle Kepler.txt). Due October 5 before class 8 points. Do not forget: Midterm Friday, October 7. (17) Let the interaction of two point particles be described by a potential which depends only on their distance: L = m 1 2 ~v 2 1 + m
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Unformatted text preview: 2 2 ~v 2 2-U ( r ) , r = | ~ r | , ~ r = ~ r 1-~ r 2 . 1. Is the energy of this system conserved (with reason) (1 point)? 2. Deﬁne the center of mass vector by ~ R = ( m 1 ~ r 1 + m 2 ~ r 2 ) /M , M = m 1 + m 2 and express ~ r 1 and ~ r 2 through ~ R and ~ r (2 points). The center of mass (cm) frame is deﬁned by ~ R ( t ) = 0. Show the following equalities in the cm system: 3. T cm = m 1 ~v 2 1 / 2 + m 2 ~v 2 2 / 2 = μ~v 2 / 2 with ~v = ˙ ~ r and μ the reduced mass. Express μ through m 1 , m 2 and M (2 points). 4. ~ L = ~ r 1 × ~ p 1 + ~ r 2 × ~ p 2 = μ~ r × ~v (2 points). Due in class. (18) Plot the eﬀective potentials corresponding to the initial conditions of the ta-ble together with the energies in the CM frames (the Potential is U ( r ) =-G m 1 m 2 /r ). Due October 14 before class 8 points....
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