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Unformatted text preview: Phys 341: Homework #4 Solutions 1. (a) Recall that the center of mass is defined such that m A a A = m B a B , so we have m A m B = a B a A = 1 . 466 = 2 . 15 Since m A /m B > 1, star A is more massive than star B. (b) Recall that the distance can be determined from the parallax angle as d = 1 pc p 1 00 1 So for p = 0 . 377 00 , we have d = 2 . 65 pc. (c) The phrase the semimajor axis of the reduced mass simply refers to the sum of the semimajor axes of the two star, a A + a B . This subtends an angle of 7 . 62 00 ; to get the corresponding length we multiply by the distance, finding a A + a B = 2 . 65 pc 7 . 62 00 180 3600 00 = 9 . 79 10 5 pc = 3 . 03 10 14 cm The last factor on the first line converts from arcseconds to radians. Keplers Third Law now tells us the sum of the masses, m A + m B = 4 2 ( a A + a B ) 3 GP 2 = 4 2 (3 . 03 10 14 cm) 3 (6 . 67 10 8 g 1 cm 3 s 2 ) (49 . 94 365 86400 s) 2 = 6 . 64 10 33 g = 3 . 34 M In the second line, the factor of 365 86400 in the denominator converts from years to seconds (86400 seconds per day, 365 days per year). Combining this with the result m B = 0 . 466 m A from part (a), we get m A (1 + 0 . 466) = 3 . 34 M m A = 2 . 28 M which then yields m B = 1 . 06 M 2. (a) For circular orbits, recall that the velocities are v A = 2 a A P = 2 P m B m A + m B a (1) v B = 2 a B P = 2 P m A m A + m B a (2) where a = a A + a B . Taking the ratio, we reproduce the result from class that the mass ratio can be determined from the velocity ratio, m A m B = v B v A = 22 . 4 km/s 5 . 4 km/s = 4 . 15 (b) From the expressions for the velocity above, we have...
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This note was uploaded on 09/08/2008 for the course PHYS 341 taught by Professor Keeton during the Spring '08 term at Rutgers.
 Spring '08
 Keeton
 Physics, Center Of Mass, Mass, Work

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