problem_set_5_sol

problem_set_5_sol - Astronomy 160: Frontiers and...

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Unformatted text preview: Astronomy 160: Frontiers and Controversies in Astrophysics Homework Set # 5 Solutions 1) Consider a binary star system consisting of two 1.5M neutron stars. Sup- pose one of the neutron stars is a pulsar with an average observed pulse period of exactly 2 seconds. If the orbit is circular and edge-on to our line of sight, what is the maximum and minimum pulse period observed? (Note: the pulse period P p obeys the same Doppler shift formulas as wave- length does). Total Mass: M = M NS + M P = 1 . 5 M + 1 . 5 M = 3 M = 6 10 30 kg, ORBITAL period: P = 8 hr = 8 hr 3600 s hr = 3 10 4 s. Now use this information to solve for the orbital velocity of the pulsar. In order to do that, we must first find the total orbital velocity of the system, where V = V NS + V P . Because V NS M NS = V P M P and M NS = M P , then V P = V/ 2. Now combine the following equations a 3 = P 2 GM 4 2 and a = V P 2 V P 2 3 = P 2 GM 4 2 V 3 P 3 8 3 = P 2 GM 4 2 V 3 = 2 GM P V = 2 GM P 1 / 3 V = 2 7 10- 11 6 10 30 3 10 4 ! 1 / 3 V = 80 10 19 10 4 ! 1 / 3 V = 80 10 15 1 / 3 V = 4 10 5 m / s therefore V P = V 2 = 4 10 5 m / s 2 = 2 10 5 m / s This isnt very fast so we can use the Newtonian Doppler formula. We are total that the average pulse period is P p = 2 s: P p P p = v c 1 P p = v c P p = 2 10 5 m / s 3 10 8 m / s 2 s = 4 3 10-...
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problem_set_5_sol - Astronomy 160: Frontiers and...

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