PS05_binary_stars_2bodytheory_solutions

Physics 341: Problem Set #5 due October 18 by 2:00 PM in PDF format on Canvas [40 pts] You are encouraged to work in groups on these problems, but each student must write up the solutions individually. You must also list your collaborators on your solutions, and cite any external sources you used (other than the course notes or textbook). Using online solutions to these or similar problems is not allowed. Show your work and explain your reasoning. The number of points for each problem is listed; partial credit will be given. Be careful with units. Questions asking about concepts require at most a few sentences to answer. Please make your answers clear and concise . 1. Which of these is a true statement about typical visual binaries and typical spectro- scopic binaries that we observe? [5 pts] A. The typical spectroscopic binaries have higher eccentricities B. The typical spectroscopic binaries have lower inclinations (closer to face-on) C. The typical visual binaries have longer periods D. The typical visual binaries have faster orbital speeds Answer: C 2. In a single-lined spectroscopic binary, where we only measure the period P and one radial velocity amplitude K 1 , what can we determine about the unseen companion? [5 pts] A. Its minimum mass B. Its maximum mass C. Its minimum radius D. Its maximum radius E. none of the above Answer: E. 3. Below is a sketch of the orbits of two stars orbiting each other about a common focal point marked by the ⋆ . The locations of star 1 are marked at times A , B , C , and D (in 1

order of earliest to latest). The distance x is 0 . 1 AU, and the eccentricity is e = 0 . 4. [15 pts] (a) In the diagram above, mark the location of star 2 at times A , B , C , and D . Star 2 must be on the opposite side of the focal point from star 1 at each time. Therefore, the locations are: (b) Sketch a plot of the kinetic and potential energies of star 1 as a function of time. On a separate plot, sketch the kinetic and potential energies of star 2 as a function of time. Indicate on the time axes where times A , B , C , and D are. You do not need to place numeric values on the axes, but you do need to indicate the location of zero on the energy axis. Star 1 has the lowest kinetic energy at time B, and the highest (least negative) 2

= 4 π 2 (4 . 27 × 10 10 m ) 3 (6 . 67 × 10 − 11 m 3 kg − 1 s − 2 )[(25 days )(86400 s/days )] 2 = 9 . 91 × 10 30 kg = 4 . 95 M ⊙ We know the mass ratio must be related to the ratio of semimajor axes: m 1 m 2 = a 2 a 1 = x/ (1 + e ) 3 x/ (1 + e ) = 1 3 . Therefore 4 . 95 M ⊙ = m 2 1 + m 1 m 2 m 2 = 3 . 34 M ⊙ 1 + 1 3 = 3 . 71 M ⊙ m 1 = 1 3 m 2 = 1 . 24 M ⊙ . 4. One of the most massive binary star systems known is called WR-20a and is located in the Large Magellanic Cloud (a small companion galaxy to the Milky Way). This system is nearly edge on, and the stars are moving in (nearly) circular orbits with observed speeds of 362.2 km s − 1 and 366.4 km s − 1 in an orbital period of just 3.686 days! Hint: assume it is exactly edge-on for parts a and b. [15 pts] (a) Calculate the masses of these two stars. Because we can observe the radial velocities of both components, this must be a double-lined spectroscopic binary. From Kepler’s 3rd law and for a circular orbit, we showed in class that the total mass is M = m 1 + m 2 = P 2 πG K 1 + K 2 sin i 3 In this case we have P = 3 . 686 day = 3 . 185 × 10 5 s K 1 = 362 . 2 km s − 1 = 3 . 622 × 10 7 cm s − 1 K 2 = 366 . 4 km s − 1 = 3 . 664 × 10 7 cm s − 1 i ≈ 90 ◦ (nearly edge-on) 4

⇒ M = 3 . 185 × 10 5 s 2 π × 6 . 67 × 10 − 8 cm 3 g − 1 s − 2 (3 . 622 × 10 7 + 3 . 664 × 10 7 ) cm s − 1 sin 90 ◦ 3 = 2 . 939 × 10 35 g = 147 . 8 M ⊙ That gives us the total mass M = m 1 + m 2 ; to get the individual masses we can use the fact that the mass ratio m 2 /m 1 = K 1 /K 2 ; so M = m 1 + m 2 = m 1 1 + m 2 m 1 = ⇒ m 1 = M 1 + m 2 m 1 = M 1 + K 1 K 2 Thus we can derive m 1 = 147 . 8 M ⊙ 1 + 3 . 622 × 10 7 cm s − 1 3 . 664 × 10 7 cm s − 1 = 147 . 8 M ⊙ 1 + 0 . 9886 = 74 . 3 M ⊙ m 2 = M − m 1 = 73 . 5 M ⊙ (b) What is the distance between the two stars? The distance between the two stars stays constant, and it is just a = a 1 + a 2 = P 2 π K 1 + K 2 sin i = 3 . 185 × 10 5 s 2 π (3 . 622 × 10 7 + 3 . 664 × 10 7 ) cm s − 1 sin 90 ◦ = 3 . 69 × 10 12 cm = 0 . 247 AU (c) More recent results have shown that we do not see the orbit exactly edge on. Would that fact increase, decrease, or not change your results for the masses and orbital separation? Explain why. If we’re not viewing the orbits exactly edge-on, then i < 90 ◦ and sin i < 1. This means that the true speeds of the stars are larger, and we just see the component along our line of sight. Because the sin i factor appears in the denominator in both the total mass and the orbital separation, our results for m 1 , m 2 , and a would all increase compared to what we derived in parts (a) and (b). 5