PS04_galactic_center_smbh_solutions

Physics 341: Problem Set #4 due October 11 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. An astronomer observes a galaxy at a wavelength of 4 cm using a 20 m radio telescope, but the observations are inconclusive. She decides she needs at least 3 times better angular resolution. How can she obtain such data? [5 pts] A. Observe the galaxy at 2 cm with the 20 m telescope B. Observe the galaxy at 2 cm with a 40 m telescope C. Observe the galaxy at 4 cm with a 50 m telescope D. Observe the galaxy at 15 cm with the 20 m telescope Answer: B (Based on using resolution goes as λ /diameter of the telescope) 2. Why is it surprising that the mass of the central supermassive black hole in a galaxy is correlated to the average random velocity of stars in the galaxy? [5 pts] A. The stars’ velocities should only depend on how far away they are from the black hole B. The central black hole does not significantly affect the motion of most of the stars C. The black hole should have swallowed up most of the stars D. The stars are individually much less massive than the black hole Answer: B. 3. The Fermi “Bubbles” are the large structures extending above and below the plane of the Milky Way emitting energy in x-rays and gamma-rays. The total energy in these 1

Bubbles has been estimated at 10 56 erg, and the event that caused them is suspected to have taken ∼ 10 5 years. [10 pts] (a) Let’s assume that the Bubbles are due to radiation emanating from material falling into our central black hole, which we’ll further assume occurred at a constant rate. Estimate the rate of mass accretion onto the black hole to create the Bubbles in solar masses per year. The total energy emitted is E = 10 56 erg, and was emitted over ∆ t = 5 × 10 5 years. Assuming that the emission was constant, the luminosity must be L = E ∆ t = 10 56 erg 10 5 years 1 year 3 . 1536 × 10 7 s = 3 . 2 × 10 43 erg/s . We derived in class that the mass accretion rate on a black hole is related to the luminosity by L = ϵ ˙ Mc 2 where ϵ is an efficiency factor that equals 1 2 in the purely Newtonian approxima- tion, and ϵ ∼ 0 . 1 including more detailed physics that is beyond the scope of this course. Taking ϵ = 1 / 2, I would calculate ˙ M = 2 L c 2 = 2 × 3 . 2 × 10 43 erg/s (3 × 10 8 m/s) 2 1 J 10 7 erg = 7 . 0 × 10 19 kg/s . The answer needs to be in solar masses per year, so ˙ M = 7 . 0 × 10 19 kg/s 1 M ⊙ 2 × 10 30 kg 3 . 1536 × 10 7 s 1 year = 1 . 1 × 10 − 3 M ⊙ / year . If I had taken an efficiency of ϵ = 0 . 1, I would have found ˙ M = 5 . 6 × 10 − 3 M ⊙ / year . (b) Assuming constant mass accretion, how long would it take to build the ∼ 4 × 10 6 M ⊙ black hole in the Galactic Center? Compare this to the age of the Universe (14 Gyr), do you think that this rate of accretion could be constant over the lifetime of our central black hole? Dividing the final mass of the black hole by the accretion rate yields the time it would take to build the black hole from constant accretion of matter: t = M ˙ M = 4 × 10 6 M ⊙ 1 . 1 × 10 − 3 M ⊙ / year = 3 . 6 × 10 9 years . This time, 3.6 Gyr, is shorter than the age of the Universe, 14 Gyr but not by many orders of magnitude. One might conclude that such a rate of accretion, 2

(b) Compute the pericenter (closest approach) and apocenter (farthest away) distances (in units of AU) for S2 and S62. The pericenter distance is r p = a (1 − e ), and the apocenter distance is r a = a (1+ e ) (Keeton Figure 3.1), so for the two stars we have: S2: r p = 1030 AU(1 − 0 . 885) = 118 AU r a = 1030 AU(1 + 0 . 885) = 1942 AU S62: r p = 747 AU(1 − 0 . 976) = 17 . 9 AU r a = 747 AU(1 + 0 . 976) = 1480 AU (c) What are the pericenter distances for the two stars in units of the radius of the black hole event horizon (the Schwarzschild radius)? The Schwarzschild radius for the black hole is R S = 2 GM c 2 = 2(6 . 67 × 10 − 8 cm 3 g − 1 s − 2 )(4 . 24 × 10 6 × 1 . 99 × 10 33 g) (3 . 0 × 10 10 cm s − 1 ) 2 = 1 . 25 × 10 12 cm = 0 . 084 AU So that means star S2 comes within 118/0.084 = 1400 R S , while star S62 gets as close as 17.9/0.084 = 213 R S . (d) Compute the speeds (in km s − 1 ) of S2 and S62 at pericenter and apocenter. What fraction of the speed of light do the stars reach when they are going their fastest? Hint: the tangential velocity is v θ = rω , and the specific angular momentum ℓ = r 2 ω is conserved throughout the orbit. Using the hint, the specific angular momentum is conserved over the orbit. From Lecture 4 slide 13 (or Keeton eqn. 3.11), ℓ = p GMa (1 − e 2 ). Also from Lecture 4 slide 12, ℓ = r 2 ω . This makes sense: because angular momentum is conserved, when the object orbits closer (lower r ) it has to go faster (higher ω ). Conversely, when the object is farther away (larger r ), it must move more slowly (smaller ω ). At pericenter, r is a minimum, and so v r = dr/dt = 0. Thus the velocity at pericenter is purely tangential ( v = v θ ). So then the speed at pericenter is | ⃗ v p | = v θ = r p ω = ℓ r p = p GMa (1 − e 2 ) a (1 − e ) = GM a 1 + e 1 − e 1 / 2 Similarly, v r = 0 at apocenter when r is a maximum, and the velocity at apocenter is also purely tangential. So the speed at apocenter is | ⃗ v a | = v θ = r a ω = ℓ r a = p GMa (1 − e 2 ) a (1 + e ) = GM a 1 − e 1 + e 1 / 2 Another way to derive this is to subtract the specific potential energy ( U/m = − GM/r ) from the specific total energy ( E tot /m = − GM/ (2 a )) to get the specific kinetic energy ( K/m = | ⃗v | 2 / 2). 4

Plugging in the numbers, for S2 we have: v p = 6 . 67 × 10 − 8 cm 3 g − 1 s − 2 × 4 . 24 × 10 6 × 1 . 99 × 10 33 g 1030 × 1 . 496 × 10 13 cm 1 + 0 . 885 1 − 0 . 885 1 / 2 = 7 . 7 × 10 8 cm s − 1 = 7700 km s − 1 v a = 6 . 67 × 10 − 8 cm 3 g − 1 s − 2 × 4 . 24 × 10 6 × 1 . 99 × 10 33 g 1030 × 1 . 496 × 10 13 cm 1 − 0 . 885 1 + 0 . 885 1 / 2 = 4 . 7 × 10 7 cm s − 1 = 470 km s − 1 For S62 we calculate v p = 6 . 67 × 10 − 8 cm 3 g − 1 s − 2 × 4 . 24 × 10 6 × 1 . 99 × 10 33 g 747 × 1 . 496 × 10 13 cm 1 + 0 . 976 1 − 0 . 976 1 / 2 = 2 . 04 × 10 9 cm s − 1 = 20400 km s − 1 v a = 6 . 67 × 10 − 8 cm 3 g − 1 s − 2 × 4 . 24 × 10 6 × 1 . 99 × 10 33 g 747 × 1 . 496 × 10 13 cm 1 − 0 . 976 1 + 0 . 976 1 / 2 = 2 . 5 × 10 7 cm s − 1 = 250 km s − 1 The stars are really zipping at perigalacticon! S2 reaches 7 . 7 × 10 8 cm s − 1 / 3 . 0 × 10 10 cm s − 1 = 0 . 026 c (2.6% of the speed of light), while S62 reaches an astounding 2 . 04 × 10 9 cm s − 1 / 3 . 0 × 10 10 cm s − 1 = 0 . 068 c , 6.8% the speed of light! 5