PS08_elliptical_galaxies_solutions

Physics 341: Problem Set #8 due November 22 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. Short answer. You should be able to answer these questions using the virial theorem without doing any detailed calculations. [5 pts] (a) For a system in virial equilibrium, is the total energy positive, negative, or zero? ⟨ E ⟩ = − ⟨ K ⟩ = ⟨ U ⟩ / 2 Because ⟨ K ⟩ is always positive, ⟨ E ⟩ must be negative. Similarly, because ⟨ U ⟩ is defined to be negative (for a bound system), ⟨ E ⟩ must be negative. (b) Suppose a system is in equilibrium. In order for the system to shrink in size, how must the total energy change? ⟨ E ⟩ = ⟨ U ⟩ / 2 = − GM 2 R If R decreases, the total energy becomes more negative. 2. An elliptical galaxy is observed to have an angular diameter of 81.4 arcseconds on the sky. We observe the hydrogen Hβ line (rest wavelength λ rest = 486 . 13 nm) in the galaxy spectrum at a wavelength λ obs = 518 . 80 nm, and we measure that the stars in the galaxy have a radial velocity dispersion σ = 255 km s − 1 . Take H 0 = 72 km s − 1 Mpc − 1 . [10 pts] (a) How far away is the galaxy (in Mpc)? 1

See LN24. The galaxy light is redshifted because of the expanding Universe. The redshift is z = λ obs λ rest − 1 = 518 . 80 nm 486 . 13 nm − 1 = 0 . 0672 Hubble’s Law says that the recession velocity is proportional to the distance, v = H 0 d , with z = v/c . So d = cz H 0 = (3 . 0 × 10 5 km s − 1 )(0 . 0672) 72 km s − 1 Mpc − 1 = 280 Mpc (b) How massive is it (in M ⊙ )? Be sure to state any assumptions you make. See LN17. Our virial mass estimate for an elliptical galaxy is M = 3 β η Rσ 2 G If we assume that the velocity dispersion is based on isotropic orbits of identical stars , then β = 1. Furthermore, if we assume an isothermal sphere model for the galaxy , η = 1 also. Finally, given the distance to the galaxy from part a, we can calculate its physical radius based on its angular radius, θ = 81 . 4 ′′ / 2 = 40 . 7 ′′ : R = θd = 40 . 7 ′′ × 1 ◦ 3600 ′′ π rad 180 ◦ × 280 Mpc = 0 . 05525 Mpc So our mass estimate is M = 3 Rσ 2 G = 3(0 . 05525 × 3 . 086 × 10 24 cm)(255 × 10 5 cm s − 1 ) 2 6 . 67 × 10 − 8 cm 3 g − 1 s − 2 = 4 . 99 × 10 45 g = 2 . 51 × 10 12 M ⊙ 3. Use the results we derived in class based on the virial theorem to estimate the mass of the Coma cluster of galaxies, which has a radial velocity dispersion of σ = 977 km s − 1 and a radius R = 3 Mpc . You can assume an isothermal sphere model for the mass distribution and that the measured radial velocity dispersion is based on isotropic orbits of identical galaxies. Would your mass estimate increase, decrease, or stay the same if you used a constant density model? [10 pts] From class we saw that the virial mass estimate is M = 3 β η Rσ 2 G 2

In class, we assumed isotropic orbits, ⟨ v αx ⟩ = ⟨ v αy ⟩ = ⟨ v αz ⟩ . For circular orbits, however, there is only motion in two directions, but the average of that velocity is equal in those two directions, e.g., in x and z (based on how we usually set up our coordinate system with x and y being in the plane of the sky). Let ⟨ v αx ⟩ = ⟨ v αz ⟩ . Thus ⟨ K ⟩ = 2 2 X α m α v 2 αz = X α m α v 2 αz We measure the velocity dispersion of stars, which is given by σ 2 = ∑ α ⟨ v 2 αz ⟩ N If we assume all the stars are the same, we can write the total kinetic energy as ⟨ K ⟩ = m X α v 2 αz = m N σ 2 = Mσ 2 We previous found that ⟨ K ⟩ = (3 / 2) βMσ 2 . Comparing to what we just derived, we can see that β = 2 / 3 for circular orbits. (c) Now use the virial theorem to get the virial mass estimate for an isothermal sphere with circular orbits in terms of σ . How are v c and σ related for circular orbits? The virial theorem is 2 ⟨ K ⟩ + ⟨ U ⟩ = 0 For an isothermal sphere with η = 1 and our newly derived ⟨ K ⟩ in the virial theorem, we have 2 Mσ 2 − GM 2 R = 0 Rearranging, we get the virial mass estimate M = 2 Rσ 2 G Comparing to (a), we see that v 2 c = 2 σ 2 for circular orbits. 4