MIDTERM #2 THURSDAY APRIL 16, 2015 AST142 1. Black hole Growth The number of quasars per unit volume peaks at a redshift of about 2. This time is thought to be the epoch of massive black hole growth. How long would it take a 106 M black hole accreting at its Eddington luminosity to increase to 109 M ? Assume the luminosity L = M˙ c2 where M˙ is the mass accretion rate onto the black hole, c is the spead of light and an efficiency of = 0.1. ———————————Solution 38 LED = 1.3 × 10 erg s −1 M M We set this equal to M˙ c2 and solve for the black hole mass as a function of time. Since M˙ ∝ M the solution is an exponential. 1.3 × 1038 erg s−1 M˙ = = 7.2 × 10−16 s−1 M M c2 ts = M/M˙ = 4.5 × 107 years M (t) = M (t = 0) exp(t/ts ) At t = 0 we have 106 M and at t later we have 109 M . The ratio of masses is 1000. ln 1000 = 6.9 = t/ts So the time is t = 6.9ts = 3 × 108 years. ———————————— 2. Correcting for Extinction A standard interstellar dust extinction law is listed in Table 1. 1 2 MIDTERM #2 THURSDAY APRIL 16, 2015 AST142 Table 1 Band Wavelength A(λ)/AV B 0.45 µm 1.29 V 0.55 µm 1.00 R 0.66 µm 0.75 J 1.2 µm 0.28 H 1.6 µm 0.19 K 2.2 µm 0.112 An F0 main sequence star has an absolute magnitude of MV = 2.7 and intrinsic color (B − V )0 = 0.30. You observe an F0 main sequence star (with spectral type identified spectroscopically) finding apparent magnitudes mV = 10.00 and mB = 10.56. (a) What is the extinction AV for this star? (b) What is the distance to this star? ———————————— Solutions Corrected or intrinsic magnitudes mV 0 = mV − AV mB0 = mB − AB = mB − 1.29AV giving mB0 − mV 0 = (B − V )0 = 0.30 mB0 − mV 0 = mB − mV − (1.29 − 1)AV = 12.56 − 12.0 − 0.29AV = 0.56 − 0.29AV Set these two together 0.30 = 0.56 − 0.29AV 0.26 = 0.29AV AV = 0.26/0.29 = 0.90 And this gives mV 0 = 10.00 − 0.90 = 9.10 and mB0 = 10.56 − 0.90 × 1.29 = 9.40 Let’s check. The difference is 0.3 as we expected! The observed V (corrected for extinction) is 9.10 and the absolute magnitude is 2.7. The distance modulus is 6.3 giving a distance d = 106.3/5 × 10 = 182pc ———————————— MIDTERM #2 THURSDAY APRIL 16, 2015 AST142 3 3. Dark matter in the Coma Cluster The Coma cluster is a nearby cluster of over 1000 galaxies. Zwicky measured the radius of the Coma cluster to be about a Mpc and a radial velocity dispersion of about vr2 ∼ 5 × 105 (km/s)2 . a. Estimate the total mass in the cluster using the virial theorem. b. Suppose the mean mass of each galaxy is 5×1010 M . What does your mass estimate say about the dark matter fraction in the cluster? (Ignore the mass in X-ray emitting gas in the intercluster medium). ——————————– Solutions Using the virial theorem kinetic energy is equal to twice the potential energy. And there is a factor of 3 giving 6vr2 r0 = 7 × 1014 M G If each galaxy has 5 × 1010 M then we are still a factor of 10 low in accounting for the matter in the cluster. Lots of dark matter. At 90%. M= ———————————— 4. Short description Discuss one of the following a. Processes affecting galaxy morphology b. Part of the astronomical distance ladder c. Active Galactic nuclei (AGNs), their diversity and unification