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SOLUTIONS TO EXERCISE 6
(a) Every year, one solar mass (2 * 10^30 kg) of matter falls onto the quasar's black hole, and 10% of that rest mass is converted into luminosity. So the luminosity of this quasar in a year is E = mc^2 = 0.1 * (2 * 10^30 kg) * (3 * 10^8 m/s)^2 = 1.8 * 10^46 J which, in Joules per second, is (1.8 * 10^46 J/yr)/(3.16 * 10^7 s/yr) = 5.7 * 10^38 Watts. Flux is given by F = L/(4 * pi * r^2), and is given to be 5 * 10^-15 Watts/m^2; so 5 * 10^-15 W/m^2 = (5.7 * 10^38 W)/(4 * 3.14 * r^2) and thus r = sqrt( (5.7 * 10^38 W)/(4 * 3.14 * 5 * 10^-15 W/m^2) ) = 9.5 * 10^25 m Which is (9.5 * 10^25 m)/(3.1 * 10^22 m/Mpc) = 3.1 * 10^3 Mpc. By the way, since 1 Mpc = 3300 light-years, the light from this quasar is approximately 3100 * 3300 = 10.2 billion years old! (b) If the Milky Way is a typical galaxy, and the Sun is a typical star, then we could estimate the typical luminosity of a galaxy to be L = (10^11 stars in the Milky Way) * (3.9 * 10^26 Watts per star) = 3.9 * 10^37 W Now, if the luminosity all galaxies in the universe is 10 times greater the luminosity of all quasars, then L(galaxy) * (number of galaxies) -------------------------------- = 10. Using L(quasar) from 2(a) above, L(quasar) * (number of quasars) (5.7 * 10^38 W) (number of galaxies) = -------------------- * (number of quasars) = 150 (3.9 * 10^37 W) * 10 |