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SOLUTIONS TO EXERCISE 6

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(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