1. In class, we showed that each radial shell contributes $n L (r)$ to the flux seen on earth, where the density $n$ and luminosity $L$ may in principle depend on $r$. Thus the flux is given by
   $$\phi = \int_0^\infty\ dr\ n L (r)\ \ .$$
   Giving $L$ a time dependence also gives it radial dependence as viewed from earth, as light received from the shell at $r$ at time $t$ was emitted at time $t - r/c$. For our simple model, $L$ is a step function $L\, \Theta (t)$ turning on at time 0, a time $t_o = 10^{10}$ years ago. Thus
   $$L(r)= \left\{ \begin{array}{ll}L &\mbox{for}\ r \le ct_o\\ 0 &\mbox{for}\ r > ct_o\end{array}\right.$$
   since signals from $r > ct_o$ have not yet had time to arrive at earth. The flux of starlight on earth is therefore
   $$\phi = \int_0^{ct_o}\ dr\ n L \ \ = \ nL ct_o\ \ .$$
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   To compare this to the brightness of the sun

   $$\phi_\odot = \frac{L}{ 4\pi\, R_\odot^2} \ ,$$
   we note that
   $$\phi = \phi_\odot \ 4\pi\, R_\odot^2\ nct_o\ \ ,$$
   where
   $$ct_o = 3 { \, \times\, 10^{8}}\ \frac{ \text{m}}{\text{s}} \ \cdo... ...mes\, 10^{16}} \ \text{m}} \quad = \quad 3{ \, \times\, 10^{9}}\ \text{pc}\ \ .$$
   Plugging in the cosmological values gives
   $$\phi = \phi_\odot \ \cdot\ 4\pi\ \cdot\ 10^{-14}\$$   pc$$^2\ \cdot\ \frac{ 10^{10}}{ 10^{18} \text{pc}^3} \ \cdot\ \quad 3{ \, \times\, 10^{9}}\ \text{pc}\quad \approx \ 10^{-12}\ \phi_\odot\ \ .$$
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   This great suppression of starlight flux in the night sky comes about because the radial cutoff $ct_o$ for starlight to have reached us since the stars first formed is 12 orders of magnitude smaller than the cutoff calculated in class, due to the opaqueness of stars.

2. The given redshift is well within the nonrelativistic limit ($z \le 0.4$). The Doppler effect thus gives a recessional velocity for the Virgo Cluster
   $$v = cz = 3 { \, \times\, 10^{8}}\ \frac{ \text{m}}{\text{s}} \ \c... ...}} \ \frac{ \text{m}}{\text{s}} \quad = 1200 \ \frac{ \text{km}}{\text{s}}\ \ .$$
   $\qedsymbol$ Hubble's law thus gives for the distance to the Virgo cluster
   $$d = \frac{v}{H_o} = \frac{1200 \ \text{km}/\text{s}} { h \,\cdot\, 100 \ \frac{ \text{km}/\text{s}}{\text{Mpc}}} \quad = 12 h^{-1} \text{Mpc}\ \ .$$
   For accepted values of $h$ ranging from $0.5$ to $0.85$, this gives $d$ from $14$ to $24$ Mpc. $\qedsymbol$

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