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\section{Stars \& Galaxies, Galaxies example class 1, answers}
\noindent{\bf Question 1.}
\begin{enumerate}
\item Spiral: disk, old+young stars, in field, contain dust vs
Elliptical spheroidal, old stars, in groups and clusters, no dust.

\item Disk (most of the light, planar, spiral arms, exponential surface
brightness), Central spheroidal bulge (r$^{1/4}$ law, most of the
remaining light), spheroidal power-law halo, containing globular
clusters.

\item A gas cloud where the hydrogem gas is photo-ionised by light from
  massive stars. They trace spiral arms because (the massive) stars are
  born in the spiral arms, and they are short lived, so cannot drift
  out of the arms.

\item The visual light from the bulge is absorbed by dust in the
  intervening interstellar medium.

\item The angular extent $\theta$ of 1AU is 1'' from 1 parsec distance,
  by definition, hence $1{\rm pc}=1{\rm AU}/\theta$. Therefore $1{\rm
  pc}={1\rm AU}/(!pi/180*3600)=3.085\times 10^{16}\,m$. (The factor
  $!pi/180*3600$ converts arc seconds to radians.


\item In the absence of absorption, the observed flux from a star with
luminosity $L$ decreases with distance $r$ as ${\rm flux}=L/(4\pi
r^2)$. Since magnitude $m=-2.5\log({\rm flux})+c$, where $c$ is a
constant, this gives $m=5\log(r)-2.5\log(L/4\pi)+c$. By definition, the
apparent magnitude $m$ and the absolute magnitude $M$ are the same,
when $r=10\pc$. Hence $M=5-2.5\log(L/4\pi)+c$. ($r$ in
pc). Substracting these two equations yields $m-M=5\log(r)-5$.


\end{enumerate}


\noindent{\bf Question 2}\\
\begin{enumerate}
\item 
\begin{eqnarray}
F &=&L_{\rm B}/4\pi r^2=6.4\times 10^{-20}\,{\rm W}\,{\rm m}^{-2}\nonumber\\
m &=&5.48-5+5\log(r)=+29.7\,.
\end{eqnarray}

\item $\sigma = 2\int_0^\infty \exp(-z/z_0)\,dz=200$~pc$^{-2}$.

\item The surface are of the square is $(r\theta)^2$, when $\theta$ is
  expressed in radians, since $\theta\ll 1$. The number of stars in the
  square is $N=\sigma\,S$, so the flux 
\begin{equation}
F=N\,L_{\rm B}/4\pi r^2=\sigma\theta^2\,L_{\rm B}/4\pi\,.
\end{equation}
is distance independent. The surface brightness is

\begin{eqnarray}
\mu &=&-2.5\log(F)+C\nonumber\\
  &=&-2.5\log(L_{\rm B}/4\pi)-2.5\log(\sigma\theta^2)+C\,,
\end{eqnarray}
with $\sigma=200$ and $\theta=1./206265$ an arc second in radians.
For solar type stars, $M_{\rm B}=5.48=-2.5\log(L_{\rm B})+5+C$.
Collecting terms gives $\mu_{\rm B}=21.3$

\item We just need to replace $n_0$ by $n_0\exp(-R/R_0)$ in the
  previous analysis, yielding $\mu_{\rm B}=21.3+2.5\log(e)\,(R/R_0)$.



\end{enumerate}



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