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\section*{Stars \& galaxies, example class 4}

\noindent{\bf Short questions}
\begin{enumerate}
\item Evidence for the presence of a very dark, massive object,
  presumably a black hole, comes from, for example, 
\begin{itemize}
\item photometric observations, suggesting a high density of stars
\item stellar kinematics, implying very high velocities near the centre
\item measurements of maser emission, suggesting a very high
  mass-density.
\item (several others are mentioned in the notes)
\end{itemize}
\item The evidence is based on proper-motion studies of stars close to
  the centre, which imply the presence of a very massive but dark
  object.
\item The presence of the clusters causes the images of background
  galaxies to be distorted, or images of the same background galaxy to
  appear several times. Modelling of the required potential constrains
  the mass of the cluster.
\end{enumerate}

\noindent{\bf Question 1.}\\
\begin{itemize}
\item Since $N(M)\,dM$ is the number of stars with mass between $M$ and
  $M+dM$, the total mass of stars in that interval is $M\,N(M)\,dM$,
  and therefore, expressing all masses in $M_\odot$, the total mass of
  stars in the cluster is
\begin{equation}
M_T = \int_{M_-}^{M_+}\, M\, N(M)dM =  \int_{M_-}^{M_+}\, M^{-1.35}\,dM
= 5.5\,N_0\,
\end{equation}
hence $N_0=1.d6/5.5$ (in which $M_-=0.1$ and $M_+=30$. The number of
SNe is
\begin{equation}
N = \int_{8}^{M_+}\,N(M)\,dM = 0.04\,N_0=6700\,.
\end{equation}

\item In solar units, the total luminosity is

\begin{equation}
L_T = \int_{M_-}^{M_+} M^2\,N(M)dM = 13.7\,N_0\,
\end{equation}
therefore the mass-to-light ratio is $M_T/L_T=0.4$.

\item In this case we need to use $1M_\odot$ for $M_+$, hence the
  luminosity is now
\begin{equation}
L_T = \int_{M_-}^{1} M^2\,N(M)dM = 1.2\,N_0\,
\end{equation}
therefore the mass-to-light ratio is $M_T/L_T=3.$.
\end{itemize}


\noindent{\bf Question 2.}\\
The equation for the angle, $\theta_E$, under which you see the ring is

\begin{eqnarray}
\theta_E &=& {R_E\over D_{OL}}\nonumber\\
         &=& ({4GM\over c^2})^{1/2}\,\left({D_{OS}-D_{OL}\over
D_{OS}\,D_OL}\right)^{1/2}
\end{eqnarray}

Substituting the numbers ($M_\odot=1.989\times 10^{30}{\rm kg}$, $1{\rm
pc}=3.086\times 10^{16}{\rm m}$, ${\rm G}=6.67\times 10^{-11}\,{\rm m}^3{\rm
kg}^{-1}{\rm s}^{-2}$, $c=3\times 10^8{\rm m}{\rm s}^{-1}$), this becomes
\begin{eqnarray}
\theta_E &=&\big[
{4\times6.67\times1.989\over 3.086\times 3.^2}\times{10^{-11}\times10^{13}\times10^{30}\over
10^3\times10^6\times 10^{16}\times 10^{16}}\times{{\rm m}^3\,{\rm
kg}^{-1}\,{\rm s}^{-2}{\rm kg}\over {\rm m}\,{\rm m}^2/{\rm
s}^{-2}}\big]^{1/2}\nonumber\\
         &=&4.37110\times 10^{-5}\,{\rm radians}\,
\end{eqnarray}
which is 9.0 arc seconds. Note how it is easy to get the order of
magnitude by just working-out the exponents, and how you can check
whether the answer makes sense by checking the units. 

The angle under which you see the galaxy is
\begin{equation}
\theta={20\kpc\over 500\Mpc}=8.25\arcsec
\end{equation}

The diameter of the ring is slightly larger than that of the galaxy.


\pagebreak
\noindent{\bf Question 3.}\\
\begin{itemize}
\item The mass due to the density distribution $\rho(r)$ is
\begin{equation}
M_\rho(<r_1) = 4\pi\rho_1\int_0^r {r_1\over r}\,
r^2\,dr=2\pi\rho_1r_1r^2=M_1 (r/r_1)^2
\end{equation}
for $r<r_1$, where $M_1\equiv 2\pi\rho_1r_1^3$.
(Note that this is found by summing the masses of spherical shells,
$M(<r)=\int_0^M dM$, where the mass of a spherical shell with radius
$r$, density $\rho(r)$ and thickness $dr$ is $dM=4\pi r^2\rho(r)dr$)

The mass in a shell between $r_1$ and $r<r_2$ is
\begin{equation}
M_\rho(r_1<r) = 4\pi\rho_1\int_{r_1}^r (r_1/r)^2\,r^2\,dr = 2M_1(r/r_1)-2\,M_1
\end{equation}

Therefore, the enclosed mass is

\begin{eqnarray}
M(<r) &=& M_0 + M_1 (r/r_1)^2 \,\,\,\,\,\hbox{for $r\le
r_1$}\nonumber\\
      &=&M_0+2M_1(r/r_1)-M_1\,\,\,\,\,\hbox{for $r_1\le
r\le r_2$}\nonumber\\
      &=& M_0 + {2r_2-r_1\over r_1} M_1 \,\,\,\hbox{for $r> r_2$}\,.
\end{eqnarray}

\item The circular velocity at distance $r$ is $V_c^2(r)=GM(<r)/r$, hence
\begin{eqnarray}
V_c^2(r) &=& {GM_0\over r} + {G\,M_1\over r_1}\, {r\over r_1} \,\,\,\,\,\hbox{for $r\le
r_1$}\nonumber\\
      &=&{G(M_0-M_1)\over r}+{2G M_1\over r_1}\,\,\,\,\,\hbox{for $r_1\le
r\le r_2$}\nonumber\\
      &=& {G (M_0 + (2r_2/r_1-1)M_1)\over r}\,\,\,\hbox{for $r> r_2$}\,.
\end{eqnarray}


\item If we neglect $M_0$ in the equation for $V_c$, then from the
previous equation, $V_c^2(r=8.5\kpc)=(220\kms)^2={GM(r<8.5\kpc)\over
r}={G(M_0-M_1)\over r}+{2G M_1\over r_1}\approx {-GM_1\over r}+{2G
M_1\over r_1}$, for $r=8.5\kpc$. Solving for $M_1$ requires
$M_1=5.98\times 10^{9}\msun$, hence $M(r<8.5\kpc)=9.56\times 10^{10}\msun$
which is $\gg M_0$ so our simplification is justified: the SMBH does
not contribute significantly to the rotation velocity at
$r=8.5\kpc$. Substituting this value for $M_1$ in $M_1=2\pi\rho_1r_1^3$
gives $\rho_1=9.5\times 10^{8}\msun/\kpc^3=6.44\times 10^{-23}{\rm
g}/\cm^3$.

\item  To find evidence for the super-massive black hole, it should
start to contribute significantly to $V_c$, which starts to happen when
$M_0/r\approx M_1 r/r_1^2$, ($GM_0/r$ is the contribution of the black
hole to $V_c^2$, $M_1 r/r_1^2$ is the contribution of the \lq bulge\rq
with density distribution $\propto 1/r$ to $V_c^2$. Hence when these
contributions are equal, 
\begin{equation}
r = ({M_0\over M_1})^{1/2}r_1 = ({10^6\over 5.98\times
10^{9}})^{1/2}\,\kpc\approx 13\pc\,.
\end{equation}

The angle $\theta$ under which you see 13\pc\ at a distance of 5\Mpc\ is
\begin{equation}
\theta = {13\over 5\times 10^6} \approx 0.53\arcsec
\end{equation}

\end{itemize}



\begin{figure*}[h]
\center\resizebox{.6\textwidth}{!}{\includegraphics{mass.pdf}}
\caption{Enclosed mass vs radius}
\label{fig:theta}
\end{figure*}


\begin{figure*}[h]
\center\resizebox{.6\textwidth}{!}{\includegraphics{vrot.pdf}}
\caption{Rotation curve, the filled circle is at $r=8.5\kpc$,
$V_c=220\kms$. There is Keplerian fall-off $V_c\propto 1/r^{1/2}$ both
very close to the black hole (where it dominates the mass), and on large
scales $r\ge r_2$, where the galaxy mass is constant. In the inner
region, where the mass is dominated by the $1/r$ density profile, the
circular velocity rises, $V_c\propto r^{1/2}$, and where the $1/r^2$
density dominates the mass, the circular velocity becomes almost
constant, $V_c\approx 220\kms$. The short dashed line is the Keplerian
fall-off due to the black hole, the long-dashed is the Keplerian
fall-off due to the $1/r$ density distribution. The open circle is
where we computed that the black hole started contributing
significantly to $V_c$.}
\label{fig:theta}
\end{figure*}

\end{document}
