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\begin{document}

\section*{Stars \& galaxies, multiple choice questions}

\noindent{\bf Problem 1.}\\

\begin{enumerate}
\item Answer {\bf a}: The distance $r=300$~\kpc\ is beyond the extent of
  the dark halo (which has radius 200~\kpc), hence equate $v_c^2=GM/r$,
  where $M=1.0\times 10^{12}M_\odot$ is the total mass of the halo, to
  find $v_c=170$ km s$^{-1}$

\item Answer {\bf b}: The $1/r^2$ density profile gives rise to a flat
  rotation curve inside the halo. Therefore the circular velocity at
  $r=100$~\kpc\ is the same as at $r=200~$\kpc\ where it is $v_h^2=GM/r$,
  with $M=2.0\times 10^{12}M_\odot$, hence $v_h=207$\kms.
\item Answer {\bf c}: The circular velocity is due to dark halo and
  bulge combined, so $v_c^2=GM_{\rm halo}(<r)/r+GM_{\rm
  bulge}(<r)/r=v_h^2+GM_{\rm bulge}(<r)/r$. Since the bulge is uniform
  density, $M_{\rm bulge}(<r)=(r/r_b)^3\,M_b$, hence
  $M_b(r<2\kpc)=(2/4)^3\,M_b$, hence $v_c=220$~\kms
\item Answer {\bf b}: The expression for Oort's constant $A$ is given
  by Eq.~(5.5) in the notes,
\begin{eqnarray}
  A &=& -{1\over 2}\,({dV_c\over dr} - {v_c\over r})\nonumber\\
    &=& -{1\over 2}\,({GM_b\over r_b^3}\,{r\over v_c}-{v_c\over
    r})\nonumber\\
    &=& 48.9~\kms\,,\nonumber
\end{eqnarray}
with $v_c=220~$\kms from above
\end{enumerate}

\noindent{\bf Problem 2.}\\
\begin{enumerate}
\item Answer {\bf b}: The 3D velocity dispersion $\sigma^2_{\rm
  3D}=\sigma^2_{1}+\sigma^2_{2}+\sigma^2_{3}$ which, if the orbits are
  isotropic, $\sigma^2_{\rm 3D}=3\sigma^2_{1D}$, hence $\sigma_{\rm
  3D}=1386$\kms.
\item Answer {\bf b}: The virial mass is given by Eq.~(9.4) in the notes, 
\begin{equation}
M={R\sigma_{\rm 3D}^2\over G}=2.2\times 10^{14}M_\odot\,.
\end{equation}
\item Answer {\bf a}: The virial temperature is given by Eq.~(9.7) in the notes,
\begin{equation}
T = {2\mu m_p\over 3\boltz}\,{GM\over R}=1.7\times 10^8\,{\rm K}\,,
\end{equation}
where the mean molecular weight per particle, $\mu=0.5$, if we assume
fully ionised hydrogen. At this temperature, the typical kinetic energy
of a particle, $\boltz T\sim 15$~keV, is much large than the binding
energy of Hydrogen, $13.6$~eV, which is why the gas is indeed highly
ionised.
\item Answer {\bf b}: Given mass and radius, we can compute the 3D, and
  hence 1D velocity dispersion of the cluster, $\sigma_{\rm
  1D}=\sigma_{\rm 3D}/\sqrt{3}$, with $\sigma_{\rm 3D}^2=GM/R$. Then
  the crossing time is
\begin{equation}
t_{\rm cr}={R\over \sigma_{\rm 1D}}=820\,{\rm Myr}.
\end{equation}
\end{enumerate}
\end{document}



