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\begin{document}

\section*{Stars \& galaxies, homework problems}

\noindent{\bf Problem 3.}\\

\begin{enumerate}
\item $T(r)$ is determined from modelling the line-strengths in the
X-ray spectrum, $\rho(r,T)$ from the X-ray emissivity due to thermal
bremsstrahlung, $\propto \rho^2 T^{1/2}$.

\item In hydrostatic equilibrium, $GM/r^2 = -(1/\rho_g)dp/dr$, where
$p=(\boltz T/\mu m_p)\rho_g$ is the gas pressure and $\rho_g$ the gas
density. This simplifies when $T$ is constant to $GM = -(\boltz T/\mu
m_p) r^2d\ln\rho_g/dr$. If $\rho_g\propto 1/r^2$, the rhs of this
equation becomes $(2\boltz T/\mu m_p) r$, or $M(<r)\propto r$. Taking
the derivative of both sides wrt $r$ shows $\rho_{\rm tot}(r)\propto
1/r^2$, where $\rho_{\rm tot}$ is the {\em total} mass density (stars,
gas, dark matter).

\item The mass is $M(<r)=(2\boltz T/G \mu m_p) r = 3.1\times
10^{12}\msun$ since $\mu=0.5$ for a fully ionised hydrogen gas.

\item Writing the density as $\rho(r)=A/r^2$, and assuming
  $\sigma(r)=\sigma_0$ is constant, the Jeans
  equation becomes
\begin{equation}
{G(M<r)\over r^2} = {2\boltz T\over \mu m_p}\,{1\over r} = {2\over r}\,\sigma_0^2\,.
\end{equation}
Hence this is a solution for $\sigma_0=575$ km s$^{-1}$.

\item Using the previous result, the circular velocity
$V_c^2=GM(<r)/r=2\sigma_0^2$. Then we use Eq.~(6.5) in the lecture
notes to find for the escape velocity $v_e$,
$v_e^2=2V_c^2(1+\ln(r_\star/r))$ or $v_e=1.5\times 10^{3}\kms$.

\end{enumerate}

\end{document}



