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\begin{document}

\def\outdate{23 Mar 2012} 
\def\duedate{Due: 2 Apr 2012}
\def\psetno{9}

 
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\noindent Astronomy 62 \hfill Ann Esin
 
\noindent Introduction to Astrophysics \hfill \outdate
 
\vskip0.3cm \hrule height1pt \vskip0.3cm
 
\noindent {\large Problem Set \psetno} \hfill  \duedate


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\noindent {\bf Please staple problems 1+2 and 3+4.}
\vskip 0.3cm

\prob 1. {\bf Planetary Nebula} 
%(based on C\&O, problem 13.6)
\vskip 0.2cm

\noindent The Helix Nebula is a planetary nebula with an angular
diameter of $15^{\prime}$ that is located approximately 120 pc from
Earth.  Like all such objects, it is powered by the UV radiation from
the compact white dwarf remnant of the original star, located at the
center of the nebula.  UV photons are absorbed by atoms in the
expanding shell, and as excited electrons cascade back down to the
ground state, many optical emission lines are produced.

\bl
\item{\bf (a)} Calculate the diameter of the Helix Nebula.

\item{\bf (b)} Assuming that the outer edge of the nebula is expanding
away from the central star at a constant velocity of $20\km\s^{-1}$
estimate its age.

\item{\bf (c)} The evolution of a planetary nebula can be described as
a free expansion, i.e. gravity is negligible, so the outward velocity
of each clump remains constant with time.  Then the material that was
ejected with higher velocity soon overtakes slower moving clumps.
Because of this, a snapshot of the nebula at any moment will show that
the radial velocity of the moving material increases linearly with the
distance from the central star.  Let's say at time $t_0$ the nebula has 
an outer radius $R_0$, thickness $\Delta R_0$, uniform number
density of UV absorbers $n_0$, and optical depth for absorption of UV
photons $\tau_0 \simeq \Delta R_0 n_0 \sigma_{UV}$.  Derive an
equation describing the time evolution of the optical depth of the
nebula at times $t>t_0$.  You can assume that the density remains
uniform throughout the expansion, while its radius and
thickness increase linearly with time.  Start with calculating $n$
as a function of time.

\item{\bf (d)} The energy balance of the nebula requires that its 
luminosity is equal to the fraction of the white dwarf luminosity that
it absorbs.  Use your answer for part (c) to derive an equation for the
total optical luminosity of the nebula as a function of the UV
luminosity of the white dwarf, $L_{\rm UV}$, and other relevant
parameters.  
\el

\prob 2. {\bf Supernova Light Curves}
\vskip 0.2cm

\noindent The emission from a regular supernova is dominated for several months
by the radioactive decay of heavy elements created in the explosion,
as a the blast wave moves through the stellar envelope
(e.g. $^{56}_{27}$Co or $^{56}_{27}$Ni).

\vskip 0.3cm
\noindent {\bf (a)} Suppose we start with $N_0$ atoms of some radiactive
material with a half-life of $\tau_{1/2}$.  Starting with Eq. (15.9) in C\&O, 
show that the amount of radioactive material remaining after time $t$ is given 
by equation:
\be
N(t) = N_0 e^{-\lambda t},
\ee
where $\lambda$ is a constant, given by 
\be
\lambda = \frac{\ln{2}}{\tau_{1/2}}.
\ee

\begin{wrapfigure}{r}{10cm}
\vskip -0.5cm
\includegraphics[width=10cm]{SNlt}
\caption{The bolometric lightcurve of SN 1987A.  The dashed lines show
the contributions expected from the radioactive isotopes produced by
the shock wave. (Figure from Suntzeff et al., 1992. Also Fig 15.12 in C\&O.)}
\vskip 0.2cm
\end{wrapfigure}

\vskip 0.2cm
\noindent {\bf (b)} Assuming that the light curve of a supernova is
dominated by the energy released in the radioactive decay of an
isotope that has a decay constant $\lambda$, show that the slope of
the light curve is given by equation:
\be
\frac{d}{d t}\left(\log_{10}{L}\right) = -0.434 \lambda.
\ee
Thus, by measuring the slope of the light curve, astronomers can verify the
presence of large quantities of a specific radioactive isotope.

Calculate the expected light curve slope from the decay of
$^{56}_{27}$Co with a half-life of 77.7 days and compare your answer
to the data shown in Figure 1

\vskip 0.3cm
\noindent {\bf (c)} The energy released during the decay of one
$^{56}_{27}$Co atom is $3.72\,{\rm MeV}$.  If 0.075$\msun$ of cobalt
was produced by the decay of $^{56}_{28}$Ni following the explosion of
SN 1987A, estimate the amount of energy released per second through
the radioactive decay of cobalt one year after the explosion and compare
your answer to the data shown in Figure 1.


\prob 3. {\bf Stellar Clusters}
\vskip 0.2cm
\noindent
Download the photometry for two clusters, 47 Tuc and M45 (Pleiades).

\bl
\item{\bf (a)} Use the data provided to construct the color-magnitude
  diagram of the two clusters.  Please include your figure in you
  homework solution. Make sure to correctly orient and label your
  axes.

\item{\bf (b)} If the distance to M45 is 135\pc, use the technique of
main-sequence fitting to estimate the distance to 47 Tuc. Do {\it not}
use just {\it one} point on the main sequence!

\item{\bf (c)} The accepted distance to 47 Tuc is around 4.5\,kpc.
  Compare thi\ s to your result in (b).  If there is a discrepancy,
  discuss its likely causes.  Hint: think of the metallicities of the
  two clusters.
\el


\prob 4. {\bf GRB Explosions}
\vskip 0.2cm

\bl
\item{\bf (a)} The neutrino fluence (the particle flux times the
  duration of the burst) from SN 1987A was estimated to be $1.3\times           
  10^{14}$ particles per m$^2$ at the location of Earth.  If the
  average energy per neutrino was approximately $4.2\,{\rm MeV}$,
  estimate the amount of energy released via neutrinos during the
  supernova explosion, if the distance to the SN is roughly 50\,kpc.

\item{\bf (b)} Estimate the gravitational binding energy of a neutron
star with a mass $1.4 \msun$ and a radius of 10\,km.  Compare your
answer with the amount of energy released in neutrinos during the
collapse of the iron core of Sk~-69~202 (the progenitor of SN~1987A).


\item{\bf (c)} A supernova in which the iron core exceeds the maximum
  allowed mass for a neutron star (believed to be about 3-5$\msun$)
  produces a Gamma-Ray Burst (GRB) event in which most of the total
  gravitational energy released in the collapse is emitted as
  $\gamma$-rays with energies $> 50 \keV$, rather than neutrinos.  If
  such an event were to occur in the vicinity of the solar system, it
  might have very real consequences for us.  A fatal dose of radiation
  for a human body is roughly equivalent to $5 \J\kg^{-1}$ of absorbed
  $\gamma$-rays per unit mass.  Estimate the total exposure per person
  ignoring the atmospheric absorption (you can assume that a
  $\gamma$-ray has a 100\% probability of being absorbed by a human
  body) if Betelgeuse were to end its life as a GRB.  Take the
  distance to Betelgeuse to be 150\pc.

\item{\bf (d)} Fortunately, the optical depth of the atmosphere in the
  $\gamma$-ray energy range is on the order of 100.  Revise your
  estimate in (c) to account for atmospheric absorption.  What are
  your conclusions?  \el

\end{document}