Sample Latex file

astrofinal.tex

\input{latex_template.tex}
\title{Astrophysics Final Exam}
\begin{document}
\maketitle

\section{Helium-burning stars}
\begin{enumerate}
\item To find the energy liberated in the reaction, we must find the mass lost
in the reaction, then use $E=mc^2$ to find the equivalent quantity of energy
released.
\begin{align*}
E_R &= c^2(4m_{\mathrm{He}} - m_\mathrm{O}) \\
E_R &= (3 \times 10^{10} \textrm{cm s}^{-1})^2 
  (4 \cdot 4.002603 m_p - 15.994915 m_p) \\
E_R &= 9 \times 10^{20} \mathrm{cm}^2 \mathrm{s}^{-2}
  \cdot 0.015497 m_p \\
E_R &= 9 \times 10^{20} \mathrm{cm}^2 \mathrm{s}^{-2}
  \cdot 2.5880 \times 10^{-26} \mathrm{g} \\
E_R &= 2.3292 \times 10^{-5} \mathrm{erg}
\end{align*}
\item To find the number of reactions, we first take the luminosity and divide
by the amount of energy released in every reaction to get the number of
reactions per unit time.
\begin{align*}
r &= \frac{L}{E_R} \\
r &= \frac{3 \times 10^{38} \,\mathrm{erg} \, \mathrm{s}^{-1}}{
  2.3292 \times 10^{-5} \,\mathrm{erg}} \\
r &= 1.288 \times 10^{43} \,\mathrm{s}^{-1}
\end{align*}
Then multiply by the amount of time passed to get the number of oxygen atoms
generated:
\begin{align*}
n_\mathrm{O} &= tr = 3.15569 \times 10^{13} \,\mathrm{s}
  \cdot 1.288 \times 10^{43} \,\mathrm{s}^{-1} \\
n_\mathrm{O} &= 4.0645 \times 10^{56}
\end{align*}
\item First, find the number of Hydrogen atoms present in the star at the start
of its lifetime by dividing its mass by the mass of a Hydrogen atom.
\begin{align*}
n_\mathrm{H} &= \frac{20 \sol{M}}{m_\mathrm{H}} \\
n_\mathrm{H} &= \frac{20 \cdot 2 \times 10^{33} \mathrm{g}}{1.67 \cdot 10^{-24}} \\
n_\mathrm{H} &= 2.3952 \times 10^{58}
\end{align*}
Then divide the number of Oxygen atoms by this value to get the number of Oxygen
atoms created per Hydrogen atom
\begin{equation*}
\frac{n_\mathrm{O}}{n_\mathrm{H}} 
  = \frac{4.0645 \times 10^{56}}{2.3952 \times 10^{58}} = 0.0169
\end{equation*}
The star generates one Oxygen atom for every 100 Hydrogen atoms.
\end{enumerate}

\section{White dwarf stars}
\begin{enumerate}
\item Assume that the mass of the electrons are negligable, and all of the white
dwarf's mass is due to the ion gas of Carbon. To find the number of Carbon atoms
in the white dwarf, divide the mass of the white dwarf by the mass of a Carbon
atom
\begin{equation*}
N_i = \frac{0.5 \sol{M}}{12 m_p} =
  \frac{0.5 \cdot 2 \times 10^{33}}{12 \cdot 1.67 \times 10^{-24}} =
  5.988 \times 10^{56}
\end{equation*}
The total amount of kinetic energy in the white dwarf star is therefore
\begin{align*}
K &= N_i \cdot 1.5kT \\
  &= 5.988 \times 10^{56} \cdot 1.5 
  \cdot 1.38 \times 10^{-16} \textrm{erg K}^{-1}
  \cdot 10^8 \mathrm{K} \\
  &= 1.2395 \times 10^{49} \mathrm{erg}
\end{align*}

\item To find the amount of time the white dwarf will take to radiate this
energy at a luminosity of $L = 10^{-4} \sol{L}$, divide the total kinetic energy
by the luminosity of the white dwarf
\begin{align*}
t &= \frac{K}{L} = \frac{K}{10^{-4} \sol{L}} \\
t &= \frac{1.2395 \times 10^{49} \mathrm{erg}}
  {10^{-4} \cdot 4 \times 10^{33} \textrm{erg s}^{-1}} \\
t &= 3.0988 \times 10^{19} \mathrm{s} = 9.826 \times 10^{11} \mathrm{yrs}
\end{align*}

\item The galaxy is not old enough for any of these white dwarves to have formed
yet.
\end{enumerate}

\section{Neutrino astronomy}
\begin{enumerate}
\item First, find the amount of energy which passed through the detector
\begin{align*}
E_D &= 20 \cdot 1.6 \times 10^{-6} \textrm{ erg} \cdot 3 \times 10^8 \\
E_D &= 9.6 \times 10^3 \textrm{ erg}
\end{align*}
Now, calculate the proportion of the energy which passed through the detector,
by comparing the surface area of the detector to the surface area of the sphere
centered at the location of the supernova whose boundary passes through the
detector. Note, this assumes that the area of the detector orthogonal to the
radius of this imaginary sphere is 1 $\mathrm{cm}^2$.
\begin{align*}
\frac{1 \mathrm{cm}^2}{4 \pi r^2} =
  \frac{1 \mathrm{cm}^2}{4 \pi (3.1 \times 10^{21} \mathrm{cm})^2} 
  = 8.2807 \times 10^{-45}
\end{align*}
If we divide the energy the detector found by this quantity, we will get an
estimate for the amount of energy released by the supernova:
\begin{align*}
E &= \frac{9.6 \times 10^{3}}{8.2807 \times 10^{-45}} \\
&= 1.1593 \times 10^{48} \mathrm{erg}
\end{align*}

\item We can calculate the approximate gravitational energy released in the
collapse by taking the absolute value of the gravitational binding energy
\begin{align*}
E_\mathrm{grav} &= \left| \frac{-GM^2}{R} \right| \\
E_\mathrm{grav} &= \left| \frac{-6.67 \times 10^{-8} \cdot
  (1.5 \cdot 2 \times 10^{33})^2}{10^7 \mathrm{cm}} \right| \\
E_\mathrm{grav} &= 6.00 \times 10^{52} \mathrm{erg}
\end{align*}

\item These results suggest that most of the energy is not released by
neutrinos, as the estimated amount of energy carried away by neutrinos is four
orders of magnitude less than the gravitational energy that must have been
released. However, there's a significant amount of error present in our estimate
for the energy carried away by neutrinos, because we have to estimate both the
energy of the neutrinos detected on Earth and the distance between us and the
supernova, so this result could be due to error.
\end{enumerate}
\end{document}