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- \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}
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