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April 11, 2012 23:07
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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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