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\def\outdate{2 March 2012} 
\def\duedate{Due: 9 Mar 2012}
\def\psetno{7}

 
\def\prob{\medskip\noindent\hskip-16pt }

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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 {\large Please staple problems 1+2 and 3+4.}
%\medskip

\prob 1. {\bf Adiabatic Atmosphere and Greenhouse Effect} 
\vskip 0.3cm

\noindent In HW\#6 we constructed an isothermal (constant
temperature) model of Earth's atmosphere.  In reality, the atmospheric
temperature decreases with height above the sea level, since it is
mainly heated by absorption of the IR radiation from the Earth's
surface.  Since conduction and radiative heat transfer are very
ineffective in our atmosphere, its temperature profile is set by
convection.

\bl
\item{\bf (a)} Combine the equations for hydrostatic equilibrium and
convective temperature gradient,  to show that the temperature drop per 
height $h$ in the atmosphere is given by
\be 
\Delta T = T(R_{\oplus}) - T(R_{\oplus}+h)= 
\frac{\mu m_H}{k} g h \left(1-\frac{1}{\gamma}\right), 
\ee 
where $g = G M_{\oplus}/R_{\oplus}^2$ is gravitational acceleration on 
Earth's surface.  ({\it Hint:} $h \ll R_{\oplus}$, so you can assume that the 
gravitational acceleration $g$ is constant.  The atmosphere also
contains very little mass, so $M_r = M_{\oplus}$.)  
Evaluate your result for $h= 1\km$, keeping in mind that air contains
mainly N$_2$, and its adiabatic exponent is roughly equal to $\gamma\simeq 
1.4$.  

\item{\bf (b)} The Earth's photospheric temperature is set by the thermal
balance with the Sun; in HW\#6 we calculated it to be $T_p =
254\K$.  If the average temperature at sea level is about $280\K$, 
calculate the height above sea level of the Earth's photosphere.

\item{\bf (c)} C0$_2$ is responsible for most of the atmospheric
opacity to IR radiation. Increasing the amount of C0$_2$ in the
atmosphere leads to an increase in the overall optical depth of the
atmosphere and a displacement of the photosphere {\it upward}
(i.e. farther from the Earth's surface).  What happens to the surface
temperature of the Earth under these circumstances?  
This result is generally known as the greenhouse effect.  
\el

\vskip 0.3cm
\prob 2. {\bf Nuclear Reactions}

\bl
\item{\bf (a)} Complete the following reaction sequences.
Be sure to include any necessary leptons.

\ba
^{27}_{14}{\rm Si} &\rightarrow& ^{?}_{13}{\rm Al} + e^+ + ? \nonumber  \\
^{?}_{13}{\rm Al} + ^{1}_{1}{\rm H} &\rightarrow& ^{24}_{12}{\rm Mg} +
^{4}_{?}{\rm ?} \nonumber \\
^{35}_{17}{\rm Cl} + ^{1}_{1}{\rm H} &\rightarrow& ^{36}_{18}{\rm Ar} + ? 
\nonumber
\ea 

\item{\bf (b)} Calculate the efficiency of energy generation from
``burning'' He through triple-$\alpha$ process.  Repeat your calculations
for C and O burning (for these two consider only lower-temperature 
reactions $^{12}_6{\rm C} + ^4_2{\rm He} \rightarrow ^{16}_8{\rm O}$
and $_8^{16}{\rm O} + ^4_2{\rm He} \rightarrow
^{20}_{10}{\rm Ne}$).  Comment on how they compare with each other
and with H burning.  These processes will become important when we
talk about stellar evolution.  You might find the following masses
useful, 
\ba m(^4_2{\rm He}) &=& 4.002603 u; \nonumber \\ 
m(^{12}_6{\rm C}) &=& 12.0 u; \nonumber \\ 
m(^{16}_8{\rm O}) &=& 15.99491 u;\nonumber \\ 
m(^{20}_{10}{\rm Ne}) &=& 19.99244 u, \nonumber 
\ea 
where $u=1.6605\times 10^{-27}\kg$ is the atomic mass unit, defined as
$1/12$ of the mass of $^{12}$C.  
\el

\vskip 0.3cm

\prob 3. {\bf Polytropes}
\vskip 0.2cm

\noindent We can solve the stellar structure equations analytically if
we make an assumption that the pressure equation of state can be
written in the form $P=K \rho^{(n+1)/n}$.  Here $K$ is a constant and
$n$ is called the polytropic index.  Stellar models computed using
this equation are called polytropes of index $n$; models with $n=1.5$ (which 
produces the adiabatic equation $P = K \rho^{\gamma}$) and
$n=3$ describe quite well purely convective and purely radiative stars,
respectively.  Only
models corresponding to $n=0,\,1,\,5$ have analytic solutions.  In this
problem we will build a model with $n=1$, i.e. $P=K\rho^2$.

\bl
\item{\bf (a)} Combine this simplified equation of state with the equations of 
hydrostatic equilibrium and mass
conservation to derive the following second-order differential equation for 
$\rho(r)$:
\be
\label{poly}
\frac{1}{x^2} \frac{d}{d x}\left[\frac{d \rho}{d x} x^2\right] = -\rho,
\ee
where $x = r/r_0$ and $r_0^2 = 2 K/(4 \pi G)$.

\item{\bf (b)} Verify that $\rho(x) = \rho_c \sin{x}/x$ is a solution
to Equation (\ref{poly}).  Sketch this function and find an expression for 
the radius of our model
star, $R$. ({\it Hint:} Think of the boundary condition for $\rho$ at the 
stellar surface.)  What parameters does $R$ depend on? 

\item{\bf (c)} In general, $K$ is a known constant (it depends only on
the physics of the equation of state).  Show that the central density, 
$\rho_c$, in our model depends only on the total mass of the star, $M$.
({\it Hint:} Integrate $\rho$ to find $M$.)  Argue that this implies that 
$\rho(r)$, $P(r)$ and $T(r)$ are completely determined by $M$ as well.  
\el

\newpage
\prob 4. {\bf Eddington Limit and Maximum Stellar Mass}

\vskip 0.3cm
\noindent In this problem you will derive the fundamental physical
limit on the luminosity of any object held together by gravity.
Consider a balance of forces on a proton-electron pair located a
distance $r > R$ from the center of a star with mass $M$, radius $R$
and luminosity $L$.  The proton in pulled inward by the star's
gravity, while the electron is pushed outward by the radiation
pressure from the star.  If the force due to radiation pressure 
overwhelms gravitational force, the pair will be blown away.   

\bl
\item{\bf (a)} Show that the momentum gain for an electron in one
photon-electron scattering is approximately $\Delta p \simeq h \nu/c$, 
where $\nu$ is the photon frequency. 


\item{\bf (b)} The force due to repeated scattering is $f_{\rm rad} = \Delta p/
\Delta t$, where $\Delta t$ is the approximate time between scatterings.  
Show that 
\be
\label{frad}
f_{\rm rad} = \frac{L}{4 \pi r^2} \frac{\sigma_T}{c}, 
\ee 
where $\sigma_T = 6.65\times 10^{-29}\m^2$ is the Thomson cross section for 
photon-electron scattering.  (We are using lower-case $f$ to avoid confusion 
between force and flux.)

\item{\bf (c)} Consider a star composed of pure hydrogen, and use 
Eq. (\ref{frad}) to show that for this star to remain stable, we must have
\be
\label{Edd}
L < \frac{4 \pi G M m_p c}{\sigma_T}.
\ee
The quantity on the right-hand side is called the Eddington Luminosity,
and is generally denoted by $L_{\rm Edd}$.  Note that it is independent 
of the distance from the star.  Stars massive
enough to have $L > L_{\rm Edd}$ will simply blow themselves apart.

\item{\bf (d)} Combine Eq. (\ref{Edd}) with the 
mass-luminosity relation, $L/L_{\odot}=(M/\msun)^{3.5}$ to estimate the 
maximum allowed mass for a main-sequence star.

\item{\bf (e)} The presence of metals in atmospheres of real stars
increases the cross section for electron scattering far above $\sigma_T$.
How would this affect your result in part (d)?
\el

\vskip 0.3cm



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