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\def\au{\, {\rm AU}}
\def\tot{{\rm Tot}}
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\begin{document}
{\it Astronomy 45: Midterm solutions}\hfill {\it Spring
1997}\newline
\hrule
\vskip12pt
\begin{center}
{\bf Astronomy 45.  Introduction to Astrophysics}
\end{center}
\medskip
\noindent{\bf Problem 1}
\smallbreak
\noindent {\bf a)} Parsec is a unit of length in astronomy defined so that
a star at a distance of 1 parsec has a parallax angle equal to 1 arc second.
The parallax baseline in this case is 1 AU.
\medbreak
\noindent {\bf b)} Standard candle is a class of astronomical objects all of
which have the same known luminosity.  Observations of flux from such 
object allows us to determine its distance using the inverse square law.
\medbreak
\noindent {\bf c)} Specific intensity $I_{\nu}$ is defined as energy per unit
frequency, area, time, and solid angle emitted by a source.
\medbreak
\noindent {\bf d)} Boson is any particle that does not obey Pauli exclusion
principle, i.e. bosons like to {\em  clump} together.  For example, photons and
gravitons are bosons.  For those of you who studied or will study quantum 
mechanics, bosons are the particles with integral spin.
\medbreak\noindent{\bf e)} Hertzsprung-Russell diagram is a plot of stellar
luminosties versus their effective blackbody temperatures or colors.

\bigskip
\noindent{\bf Problem 2}
\smallbreak
\noindent {\bf a)} $a = \sqrt{2} R_{\odot} = \sqrt{2}\times 7\times 10^{10}\,\cm/1.5\times 10^{13}\,\cm/\au = 0.0066\,\au$.

\medbreak
\noindent {\bf b)} Kepler's law gives us $P^2 = a^3$, where $P$ is measured in
years, and $a$ is measured in $\au$.  Then 
$$
P = 365\,{\rm days/yr}\times (0.0066)^{3/2}\, {\rm yr} = 0.196\,{\rm days}.
$$

\medbreak
\noindent {\bf c)} 
The key here is to realize that the angular diameter of the Sun is 
$90^{\circ}$.  Then
$$
\Omega = \int{d \Omega} = \int_0^{2 \pi} { \phi\, 
\int_0^{\pi/4}{d \theta\, \sin{\theta}}} = 2 \pi (1-1/\sqrt{2}) \simeq 1.84\,{\rm sterradian}.
$$

\medbreak
\noindent {\bf d)}
Energy per unit time absorbed by the planet is $F_{\odot}\, \pi R^2$, where
$R$ is the planet's radius. Energy per unit time radiated by a planet is
$4 \pi R^2 \sigma T^4$.  The two must be the same at equilibrium, so
$$
F_{\odot} = \frac{L_{\odot}}{4 \pi D^2} = 4 \sigma T^4,
$$
where $D$ is the planet's distance from the Sun.

Now there are two ways to compute $T$.  You can just substitute the 
value for $L_{\odot}$ and $D$ and compute $T$ directly.  Another way is to 
scale from a value for the equilibrium temperature of Earth derived in class.
$T_{\oplus} \approx 300\K$, so 
$$
T \simeq T_{\oplus} \left(\frac{D_{\oplus}}{D}\right)^{1/2} = 
300\K\times (0.0066)^{-1/2} \simeq 3700\K.
$$
\bigskip
\noindent{\bf Problem 3}
\smallbreak
\noindent {\bf a)} We know that 
$$
V-M_V = 5 \log{\left(\frac{D}{10\pc}\right)}.
$$
Then $D = 10\pc\, 10^{(V-M_V)/5} = 10^5\pc$.
\medbreak
\noindent {\bf b)} Again, by the definition of absolution magnitude
$$
M_V-M_{V,\odot} = -2.5 \log{\left(\frac{F_V}{F_{V,\odot}}\right)} =
-2.5 \log{\left(\frac{L}{L_{\odot}}\right)}.
$$
We replaced fluxes by luminosities, since both stars are at the same distance 
(10\pc).  Note that we also assumed that luminosities scale as the 
V-band fluxes.  That is not true in general, but good enough for our 
purposes.  

Then 
$$
L = L_{\odot}\,10^{-(M_V-M_{V,\odot})/2.5} \simeq 230 L_{\odot} = 
9 \times 10^{35}\,{\rm ergs/s}.
$$
\medbreak
\noindent {\bf c)} The observed flux is just  
$$
F = \frac{L}{4 \pi D^2} = \frac{9 \times 10^{35}\,{\rm ergs/s}}{4 \pi\,(
10^5\pc\times 3\times 10^{18}\,{\rm cm/pc})^2} = 8 \times 10^{-13}\,
{\rm ergs/s/cm^2}.
$$

\medbreak
\noindent {\bf d)} The average wavelength of the visible light is $\lambda 
\sim 5500\AA = 5.5\times 10^{-5}\cm$. 
Assuming that the entire flux is radiated in visible energy 
band, the number of photons observed is
$$
n_{\gamma} = \frac{F}{h \nu} = \frac{F \lambda}{h c} = 
\frac{8 \times 10^{-13}\,{\rm ergs/s/cm^2}\ 5.5\times 10^{-5}\cm}
{6.63 \times 10^{-27}\,{\rm ergs\, s}\ 3\times 10^{10}\,{\rm cm/s}} = 
$$
$$
n_{\gamma} = 0.22\ {\rm photons/s/cm^2}.
$$

\bigbreak\noindent{\bf Problem 4}
\smallbreak
\noindent {\bf a)} The total incident flux is equal to the total outgoing flux,
so 
$$
I = x + y.
$$
The same requirement has to hold for the Earth's surface, so
$$
I + w = z,
$$
and for the greenhouse layer,
$$
z = x + y + w.
$$
Finally,  since the greenhouse layer absorbs only 50\% and emits equal amounts
up and down, we must have $z = 2 x$ and $y=w$.

Now we just have to solve this system of equations.  It is not hard to see 
that 
$$y=w = I/3;\ \ \ x = 2 I/3;\ \ \ z = 4 I/3.$$

\medbreak
\noindent{\bf b)} The Earth must emit $4 I/3$ as compared with $2 I$, when the
greenhouse layer is completely absorbing.  Since emission scales as $T^4$, the
new temperature is smaller by a factor of $(2/3)^{1/4}$, so $T = 333\K\ 
(2/3)^{1/4} \simeq 301\K$.

\medbreak
\noindent{\bf b)} In the IR frequency range there are some transparent 
windows, as well as almost completely opaque absoption bands.  So the average
transparency must be somewhere between 0\% and 100\%.

\end{document}