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\begin{document}
{\it Astronomy 45 solutions to problem set 1}\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)} Let $\w$ and $\w_\oplus$ be the orbital angular
velocities of a planet and of Earth respectively. Then the angular velocity
of the planet in the frame of reference in which Earth is stationary is
simply
$$
\w' = |\w -\w_\oplus |\;.
$$
By definition of the synodic period, $S={2\pi\over \w'}$, and of course
$P={2\pi \over \w}$ and $P_\oplus = {2\pi \over \w_\oplus}$.  Then we get
$$
{1\over S} = \l| {1\over P}-{1\over P_\oplus}\r|\;.
$$

For {\it inferior} planets, $P<P_\oplus$, so  \fbox{${1\over S}={1\over
P}-{1\over P_\oplus}$}.
\smallbreak
For {\it superior} planets, $P>P_\oplus$, so \fbox{$\frac1S =
\frac1{P_\oplus}-\frac1P$}.
\bigbreak\noindent{\bf b)} One way to solve this problem is of course to
compute the synodic period for every planet.  However, one can do it in a
more general way.  From part (a) we know that $S={P_\oplus P\over
|P-P_\oplus|}$.  We can plot this relationship to see how $S$ changes with
$P$.
\figin{1}{1000}

For $P\gg P_\oplus$, $S\to P_\oplus$.  For $P\ll P_\oplus$, $S\to P$.

From this it is clear, that interior planets with sidereal periods less than
${P_\oplus \over 2}$ have the shortest synodic periods.  Only Mercury
satisfies this criterion, so its synodic period is the shortest
\medbreak
\centerline{\fbox{$S_M = 0.317$ year $= 115.9$ days} }
\medbreak
 

Note that Venus, the next planet after Mercury has $P=0.62 \yr$ so if
Mercury did not exist, computing the synodic period for every planet would
be the only way to solve this problem.

\bigbreak\noindent{\bf Problem 2}
\medbreak\noindent{\bf a)} Let's do this problem in terms of angles and
angular velocities.
\figin{2}{1000}

Angular velocity of Venus in Earth's reference frame is
$$
\w'_V = \w_V - \w_\oplus = \underbrace{2\pi \over P_V}_{\hbox{sidereal
period}\atop\hbox{ of Venus} } - \underbrace{2\pi\over
P_\oplus}_{\hbox{sidereal period}\atop\hbox{of Earth}} \;.
$$
We know that 
\begin{eqnarray*}
P_V = 0.615\yr  &=& 0.615\yr \times 365.26 \,\hbox{days/yr} \times
24\, \hbox{hours/day}\\
&=& 53913\  \hbox{hours}\end{eqnarray*}
and
\begin{eqnarray*}
P_\oplus &=& 1\yr \times 365.26 \,\hbox{days/yr} \times 24\,
\hbox{hours/day}\\
\Rightarrow \w'_V&=& 4.49\times 10^{-4}\,
\hbox{rad/hour}\;.\end{eqnarray*}
Now, the angular distance Venus has to travel is just
$$
x=2R_\odot \cos 42^\circ = D_\odot \cos 42^\circ = 32' \cos 42^\circ = 23'78
$$
or
$$
x=6.92\times 10^{-3} \, \hbox{rad}\;.
$$
Therefore, the crossing time is
$$
\Delta t = {x\over \w'_V } = {6.92\times 10^{-3} \rad \over 4.49\times
10^{-4} \radhr} = \hbox{\fbox{15.4\ hours}}
$$
\medbreak\noindent{\bf b)} This is a very trigonometry-intensive problem. 
To get a general picture of what is happening, consider the following figure:
(the figure is of course not to scale) 
 \figin{3}{700}
\noindent So the southern observer will actually see Venus crossing more
to the North of the Sun, at some latitude $(42^\circ + \theta)$.  And our
object is to find $\theta$.

Now, the radius of the Sun is very small compared with $1\au$
($\frac{R_\oplus}{1\au} = 0.0047<1\%$), so we can consider that the line
from Earth to Venus to the Sun is actually parallel to the ecliptic.

\figin{4}{1000}
\vglue-36truept
The distance $x$ is easily computed:
$$
{8000\km\over (1-0.72)\au} = {x\over 0.72\au}\ \ \ \Rightarrow\ \ \  x=20571\km
$$
Now we assume that $\theta$ is very small, so that the segment $AB$ is
roughly perpendicular to $AO$, and the angle $BAC$ is $42^\circ$.  Then 
$AB={x\over \cos 42^\circ}=27682\km$.  Again, since $\theta$ is small,
the arclength between the points $A$ and $B$ is roughly equal to $AB$, so
$$
\theta \simeq {AB\over R_\odot} = {27682\km\over 6.96\times
10^{10}\cm} = {2.77\times 10^9\cm \over 6.96 \times 10^{10}\cm} =
0.0398\rad = \fbox{$2^\circ.28$}
$$
So $\theta$ is indeed small and our approximations are valid.  Thus, the $S$
observer sees Venus cross at \fbox{$44.28^\circ N$.}

\medbreak\noindent{\bf c)} We assume that both observers see Venus cross
the mid line of the Sun at the same time.  Then the difference in the time
when Venus touches the Sun's disk is just
\begin{eqnarray*}
\Delta t  =  {x_N-x_S\over 2\w'_V}& =& {2R_\odot \cos 42^\circ - 2R_\odot
\cos 44^\circ.28\over 2\w'_V} = {R_\odot\over \w'_V} (\cos 42^\circ-\cos
44^\circ.28) \\
&=& {16'\over 4.49\times 10^{-4}\radhr} \times 0.0272=0.282\hr = 16\min
55.32\sec\;.\end{eqnarray*}
\figin{5}{700}
Thus, the $S$ observer sees Venus touch the sun at \fbox{11:16:55.32}.

\bigbreak\noindent{\bf Problem 3}
\medbreak\noindent{\bf a)} 

The parallax angle of Mars is $\theta=33''.6=1.63\times 10^{-4}\rad$.  
Since $\theta$ is small,
\begin{eqnarray*}
D&\simeq& {D_\oplus\over \theta} = {2\times 6.38\times 10^8\cm\over
\theta} = \fbox{$7.83\times 10^{12}\cm$} = 0.52\au\;.
\end{eqnarray*}
\figin{6}{1000}
\noindent{\bf b)} Let's first show that Earth's rotation can be ignored. 
Orbital velocities of Earth and Mars are
\beq
v_\oplus = \w_\oplus a_\oplus\;, && a_\oplus = \hbox{distance from Earth
to the Sun}\\
v_M=\w_Ma_M\;, && a_M = \hbox{distance from Mars to the Sun}
\eeq
But
\beq
a_\oplus &=& 1\au = 1.496\times 10^{13}\cm\\
a_M &= &a_\oplus +D=1.496\times 10^{13}\cm + 7.83 \times 10^{12}\cm =
2.28\times 10^{13}\cm \eeq
and
\beq
\w_\oplus& = &{2\pi \over P_\oplus} = 1.99\times 10^{-7} \rads\\
\w_M &= &{2\pi\over P_M} = {2\pi \over 1.8809\yr} = 1.06\times
10^{-7}\rads\eeq
so
\beq
v_\oplus &= &2.977 \times 10^6 \cms = 29.77\kms \\
v_M&=& 2.41\times 10^6\cms = 24.13\kms \;.\eeq
The relative velocity between Earth and Mars-to-Sun line (we are assuming
that mars is fixed and Earth is flying past it) at the time of opposition is
$$
v_R=v_\oplus - v_M = 5.64\kms.
$$  
The velocity of a point on
Earth's equator due to Earth rotation is 
$$
v= R_\oplus {2\pi \over 1\hbox{day}} = 6.38\times 10^8 \cm \times 7.27 
\times 10^{-5} \rads = 0.46\kms, 
$$ 
where $R_\oplus$ is the Earth's radius.  So $v$ is
$\sim 12$ times smaller than $v_R$ and since we are looking for a 10\%
accuracy, it is negligible.

Now let's comput how closely the clocks must be synchronized.  The
astronomers in 1672 would assume that they know the baseline
$(D_\oplus)$ well.  So to measure $D$ to within 10\%, we must measure
$\theta$ to within 10\%.  Since we know what the distance to Mars really
is, we can compute what that means for the time synchronization.
$ \theta = {  \hbox{Baseline}\over D}$, where Baseline $= D_\oplus \pm
v_R\Delta t$, where $\Delta t$ is the difference between the observer's
clocks.  So 
$$
\Delta \theta={v_R\Delta t\over D}< 0.1\theta,\ \ \ \ \mbox{\rm and}
\ \ \ \ \Delta t<{ 0.1\theta D\over v_R}\;.
$$
Assuming that values in part (a) for $\theta$ and $D$ are correct, we get
$$
\Delta t <{0.1\times 1.63\times 10^{-4} \times 7.83 \times 10^7 \km \over
5.64 \kms} = 226\sec = \fbox{$3.8\min$}
$$
So the observers' clocks had to by synchronized to $\sim 4 \min$!

\bigbreak\noindent{\bf Problem 4}
\medbreak\noindent{\bf a)}

\begin{tabular}{l|l|l}
& $\alpha$ & $\delta$ \\
\hline
$\alpha$ Centauri & $219^\circ.900$ & $-60^\circ.83$\\
Prox.\ Centauri & $216^\circ.575$ & $-62^\circ.467$ \end{tabular}
(We converted $\alpha$ to degrees.)
\medbreak
Remember that the coordinates on a unit sphere are:
\beq
z & = & \sin \delta\\
x&=& \cos \delta\cos \alpha\\
y&=& \cos \delta \sin \alpha\eeq
Let $\theta$ be the angle between the two stars.  Then
$$
\tan \l({\theta\over 2}\r) = \l| {\hat r_1 - \hat r_2\over \hat r_1 + \hat
r_2}\r|\;,
$$
where $\hat r_1$ and $\hat r_2$ are the position vectors for the two
stars.  Then
$$
\tan \l( {\theta\over 2}\r)  =  \left[{ (x_1-x_2)^2 + (y_1-y_2)^2 + (z_1 -
z_2)^2\over (x_1+x_2)^2 + (y_1+y_2)^2 + (z_1+z_2)^2}\right]^{1/2}
=$$ 
$$ \tan \l({\textstyle {\theta\over 2}}\r)  =  \left[ {\textstyle {(\cos
\delta_1 \cos \alpha_1 -
\cos\delta_2\cos
\alpha_2)^2 + (\cos
\delta_1 \sin \alpha_1-\cos \delta_2\sin\alpha_2)^2+ (\sin \delta_1-\sin
\delta_2)^2\over
(\cos \delta_1 \cos \alpha_1+\cos \delta_2 \cos\alpha_2)^2 + (\cos
\delta_1\sin \alpha_1 + \cos \delta_2\sin \alpha_2)^2 + (\sin \delta_1 +
\sin \delta_1)^2}}\r]^{1/2} \;.
$$
Expanding the squares and using basic trigonometry laws ($\cos^2 x + \sin^2 x
=1$;  $\cos x \cos y + \sin x \sin y = \cos[x-y]$), we get
$$
\fbox{$
\tan \l({\theta\over 2}\r) = \l[ {1-\sin \delta_1 \sin \delta_2 - \cos
\delta_1\cos \delta_2 \cos(\alpha_1-\alpha_2)\over 1+\sin \delta_1\sin
\delta_2+\cos \delta_1\cos\delta_2\cos(\alpha_1-\alpha_2)}\r]^{1/2}$}
$$
Substituting the values for  $\delta_1,\delta_2$ and $\alpha_1,\alpha_2$,
we finally obtain
$$
\tan \l({\theta\over 2}\r) = 0.0198
$$
and
$$
\theta= 2\arctan (0.0198) = \fbox{$2^\circ.27 = 0.04\rad$.}
$$
\medbreak\noindent{\bf b)} Projected distance (distance in the plane of
Proxima Centauri) is
\beq
x&=& \theta \cdot D_{\hbox{Pr.\ Cen}} = 0.04\times 4.3\lyr =
\fbox{$0.17\lyr$} \\
y &=& 4.4\lyr - 4.3\lyr = 0.1\lyr.\eeq
Then 
$$
D_\ast = \sqrt{x^2+y^2} = \fbox{$0.2\lyr$}
$$
\figin{7}{1000}
\medbreak\noindent{\bf c)} By definition of the apparent magnitude:
$$
m_{\rm tot} = {\rm const} - 2.5\log F_{\rm tot}$$
where
$$
F_\tot = F_1+F_2 +F_3 = \hbox{sum of the fluxes}
$$
But
\beq
m_1 &=& \const - 2.5\log F_1; \Rightarrow F_1 = 10^{(\const-m_1)}/2.5\\
m_2 &=& \const - 2.5\log F_2; \Rightarrow F_2 = 10^{(\const - m_2)}/2.5\\
m_3 &=& \const - 2.5\log F_3; \Rightarrow F_3 = 10^{(\const -
m_3)}/2.5\; . \eeq
So
\beq
m_\tot &=& \const - 2.5 \log \l[ 10^{\const/2.5} \l(10^{-m_1/2.5} +
10^{-m_2/2.5} + 10^{-m_3/2.5}\r)\r] \\
&=& \hskip10pt \big/ \hskip-15pt\const - 2.5
\hskip20pt \Big/\hskip-20pt{\const\over 2.5} - 2.5
\log
\l[ 10^{-m_1/2.5}+10^{-m_2/2.5} + 10^{-m_3/2.5}\r] \\
&=& -2.5\log \l[ 10^{-m_2/2.5} + 10^{-m_2/2.5} +10^{-m_3/2.5}\r] \;.\eeq
We know that $m_1=0$, $m_2=1.3$, and $m_3=11.0$  Then \fbox{$m_\tot =
-0.287$}.

The total flux in the visual band can be computed using formula on p.~17 in
the notes.
$$
F_C = 10^{-4.42} \times 0.098\mu\times 10^{-0.4(-0.287)} {\erg\over
\cm^2\s\mu} = \fbox{$4.4\times 10^{-6} {\erg\over \cm^2\s } $}
$$
\medbreak\noindent {\bf d)} $M= m-5\log \l( {D\over 10\pc}\r)$
\medbreak
\begin{tabbing}
\={\bf Proxima Cen}: \= $M=11.0, D=4.3\lyr = 1.32\pc$\kill
\>$\alpha$ {\bf Cen}:\>$m=0.0, D=4.4\lyr = 1.35\pc$\ \ \ $\Rightarrow$ \fbox{$M=4.35$}\\
\>{\bf Proxima Cen}:\>$m=11.0, D=4.3\lyr = 1.32\pc$\ \ \ $\Rightarrow$ \fbox{$M=15.4$}\\
\>{\bf Third star}:\>$m=1.3, D\simeq 4.35\lyr = 1.33\pc$\ \ \ $\Rightarrow$ \fbox{$M=5.67$} 
\end{tabbing}
\medbreak
To get the spectral types and masses of these stars, we use the table on
page~18 of the notes.

$\alpha$ Cen has spectral type between $F$ and $G$, and mass $\sim
1.2M_\odot$.

Proxima Cen has spectral type $M$ and mass $\sim 0.2M_\odot$.  

The third companion has spectral type between $G$ and $K$ and mass $\sim
0.9 M_\odot$.

This part might not make too much sense to you now, but later you will
learn more about it.
 \end{document}

