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\hspace*{3.0in}Name:{\hrulefill} \\
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\begin{Large}
\section{\bf Discovering Exoplanets}
\end{Large}

\subsection{Introduction}

One of the most exciting discoveries in Astronomy over the last twenty years 
was the conclusive detection of planets orbiting other stars. At last count,
we are closing-in on having discovered {\it two thousand} planets orbiting
other stars. Planets orbiting other stars are called ``exoplanets.'' These
exoplanets range in size from similar to the Earth, to larger than Jupiter.
With much hard work, we now know that small
exoplanets are much more common than big exoplanets, and some astronomers
believe that Earth-sized planets orbit nearly every normal star. The current
goal of astronomers is to find exoplanets that are most similar to Earth
(same mass, radius, orbiting their host star at 1 AU, etc.). With improvements 
in technology, we will one day be able to determine whether such exoplanets
support life. In the distant future, maybe we will be able to send a space
probe to those exoplanets to investigate the life found there.\\

Astronomers have been studying the sky with advanced instruments for more than
100 years, but it was only in the early 1990's that the first real
exoplanets were found. Why did it take so long? The answer is that compared
to their host stars, exoplanets are tiny, and hard to see. We will quantify how
hard it is to see them shortly. First though, how might we discover such 
objects? There are three main techniques: direct imaging, transits 
(mini-eclipses), and ``radial velocity'' measurements. As its name suggests, 
direct imaging is simply taking a picture of a star and looking for its 
planets. The big problem is that the star is very bright (it generates its 
own energy), while an exoplanet shines by reflected light from the star. This 
is by far the hardest method to find exoplanets. To be effective, we will need 
to launch special telescopes into space where our image-disturbing atmosphere 
does not exist, allowing us to see much, much more clearly.\\

 The transit method 
is much easier in that what we monitor is the light output from a star, and if 
an exoplanet crosses in front of the star, the light briefly dims. As we will 
learn, this technique also tells us the {\it diameter} of the exoplanet. The 
radial velocity method uses the Doppler effect to detect the orbital motion of 
the planet. The radial velocity technique allows us to determine the 
{\it mass} of the exoplanet.  If we can combine the transit and radial velocity 
techniques, we can get the size and mass of a planet, and thus measure its 
{\it density}, and therefore constrain its composition. We will investigate all 
three methods in this lab, and then learn how we can characterize the 
properties of these objects.\\

\section{Why are Exoplanets so hard to see?}

In our first experiment, we are simply going to demonstrate how hard it is
to directly see an exoplanet. First, however, a diagram to remind you how
small the Earth and Jupiter are compared to the Sun (Figure \ref{suncomp}).\\

\begin{figure}[htb]
\centerline{{\includegraphics[width=7cm]{sunearthjupiter.jpg}}}
\caption{Comparison of the size of the Earth and Jupiter to the Sun.}
\label{suncomp}
\end{figure}

Let's look at some numbers. The radius of the Sun is 695,550 km, the radius
of Jupiter is 69,911 km, and the radius of the Earth is 6,371 km. Note that
these objects are all spheres, and thus when we look at them from space, they
all appear to be circles (``disks''). What is the area of a circle? 
A$_{\rm circle}$ = $\pi$R$^{\rm 2}$.\\

\begin{flushleft}
{\bf Exercise \#1:} Calculate the areas of the circular disks of the Sun, 
Jupiter, and Earth (if you want make the calculation simpler, just set 
R$_{\rm Sun}$ = 700,000 km, R$_{\rm Jupiter}$ = 70,000 km, and 
R$_{\rm Earth}$ = 6000 km).
{\bf (3 points)}.\\
\vspace{0.2in}
(Area of Earth) = {\underline{\hskip 2in{}} } km$^{\rm 2}$\\ %4.06e7
\vspace{0.2in}
(Area of Jupiter) = {\underline{\hskip 2in{}} } km$^{\rm 2}$\\ %4.89e9
\vspace{0.2in}
(Area of Sun) = {\underline{\hskip 2in{}} } km$^{\rm 2}$\\ %4.84e11
\vspace{0.2in}
\end{flushleft}
If we are going to take a picture of a exoplanet around a star, we have two
problems: how much light will the exoplanet reflect compared to its star,
and how close-in is it? Let's tackle the first question.\\

\begin{flushleft}
{\bf Exercise \#2:} We are going to keep everything very simple, and just
estimate how much sunlight the Earth or Jupiter would reflect compared to
that emitted by the Sun. We will assume that these planets reflect 100\%
of the light that hits them, and we are going to ignore the fact that the amount
of sunlight at the orbits of each of these planets is less than at the surface 
of the Sun (remember, the amount of light passing through a sphere surrounding
a light source drops off as 1/R$^{\rm 2}$). In this unrealistic scenario,
the maximum amount of light that a planet can reflect is simply the ratio of
its area to that of the star it orbits. Calculate the following:
{\bf (2 points)}.\\
\vspace{0.2in}
(Area of Earth)/(Area of Sun) = {\underline{\hskip 2in{}} }\\ %8.39e-5 
\vspace{0.2in}
(Area of Jupiter)/(Area of Sun) = {\underline{\hskip 2in{}} }\\ %0.01
\vspace{0.2in}
\end{flushleft}

These already small numbers are actually way too big. As we noted, Earth
can only reflect the amount of light it intercepts at the distance it
is from the Sun. In fact, the Earth only intercepts 1.67 $\times$ 10$^{\rm -9}$
of the Sun's light output, and the amount of visible light it reflects
(its ``albedo'') is 40\%. So, seen from distant space, the Earth
is only {\it one billionth} as bright as the Sun! Jupiter is obviously
much bigger than the Earth, but remember, Jupiter is at 5.2 AU, so it
actually receives 1/27$^{\rm th}$ the amount of sunlight as does the Earth.
Thus, to an observer outside our solar system, Jupiter is only 4.4 times more 
luminous than the Earth. \\

Directly detecting exoplanets is going to be hard, besides being very faint,
they are located very close to their host stars. We need a way to ``turn off''
the star. One way to do this is to block its light out with a small,
opaque metal disk. As shown in Figure \ref{exoimage},
we now have the capability to do this, but only for finding big planets located
far from their host stars (in fact, to date, only Jupiter-sized planets
located at large distances from their host stars have been directly imaged).
There is a more complex technique called ``nulling interferometry'' where
you use the star's own light to cancel itself out, but not its planets,
that lets astronomers search for planets closer to the host star. While
it can be done from the ground, it is better from space. You 
can more read about this method by searching for the canceled NASA mission
``Terrestrial Planet Finder'' on the web (it was killed due to budget cuts).\\
\begin{figure}[htb]
\centerline{{\includegraphics[width=9cm]{Fomalhaut_B.jpg}}}
\caption{A planet orbiting the star Fomalhaut (inside the box, with
the arrow labeled ``2012''). This image was obtained with the Hubble
Space Telescope, and the star's light has been blocked-out using a small metal 
disk. Fomalhaut is also surrounded by a dusty disk of material---the broad 
band of light that makes a complete circle around the star. This band of dusty 
material is about the same size as the Kuiper belt in our solar system. The 
planet, ``Fomalhaut B'', is estimated to take 1,700 years to orbit once around 
the star. Thus, using Kepler's third law (P$^{\rm 2}$ $\propto$ a$^{\rm 3}$),
it is roughly about 140 AU from Fomalhaut (remember that Pluto orbits at 
39.5 AU from the Sun).}
\label{exoimage}
\end{figure}

\section{Exoplanet Transits}

The exoplanet transit method of discovery is simple to envision, and
the easiest to carry-out. As shown in Figure \ref{transit}, a transit occurs
when an exoplanet crosses the disk of its host star as seen by observers on
Earth. Since the planet does not emit any light (we are looking at the 
``nighttime side''), it is completely dark. Thus, the amount of light 
from the star will dim as the planet blocks out (``eclipses'') a small portion 
of the star's light-emitting disk. The plot of the brightness of a star versus 
time is called a ``light curve''. The light curve of the transit is shown below
the cartoon of the star and exoplanet in Figure \ref{transit}.\\
\begin{figure}[htb!]
\centerline{{\includegraphics[width=7cm]{exotransit.jpg}}}
\caption{The diagram of an exoplanet transit. The planet, small, dark 
circle/ disk, crosses in front of the star as seen from Earth. In the process, 
it blocks out some light. The light curve shown on the bottom, a plot of 
brightness versus time, shows that the star brightness is steady until the 
exoplanet starts to cover up some of the visible surface of the star. As it 
does so, the star dims. It eventually returns back to its normal brightness 
only to await the next transit.}
\label{transit}
\end{figure}

\begin{flushleft}
{\bf Exercise \#3: Simulating an Exoplanet transit}\\
\vspace{0.1in}
As part of the materials set out for you to use during today's lab is a
device to simulate an exoplanet transit. Take a look at the wooden device.
It has a light meter attached to the back, and three rods that dangle
down in front of the light meter. We will use the desk lamp as our light 
source (``star''), and move the rods across the light meter. Note that the 
dowel rod on top has five notches. The two furthest from the center represent
where we will be at the start and end of our simulation. At these positions, 
no light is blocked (similar to position \#1 in Figure \ref{transit}). Note also 
that we have two planets, one big, one small, and a bare rod without a planet. 
What do you think the latter is used for? Yes, our planets need to be attached 
to {\it something} to allow us to perform this experiment. Thus, this 
planet-less rod allows us to measure how much light just the bare rod blocks
out.  We will have to take this into account when we plot our light curves.\\
\vspace{0.1in}

The light meter itself is rather simple, it has power, mode, and hold buttons. 
We only will use the power and mode buttons. Hit the power button (note
that to extend battery life, the device automatically shuts off after 45
seconds).  At the bottom of the device display window there will either be a
``LUX'' or ``FC'' displayed. We want the unit to be in LUX, so click the mode 
button until LUX is displayed.\\
\vspace{0.08in}
{\bf Setting things up:} Move all of the metal rods to the left end of the 
dowel rod so that nothing is blocking the lamp light from illuminating the 
white circle. Turn on the light meter. Note that the number bounces 
around--this is due to electronic noise. Every electronic device has this type 
of noise, and it takes much hard work (and expense) to minimize this noise 
(one way is to chill the device to low temperatures). Here we have to live 
with it, but this is just like what an astronomer would have to deal with in a 
real observation. You are going to have to make a mental average of the values 
at each measurement point. For example, in five seconds, if the numbers are 
78, 81, 79, 82, and 78, we would just estimate the count rate as ``80''. Note: 
the light meter is very sensitive, so you must keep yourself and your hands 
well away from the front of the device when making a measurement (the meter 
will detect light reflected off of {\it you}, making it hard to figure out what
is going on!).\\
\vspace{0.08in}
With the room lights turned off, set the desk lamp about two feet in front
of the transit device. Power on the light and the light meter. With no rods
in front of the glass disk, adjust the {\it height and direction} of the desk 
lamp to maximize the number of counts. Make sure the light bulb in the lamp is 
at roughly the same height as the round, glass disk in front of the light 
meter. One way to do this is move the big planet in front (putting the rod in 
the centermost notch) and make sure its shadow hits the center of the glass 
disk. Move all of the rods out of the way, and then move the transit device 
closer to the lamp until it gives a reading above 200 counts. \\
\vspace{0.1in}
Now we are simply going to move each of the three rods (Bare Rod, Small Planet,
Large Planet) into the five notches on the top dowel rod, and write down the 
average value of the light meter measurement at each position into Table 
\ref{measurement}. We do this one rod at a time. Once done, move that rod
to the far right side of the dowel rod to start the process for the next
rod. The rods may swing around a bit, just let them stop moving, back away
from the front of the device, and take your measurement. It sometimes takes
a few seconds for the light meter to settle to the correct value, so give it
a few seconds, and then make a estimate of the average light value at this
position. Note: if you accidently bump the lamp or transit device you
have to start over! Small changes in the separation or lamp height will
result in bad data.\\
\vspace{0.1in}

\begin{table}[htb]
\begin{center}
\caption{Exoplanet Transit Data}
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
{\bf Position} &{\bf Bare Rod} & $~ ~ ~\delta$~ ~ ~&{\bf Small P.}&~ ~{\bf S.P.$+$$\delta$}~ ~&{\bf Large P.}&~ ~{\bf L.P.$+$$\delta$}~ ~\\
\hline
\#1& &0.0 & & & & \\ \hline
\#2& & & & & & \\ \hline
\#3& & & & & & \\ \hline
\#4& & & & & & \\ \hline
\#5& &0.0 & & & & \\ \hline
\end{tabular}
\label{measurement}
\end{center}
\end{table}

Now we have to account for the dimming effect of the rod. First, add the
bare rod measurements at positions \#1 and \#5 together and divide by 2 to
create the average unobscured value. Now, in the column labeled ``$\delta$'',
fill in the differences between the average unobscured value you just 
calculated, and your bare rod measurements at positions 2 through 4 
($\delta$ = Ave.  $-$ \#2, etc.). Then in columns 5 (S.P. $+$ $\delta$) and 7
(L.P. $+$ $\delta$), add the value in column 2 to the measurements in columns 
4 and 6, respectively for all five measurements [obviously, you add
the value of $\delta$ at position \#2 to the value of Small P. at position
\#2 to get the value of (S.P. $+$ $\delta$) at position \#2]. {\bf (14 points)}
\\
\vspace{0.1in}
{\bf Making Light Curves}

Now we want to plot the data in Table \ref{measurement} to make a light
curve for our two planets. Plot your data on the graph paper in the next
two windows. We have filled-in the X axis with notation for the five
positions you measured. You will have to put values on the Y axis that 
allow the entire light curve to be plotted. For example if the unobscured
value was near 285 (positions \#1 or \#5), the top Y axis grid line might
be set to 300. If the value at position \#3 was 223, the bottom of the
Y axis could have a value of 200. It depends on your light meter, and how
bright the light source was. You will have to decide how to label the Y axis!
Plot the data for both planets in Figures \ref{smalltransit} and 
\ref{bigtransit}. {\bf (8 points)}.\\
\end{flushleft}

\begin{figure}[htb]
\centerline{{\includegraphics[width=7.5cm]{exograph.jpg}}}
\caption{The light curve of the transit of the small planet.}
\label{smalltransit}
\end{figure}
\begin{figure}[htb]
\centerline{{\includegraphics[width=7.5cm]{exograph.jpg}}}
\caption{The light curve of the transit of the large planet.}
\label{bigtransit}
\end{figure}

\vspace{0.2in}
\subsection{Real exoplanet transits}

Now that we have seen how one might observe a real exoplanet transit, and 
construct its light curve. We now want to examine how hard this really is.
You probably have already found the dimming signal due to the small planet was 
quite small. Let's calculate how much the light dimmed in our simulations so we 
can compare them to real exoplanet transits. First we need to find the 
difference between the unobscured value, and the value at position \#3 for 
both planets: {\bf (2 points)}\\
\vspace{0.2in}
\begin{flushleft}
Total dimming small planet = (Position \#1) $-$ (Position \#3) = {\underline{\hskip 0.52in{}} } counts\\ 
\vspace{0.2in}
Total dimming large planet = (Position \#1) $-$ (Position \#3) = {\underline{\hskip 0.52in{}} } counts\\ 
\vspace{0.2in}
\end{flushleft}
Now let's put this in the fractional amount of dimming (``$\Delta$F/F''): \\

\begin{flushleft}
Fractional dimming small planet = (Total dimming small planet)/(Position \#1) = {\underline{\hskip 0.52in{}} } \\ 
\vspace{0.2in}
Fractional dimming large planet = (Total dimming large planet)/(Position \#1) = {\underline{\hskip 0.52in{}} } \\ 
\vspace{0.2in}
\end{flushleft}
How does this compare to the real world? You actually already calculated
the percent dimming for the Earth and Jupiter in Exercise \#2. In that 
exercise we calculated the ratio of the areas of the planets relative to
the Sun---this ratio is in fact how much the light from the Sun would dim
(in fractional terms) when the Earth or Jupiter transited it as seen from a 
very distant point in space (or as some alien would measure watching those
crazy exoplanets transit the star we call the Sun!).\\

\begin{flushleft}
{\bf Questions:}\\
\vspace{0.1in}
1) Compare the percent dimming of our simulated exoplanets to the values for
the Earth and Jupiter found in Exercise \#2. Was our simulation very
realistic? {\bf (2 points)}\\
~\\
\vspace{2.5in}
2) Let's imagine an alien pointed their telescope at our Sun to watch a transit
of the Earth. If his light meter was measuring 25,000 counts from the Sun
before the Earth transited (i.e., Point \#1), what would it read at 
mid-transit (i.e., Point \#3)? Show your math. [Hint: remember the dimming is 
very small, so the mid-transit number will be very close to the unobscured 
value.] {\bf (2 points)}\\
\vspace{1.0in}
\end{flushleft}
As you have now seen, detecting planets around other stars is very hard.
The amount of dimming during a transit is only about 1\% for a Jupiter-sized
exoplanet that orbits another star. To make such high precision measurements,
especially to see Earth-sized planets, requires us to get above the Earth's
atmosphere and use special detectors that have very low noise. Note that
we also have to observe for a very long time---the Earth only has one
transit per year! Jupiter would have one every 12 years! These events only
last a few hours, so we also have to observe the star continuously so we
do not miss the transit. This requires a dedicated instrument, and this
need was the genesis of the Kepler mission launched by NASA several years
ago. Kepler detected over 1,000 transiting exoplanets during its four
year mission. Unfortunately, Kepler is no longer fully functional, and it 
will not be able to continue searching for Earth-like planets.\\

Before we leave the subject of transits behind, we want to talk a little more
about how we can use light curves to get actual information on the exoplanet. 
In Figure \ref{bigexoplanet} is plotted
an exoplanet transit and light curve, with all of the math (scary, eh?) that 
needs to be taken into account to decipher exactly what is going on (actually
the math is not real scary, as it is derived from Kepler's laws). In the
preceding  we have assumed that the planet crosses the center of the 
star---but this almost never happens. The orbit is tilted a little bit, so the 
transit path
is shortened. There are ways to figure all of this out, as demonstrated by
the many math equations in this figure. But we want to focus your attention
on the most important result that a transit tells you: the radius of the
exoplanet.  In the top corner of Figure \ref{bigexoplanet} there is a simple
equation: $\Delta$F/F = (R$_{\rm p}$/R$_{*}$)$^{\rm 2}$. As we have
calculated above, the depth of the eclipse, $\Delta$F/F, allows you to
determine the radius of an exoplanet. As all of the math in this figure
shows you, if you can estimate the stellar parameters (R$_{*}$, M$_{*}$),
you can also determine other characteristics of the exoplanet orbit 
(semi-major axis, orbital inclination). It is fairly simple to estimate 
R$_{*}$ and M$_{*}$. In fact, if we measure the period of the orbiting
planet, we can measure the mass of the host star using Kepler's laws
(the P$^{\rm 2}$ = 4$\pi^{\rm 2}$/GM$_{*}$ equation). Thus observing transits
provides much insight into the nature of an exoplanet, its orbit, and the host
star.\\
\begin{figure}[htb]
\centerline{{\includegraphics[width=12.cm]{Transit.jpg}}}
\caption{An exoplanet transit light curve (bottom) can provide
a useful amount of information. As we have shown, the most important
attribute is the radius of the exoplanet. But if you know the mass and
radius of the exoplanet host star, you can determine other details about the
exoplanet's orbit. As the figure suggests, by observing multiple transits
of an exoplanet, you can actually determine whether it has a moon! This
is because the exoplanet and its moon orbit around the center of mass
of the system (``barycenter''), and thus the planet appears to wobble back 
and forth relative to the host star. We will discuss center of mass, and the
orbits of stars and exoplanets around the center of mass, in the next section.}
\label{bigexoplanet}
\end{figure}
\section{Exoplanet Detection by Radial Velocity Variations}

The final method we are going to investigate today, the technique of
radial velocity variations, is the most difficult to understand as we
have to talk about ``center of mass'', the Doppler effect, and spectroscopy.
You have probably heard about all three of these during the lectures over
this past semester, but we are sure you need to have a review of these
topics.\\

You are certainly aware of the concept center of mass, even if you never knew
what it was called. Take the pencil or pen that you have with
you today and try to balance it across the tip of your finger. The point on
the pencil/pen where it balances on your finger tip is its center of mass.
A teeter totter is another good way to envision the center of mass. If a
small kid and a big kid are playing on the teeter totter, the balance is
not good, and it is hard to have fun. You need to either adjust the balance
point of the teeter totter, or have two kids with the same weight use it.\\

A diagram for defining the center of mass for two objects with different
masses is shown in Figure \ref{com}. If the two objects had the same mass,
the center of mass would be halfway between them. If one object has a much
bigger mass, than the center of mass will be located closer to it. You have
a device today that clearly demonstrates this type of system.

\begin{figure}[htb]
\centerline{{\includegraphics[width=7.cm]{com.jpg}}}
\caption{Center of mass, ``x$_{\rm CM}$'', for two objects that have unequal 
masses. The center of mass can be thought as being the point where the
system would balance on a ``fulcrum'' if connected by a rod.}
\label{com}
\end{figure}

\begin{flushleft}
{\bf Exercise \#4: Defining the Center of Mass for a Two Body System}\\
\vspace{0.2in}
As part of the materials for today's lab, you were given a center of mass
demonstrator. It consists of a large black mass connected to a small white
mass by a long rod. There is also a wooden handle with a small pin at one
end.\\
\vspace{0.15in}
Remove the wooden handle from the long rod. Using the meter stick,
estimate the length of the entire device, from the {\bf center} of the black 
sphere (we will call it ``M$_{\rm 1}$''), to the {\bf center} of the white 
sphere (``M$_{\rm 2}$''). What is this number in cm?  {\bf (1 point)}:\\
\clearpage
Ok, now find the halfway point from the center of one ball to the next. You
need to divide the length you just measured by two, and measure in from
one of the balls and note its location (if necessary, use a piece of tape).
Is there a hole there? If you try to balance the device on the tip of your
finger at this center point/hole, what happens? {\bf (2 points)}\\
\vspace{0.9in}
Now put the device on the tip of your finger and find the balance point of the 
device. There is also a hole there. Use the meter stick to estimate (and
write down) the distance between the center of the black ball to this point
(we will call this ``X$_{\rm 1}$''), and the distance between the center of 
the white ball and this point (we will call this ``X$_{\rm 2}$''). This
exercise is best done by two people.  {\bf (2 points)}\\
\vspace{0.2in}
X$_{\rm 1}$ = {\underline{\hskip 2in{}} } cm \\
\vspace{0.2in}
X$_{\rm 2}$ = {\underline{\hskip 2in{}} } cm \\
\vspace{0.2in} 
This spot on the rod is ``the center of mass''. The center of mass point is 
important, as it allows us
to determine the ``mass ratio'', and if we know the mass of one of the 
objects, we can figure out the mass of the other object. The equation for
center of mass is this:
\vspace{0.2in}
\end{flushleft}
\begin{center}
{\Large M$_{\rm 1}$X$_{\rm 1}$ = M$_{\rm 2}$X$_{\rm 2}$}\\
\end{center}
\begin{flushleft}
\vspace{0.2in}
and the mass ratio is:\\
\end{flushleft}
\vspace{0.15in}
\begin{center}
{\Large M$_{\rm 1}$/M$_{\rm 2}$ = X$_{\rm 2}$/X$_{\rm 1}$}\\
\end{center}
\begin{flushleft}
Determine the mass ratio for the center of mass device. {\bf(2 points)}\\
\vspace{0.3in}
M$_{\rm 1}$/M$_{\rm 2}$ = \\
\vspace{0.3in}
If M$_{\rm 1}$ = 250 grams, what is the mass of M$_{\rm 2}$? {\bf(2 points)}\\
\vspace{0.3in}
M$_{\rm 2}$ = {\underline{\hskip 2in{}} } grams\\
\vspace{0.4in}
Now that we have explored the concept of center of mass, let's see how
it applies to objects that orbit each other. Inserting the pin on the
wooden handle into the center point of the rod (not the center of mass hole!),
hold the wooden handle and try to spin the device. Now, move the wooden
handle to the center of mass hole. Spin the device. Explain what happened
at both locations: {\bf (2 points)}
\vspace{2in}
\end{flushleft}
Any two objects in orbit around each other actually orbit the center of mass
of the system. This is diagrammed in Figure 
\ref{com2}\footnotemark[1]\footnotetext[1]{An animation of this can be found at http://astronomy.nmsu.edu/tharriso/ast105/Orbit3.gif}.
Thus, the Earth and Sun orbit each other around their center of mass, and
Jupiter and the Sun orbit each other around their center of mass, etc. In
fact, the motion of the Sun is a complex combination of the orbits of all
of the planets in our solar system. For
now, we are going to ignore the other planets, and figure out where the
center of mass is for the Sun$-$Earth system.\\

\begin{flushleft}
The Sun has a mass of M$_{\rm Sun}$ = 2.0 $\times$ 10$^{\rm 30}$ kg, while
the Earth has a mass of M$_{\rm Earth}$ = 6.0 $\times$ 10$^{\rm 24}$ kg.
We will save you some math and just tell you that the approximate mass ratio 
is:\\
\vspace{0.2in}
M$_{\rm Sun}$/M$_{\rm Earth}$ = 330,000 \\
\vspace{0.2in}
\end{flushleft}
\begin{figure}[htb]
\centerline{{\includegraphics[width=7.cm]{center-of-mass.jpg}}}
\caption{If two stars are orbiting around each other, or a planet is orbiting
a star, they both {\it actually} orbit the center of mass. If the two objects 
have the same mass, the center of mass is exactly halfway between the two 
objects. Otherwise, the orbits have different sizes.}
\label{com2}
\end{figure}
To determine where the center of mass is for the Earth-Sun system, we have
to do a little bit of algebra. Remember that the mean distance between
the Earth and the Sun is 1 AU. Thus, using our notation from above:\\
\vspace{0.2in}
\begin{center}
{\Large 1 AU = X$_{\rm 1}$ + X$_{\rm 2}$} \\
\end{center}
\vspace{0.2in}
Therefore, \\
\begin{center} 
{\Large X$_{\rm 1}$ = 1 AU $-$ X$_{\rm 2}$}~ ~ ~ ~ ~ ~ (Equation \#1)\\
\end{center}
\vspace{0.2in}
Does that make sense to you? X$_{\rm 1}$ and X$_{\rm 2}$ are the distance
from the Sun to the center of mass, and the Earth to the center of mass,
respectively. As the center of mass demonstration device shows you, the
center of mass is located somewhere on the line that connects the two objects. 
Thus, X$_{\rm 1}$ + X$_{\rm 2}$ = distance between the two masses. For the
Earth and Sun, X$_{\rm 1}$ + X$_{\rm 2}$ = 1 AU. Now, going back to our
center of mass equation:\\
\vspace{0.2in}
\begin{center}
{\Large M$_{\rm 1}$X$_{\rm 1}$ = M$_{\rm 2}$X$_{\rm 2}$} ~ ~ ~ ~ ~ ~(Equation
 \#2)\\
\end{center}
\vspace{0.2in}
We can substitute the result for X$_{\rm 1}$ in equation \#1 into Equation 
\#2:\\
\vspace{0.2in}
\begin{center}
{\Large M$_{\rm 1}$(1 AU $-$ X$_{\rm 2}$) = M$_{\rm 2}$X$_{\rm 2}$}~ ~ ~ ~ ~ ~ (Equation \#3)\\
\end{center}
\vspace{0.2in}
Dividing both sides of Equation \#3 by M$_{\rm 1}$ gives:\\
\vspace{0.2in}
\begin{center}
{\Large 1 AU $-$ X$_{\rm 2}$ = (M$_{\rm 2}$/M$_{\rm 1}$)X$_{\rm 2}$}~ ~ ~ ~ ~ ~ (Equation \#4)\\
\end{center}
\vspace{0.2in}
But (M$_{\rm 2}$/M$_{\rm 1}$) = 1/330,000 for the Earth-Sun system, and now
we can solve to find X$_{\rm 2}$:\\
\vspace{0.2in}
\begin{center}
{\large 1 AU  = (M$_{\rm 2}$/M$_{\rm 1}$)X$_{\rm 2}$ $+$ X$_{\rm 2}$ =
(1/330,000 $+$ 1)X$_{\rm 2}$}\\
\vspace{0.2in}
\end{center}
Thus, \\
\vspace{0.2in}
\begin{center}
{\Large X$_{\rm 2}$ = 1.0/(1 + 1/330,000) = 0.999997 AU}\\
\end{center}
\vspace{0.2in}
Essentially, the Earth is 1 AU from the center of mass, how far away is
the Sun from the Earth-Sun center of mass? Go back to equation \#1: \\
\vspace{0.2in}
\begin{center}
{\Large X$_{\rm 1}$ = 1 $-$ 0.999997 = 0.000003 AU}\\
\end{center}
\vspace{0.2in}
The Sun is very close to the center of mass of the Earth-Sun system.\\
~ \
\begin{flushleft}
{\bf Exercise \#5: Determining the Size and Velocity of the Sun's ``Reflex 
Motion''} \\
\vspace{0.2in}
We are going to calculate the size of the Sun's orbit around the center
of mass for the Sun-Earth system, and then determine how fast the Sun
is actually moving. The motion of the Sun (or any star) due to an orbiting
planet is called the ``reflex motion''. Like the name suggests, it is the
response of the star to the gravitational pull of the planet. Since AU per 
year is not a normal unit with which to measure velocity, we need to convert 
the numbers we 
have just calculated to something more useful.\\
\vspace{0.2in}
1 AU = 149,597,871 km. How far from the center of mass is the Sun in km? 
{\bf (1 point)}:\\ %448.794 km
\vspace{0.2in}
{\large X$_{\rm 1}$ (km) = X$_{\rm 1}$ (AU) $\times$ 149,597,871 (km/AU) = {\underline{\hskip 1.5in{}} } km} \\
\vspace{0.2in}
Hopefully, you noticed how the units of length canceled in the last equation.
\\
~\\
So, we now have the
distance of the Sun from the center of mass. Note that this number puts
the center of mass of the Earth-Sun system well inside the Sun (actually
very close to its core). We now want to figure out what the length of the orbit
is that the Sun executes over one year (remember, the Earth takes one year
to orbit the Sun, so the ``orbital period'' of the Sun around the Earth-Sun
center of mass will be one year). Referring back to the center of mass
device, if you put the handle in the center of mass hole and spin the
system, what path do the masses trace? That's right, a circle. Do you
remember how to calculate the circumference of a circle? C = 2$\pi$R, where R 
is the radius of the circle and $\pi$ = 3.14. \\
~\\
What is the circumference of the orbit circle (in km) that is traced-out by 
the Sun?  {\bf (2 points)}:\\ %2819.85
\vspace{0.5in}
This is how far the Sun travels each year, thus we can turn this into a
velocity (km/hr = kph) by dividing the distance traveled (in km) by the 
number of hours in a year. Show your math {\bf (2 points)}:\\ %0.322 kph
\vspace{0.2in}
{\large V$_{\rm Sun}$ (km/hr) = C (km) $\div$ (\# hours in year) =  ???}\\
\vspace{1.5in}
1) Comment on the size of the reflex velocity (V$_{\rm Sun}$) of the Sun. Note 
that the Earth travels much, much further during the year, so its velocity is 
much, much higher: 107,000 km/hr! {\bf (3 points)}:\\
\vspace{2in}
Because all of the math above involved simple, ``linear'' equations, we can
quickly estimate the reflex velocity of the Sun if we replaced the Earth
by something more massive. For example, if we put an object with 10
Earth masses in an orbit with R = 1 AU, the reflex velocity of the Sun would
be 10 times that which you just calculated for the Earth.\\
\vspace{0.2in}
2) Jupiter has a mass that is 318 times that of the Earth. If Jupiter orbited
the Sun at 1 AU, what would the reflex velocity of the Sun be? {\bf (2 points)}:\\
%1=(1/1037.7 +1)X2 = 0.99904, 0.00096=x1 =144,019 km => 9.05e5km=>103.3 kph
\vspace{1.5in}
\end{flushleft}
Since Jupiter is at 5.2 AU, and its orbital period is 11.9 yr, the reflex
motion of Jupiter is actually: V$_{\rm Jupiter}$ = 318 $\times$ V$_{\rm Earth}$ 
$\times$ 5.2 $\div$ 11.9 $\approx$ 45 km/hr.\\
%5.2=(1/1037.7 +1)X2=5.195 x1=0.005 7.49e5=> 4.71e6 km => 45.18 kph
~ \\
\begin{flushleft}
{\bf Exercise \#6: Understanding the Sizes of the Reflex Motions} \\
~ \\
For the final exercise of today's lab, we want to demonstrate how big
these reflex motions are by comparing them to the velocities that you can
generate. To do so, we are going to be using radar guns just like those
used by the police to catch speeders. These devices are very expensive,
so please be extremely careful with them. The radar guns are a bit technical
to set-up, so your TA will put them in the correct mode for measuring
velocities in km/hr.\\
~ \\
Your lab group should head out of the classroom, and into the hallway (or
outside) to
get a long enough path to execute this part of the lab. The idea is to
have one of the lab members move down the hallway, and act as the ``speeding
car''. Note that if there are other people moving around in the hallway,
the radar gun might get a confusing signal and not read correctly. So, make
sure only one person is moving when doing this.\\
~ \\
1) One lab member hold the radar gun, have another lab member walk towards
the radar gun. Hold down the trigger a few seconds and then let go. Do
this several times to get a good reading. What is the average velocity
of the walking speed of this lab member? {\bf (2 points)}:\\
\vspace{2 in}
2) Now, we are going to measure the running speed. {\bf BE CAREFUL!}. Have
everyone participate, and see who can run the fastest. What are the
velocities for the various lab members? {\bf (2 points)}:\\
~ \\
~ \\
~ \\
\vspace{1.5 in}
3) Compare your walking and running velocities to the Sun's reflex velocity 
caused by the Earth that you calculated above. How massive a planet (in Earth 
masses) would it take to get your walking reflex motion to be executed by the 
Sun? How about your running reflex motion?
{\bf (5 points)}.
\vspace{3 in}
\end{flushleft}
\section{Radial Velocity and the Doppler Effect}

Earlier we called this final exoplanet discovery technique ``the radial 
velocity'' method. What do we mean by this term? The radial velocity is a 
measurement of how fast something is coming towards you, or going away from 
you. If an object is moving across your line of sight (like the cars on the 
road as you wait to cross a street at the pedestrian crossing), it has no 
radial velocity (formally, they would have a ``tangential velocity'' only).
If we were an alien watching the Sun, the Sun would sometimes have a 
radial velocity coming towards us (normally defined to be a negative number), 
and a radial velocity going away from us (normally defined as a positive
number), due to the reflex motions imparted on it by the planets in our
solar system. This gives rise to something called a ``radial velocity curve''.
\\

So how do we detect the radial velocity of a star? We use something called
the Doppler effect. The Doppler effect is the change in frequency of
a sound or light {\it wave} due to motion of the source. Think of an ambulance.
When the ambulance is coming towards you, the siren has a high pitch. As
it passes by you, the pitch drops (for audio examples, go here:
http://www.soundsnap.com/search/audio/doppler/score?page=1"). This is
shown in Figure \ref{doppler}. The radar guns you just used emit microwaves
that are Doppler shifted by moving objects. Stars are too far away to use
radar. Fortunately, the same process happens with all types of 
electromagnetic radiation. Astronomers use visible light to search for
Exoplanets. In a source coming towards us the light waves get
compressed to higher frequency. When it is receding the light waves are 
stretched to lower frequency. Compressing the frequency of light adds energy, 
so it ``blueshifts'' the light. Lowering the frequency removes energy, so
it ``redshifts'' the light. For an object orbiting the center of mass,
sometimes the light is blueshifted (at point \#4 in Figure \ref{radvel}),
sometimes it is redshifted (at point \#2 in Figure \ref{radvel}).
\begin{figure}[htb]
\centerline{{\includegraphics[width=10cm]{doppler.jpg}}}
\caption{For a stationary vehicle emitting sound, there is no Doppler effect.
As the vehicle begins to move, however, the sound is compressed in the
direction it is moving, and stretched-out in the opposite direction.}
\label{doppler}
\end{figure}
~ \\
\begin{figure}[htb]
\centerline{{\includegraphics[width=10cm]{RadVel.pdf}}}
\caption{A radial velocity curve (left) for a planet with a one year
orbit like Earth, but that imparts a reflex velocity of 1.5 km/hr on its
host star. When the motion is directly away from us, \#2, we have the maximum 
amount of positive radial velocity. When the motion of the object is directly
towards us, \#4, we
have the maximum negative radial velocity. At points \#1 and \#3, the 
object is not coming towards us, or going away from us, thus its radial velocity
is 0 km/hr. The orbit of the object around the center of mass (``X'') is shown 
in the right hand panel, where the observer is at the bottom of the diagram.
The numbered points represent the same places in the orbit in both panels.}
\label{radvel}
\end{figure}
This is how astronomers discover exoplanets, they monitor the spectrum of a
star and look for a changing radial velocity like that shown in Figure 
\ref{radvel}.
What they see is that the absorption lines in the spectrum of the exoplanet
host star shift back and forth, red to blue to red to blue. Measuring the 
shift gives them the velocity. Measuring the time it takes to go from maximum
blueshift to maximum redshift and back to maximum blueshift, is the 
exoplanet's orbital period. Remember, the exoplanet is too faint to detect 
directly, it is only the reflex velocity of the host star that can be observed. 
And, now you should understand how we measure the mass of the exoplanet. The 
amount of reflex velocity is directly related to the mass of the exoplanet and 
the size of its orbit. We can use the orbital period and Kepler's laws to
figure out the size of the exoplanet's orbit. We then measure the
radial velocity curve, and if we can estimate the host star's mass, we
can directly measure the mass of the exoplanet using the techniques you
have learned today.\\
~ \\
Here is how it is done. To determine the mass of an exoplanet, we
first must figure out the semi-major axis of its orbit (for the Earth,
the semi-major axis = R = 1 AU). We return to Kepler's laws:

\begin{displaymath} R^3 = \frac {GM_{star}}{4\pi^2} P^2  ~ ~ ~ ~ (Equation ~ \#5)
\end{displaymath}

In this equation, ``G'' is the gravitational constant. P is the orbital
period. In physics equations like these, the {\it system} of units used
must be the same for each parameter. Such as centimeter-gram-second, or
meter-kilogram-second. We call these the ```cgs'' and ``mks'' systems,
respectively. You cannot mix and match. Thus, there have to be two flavors of 
G for this equation: G$_{\rm cgs}$ = 6.67 $\times$ 10$^{-8}$, and 
G$_{\rm mks}$ = 6.67 $\times$ 10$^{-11}$. The equation above is just 
Kepler's third law P$^{\rm 2}$ $\propto$ a$^{\rm 3}$ you learned about at the 
beginning of the semester. What Isaac Newton did was figure out what is needed 
to change the ``$\propto$'' into the ``='' sign. If we know ``R'' and
the exoplanet host star mass (M$_{star}$) we can figure out the exoplanet's 
mass. So using equation \#5 above, we find R. Since we know the orbital period 
(P), we can estimate the exoplanet's orbital velocity:

 \begin{displaymath} V_{pl} = \frac {2\pi R}{P} ~ ~ ~ ~ (Equation ~ \#6)
\end{displaymath}
The mass of the planet is simply:

\begin{displaymath} M_{pl} = \frac {M_{star}V_{star}}{V_{pl}}
\end{displaymath}
In this equation V$_{\rm star}$ is the host star reflex velocity like
those we calculated above for the Earth-Sun, and Jupiter-Sun systems. 
The biggest unknown when making such mass measurements is estimating
the host star mass. There are ways to do this, but they are beyond the scope
of today's lab. We will use these equations in the take-home part of this
lab, so make sure you understand what is going on here before leaving 
today.
\clearpage
\hspace*{3.0in}Name:{\hrulefill} \\
\hspace*{3.25in}Date:{\hrulefill}\\

\subsection{Take-Home Exercise (35 points total)}

\begin{flushleft}
1) Discuss what you have learned about Exoplanets today. How hard are they
to detect, and what are the main techniques astronomers use to find them?
{\bf (10 points)}\\
\vspace{3.0in}
2) You have obtained a radial velocity curve for a transiting exoplanet
that orbits the star 18 Scorpii, which is referred to as a ``solar twin''
(that is, identical in every way to the Sun). The planet has an orbital
period of 400 days. The maximum reflex velocity of the star 18 Scorpii is 
1 m/s. What is the mass of this exoplanet? You are going to need to use
equations \#4 and \#5, and the mass (listed in Exercise \#4) and radius (from 
Exercise \#1) of the Sun. Remember that you must use a consistent set
of units. In equation \#4 ``{\bf G}'' is listed in ``mks''. Thus, the period 
of the exoplanet must be converted from days to seconds, and the mass and 
radius of the Sun must be in kilograms and meters, respectively, to correctly 
use equations \#4 and \#5. Compare the mass of this exoplanet to the mass of 
the Earth (6.0 $\times$ 10$^{\rm 24}$ kg). Show your work. {\bf (10 points)}\\
%radius of sun=695,550 km = 6.96e8m, mass of sun=2e30g, P=400d=3.46e7 s
%G=6.67e-11  Rcubed= 4.05e33 R=1.59e11 m, Vpl=2.89e4 m/s
%Mp = 2e308*1/2.89e4 = 6.93e25 kg  (mass of Earth = 6e24kg), so 11.6 Mearth
~\\
\vspace{3.8in}
~\\
~\\
3) As we noted, the exoplanet around 18 Scorpii is also a transiting
system, and $\Delta$F/F = 4.8 $\times$ 10$^{\rm -4}$. Calculate the
radius of this planet (like those in Exercise \#3). Compare it to the
radius of the Earth. What is the density of this planet in kg/m$^{\rm 3}$? 
[Hint: density = mass/volume. What is the volume of a sphere?] Compare
this to the density of the Earth (5,511 kg/m$^{\rm 3}$), and Jupiter
(1,326 kg/m$^{\rm 3}$). Is ``18 Scorpii B'' more like a Terrestrial planet,
or a Jovian planet? Show your work. {\bf (15 points)}\\
%Rp=  1.52e7 m, Vpl=1.48e22 m3. rho= 4675 I wanted this to be ambiguous
%to see what they answer. The mass is like Uranus/Neptune, but the
%orbital period is like Earth and Mars, as is the density.
\vspace{4in}
\end{flushleft}
\subsection{Possible Quiz Questions}
\begin{flushleft}
1) What is an Exoplanet?\\
2) Name one of the techniques used to find Exoplanets\\
3) Why are Exoplanets so hard to discover?\\
\end{flushleft}
\subsection{Extra Credit (make sure you get permission from your
TA before attempting, 5 points)}

Using the web, search for an article on an ``Earth-like exoplanet'' and
write a one page discussion of this object, and what makes it ``Earth-like''.
Note that there are quite a few such objects, just pick the one you find
most interesting (and one that has sufficient discussion to allow  you
to write a short paper).

\end{document}
