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\hspace*{3.0in}Name:{\hrulefill} \\
\hspace*{3.25in}Date:{\hrulefill}

\begin{Large}
\section{\bf The Volcanoes of Io}
\end{Large}

\subsection{Introduction}

During this lab, we will explore Jupiter's moon Io, the most volcanically active 
body in the Solar System. The reason for Io's extreme level of volcanic
activity is due to the intense tidal `stretching' it experiences 
because of its proximity to Jupiter, and due to its interaction with the
moons Europa and Ganymede. The regions of the surface where molten lava 
from the interior comes up from below are very hot, but in general the rest of the
surface is quite cold (about $-$172$^{o}$C = $-$279$^{o}$F) since Io is 5.2 AU from 
the Sun. Regions of different surface temperatures emit different amounts
of thermal (blackbody) radiation, since the amount of thermal energy emitted 
is proportional to the temperature raised to the 4th power: T$^4$. We will use 
{\it infrared} observations, obtained with the Galileo spacecraft in the late 
1990's, to determine the temperatures of some of the volcanic regions on Io,
and estimate the total amount of energy being emitted by the volcanoes on
Io.

\begin{flushleft}
Supplies:
\begin{enumerate}
\item Exercise squeeze balls and thermometers
\item Visual and thermal images of regions on Io
\item  A map of Io with various features identified by name
\item  A transparency sheet for temperature fitting of blackbodies
\end{enumerate}
\end{flushleft}

\subsection{Introduction to Io}

Io (pronounced eye-Oh) is one of the four large moons of Jupiter discovered
by Galileo. Images of these four moons (Io, Europa, Ganymede, and Callisto)
are shown in Figure \ref{galilean}. Io, Ganymede and Callisto are all larger
than the Earth's moon, while Europa is slightly smaller. It is clear from
Figure \ref{galilean} that Io appears to be quite different from the other
Galilean satellites (especially when viewed in color!): it has few obvious
impact craters, and has a mottled surface that is unlike any other object
in the solar system. Even before the two Voyager probes first flew past Io
back in the late 1970's, it was already known that it was an unusual object. 
The Voyager images of Io certainly suggested that it was covered with 
volcanoes and lava flows, but it was not until 
an image showing an erupting volcano, also shown in Figure \ref{galilean}, that the 
case was clinched. From the imaging data, astronomers estimate that there may be as
many as 200 volcanoes on Io! \\

\begin{figure}
\centering
\mbox{\subfigure{\includegraphics[width=3in]{galilean.jpg}
\quad
\subfigure{\includegraphics[width=3.28in]{io_volcano_big.jpg} }}}
\caption{Left: The four Galilean moons of Jupiter. Right: An erupting volcano on 
Io seen in a Voyager image.}
\label{galilean}
\end{figure}

Why does Io have so many volcanoes? It has to do with a process called ``tidal
heating''. As you have learned in the lectures this semester, the gravitational
pull on one body by a second massive body raises tides---an example are those caused
by the Moon upon the Earth's oceans. As we have also found this semester, the orbits
of objects in the solar system are not perfect circles, but ellipses. That means
the distance of an object orbiting a larger body (planet around the Sun, or
moon around a planet) is constantly changing. In the case of Io, we have an
object that has about the same mass as the Earth's moon, but it orbits Jupiter,
an object that has 300 times the mass of the Earth! We have learned that the force
of gravity is directly proportional to the mass of an object, Newton's second 
law: F = m$a$. For gravity, Newton's second law is F = 
({\it G}m$_{\rm 1}$m$_{\rm 2}$)/r$^{\rm 2}$ (``{\it G}'' is the ``gravitational
constant''). Thus, even a slightly eccentric 
orbit, as demonstrated in Figure \ref{tidal}, means that large changes in 
tidal force are felt as Io goes around Jupiter (the 1/r$^{\rm 2}$ term in
the equation). In fact, the surface of Io rises and falls by about 100 meters 
over an orbit! This should be compared to the approximate 0.3 meter rise and fall of the 
Earth's surface due to the Moon's pull.\\

The reason that Io's orbit is so eccentric is due to the gravity of
Europa and Ganymede. First, let's look at the orbital periods (i.e., the
time it takes the moon to orbit Jupiter a single time) of these three moons: P$_{\rm Io}$ = 
1.769 days, P$_{\rm Europa}$ = 3.551 days, and P$_{\rm Ganymede}$ = 7.155 days.
It we take the ratios of these orbital periods we get the following answers:
P$_{\rm Europa}$/P$_{\rm Io}$ = 2.0, P$_{\rm Ganymede}$/P$_{\rm Io}$ = 4.0.
What does this mean? Well, it tells you that every 3.551 days Europa and
Io will be in the same exact location (relative to each other), and that every 
7.155 days Ganymede, Europa {\it and} Io will be in the same relative places! 
A diagram of this is shown in Figure \ref{tidal}. The term
astronomers use for such an arrangement is ``orbital resonance''. Because of these
orbital resonances, the gravitational tug on Io is amplified, as it and Europa
(and it and Ganymede) make close approaches on a regular, and repeating basis. Thus, 
Europa and Ganymede continually pull on Io, making its orbit more eccentric.
[Note that we believe that Europa also has considerable tidal heating, and this
heating may mean that below its frozen surface, there is a large ocean of
liquid water that could support primitive life. This might even be happening
on Ganymede.] The tidal heating causes the interior of Io to become molten,
and this liquid rises to the surface, where it erupts in volcanoes. We will
return to Io later in this lab, but before we do so, we must cover several
complicated topics that will allow us to better understand what is happening on Io.\\

\begin{figure}
\centering
\mbox{\subfigure{\includegraphics[width=3in]{IoTidal.jpg}
\quad
\subfigure{\includegraphics[width=2.5in]{IoResonance.pdf} }}}
\caption{Left: Because Io's orbit around Jupiter is an ellipse, the distance is
constantly changing, and so is the gravitational force exerted on Io by
Jupiter (note that this figure is not to scale, and the ellipticity of the
orbit and the shape of Io have been grossly exaggerated to demonstrate the
effect). This changing force causes Io to stretch and relax over each orbit.
Right: The tidal forces exerted by Europa and Ganymede distort the
orbit of Io because the orbits of all three moons are in ``resonance'': 
for every four trips Io makes around Jupiter, Europa makes two, and Ganymede makes 
one. This resonance enhances the gravitational forces of Europa and Ganymede, as 
these three moons keep returning to the same (relative) places on a regular basis. 
This repeated and periodic tugging on Io causes its orbit to be much more 
eccentric than it would be if Europa and Ganymede did not exist. }
\label{tidal}
\end{figure}

\subsection{The Electromagnetic Spectrum}

Before we begin today's lab, we have to review what is meant by the term
``spectrum'', and ``wavelength''. As we have discussed in class, light is
an energy wave that travels through space. For now, we can use the analogy
that waves of light are like waves of water: they have crests, and troughs.
The ``wavelength'' is the distance between two crests, as shown in Fig.
\ref{wavelength}. The energy contained in light is directly related to
the wavelength: low energy light has long wavelengths, while high energy light
has short wavelengths. Thus, scientists have constructed several categories
of light based on wavelength, and which you have certainly heard about:
Gamma-ray, X-ray, Ultraviolet, Visible, Infrared, Microwave and Radio. Gamma- 
and X-rays have very short wavelengths and have lots of energy, so they 
penetrate through
materials, and often damage them as they pass through. Ultraviolet light
causes sunburns and skin cancer. Visible light is what our eyes detect. We feel
intense infrared light as ``heat'', microwaves cook our food, while radio waves 
allow you to listen to music and watch television. The common textbook plot 
of the electromagnetic spectrum is shown in Fig \ref{em}. When we break-up
light and plot how much energy is coming out at each wavelength, we construct
a ``spectrum''. A spectrum of an object supplies a lot of information, and is
the main tool astronomers use to understand the objects they study.\\

We can also think
of the electromagnetic spectrum as a way to represent temperature. For example,
objects that emit X-rays are at temperatures of millions of degrees, while
objects that emit visible light have temperatures of thousands of degrees (like
the Sun), while infrared sources have temperatures of 100's of degrees. To 
understand this concept, we must talk about ``blackbody'' radiation.\\

\begin{figure}[htbp]
\centerline{\includegraphics[width=2.0in]{wave_crest.jpg}}
\caption{The wavelength is the distance between two crests.}
\label{wavelength}
\end{figure}
\begin{figure}[htbp]
\centerline{\includegraphics[width=6.0in]{EM_spectrum.jpg}}
\caption{The electromagnetic spectrum.}
\label{em}
\end{figure}


\subsection{Blackbody Radiation Review}

Let us review the properties of \underline{\bf blackbody radiation}.  A 
blackbody is an object that exactly satisfies the Stefan-Boltzmann law (named
for the two scientists who first figured it out), and has a spectrum that is
always the same shape, no matter what temperature the source has, as shown in 
Fig.  \ref{bbody}. While real objects do not exactly behave like this, many 
objects come very close and in general we assume that most solar system
objects (including Io) are blackbodies.\\

\begin{figure}[htbp]
\centerline{\includegraphics[width=6in]{blackbody.jpg}}
\caption{The spectra of blackbodies always have the same shape, but the
wavelength where the {\it peak emission} occurs depends on temperature, and can
be calculated using the ``Wien displacement law'' (since Wien is a German name, it 
is properly pronounced ``Veen''). In this particular plot the unit of wavelength is 
the micrometer, 10$^{\rm -6}$ meter, symbolized by ``$\mu$m.'' Note also that
the x-axis is plotted as the {\it log} of wavelength, and the y-axis is the
{\it log} of the radiant energy. We have to use this type of ``log-log'' plot since 
blackbodies cover a large range in radiant energy and wavelength, and we
need an efficient way to compress the axes to make compact plots. We will be
using these types of plots for the volcanoes of Io.}
\label{bbody}
\end{figure}
The Stefan-Boltzmann law states that {\it the total
amount of energy at all wavelengths emitted by a blackbody at temperature T
is proportional (``$\propto$'') to the fourth power of its temperature}, which 
can be written in equation form as:

\begin{equation}
E \propto T^4 .
\end{equation}

Here {\it E} is the amount of energy emitted by {\it each square meter} of the
object each second.  You might be wondering to yourself why we write
E $\propto$ T$^{\rm 4}$, instead of E = T$^{\rm 4}$. In fact, the real 
blackbody equation is  E = $s$T$^{\rm 4}$, where ``{\it s}'' is the
``Stefan-Boltzmann constant.'' The Stefan-Boltzmann constant is a special
number that makes the equation work, and insures that the output energy is in 
Watts (or another appropriate energy unit), instead of $^{\rm o}$F$^{\rm 4}$. 
You measure the energy of a light bulb in Watts, not the fourth power of 
degrees Fahrenheit. The actual value of $s$ is 5.6703 $\times$ 10$^{\rm -8}$. 
This is a horrible number to deal with, so we will use a technique that does 
not require us to remember it!\\

As noted in Fig. \ref{bbody}, the Wein displacement law relates the temperature of a blackbody,
and the wavelength ($\lambda$) of its maximum emission: $\lambda_{\rm max}$ $\times$ T = 3670,
where 3670 is the value of ``Wien's constant'' when wavelength is measured in 
micrometers, and radiant energy in Watts/m$^{\rm 2}$ (as we will use in this lab).\\

\begin{flushleft}
{\bf Definition of Temperature}
\end{flushleft}

Before we go any further in understanding blackbodies, we must define the 
temperature scale that is used in the Stefan-Boltzmann formula, and in Wien's
law. In the United States, our weather forecasts use the 
Fahrenheit scale. This scale was developed around the idea that in our everyday 
experience, a big number like ``100$^{\rm o}$ F'' would be ``hot'', and 
``0$^{\rm o}$ F'' would be ``very cold.'' On this scale water boils at 
212$^{\rm o}$ F, and freezes at 32$^{\rm o}$ F. The Fahrenheit scale is not 
very easy to work with, in that it has 180$^{\rm o}$ F between the
boiling and freezing point of water (two processes that are easy to observe,
allowing accurate calibration). With the development of the metric system, 
based on powers of 10, a temperature scale was developed where the freezing 
point of water was defined to be 0$^{\rm o}$, and the boiling point was set 
to 100$^{\rm o}$. This is the ``Celsius'' scale
(denoted by ``$^{\rm o}$ C''), predominantly used outside the United States.\\

Both the Fahrenheit and Celsius scales, however, cannot be used with the
blackbody energy equation. Why? Because both scales have ``zeroes'' and 
negative temperatures. Even in Las Cruces, the temperature often goes to 
0$^{\rm o}$ C or below on the Celsius scale during winter (and once in a while,
as in 2010, it goes below zero on 
the Fahrenheit scale!). Look at our equation again, E $\propto$  T$^{\rm 4}$.
If the temperature changes from 3$^{\rm o}$ C to 0$^{\rm o}$ C, the amount
of energy emitted by a blackbody {\it would go from positive to zero.} If this object
got colder and colder, however, its emitted energy would increase! For example, if
its temperature had now dropped to $-$3$^{\rm o}$ C, the emitted energy would be 
the {\it same} as it was at $+$3$^{\rm o}$ C: E = $-$3 $\times$ $-$3 $\times$ $-$3 
$\times$ $-$3 = 81 = 3 $\times$ 3 $\times$ 3 $\times$ 3. Do you 
see why this is? The fourth power (or any even power in the exponent) means that a 
negative number will turn out positive: (10)$^4$ = ($-$10)$^4$ = 10,000, because
every time you multiply two negative numbers together, the result is a positive
number. \\

If we were to use the Fahrenheit or Celsius temperature scales, our equation
would produce nonsensical answers, since it is obvious that a hotter object 
has more energy than a colder one. Thus, scientists use a scale that has no 
negative numbers, the ``Kelvin'' scale. On the Kelvin scale, the temperature at which water freezes is 
273 K, and it boils at 373 K (Kelvin has the same size degrees as the Celsius 
scale, and note also that the little degree symbol, ``$^{\rm o}$'', is not used 
with Kelvin). In our example, 3$^{\rm o}$ C = 276 K, and 0$^{\rm o}$ C = 273 K. 
Now, a drop in temperature by 3 degrees does not cause the emitted energy to go 
from positive to zero, the energy simply decreases. There is a 0 K, but that 
temperature is so cold that any object with that temperature {\it would} emit 
zero energy (that, in fact, is the definition of 0 K!).\\
\vspace{0.05in}
\begin{flushleft}
{\bf Working with the Stefan-Boltzmann Law}
\end{flushleft}

An equation like the Stefan-Boltzmann law is scary to many Astronomy 105
students. Nearly all of you have heard about ``squares'', such as the area
of a circle being $\pi$R$^{\rm 2}$. But, there are many equations in
science when the exponent is larger than 2. All an exponent says is that you
must multiply the number by itself that many times: R$^{\rm 2}$ = R $\times$ R.
Or, R$^{\rm 5}$ = R $\times$ R $\times$ R $\times$ R $\times$ R. Other than
the large numbers that come out of the Stefan-Boltzmann law (it is astronomy
after all!), there is nothing difficult about understanding how to deal
with T$^{\rm 4}$.\\

Ok, let's see how to use equation (1) so we can compare the energy emitted by 
{\it each square meter} of the surface of two different objects, A and B. We 
will construct the ratio so we do not have to worry about the value of the
Stefan-Boltzmann constant:\\

\begin{equation}
\frac{E_A}{E_B} = \frac{sT_A^4}{sT_B^4} = \left(\frac{T_A}{T_B}\right)^4
\end{equation}

Do you understand what happened? We had an {\it s} on the top and bottom of our
equation, but {\it s} = {\it s}, so it cancels out! We also use the property
where $T_{A}^{4}$ $\div$  $T_{B}^{4}$ = ($T_{A}$/$T_{B}$)$^{\rm 4}$
(in math this is called the ``Power of a Quotient property''). \\

Let's work an example. Object P has a temperature of 43 K, and object Q
has a temperature of 33 K. The objects have the same area. How many times 
greater is the energy emitted by P
compared to the energy emitted by Q? Set-up the equation:\\

\begin{equation}
\frac{E_P}{E_Q} = \frac{s(43)^4}{s(33)^4} = \left(\frac{43}{33}\right)^4 =
(1.3)^4 = 1.3 \times 1.3 \times 1.3 \times 1.3 = 2.86
\end{equation}

\begin{flushleft}
Now it is your turn:\\
\end{flushleft}
 
\begin{enumerate}
\item Assume that $T_A$, the surface temperature of Object A, is 200 K, and 
$T_B$, the surface temperature of Object B, is 100 K. The objects have
the same area. How many times greater is the 
energy emitted by A compared to the energy emitted by B? ({\bf 2 points})\\
\vspace{2.7cm}

\item Object R and Object S have the same temperature. But object R
has an area of 4 square meters, and object S has an area of 2 square meters. 
How much more energy does object R emit compared to Object S? ({\bf 2 points})\\
\vspace{2.3cm}

\item  Now we are going to go backwards (much harder!): assume that we receive 
81 times more energy from Object X than from Object Y. Object X and Y have
the same areas. How many times hotter is the 
surface of X compared to the surface of Y? [Hint: what number multiplied
by itself 4 times = 81?] ({\bf 2 points})\\
\vspace{2.3cm}
\end{enumerate}
We know that the last problem was hard! How does one solve such equations?
The key to understanding this is to realize that for every mathematical
operation that uses exponents, there is the reverse process
of ``taking the root''. For example, two squared: 2$^{\rm 2}$ = 4. What is the
square root of 4? $\sqrt{4}$ = 2. The square root can also be written as
a fractional exponent: (4)$^{\rm 1/2}$ = 2. This is how we solve the problem
above. Here is an example: What is Q, if Q$^{\rm 4}$ = 6561? On a fancy
scientific calculator, we just enter this: (6561)$^{\rm 1/4}$ = 9. But the
fourth root is really just two {\it successive} square roots: $\sqrt{6561}$ = 81, 
$\sqrt{81}$ = 9 = (6561)$^{\rm 1/4}$. So you do not need a fancy calculator, got it?\\
\vspace{0.05in}
\begin{flushleft}
{\bf Working with Wien's Law} \\
\end{flushleft}
Unlike the Stefan-Boltzmann law, Wien's Law is very simple. So simple we do not 
think you need an example on how to use it! [Here is Wien's law again:
$\lambda_{\rm max}$ $\times$ T = 3670]\\
\begin{enumerate}
\setcounter{enumi}{3}
\item If the temperature of a black body is 1000 K, at what wavelength ($\lambda_{\rm max}$) does it
emit its peak amount of energy? (Remember to include the wavelength unit!) 
({\bf 2 points})
\vspace{1.0in}
\item An object is observed to have a blackbody spectrum that peaks at 
$\lambda_{\rm max}$ = 37 $\mu$m, what temperature is this object? (Remember to 
include the temperature unit!) ({\bf 2 points})
\vspace{1.0in}
\end{enumerate}
%\item  How does the total amount of radiant energy (or flux) emitted by
%a blackbody depend on its temperature ? How does the wavelength at which
%most of the energy is emitted depend on its temperature? ({\bf 5 points})\\
%\vspace{3cm}
\subsection{Simulating Tidal Heating}

As we noted above, the process of tidal heating is what causes Io to be
covered in active volcanoes. In this exercise we are going to simulate
tidal heating, where {\it you} are the source of the energy input. First off,
however, have you ever tried to break a piece of wire with your hands? You cannot
simply pull it apart with your hands, it is too strong. But we can break it
by adding heat. We do this by first folding the wire to create a kink, and then
rapidly bending the wire back and forth. The wire becomes very, very hot at
the kink, and will eventually snap. What you have done is transfer
energy your body generates and focused it on a tiny region of the wire. The
intense heat weakens the wire and it snaps (you should try this with a paper clip). 
This process is what is going on in Io, a 
stretching/bending of the rock that generates heat.\\

\begin{flushleft}
{\bf Exercise \#1:}
\end{flushleft}

Io is not a wire, it is a sphere! While the repeated bending of a wire is 
{\it exactly} like the process that is heating Io, it is not very realistic.
Let's take this concept to a slightly more realistic level by ``stretching'' 
a sphere.  Among the materials you were given were two, small exercise squeeze 
balls and a digital thermometer. We will now use these. To start this 
experiment, insert the thermometer into each of the balls and record
the {\bf Start Temperature}. Make sure the tip of the metal probe reaches
the center of the ball (and no further!). Note that it also takes a certain
amount of time for the temperature to stabilize at the correct value.
Enter these values into Table \ref{exercise}. \\

 Now, one member of your group should take a ball in each hand.  One of these
will be the ``control ball'', let's call that Ball \#1. You will not do anything
to Ball \#1, except hold it in your hand. But for Ball \#2, repeatedly, as
rapidly as possible, squeeze this ball as tightly as possible, release, and
repeat. {\it Do this for four straight minutes (one group member needs to be
the time keeper!).} At the end of four minutes,
as quickly as you can, insert the thermometer into the ball you have been
squeezing and record the temperature. Note that it takes quite a few seconds
for the temperature to read the correct value, continue to squeeze this ball
{\it with} the thermometer inserted, until the temperature no longer rises. 
Record
this value in the ``End Temperature'' column for Ball \#2. Now, do the same
for Ball \#1, but do not squeeze, simply continue to quietly hold it in your
hand while the thermometer rises to its maximum temperature. Put this value in 
Table \ref{exercise}. [If you cannot repeatedly squeeze Ball \#2
for four straight minutes in one hand, go ahead and switch hands, as long
as the same ball is the one that continues to get squeezed.]\\

\begin{flushleft}
Take the difference between the End and Start temperatures and enter it into
the final column of Table \ref{exercise}. ({\bf 6 points})\\
\vspace{0.15in}
\begin{table}[htb]
\begin{center}
\caption{Exercise Ball Temperatures}
\begin{tabular}{|c|c|c|c|}\hline
 &{\bf Start Temperature} & {\bf End Temperature}& {\bf Change in Temperature}\\
\hline
Ball \#1 & & & \\
\hline
Ball \#2 & & & \\
\hline
\end{tabular}
\label{exercise}
\end{center}
\end{table}

{\bf Answer the following questions:} Are the start and end temperatures for both 
balls different? Why do you think we had you hold onto Ball \#1 the entire time 
you were squeezing Ball \#2?  Which ball showed the greater temperature rise? Why 
did this happen, and where did this energy come from? ({\bf 6 points})
\vspace{2.5in}

\end{flushleft}


\subsection{Investigating the Volcanoes of Io}

Now to the main part of today's lab, the volcanoes of Io. Along with the
other lab materials, we have supplied you with a three ring binder
containing images of Io, along with a large laminated map
of Io. Please do not write on any of these items! The first section contains
some images of Io taken with the Galileo spacecraft. Just page through
them to get familiar with Io (including color versions of the Figures in the 
introduction of this lab). Io is an unusual place!\\

Today we are going to look at images and data obtained with three different
instruments of the Galileo spacecraft: the Solid State Imager (SSI), the
Near-Infrared Mapping Spectrometer (NIMS), and the Photopolarimeter-Radiometer
(PPR). The SSI is simply a (``0.6 megapixel'') digital camera not unlike the 
one in your smart phone, and only can detect visible light (technically 
wavelengths from 0.4 to 1.1 $\mu$m). NIMS is also an imager, but it detects 
near-infrared light, having wavelengths from 0.7 to 5.2 $\mu$m (your TA
will demonstrate a version of this type of infrared camera during lab). The PPR 
measures the heat output of objects (not really an imager, 
though you could make coarse pictures with it), and could detect light with 
wavelengths from 17 to 110 $\mu$m. \\

Let's go back and look at Fig. \ref{bbody}. Do you understand why these
instruments were included on a mission to Jupiter? The Sun has
a blackbody temperature of about 6,000 K, what is the wavelength of
peak emission for such a blackbody? This is the light that illuminates
the Earth during the day, and all of the other objects in our solar system.
Thus, to see these objects, we only need a regular camera (the SSI).
But Jupiter is very far from the Sun, and thus it is very cold place. For
example, at the surfaces of the Galilean satellites, the temperatures
are about 100 K. To measure such cold objects, we need an instrument like
the PPR. If there are hot spots on Jupiter or any of its moons (like Io!),
they might have temperatures between 500 and 2000 K, and we will need
a ``near-infrared'' camera like NIMS to detect this light. \\

In the second section of the three ring binder are some NIMS images. The 
first set of images shows a color picture of Io obtained with the SSI,
and two images obtained with NIMS (at 1.593 and 4.133 $\mu$m). Note that
in the SSI image there are bright and dark regions all over Io. In the
NIMS images, however, Io begins to look quite different. In image \#5a,
at 1.593 $\mu$m there is still {\it some} reflected sunlight (since this is a 
daytime NIMS image), but by 4.133 $\mu$m thermal (blackbody) emission from 
Io is now strong. \\

\begin{flushleft}
{\bf Exercise \#2}
\end{flushleft}

\begin{enumerate}
\setcounter{enumi}{5}

\item In image \#5a, we see that at 4.133 $\mu$m there are many bright
spots. Returning to Fig. \ref{bbody} (above), estimate the temperature of these
bright spots. [Hint: can you see the bright spots at 1.593 $\mu$m? What
is the hottest blackbody in this figure that has a lot of emission at 4.133 
$\mu$m, but (almost) none at 1.593 $\mu$m?] ({\bf 4 points})\\
\vspace{1.5in}
\item In image \#5b, around the dark spot near the center of the 1.3 $\mu$m 
image, there is a bright ring. But this ring is very dark at 4.2 $\mu$m,
suggesting it is very cold. How can it be bright at 1.3 $\mu$m, and dark
at 4.2 $\mu$m? These are daytime images. Can you explain this feature? [Hint: 
think about snow] ({\bf 4 points})\\
\vspace{2.0in}
\item In Fig. \ref{2bbody}, below, are plotted two blackbodies (energy emitted
in Watts vs. wavelength in micrometers). Using Wien's law, what are the
approximate temperatures of each of these blackbodies (one {\bf solid} line,
one {\bf dashed}) in ``K''? Which one is emitting more total energy? How
do you explain this? ({\bf 4 points})\\
\vspace{1.5in}
\end{enumerate}
\begin{figure}[htbp]
\centerline{\includegraphics[width=4.0in]{iolabbbfig.pdf}}
\caption{The energy vs. wavelength, the ``spectra'' (spectra is plural
of spectrum), produced by two blackbodies with
different temperatures.}
\label{2bbody}
\end{figure}

\begin{flushleft}
{\bf Exercise \#3}
\end{flushleft}

In section 3 of the binder, we have some NIMS images of active
regions on Io. On these images are some small, numbered boxes, we will 
be looking at the NIMS + PPR spectra of some of these boxed regions to 
determine 
their temperatures. The names of the features on Io are from a variety of
mythologies that have to do with deities of fire, volcanoes, the Sun, thunder 
and characters and places from Dante's Inferno. Named mountains, plateaus, 
layered terrain, and shield volcanoes are given the terms mons, mensa, planum, 
and tholus, respectively. The term ``Patera'' (plural = Paterae) means a bowl, 
and brighter, whitish regions go by the name ``Regio''.

\begin{enumerate}
\setcounter{enumi}{8}

\item Region \#1 (Image \#6) is a night time NIMS image of a region on Io. In 
this image, you can see lines of longitude and latitude. It basically runs from 
125$^{\circ}$W to 132$^{\circ}$W in longitude, and from $+$59$^{\circ}$ to
$+$71$^{\circ}$ in latitude. Using the big map of Io, what is the name of this
active region? [Note: an SSI image of this region is shown in binder 
image \#4!] ({\bf 2 points})
\vspace{1.5in}

\end{enumerate}

\begin{enumerate}
\setcounter{enumi}{10}
\item It is clear the NIMS instrument does not make very pretty pictures,
it has ``poor resolution''. When this camera was built, infrared imaging
technology was just becoming possible. The infrared camera that your
TA has demonstrated today in class is as good, or better than NIMS! In these 
NIMS images, redder colors
mean hot, and bluer colors mean cool. Compare the Region \#1 NIMS image
to Image \#4 in the binder from the SSI (they have totally different
orientations!!!). Can you figure out what is happening? 
Can you figure out which boxes in the NIMS image cover the hot, glowing
lava feature in the SSI image?  ({\bf 6 points})
\vspace{1.9in}

\item The NIMS image of Region \#2 is shown as Image \#7. Using the
large map, what is the name of this region?  ({\bf 2 points}) 
\vspace{1.2in}
\item In fact, the NIMS image of Region \#2 does not cover all of this large
feature, does it? In Fig. \ref{loki} we present the NIMS + PPR spectra of
the six boxes shown identified in Image \#7. Using the plastic blackbody
overlay, measure the temperatures for {\it only} boxes 1 and 4. [If you
are having trouble doing this, ask your TA for help.] ({\bf 4 points})
\vspace{0.5in}

\begin{table}[htb]
\begin{center}
\caption{Region \#2 Box Temperatures}
\begin{tabular}{|c|c|c|}\hline
{\bf Box} & {\bf Maximum Wavelength ($\mu$m)}& {\bf Temperature (K)}\\
\hline
Box \#1 & & \\
\hline
Box \#4 & & \\
\hline
\end{tabular}
\label{region2tab}
\end{center}
\end{table}
\item The radius of Io is 1,821.3 km, that means that the circumference of
Io is (C = 2$\pi$R) 11,443.6 km. Since there are 360$^{\circ}$ in a circle,
each degree of {\it latitude} represents 31.79 km. Assuming the northern half
of this glowing ring has the same size as the southern half, what is the
total area covered by the hot material of this feature? [Hint: The latitude
increases from the bottom to the top of the image (approximately the y-axis of
the figure), while the horizontal (x-axis) direction is longitude.
Note that the white grid lines are identical in size in the vertical
and horizontal directions, thus you can measure both sides of the box
in degrees of latitude (note that degrees of longitude only equal degrees
of latitude at the equator, and this region is not at the equator!).
The degrees of latitude are the small white numbers that run from
9 to 13.]\\

The area of a square is simply {\it side $\times$ side} = $s^{\rm 2}$. Calculate
the area in square kilometers of one white grid box (not the tiny little boxes 
you measured the temperatures for!). Next, estimate the number of such grid
squares {\it fully} covered by the ``hot'' reddish regions for the southern 
half of this feature (this will be a fraction of a grid box for some spots). 
The total area in square kilometers is the number of boxes covered times 
the area of one box---find this number.
Multiply that result by two, and you have the approximate area of the entire
feature. ({\bf 6 points})
 \vspace{2.0in}
\item Now we want to figure out the total energy output of all of the volcanoes
on Io. Step 1: In the large map of Io, the paterae are the brown regions. You 
can see that the volcano you just measured is just about the largest such 
feature on Io. The average patera appears to have about 5\% (= 0.05) the area
of this feature.  Estimate the total area covered by {\it all of the paterae} 
on Io. [Hint: note what we said in the introduction about the estimated number 
of volcanoes on Io.] ({\bf 4 points}) \\
\end{enumerate}
\begin{flushleft}
\vspace{0.25in}
{\bf Total Volcano Area} = Average area $\times$ number of volcanoes = \hspace*{0.65in} \underline{????} km$^{\rm 2}$ \\
\vspace{0.25in}
{\bf Total Volcano Area} = \hspace*{0.9in} $\times$ \hspace*{1.5in} = \hspace*{1.0in} km$^{\rm 2}$ \\
\end{flushleft} 

%\begin{figure}[htbp]
%\centerline{\includegraphics[width=7.0in]{region1fig.pdf}}
%\caption{The blackbody ``spectra'' (the plural of spectrum) of the six boxes shown in Image \#6. Be careful, these plots have {\it log} wavelength on the x-axis.}
%\label{region1}
%\end{figure}
\begin{figure}[htbp]
\centerline{\includegraphics[width=7.0in]{region2fig.pdf}}
\caption{The blackbody spectra of the six boxes shown in Image \#7.  Be careful, these plots have {\it log} wavelength on the x-axis.}
\label{loki}
\end{figure}
\vspace{0.8in}
\begin{enumerate}
\setcounter{enumi}{15}
\item Step 2: Figure out the total area of Io. The area of a sphere is
4$\pi$R$^{\rm 2}$. ({\bf 3 points})
\vspace{1.0in}
\item Step 3: We will assume that the average surface temperature of the
non-volcanic regions on Io is the same as that of box \#4 on Image \#7 that you
found above. We will assume that the average temperature of the paterae
is the same as that of box \#1 on Image \#7 that you found above. Now, we
are going to use the Stephan-Boltzmann law to calculate how much energy
the volcanoes on Io put out compared to the rest of Io. Remember, the 
Stephan-Boltzmann law was the amount of energy output per unit area (m$^{\rm 2}$):
\end{enumerate}
\begin{equation}
\frac{E_A}{E_B} = \frac{sT_A^4}{sT_B^4} = \left(\frac{T_A}{T_B}\right)^4
\end{equation}
\vspace{0.3in}
\begin{flushleft}
Since in this problem we have two different emitting areas (total Io area, and area
covered by volcanoes), we have to modify this law to explicitly include the area 
terms:

\begin{equation}
\frac{(Total~Emitted~Energy)_A}{(Total~Emitted~Energy)_B} = \frac{Area_A}{Area_B} \times \left(\frac{T_A}{T_B}\right)^4
\end{equation}
So, \\
\begin{equation}
\frac{(Total~Emitted~Energy)_{Volcano}}{(Total~Emitted~Energy)_{Io}} = \frac{(Area)_{Volcano}}{(Area)_{Io}} \times \left(\frac{T_{\#1}}{T_{\#4}}\right)^4
\end{equation}
~ \\
\begin{equation}
\frac{(Total~Emitted~Energy)_{Volcano}}{(Total~Emitted~Energy)_{Io}} =
\end{equation}
~ \\
\vspace{0.9in}
The volcanoes on Io put out how much more energy than the total {\it for all of}
 Io? Do you find this surprising? Note that the Sun is far away (5.2 AU),
and cannot heat-up Io very much. Thus, gravitational heating can be
very important. This process is probably going on elsewhere in the
solar system (such as with the moons of Saturn). What does this mean
for the possibility of life existing on/inside these moons? ({\bf 4 points})
\end{flushleft}
%The figure below makes template blackbody SEDs for measuring the
%temperature of the box. Due to Latex's scaling issues, this has
%to be identical to the previous figure. So, if you change that caption
%or size, you need to do the same to this graphic to insure it works!
%\begin{figure}[htbp]
%\centerline{\includegraphics[width=7.0in]{newtemplatebb.pdf}}
%\caption{This plot allows you to make template blackbody curves so that you
%can photocopy them to onto transparencies. Since
%this is latex, the captions must be two lines long.}
%\label{region1}
%\end{figure}
%for scaling
\newpage

\hspace*{3.0in}Name:{\hrulefill} \\
\hspace*{3.25in}Date:{\hrulefill}\\

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

\begin{enumerate}
\item  In the graph below, draw two curves indicating the blackbody 
curves (energy as a function of wavelength) emitted by i) a hot object
(T = 6,000 K), and ii) a cool object (T = 1,000 K). Both objects  have
the same area. You will be graded on the {\it relative positions} of these two 
curves with respect to one another, as well as which one emits more
energy. Label the y-axis with the appropriate numbers, and identify the 
blackbody curves! ({\bf 10 points})

\begin{figure}[htbp]
\centerline{{\includegraphics[width=16cm]{ioplot.jpg}}}
\end{figure}
\clearpage
\item  If Europa and Ganymede were further from Jupiter (had larger orbits),
but Io remained where it is, would you still expect Io to experience 
volcanism?  Explain.  ({\bf 10 points})
\vspace{2.75in}

\item  The colorful volcanic features we have studied in this lab involve the 
chemical element sulfur. It is not expected that molten sulfur gets any hotter 
than $\sim$ 350 Kelvin or so on Io's surface.  As you have found out, however, 
many spots on Io's surface have been determined to possess temperatures that
are much hotter, some as hot as 1800 K! It is believed that such regions 
must consist of molten rock (silicates, like lava here on Earth) and not molten 
sulfur.\\
 \\
a) How many times greater would the flux from such a rock-lava region be 
compared to the flux emitted by the colder regions of Io (such as you
measured in Exercise \#3, question \#13). ({\bf 3 points})\\
\vspace{1.75in}\\
b) At what wavelength would the maximum (peak) energy emission occur from
this 1,800 K region? ({\bf 2 points})\\
\clearpage
c) Returning to Figure \ref{bbody}, would this very hot lava be detectable
with the SSI? Explain. ({\bf 5 points})\\
\vspace{1.75in}
\item  Jupiter has several moons that are much, much smaller than Io and that 
orbit even closer to Jupiter than Io.  Give a brief explanation of why  
these moons do NOT show evidence of volcanism [Hint: think of a man-made
satellite in Earth orbit, even a {\it big} one such as the International Space 
Station].  
({\bf 5 points})
\vspace{3cm}
\end{enumerate}
\subsection{Possible Quiz Questions}
\begin{enumerate}
\item Why does Io have volcanoes?
\item What does the term ``orbital resonance'' mean?
\item What is a ``blackbody''?
\item What is Wien's law?
\item What does the term ``patera'' mean?
\end{enumerate}
\subsection{Extra Credit (ask your TA for permission before attempting,
5 points)}

Orbital resonances are found elsewhere in the solar system. For example,
the shaping of Saturn's ring system, or the relationship between Neptune and 
Pluto. Type-up a one page discussion of
how orbital resonances affect the appearance of Saturn's rings, or how
the Neptune-Pluto orbital resonance gives us insight into the processes
that shaped the formation of our solar system.
%\end{document}
