the cosmologist's
toolbelt
Introduction to Stars, Galaxies, and Cosmology
microtask workbook
V .2
Kei h Kilo Wa M S M A
All righ s reser ed
Tool #1: Calculators
Science is a way of describing the world. We can do this using words, for example:
“As the temperature of an object goes up, the wavelength of light it emits gets smaller.”
This is fine and is often all we need. In fact, it’s far more important that you understand these
relationships in words than it is to do the math they imply. But sometimes we need to know exactly how
things are changing. To get exact values, we need to do some math. The statement above can be
written as a math equation:
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ =
2,897,768.5
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
As the temperature gets bigger, the value of the entire fraction gets smaller (to convince yourself of this,
consider the fractions ½ and ¼ -- as the bottom number gets bigger, the value of the fraction gets
smaller).
Scientists are basically lazy (okay, technically we’re efficient), so we want to write out as little as
possible. We traditionally use the Greek letter λ (lambda) to represent wavelength and T to represent
temperature. It actually doesn’t matter what symbol you use or what you call them – you can call λ
“chicken foot” if you want, or you can use a smiley face for wavelength instead. The symbol just
represents “wavelength”, and as long as everyone understands what you mean, you can use whatever
symbol you want. Using “standard” symbols makes it easier for everyone to know what you’re talking
about (“chicken foot” might get you some funny looks). A scientist might right the above equation this
way:
λ=
2,897,768.5
𝑇𝑇
All it means is “A hotter object emits wavelengths of light that are smaller.” Plugging in numbers just lets
you get details. If you want the details – how small is the wavelength of the light? – then you’re going to
have to use your calculator to figure them out. The math we do isn’t like an algebra class, it’s more like
learning to enter the numbers properly in your calculator.
The first thing we need to learn is the concept of significant digits. When you do a mathematical
operation on your calculator, it’s going to give you a long, long string of numbers. You don’t need to
copy down all those numbers. Suppose you have $113 in your bank account. If you tell someone you
have “about $100”, you are being accurate, but you aren’t being precise. The number 100 is precise to
one significant digit. You could be more precise if you wanted to be. You could tell someone you have
“about $110”, then you are still accurate, but now you’re being more precise – the number 110 has two
significant digits. The zeroes are just “placeholder” zeroes that are there to make sure you’re still
accurate – if you tell someone that you have $1 when you actually have $113, then you are being just as
precise – one significant digit – but you are no longer being accurate. The zeroes at the end are just
there to keep the number accurate, so they are not significant. Interior zeroes, such as in “101”, are
always significant – you only have to worry about whether zeroes at the end of a whole number are
significant
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If you wanted to be very precise, you could tell someone that you have “$113.10” – five significant
digits. But wait! Aren’t zeroes not significant? Think about it, 113.1 and 113.10 are exactly the same
number. You don’t need the extra zero on the end to keep yourself accurate. The only reason to include
that zero is so you can say, “Hey! I measured this more precisely! I have $113.10 and not $113.11!”
Since that last zero is showing precision rather than accuracy, it’s a significant digit.
For numbers less than one, the same concept applies. The value “.0134” has three significant digits. The
first zero is just there for accuracy -- .134 and .0134 aren’t the same number! Because that zero is there
for accuracy and not for precision, it isn’t a significant digit. It’s important to remember: Significant
digits are NOT the number of decimal places to keep!
Report the number of significant digits in each value below:
VALUE SIG. DIGITS
VALUE
SIG. DIGITS
132
10
5.40
5.4
.101
.011
The easy way to avoid having to worry about significant digits is to use scientific notation. In properly
written scientific notation every digit is significant. Proper scientific notation for 113 looks like this:
1.13 x 102
We always have one digit in front of the decimal. The exponent of the 10 (the “2” in this case) tells you
how many places to move. If the exponent is positive, move the decimal so that the result is bigger than
one. If the exponent is negative, move the decimal so that the result is less than one. For example, .0134
in scientific notation would be:
1.34 x 10-2
In both cases, we have three digits in the main part of the number, so we have three significant digits.
Here are both numbers written to two significant digits:
110 = 1.1 x 102 and .013 = 1.3 x 10-2
What does this have to do with calculators? You never have to report all those numbers that your
calculator gives you as an answer. You only need to report three significant digits. For example, if your
calculator tells you the answer is:
66.4783284128
You only need to report:
66.5
(notice that the 4 got rounded up to 5 – do you see why?)
One of the most important things to keep in mind: You must use the scientific notation button on your
calculator! In other words, you should never enter “1.1 x 10 ^ 2” in your calculator – in certain cases, this
gives you the wrong answer! Your scientific notation button might be EE, or EXP, or x10x – you’ll just
have to find it on your particular calculator. You don’t press the “times” button at all. Instead, you enter
the number in as “1.1 EE 2”. It’s not only faster, it will always give you the right answer!
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Tool #1: Calculators
Practice Problems:
1. Write each of the following numbers to three significant digits. You may use scientific notation if
you wish.
VALUE
3 SIG DIGITS
2,897,768.5 2.90 x 10^6
4.145 x 103 4.15 x 10^-6
VALUE
3 SIG DIGITS
.003643 3.64 x 10^-3
5.4362821 5.44
2. Convert each of the following numbers into standard notation using three significant digits.
VALUE
STANDARD
2.99 x 105 2,990
1.50 x 108 0.0150
VALUE
STANDARD
2.74 x 10-3 27.4
6.40 x 101 64,000
3. Use the numbers provided to find the value of the equation below. YOU MUST USE YOUR
SCIENTIFIC NOTATION BUTTON!
𝐹𝐹
𝐿𝐿𝑠𝑠 = (9.94 𝑥𝑥 106 ) 𝜋𝜋 2
𝑝𝑝
F = 3.839 x 1026
p = 3.26 x 10-2
π = 3.1415926
3.5 x 10^9
𝐿𝐿𝑠𝑠 = __________________
4. Use the numbers provided to find the value of the equation below. Don’t forget that there are
implied parentheses in the expression below.
𝑅𝑅𝑠𝑠 =
L= 7.239 x 1026
𝐿𝐿1/2
𝑇𝑇 2
)
5800
(
T = 7500
2.08 x 10^3
𝑅𝑅𝑠𝑠 = __________________
5. Convert from parsecs to light-years if 1 parsec = 3.26 light-years
2.86 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑥𝑥
3.26 𝑙𝑙𝑙𝑙𝑙𝑙ℎ𝑡𝑡−𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
1 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑐𝑐
9.32
= ____________________
𝑙𝑙𝑙𝑙𝑙𝑙ℎ𝑡𝑡 − 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
Notice that there’s nothing special about conversions – just set up your multiplication so that
you cancel out what you don’t want. In this case, you want parsecs to cancel, so put it in the
bottom of your fraction.
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Tool #2: Conversions
Throughout this course, and often in your everyday life, you will need to convert a measurement in one
unit to a different unit. For example, if I wanted to run a 5K race, I will need to practice on the treadmill
at the gym. But when I step onto the treadmill, the display measures my distance in miles not
kilometers. I will need to convert the 5 kilometers into miles to see how long I should run to practice for
my race.
Conversions always require two things: a measurement that you need to convert and a conversion
factor. The conversion factor is a relation between the unit you have and the unit you want. It usually
has the form
1 of some unit = ### of another unit
The conversion factor is usually given to you. In our example, the conversion factor is
1 km = 0.621 miles
So in this example we have identified our “number to convert” and our “conversion factor.”
Number to convert: _______5 kilometers_________
Conversion Factor:
1 km = 0.621 miles______
Now to set up our conversion we need to cancel out the unit we have and leave us with the unit we
need. And we want to do this in a way that we will always know what to multiply and what to divide.
When doing this, we will need to split the conversion factor across the equal sign. We will place the
“1km” side in one place and the “0.621 miles” side in another place.
NOTE: Each side of the conversion factor is glued together!!! You cannot switch the unit and the
number! You will never place “1 miles” and “0.621 km” in a conversion because these two values are
NOT equal to each other!
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Using the guide grid below, place the values the 3 corners such that the unit of your number to convert
will be canceled out when we divide by the bottom values.
Number to convert: 5 km
Conversion Factor part with units we want: 0.621 miles
Conversion Factor portion with the units to cancel: 1 km
Now do a reality check.
How do you cancel the kilometers? MULTIPLY THEM TOGETHER (km * km)
Given how you set up the box, will the kilometers cancel out?
YES
DIVIDE THEM (km / km)
NO
If you answered no, go back and fix your conversion factor boxes so that the units cancel. If you
answered yes, multiply across the top and divide by each number in the bottom to find out how many
miles I need to run to practice for my 5K race.
How many miles will I have to run? ____________________________miles
Conversions and unit analysis are some of the most useful tools in science. A unit analysis will tell you if
you have set up an equation right by canceling out the units you don’t want and leaving the units you do
want.
Through the rest of the course, you will need to convert properties you’ve learned to calculate into a
different set of units to compare planets. Below are the most common conversion factors you will use
in this course. The practice worksheet will have you set up our most common conversions.
1 𝑀𝑀𝐸𝐸 = 5.97 × 1027 𝑔𝑔
1 𝑅𝑅𝐸𝐸 = 6.37 × 108 𝑐𝑐𝑐𝑐
1 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 = 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
1 𝐿𝐿𝑠𝑠 = 3.827 × 1026 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊
1 𝐴𝐴𝐴𝐴 = 1.496 × 1011 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
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Tool #2: Conversions
Practice Problems:
Complete the following conversions. Conversion factors are given below. Remember all answers need
to be in 3 significant figures.
1.
1 𝑀𝑀𝐸𝐸 = 5.97 × 1027 𝑔𝑔
1 𝑅𝑅𝐸𝐸 = 6.37 × 108 𝑐𝑐𝑐𝑐
1 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 = 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
1 𝐿𝐿𝑠𝑠 = 3.827 × 1026 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊
1 𝐴𝐴𝐴𝐴 = 1.496 × 1011 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
Convert 750 days into years.
Number to convert: ______________________
Conversion Factor: _________________________
= _________________________ years
2. Convert 1.5 MEarth into grams.
Number to convert: ______________________
Conversion Factor: _________________________
= ______________________________ grams
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3.
Convert 3.2 REarth into cm.
Number to convert: ______________________
Conversion Factor: _________________________
= _______________________________ cm
4. Convert 5.4 LSun into Watts.
Number to convert: ______________________
Conversion Factor: _________________________
= ________________________________ Watts
5.
Convert 3.5 AU into meters.
Number to convert: ______________________
Conversion Factor: _________________________
= _____________________________ meters
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Tool #3: Position in the sky
The sky outside appears to be a dome over our heads. The diagram below shows how the sky might
appear to you when standing outside in the United States. Notice that to you, the ground seems
horizontal – you don’t feel like you’re sticking out from the Earth (even though you are!). The shaded
oval in the diagram shows the horizon. The Earth’s axis is shown on the diagram; all the stars seem to
rotate around this celestial north pole (even though it is actually Earth that is turning).
If we wanted to tell someone how to find Polaris (in this example), we could easily tell them to look a
certain number of degrees above the horizon. The altitude is the distance of the star above the horizon
in degrees. We always start from zero on the horizon and go directly up to 90 degrees directly overhead.
In this case, Polaris’ altitude is about 45 degrees above the horizon.
But the horizon is a circle all around us and Polaris is not 45 degrees above all parts of that circle. We
also need to specify which part of the horizon the star is above. Polaris is the North Star, so in the
diagram is it above the northern horizon. But if we went up 45 degrees from the southern horizon, we
wouldn’t find the star! The arrow in the diagram below shows the problem.
We need another coordinate to tell us where on the ground to start moving up from. We start counting
from zero degrees at due north, to 90 degrees at due east, to 180 degrees at due south, to 270 degrees
at due west, and finally back to 360 degrees (or zero degrees) at due north again. Test yourself by filling
in the degrees for each of the four boxes in the diagram below. (Don’t peek at the next page just yet!)
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The azimuth is the distance to the star (actually the point on the ground just below the star) around the
horizon in degrees, starting from north. With these two coordinates, we can point to anything on the
sky, and everyone will know where to look. It’s simple and intuitive! These coordinates make up the
horizon coordinate system.
But there are two problems with the horizon coordinate system. First, the Earth is turning from west to
east on its axis. That means that stars appear to rise in the east and set in the west. If the star is rising,
then its altitude must be increasing. Similarly, if a star is setting, its altitude must be decreasing. It’s
azimuth is changing as well – the star might rise in the east (90 degrees), reach its highest point in the
south (180 degrees), and then set in the west (270 degrees).
This points out the first of the two additional pieces of information we have to have in order to use the
horizon coordinate system: We must know the time of the observation! If we know that the star will be
at 45 degrees altitude and 100 degrees azimuth at 9:00 PM, then we know where to look. A few hours
later, the star might be at 60 degrees altitude and 180 degrees azimuth. The diagram below shows the
star at these two times (remember that the star stars on the surface of the dome of the sky):
Remember, at 9:00 (for example) start at 100 degrees on the horizon and go up the curve of the dome
of the sky towards the point directly overhead (which is 90 degrees altitude). Since the star is about
halfway between the horizon and overhead, it must be at about 45 degrees altitude. At 11:00, the star is
above 180 degrees (south, in other words), and you go about 2/3 the way up from the horizon towards
the point directly overhead. Try to visualize this in 3D – you’ll get to work with this more on the practice
worksheet.
The other problem is that when it is night time at your location, it will be the middle of the day on the
other side of the Earth. Even if we are viewing at exactly the same instant (for example, if we call a
friend in Europe on our cell phone), the star will be in a different position (or below the horizon entirely)
and so will have a different altitude and azimuth. This is the second piece of information we need in
order to use the horizon coordinate system: We must know the place on Earth! The diagram below
shows the problem. As you can see, even though we can see the star in the United States, it’s below the
horizon from Europe. If we know these two pieces of information, though, the horizon coordinate
system is always available and easy to use.
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Tool #3: Position in the sky
Practice Problems:
1. On the diagram, draw a curved arrow to show where azimuth is measured.
2. On the diagram, draw a curved arrow to show where altitude is measured.
3. For each of the positions below, give an estimate of the altitude and azimuth of the star at that
position (write “Below the horizon” if the star is not visible):
Position
Alt
A1
A2
A3
A4
Polaris
B1
B2
B3
B4
11
Az
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Tool #4: Motions of the sky
The horizon coordinate system is perfect for pointing to an object in the sky, but the altitude and
azimuth change because the Earth is moving. In fact, Earth is moving in three different ways all at once!
In order to understand how the sky moves, we need to understand how the Earth moves.
The Earth’s rotation is its spinning on its own axis. Earth takes 23 hours and 56 minutes to spin one time
on its axis – this defines the sidereal day. Because Earth’s axis is currently pointed towards Polaris (our
current North Star), the sky seems to rotate around Polaris, as you can see in the photo below.
The key effect of the Earth’s rotation is that it is what causes the Sun, the Moon, the stars, the planets –
everything in the sky – to rise in the east and set in the west.
The Earth’s revolution is the Earth travelling around the Sun. Earth takes 365.25 days to go around the
Sun one time – this defines the sidereal year. The Earth’s axis stays pointed towards Polaris throughout
the year, so the northern hemisphere is tilted towards the Sun in the summer and away from the Sun in
the winter. The diagram below shows what is happening.
Also, Earth’s revolution is the reason why some constellations are only visible in the winter, while others
are only visible in the summer. Technically, all the constellations are in the sky all year round, but during
the winter (for example), the summer constellations are on the day side of the Earth, so we can’t really
see them. We really can only see stars when they are on the side of the Earth facing away from the Sun.
Therefore, we have different constellations viewable at different times of the year.
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Incidentally, Earth’s revolution is why our solar day is exactly 24 hours instead of 23 hours and 56
minutes. The solar day is defined to be the time from when the Sun is directly overhead on one day until
it is again directly overhead on the next day. The Earth spins 360 degrees in 23 hours and 56 minutes,
but in that same time, it has moved a little bit further around the Sun in its orbit. It has to turn just a
little further – four minutes further – in order for the Sun to be directly overhead again. If we didn’t
revolve around the Sun, our solar day would be the same as our sidereal day – 23 hours and 56 minutes.
The diagram below shows what is happening.
The third motion of the Earth may not be as familiar to you as the other two. The Earth’s precession is
the wobble of the Earth on its axis. Just like a top wobbles, the Earth does too. Our axis always keeps the
same angle – 23.5 degrees – but it’s not always pointed towards Polaris. Precession means that our
North Star slowly changes over time. Take a look at the diagram below:
It takes about 26,000 years for the Earth’s axis to wobble in one complete circle. That means that in
13,000 years, Vega will be our North Star! There are 13 constellations of the zodiac (Taurus, Gemini,
Cancer, etc.), so the Earth’s axis precesses by one constellation about every 2,000 years. Test yourself!
In the diagram below, Vega is now the North Star. Fill in each box with either winter or summer.
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Tool #4: Motions of the sky
Practice Problems:
The diagram on the left shows the view of the sky as seen from Earth. The diagram on the right shows
the sky as seen from an observer out in space. The path Star A takes over the course of a night is shown
on the left diagram. The numbers are the order of the observations of the star.
1. Which direction (north, east, south, or west) are you facing in the left diagram? How do you
know?
2. On the left diagram, fill in the boxes with the numbers that correspond to Star A’s positions in
the right diagram.
3. On the left diagram, draw an arrow on the dashed circle showing the direction that Star A
appears to move.
4. The point directly overhead is the zenith. At what numbered position will Star A be at zenith?
5. Suppose Star A is at position 3 at 6 PM. Label each position in both diagrams with the time at
which Star A will arrive there.
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Tool #5: Star Maps
Because of the motions of the Earth, the positions of objects in the sky change over time: stars (and the
Sun) rise and set, some stars are visible only during the winter, and some stars are visible only during the
summer. Since the altitude and azimuth of the stars are always changing, we need a way to find them
using coordinates that don’t change – we need a coordinate system for star maps. To accomplish this,
astronomers invented a system called “equatorial coordinates”.
Imagine that all the stars are painted on a giant invisible sphere that surrounds the Earth. The diagram
below shows this “celestial sphere”.
The stars stay fixed on the sphere, while the Earth turns within the sphere (which, of course, is what’s
really happening). Polaris is on the celestial sphere, but Earth’s axis happens to pass almost exactly
through it. As the Earth turns, Polaris appears to always be in the same place – this is why the sky seems
to turn around the North Star.
Suppose you are standing on the North Pole. Where would Polaris be in the sky? Select your answer.
____OVERHEAD
____ON THE HORIZON
____BELOW THE HORIZON
____ HALFWAY BETWEEN THE HORIZON AND OVERHEAD
Suppose you are standing on Earth’s equator. Where would Polaris be in the sky?
____OVERHEAD
____ON THE HORIZON
____BELOW THE HORIZON
____ HALFWAY BETWEEN THE HORIZON AND OVERHEAD
Suppose you were in the United States (about 45° latitude). Where would Polaris be in the sky?
____OVERHEAD
____ON THE HORIZON
____BELOW THE HORIZON
____ HALFWAY BETWEEN THE HORIZON AND OVERHEAD
Here’s a way to help figure out your location on Earth: The altitude of Polaris is always exactly equal to
your latitude. So, if Polaris is 30° above the horizon, then your latitude is 30° North.
Because stars don’t move on the celestial sphere, we can use it to make star maps. The celestial
equator divides the sphere in two (and is directly above Earth’s equator, as you can see in the diagram).
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We can draw circles parallel to the celestial equator, and we can draw lines perpendicular to the
celestial equator:
The lines parallel to the celestial equator (they look like lines of latitude on Earth) are called declination,
and they tell us the star’s distance above or below the celestial equator. Declination (DEC) is measured
in degrees just like latitude: 0° is the equator, the celestial north pole is at +90°, and the celestial south
pole is at -90°. The lines perpendicular to the celestial equator (they look like lines of longitude on
Earth) are called right ascension, and they tell us the star’s distance around the celestial equator. Right
ascension (RA) is measured in hours, but these are hours of distance, not hours of time – don’t get them
confused! By tradition (thousands of years of tradition), the 0 hour line goes through wherever the Sun
is on the first day of spring, March 21st. The hour lines increase eastward (to the right in the diagram)
until they get back to the 24 hour line, which is the same as the 0 hour line.
When we put these two sets of lines together, we get a grid on the celestial sphere.
Since the star is painted on the celestial sphere, its position on the grid never changes. When we specify
the declination (DEC) and right ascension (RA) for a star, they will always be the same.
Remember that equatorial coordinates are only used for star maps, but there are some useful rules of
thumb we can use:
•
•
DEC near 90° will be near Polaris in the north and will be visible all night.
RA near 0 hrs will be overhead at midnight in December.
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Tool #5: Star Maps
Practice Problems:
Shown below is a map of Orion’s Belt, one of the more obvious constellations in the sky. The three stars
in a line are very easy to spot in the sky. Hours and minutes of right ascension are shown at the top and
bottom of the map and go right to left (they go the opposite way because we are inside the celestial
sphere!). There are 60 minutes in each hour of right ascension. Declination is shown on the left and
right sides of the map.
1. What is the approximate RA of Mintaka? _______ hours _________ min
2. What is the approximate DEC of Mintaka? ___________ degrees
3. Draw a dot on the map at 5 hours 52 min of RA and -5 degrees of DEC.
4. Will Orion’s Belt be visible in the winter or in the summer? ___________________
5. Is Orion’s Belt visible in the north (near Polaris) or in the south? ________________
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Tool #6: Telescopes
When many people think of astronomy, they think of telescopes. After all, things in space are far away,
so we need to make them closer, or at least bigger, right? Well, not quite. The only purpose of a
telescope is to gather light. Your retina needs a certain amount of energy to be stimulated. When it is
stimulated, it sends a signal up the optic nerve to the brain – and it’s your brain that actually sees. Only
the closest and brightest stars send enough energy to your retina to stimulate it. For fainter stars, we
need to gather as many light rays as we can and concentrate that energy on the retina. That’s what a
telescope does.
Glass bends light through refraction. The greater the curvature and thickness of a glass lens, the more
the light gets bent. In the diagram below, you can see how the lens is bending the light. Extend the light
rays until they meet in one place.
The point you just found is the focal point of the lens, or just the focus. When you focus your camera,
you are adjusting the focal point so that it falls directly on the film. When your eyes focus, you are
changing the lens of your eye (usually by changing the curvature using your eye muscles) so that the
focal point falls directly on your retina. If your eye muscles can’t pull or relax enough to get the focus
onto the retina, then you’ll see a blurry image – and you probably need glasses! If you’re near-sighted
(myopia), your lens is bending the light too much. Glasses “pre-bend” the light in the opposite direction
so that when it passes through the lens of your eye, the focal point Is in the right place. You could also
have surgery (LASIK, for example) to change the thickness of your eye – a thinner lens won’t bend the
light as much, so the focal point moves back onto the retina.
Telescopes work the same way. The distance from the lens to the focal point is the focal length of the
lens, which one of the two important properties of a lens. The longer the focal length, the farther back
the light will focus. Remember that a telescope only gathers light; it does not make things bigger. Things
look bigger through a telescope because we put a secondary lens – called the eyepiece – at the focal
point of the primary lens – called the objective lens. It’s the eyepiece that makes things bigger. The
formula for magnification is an important tool:
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 =
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ℎ 𝑜𝑜𝑜𝑜 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ℎ 𝑜𝑜𝑜𝑜 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
The telescope is the objective lens – you can’t generally change it. You can change eyepieces all you
want, however – and most astronomers have a kit of several eyepieces. Assuming the objective lens
doesn’t change, circle the appropriate responses:
Characteristic
Larger F.L. Eyepiece
Smaller F.L. Eyepiece
Magnification
Decreases
Decreases
Increases
Increases
21
Stays the same
Stays the same
Notice that the diameter, the second important property of a lens, has no effect on magnification at all!
Look at the diagrams below and circle the appropriate responses.
Characteristic
Larger diameter
lens
More light rays
gathered
More energy
gathered
Brighter image
seen
Analysis
Lens A
Lens B
Same
Lens A
Lens B
Same
Lens A
Lens B
Same
Lens A
Lens B
Same
You may wonder why we don’t just use the largest telescope we can all the time. Telescopes can make
faint objects (like distant stars) look much brighter, but what if the object is already bright – like the
Moon or, worse, the Sun? If the object is too bright, you can actually fry your detector (like a camera or
– oops – your eye). We had to be careful not to let the Hubble Space Telescope even cross over the
Moon or the Earth – they are so bright, that they would destroy the telescope’s sensitive detectors.
Okay, so what about magnification? Why don’t we use the highest magnification possible all the time?
The problem here is that you don’t get something for nothing. Look at the diagrams below and circle the
appropriate responses.
Characteristic
Higher
magnification
Larger
field-of-view
Longer F.L.
eyepiece
Analysis
Eyepiece A
Eyepiece B
Same
Eyepiece A
Eyepiece B
Same
Eyepiece A
Eyepiece B
Same
If you want a wide field of view, you need to use a lower magnification eyepiece. Based on the
magnification formula, if you want a wide field of view, do you want a longer or shorter focal length
eyepiece? Logic through the steps and write your answer here: ____________________________
Refracting telescopes, or just refractors, use lenses. Reflecting telescopes, or reflectors, use curved
mirrors to gather light too. Obviously, as you can see below, since the focus is directly in front of the
mirror, we need to reflect that focused image out of the telescope. A Newtonian telescope reflects the
focused image to an eyepiece on the side of the telescope, while a Cassegrain telescope drills a tiny hole
in the mirror and reflects the focused image to an eyepiece in the back of the telescope.
22
Tool #6: Telescopes
Practice Problems:
Here is a table of characteristics for the two types of telescopes:
CHARACTERISTIC
Light Gathering Device
Maximum Diameter
Focal Length of Objective
Stability
Expense
REFRACTOR
Lens
Smaller
Longer
Fragile
Expensive
REFLECTOR
Mirror
Larger
Shorter
Sturdy
Cheap(er)
Also, remember the formula for magnification:
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 =
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ℎ 𝑜𝑜𝑜𝑜 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ℎ 𝑜𝑜𝑜𝑜 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
Finally, remember the relationship between magnification and field of view:
As magnification increases, field of view decreases.
Based on the above, circle the answer for the device you would choose:
REFRACTOR
REFLECTOR
2. Suppose you want to observe a very bright object. REFRACTOR
REFLECTOR
1. Suppose you want to observe a very faint object
3. Suppose you want to observe a very tiny object.
LONG F.L. EYEPIECE
SHORT F.L. EYEPIECE
4. Suppose you want to scan a wide area of the sky.
LONG F.L. EYEPIECE
SHORT F.L. EYEPIECE
5. Suppose you want to observe small details on the surface of the Moon.
REFRACTOR AND LONG F.L. EYEPIECE
REFRACTOR AND SHORT F.L. EYEPIECE
REFLECTOR AND LONG F.L. EYEPIECE
REFLECTOR AND SHORT F.L. EYEPIECE
23
24
Tool #7: Mounts and Detectors
Once you have chosen a telescope, your decisions are not over. You also need to decide what kind of
mount you want to put your telescope on. Once you’ve made that decision, you then have to decide
what you want to use to view celestial objects with: a camera, a scientific instrument, or your Mark I
Eyeball. As with telescopes, there is no one right answer to these questions – it depends on what you
want to do.
The choice of mounts is fairly simple. There are two kinds of mounts:
An alt-az mount is the simplest, cheapest, most versatile, and rugged mount you can buy. It swings the
telescope to point up and down from the horizon to overhead (altitude) and left and right around the
horizon (azimuth). There is nothing really to learn with this mount, you simply point it at the object you
want to observe and you’re done! Some typical alt-az mounts are shown below.
The alt-az mount has a major disadvantage, however. If you want to take a photograph of a celestial
object, you’ll need to leave the shutter open on the camera for as much as five or ten minutes in order
to gather as much light as possible. The stars (and everything else in the sky) rise in the east, reach a
maximum height above the horizon, and then set in the west. That means that you have to continually
move the telescope to follow the star. If you don’t do this absolutely precisely, then your beautiful image
of the sky will be nothing but a smear. Unfortunately, trying to track the star in both altitude and
azimuth (up/down and left/right) is like trying to draw a diagonal line with an Etch-a-Sketch – it’s nearly
impossible to do for ten minutes without stopping. Alt-az mounts, therefore are the best mount to use
unless you want to take a photo or need to follow the object for a long period of time.
If you do want to take a picture, you need the other kind of mount. An equatorial mount is designed so
that the mount only has to turn on one axis to follow the star across the sky. It usually has a motor,
called a clock drive, that turns the mount at exactly the right speed to keep the star exactly centered in
the telescope. Taking a long exposure photo is easy because the star never moves in the telescope’s
field of view. Some typical equatorial mounts are shown below. Equatorial mounts are very expensive,
(if you don’t need a picture put your money in a bigger telescope!) and must be carefully aligned to the
North Star (to make sure they turn in the correct direction) every time you use them.
25
In the spaces below, list one advantage and one disadvantage of each of the two types of mounts.
Alt-Az Mount
Equatorial Mount
Advantage
Disadvantage
Once you’ve selected a mount, you still need to decide on your detector. If you just want to go
stargazing, your detector is simply your eye! On the other hand, if you want to put a nice picture on your
wall, you’ll need a digital camera (technically called a “CCD camera”). If you want to do science, you
might need to measure the color of the object (using specially calibrated light filters), the composition of
the object (by looking at its spectrum, called “spectroscopy” – more on that later in the course), or the
brightness of the object and how that brightness changes over time (called “timing measurements”).
Using this information, match the detector with its characteristics:
All CCD cameras are greyscale cameras – even the digital camera in your phone! In order to take a color
photo, we need three CCD images: one taken through a red filter (only red light passes through), one
taken through a green filter (only green light passes through), and one taken through a blue filter (only
blue light passes through). Your digital camera colors the image from the red filter from black to red
(whiter parts of the picture are redder). It colors the image from the green filter from black to green,
and the image from the blue filter from black to blue. The three colored images are then combined into
a full color image. Your digital camera is actually three cameras all operating at the same time!
You need to take all three color images at the same time if you are photographing a three-year-old
(because three-year-olds wiggle faster than the local speed of sound). As long as you are using an
equatorial mount, objects in space won’t move in the field of view, so you can take the three pictures
one after the other. Most astronomical CCD cameras are therefore a single greyscale camera with three
color filters that can be cycled through.
26
Tool #7: Mounts and Detectors
Practice Problems:
Choose the best mount for the task:
1. Suppose you want to take a photograph of a comet.
ALT-AZ
EQUATORIAL
2. Suppose you want to go stargazing.
ALT-AZ
EQUATORIAL
3. Suppose you don’t have much money for a telescope.
ALT-AZ
EQUATORIAL
1. Suppose you want to take a photograph of a comet.
CCD
SPECTROSCOPE
2. Suppose you want to measure the color of a star.
EYE
CCD
3. Suppose you want to measure how the brightness
of a star changes when a planet moves in front of it
SPECTROSCOPE
TIMING
4. Suppose you want to determine the composition
of a star (what it is made out of)
SPECTROSCOPE
EYE
Choose the best detector for the task:
Suppose you want to take a picture of a faint distant galaxy in order to create an advertisement for a
science fiction movie. Circle the mount and detector you should use for this task:
MOUNT
DETECTOR
ALT-AZ
EYE
EQUATORIAL
CCD
SPECTROSCOPE
TIMING
27
28
Tool #8: Finding Distance
One of the most basic properties of a star, and one which allows us to determine other important
properties, is a star's distance from Earth. There are several ways to determine distance, but the
method we will be using here is called parallax.
HOW IT WORKS
1. Using a straight edge on Figure 1, draw a line from the Earth's position in January through the
nearby star to determine the apparent location of the nearby star with respect to the distant
stars as seen from Earth in January.
2. Using the same method, determine which distant star the nearby star will appear close to as
seen from Earth in July.
3. The angle between the two lines you just drew is the apparent motion of the star as seen from
Earth over the course of a year. But the parallax angle used to determine distance is only half of
this angle. Draw a line from the Sun to the nearby star and extend it to the distant stars.
4. The parallax angle is the angle between the line from the Sun to the star and one of the lines
from the Earth to the star. Label this angle p1.
5. Now draw another "nearby" star that is approximately twice the distance from Earth of the first
"nearby" star. Draw the same lines as you did in instructions 2 and 3 and label the parallax
angle p2.
6. How do the parallax angles compare for these two stars?
7. Will a more distant star have a larger or smaller parallax angle than these two stars?
8. Are the distances of stars directly proportional or inversely proportional to the parallax angles?
29
HOW WE SEE IT
Figure 2 is a series of images of a star field taken every month over the course of a year. Most stars are
so distant that they don't appear to move, but one star is "nearby" and moves from month to month.
1. Locate and circle the "nearby" star in each of the monthly images.
2. During which two months is the star at its extreme positions? This is the furthest along its path
before it turns around and goes the other direction.
3. On the diagram below (which does not include our "nearby" star) draw in the position of the
nearby star at its two extremes. Connect the two positions by a line.
4. Remember that this is the full motion of the star as Earth orbits the Sun. How does the parallax
angle (as defined on the previous page) compare to the length of this line? Indicate the length
of the parallax angle on the diagram below.
0.25"
Since stars are so distant, parallax angles are often measured in arcseconds (") which is the
equivalent of 1/3600th of a degree on the sky. The scale for 0.25" is indicated on the image
above.
5. Using the scale, estimate the size of the parallax angle you have indicated for the "nearby" star?
The distance to a star is inversely proportional to the measured parallax angle. As the distance
to a star increases, the measured parallax angle decreases. Astronomers calculate the
distances to stars using the relationship
1
𝑑𝑑 =
𝑝𝑝
where p is the parallax angle in arcseconds and d is the distance to the stars in a unit called
"parsecs." A parsec is defined to be the distance of a star with a parallax of one arcsecond. One
parsec is equal to 3.26 lightyears.
30
6. Use the parallax equation to calculate the distance to the star identified in the star field.
FIGURE 1
Drawing not
to scale
Earth in
July
Earth in
January
31
January
FIGURE 2
July
February
August
March
September
April
October
May
November
June
December
32
Tool #8: Finding Distance
Practice Problems:
Calculate the distance to the following stars using the parallax values given.
PARALLAX (")
MIRFAK
0.0055
ALMACH
0.0093
KOCHAB
0.025
REGULUS
0.0423
DISTANCE (PC)
What is the relationship between parallax and distance (circle one):
•
Closer stars have a larger parallax.
•
Distant stars have a larger parallax.
33
DISTANCE (LY)
34
Tool #9: Finding Luminosity
In order to determine how much energy a star is emitting, we need to measure the star’s luminosity.
Unfortunately, there’s no way for us to directly measure the luminosity of the star, but we can measure
the amount of energy from the star that we receive at Earth – the flux – as well as the distance to the
star. With these two observations, we can figure out just how much the star is sending to any planets
that are in orbit around it.
The luminosity is as intrinsic property of the star – it’s how bright the star really is. The flux is how bright
the star seems from Earth. The only reason the two numbers are different is because the star is very far
away – and as you know, the farther away a light is, the dimmer it appears.
Consider the following stars. Use the table to answer the following questions.
ACANASI
BADE
CERIERO
LUMINOSITY
(LSUN)
1
1
0.25
DISTANCE
(LIGHT-YEARS)
10
5
5
1. Which star is actually brighter: Acanasi, Bade or neither? Explain your reasoning.
2. Which star is seems brighter from Earth: Acansi, Bade, or neither? Explain your reasoning.
3. Which star is seems brighter from Earth: Bade, Ceriero, or neither? Explain your reasoning.
The flux we receive at Earth from a star gets smaller if the distance is greater (farther objects seem
dimmer). The flux we receive at Earth from a star gets larger if the luminosity is greater (more luminous
objects seem brighter no matter how far away they are). We can write that relationship as:
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 = (𝑎𝑎 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 2
See if this matches what you know: As the luminosity increases, the value of the flux increases. As the
distance increases, the value of the flux decreases. (The distance is squared because of the physical laws
that govern light.)
35
We can’t actually measure the distance to the star directly. We have to measure the parallax angle of
the star instead. Remember from Tool #1 (Finding Distance) that the parallax is related to the distance
as:
1
𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 =
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
If we put this relationship into our flux relationship, we have:
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 = (𝑎𝑎 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 ∗ 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 2
The parallax is now in the top, because when you divide by a fraction, you flip it and multiply. Check to
see if this still makes sense: a small parallax means the distance is large. A large distance means the flux
we receive at Earth is small.
We’re now (finally!) ready to write our relationship in a form that let’s us find the luminosity of the star
using values we can measure from Earth – which was our whole goal.
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 = (𝑎𝑎 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏ℎ 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 2
We aren’t worried about figuring out what the actual constants are – they will always be given to you –
we just want you to understand the relationships between the values. Putting in those constants gives
us our final equation for finding the luminosity of a star:
𝐹𝐹
𝐿𝐿 = (9.95 𝑥𝑥 106 ) 𝜋𝜋 2
𝑝𝑝
L is the luminosity in solar luminosities, F is the flux in Watts/m2, and p is the parallax angle in
arcseconds. π is 3.1416.
Don’t forget to use the scientific notation button on your calculator: Enter the constant as 9.95[EE]6 not
as 9.95[x]10[^]6!
36
Tool #9: Finding Luminosity
Practice Problems:
The following table lists 6 bright stars in our night sky with their fluxes as measured at Earth and their
measured parallax angles. Use the appropriate equations to fill in the table, then use the results to
answer the following questions. (Remember that 1 parsec = 3.26 light-years.)
SIRIUS
FLUX
(W/m2)
1.00 x 10-7
PARALLAX
(ARCSECONDS)
0.379
ARCTURUS
2.79 x 10-8
0.089
VEGA
2.59 x 10-8
0.129
BETELGEUSE
1.75 x 10-8
0.0076
ALTAIR
1.32 x 10-8
0.194
FORMALHAUT
9.07 x 10-9
0.130
STAR NAME
DISTANCE
(PC)
DISTANCE
(LY)
LUMINOSITY
(LSUN)
1. Without doing any math, which star is brightest as seen from Earth? Explain your
reasoning.
2. Without doing any math, which star is closest to Earth? Explain your reasoning.
3. Which star is the most luminous?
4. Is the brightest star in the sky the same as the most luminous star? Why or why not?
37
38
TOOL #10: Light and Energy
When most people think of light, they think of visible light. This is natural, as it happens to be the type
of light that our eyes have evolved to see. Visible light is not the only kind of light, however. You may
be familiar with infrared or ultraviolet light. Infrared is exactly the same as visible light, it’s just a color
so red that your eyes aren’t sensitive to it. Similarly, ultraviolet is a color so blue (purple, really) that
your eyes aren’t sensitive to it. In order to get the maximum information from light, then, we can’t just
restrict ourselves to visible light. The full range of all types of light is called the electromagnetic
spectrum.
All light is a wave (it’s also a particle called a photon), so it has three primary characteristics: wavelength,
frequency, and energy. Wavelength is distance between wave peaks. The wavelength of light is what
we perceive as color, so in some sense color and wavelength are two ways of expressing the same thing.
Examine the two diagrams below.
Which of the two types of light shown above have the longest wavelength? RADIO
GAMMA RAYS
The frequency of light is the number of waves per second that reach you. Here’s an important fact to
remember:
ALL TYPES OF LIGHT TRAVEL AT THE SAME SPEED!!!
Because all light travels at the same speed, there is more time between the peaks of a long wavelength
light than there is between the peaks of a short wavelength light. The long wavelength light therefore
has a lower (smaller) frequency than the short wavelength light.
Which of the two types of light shown above have the highest frequency?
RADIO
GAMMA RAYS
As the wavelength of the light increases, what happens to the frequency? INCREASES
DECREASES
Based on what you have observed, you should be able to write the relationship in mathematical form.
In math form, the relationship looks like this:
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 =
(𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)
𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ
39
In this case, the constant is the speed of light:
𝑓𝑓 =
𝑐𝑐
𝜆𝜆
λ is the wavelength in meters, f is the frequency in Hertz, and c is 3.00 x 108 m/sec
Using this tool, you could convert between frequency and energy, but for our purposes, the key is to
remember that wavelength and frequency are inversely proportional – as one goes up, the other goes
down.
The third property of light is its energy. The relationship between energy and frequency is very simple:
The higher the frequency, the higher the energy. If each wave peak is like a ping pong ball, a higher
frequency means you’re getting hit with more ping pong balls per second, so you are getting hit
“harder”. To convert between frequency and energy we would multiply by a special constant called
Planck’s Constant, which we represent with h. The tool therefore looks like this:
𝐸𝐸 = ℎ 𝑓𝑓
E is the energy in electron-Volts (a standard measure of energy for light), f is the frequency in Hertz, and
h is 4.60 x 10-20 eV-secs
Again, for our purposes, the key is to remember that energy and frequency are directly proportional –
as one goes up, the other goes up as well. It’s important to remember which types of light has the most
energy and which types of light have the least energy.
Now that you know the characteristics of light, we arrange all the different types of light in order, based
on their energy:
RADIO
MICROWAVE
INFRARED
LOW ENERGY
________ FREQUENCY
________ WAVELENGTH
VISIBLE
(ROYGBV)
ULTRAVIOLET
GAMMA
RAYS
HIGH ENERGY
________ FREQUENCY
________ WAVELENGTH
X RAYS
Based on what you’ve learned, fill in the blanks above with either LONG, SHORT, HIGH, or LOW.
Which type of light is the “reddest”? _________________________
Which type of light is the “bluest”?
_________________________
One last note to remember: Astronomers generally only talk about the “red end” of the spectrum and
the “blue end” of the spectrum. The red end is the end closest to the color red in the visible part of the
spectrum. The blue end is the end closest to the color blue in the visible part of the spectrum.
MEMORIZE the visible wavelength range!
Red visible = 750 nm; violet visible = 400 nm
40
TOOL #10: Light and Energy
Practice Problems:
Which type of light has the highest energy:
BLUE VISIBLE
ULTRAVIOLET
INFRARED ALL SAME
Which type of light has the longest wavelength: RED VISIBLE
BLUE VISIBLE
X RAYS
ALL SAME
GREEN VISIBLE MICROWAVE
RADIO
ALL SAME
X RAYS
ALL SAME
Which type of light has the highest frequency:
Which type of light moves the fastest:
RADIO
GAMMA RAYS
Match the types of light listed below to their wavelengths:
TYPE OF LIGHT
WAVELENGTH (m)
9.50 x 10-14
8.54 x 10-9
1.50 x 10-7
4.00 x 10-7
7.50 x 10-7
9.97 x 10-6
1.17 x 10-2
5.77 x 103
VIOLET VISIBLE
ULTRAVIOLET
RADIO
MICROWAVE
INFRARED
GAMMA RAYS
RED VISIBLE
X RAYS
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42
Tool #11: THE EM SPECTRUM
Light is a vibrating electromagnetic field – when you wiggle a magnet up and down, you’re creating light!
Now, it’s not visible light, certainly, but visible light is just a tiny fraction of all of the types of light that
exist (for the curious, when you wiggle a magnet, you produce very low-frequency radio waves). When
we discovered that objects in space emit more than just visible light, it literally opened a new window
into the universe.
Types of Light
Using the information below, fill in the table below with the types of light and one use of that light.
VISIBLE
(ROYGBV)
Visible
Visible light is special to us because it happens to be the type of light that
your eyes are attuned to detect. The reason for this is that our sun puts out
most of its light in the visible range, so your eyes evolved to take advantage
of that. Visible light is therefore what you use to see the world around you. If
you’ve ever seen a rainbow (and hopefully you have!), there are six main
colors of visible light: red, orange, yellow, green, blue, and violet.
Infrared
Infrared (or IR) is a color that is so red, your eye can’t perceive it. You do
detect it, however, as heat. Low-power IR is often used to send a signal from
your remote to your TV. Higher-power IR is used in heat lamps.
Microwave
Even redder than IR is microwave light. Microwaves are just the right
wavelength to make water molecules in your food rotate faster. Faster
movement of molecules means more temperature. This is how the microwave
cooks your food – it’s basically boiling the water in it.
Radio
The longest wavelength (and therefore the reddest) is radio. Radio is used
for, well, radio transmissions! The name of an FM radio station (98.6,
95.9) is the frequency of the radio light in MHz (millions of waves per
second). From this you can determine the wavelength of the light. Typical
FM stations have a wavelength about the size of a mountain. FM radio
can be blocked a lot more easily than an AM radio station, since AM
waves are much longer.
Which has more energy, radio or visible light?
RADIO
VISIBLE
Given your answer, which is more likely to damage to your body?
RADIO
VISIBLE
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Ultraviolet
Ultraviolet (or UV) is a light that is so blue that your eye can’t detect it.
You have a perfectly good UV detector in your body, however: your skin!
UV light is what causes sun burns, and it often leads to skin cancer. If
you’ve ever seen a “black light” at a party, that’s a UV light – certain
materials will glow when exposed to UV. The reason it is so dangerous is
that the energy of UV light is higher than that of visible light, so it is
possible for it to damage your cells over a long period of time. This is
why you should always wear sunscreen and NEVER get in a tanning bed!
X-Rays
Even bluer than UV light is x-rays. X-rays have so much energy they can actually
go through your skin. They can’t normally penetrate bone (which is much
more dense), so a x-ray photo is actually the shadow cast by your bones as the
x-ray light shines through your body. If the doctor wants to look at soft tissues
in your body (like your lungs), she has to lower the energy of the x-ray so that it
doesn’t penetrate quite as well. While getting a dental x-ray a couple of times a
year is safe, it takes much less exposure to x-rays to cause damage than does
exposure to UV. This is why they always give you a lead apron – which blocks
all
the x-ray light – whenever you get an x-ray.
Gamma Rays
Gamma rays are the bluest light of all and have the most energy.
Gamma rays are generally produced in nuclear reactions. When most
people talk about “radiation” they actually mean gamma rays, but
there’s no difference between gamma rays and visible light other than
their energy and wavelength! Gamma rays have so much energy that
they can penetrate almost anything (other than really dense materials
like lead). Their energy is so high, that they can damage your cells
almost immediately. The damage can be so bad that the cells in your
organs die, which causes the organs to die, and then causes you to die.
This is radiation poisoning. Like UV, it can also damage the cells in such as way as to cause cancer.
Gamma rays are used to heat water in nuclear power plants (which turns to steam, which turns a
turbine, which produces electricity). They are also used in medical applications – if they are targeted
carefully, they can kill cancer cells without (hopefully) killing the patient.
It’s important that you remember: ALL SEVEN TYPES OF LIGHT ARE JUST LIGHT! The only difference
between them is their wavelength, frequency, and energy. Objects in space produce all seven kinds of
light through different processes – this gives us clues as to what is happening in those objects.
Which kind of light is most harmful?
RADIO
VISIBLE
UV
GAMMA
Which of the three properties of light is responsible for this damage?
WAVELENGTH
FREQUENCY
44
ENERGY
Tool #11: THE EM SPECTRUM
Practice Problems:
Draw lines to match the types of light to the characteristics shown.
Which type of light travels the fastest? ________________________________________________
45
46
Tool #12: Finding Temperature
White light is actually made from all of the colors of the rainbow. We can see those colors by passing
the light through a prism – or raindrops. The rainbow that is produced is called a continuous spectrum:
We can also represent the continuous spectrum as a graph, showing how much of each color we
receive. This type of spectrum is usually more useful than the rainbow image because we can get actual
numbers for each of the colors. We will be using this type of spectrum graph for this task.
The star’s temperature determines the shape, size, and peak location of the continuous spectrum curve.
Wien’s Law allows us to calculate the temperature of a star by measuring the peak wavelength.
𝑇𝑇 =
2.9 𝑥𝑥 106
𝜆𝜆𝑚𝑚𝑚𝑚𝑚𝑚
where 𝜆𝜆𝑚𝑚𝑚𝑚𝑚𝑚 is the peak wavelength in nanometers and 𝑇𝑇 is the temperature in Kelvin.
As the wavelength goes down (shorter wavelengths), the temperature goes up. In other words, we can
also state Wien’s Law as: “Hotter objects are bluer.”
Examine the continuous spectrum below and answer the questions.
Energy Output
Roza
Obeton
V B G Y O R…
What color does Roza appear?
REDDISH
BLUISH
What color does Obeton appear?
REDDISH
BLUISH
Which star is hotter?
ROZA
OBETON
Using the Wien’s Law equation, calculate the temperature of both stars.
Roza
𝜆𝜆𝑚𝑚𝑚𝑚𝑚𝑚 [nm]
Obeton
725
T [K]
414
47
Analyze the continuous spectra shown below and circle the appropriate responses.
Energy OUtput
Roza
Gianfar
VBGYOR
Characteristic
Analysis
Roza
Gianfar
They peak at the same wavelength
Has a lower temperature
Roza
Gianfar
They have the same surface temperature
Appears reddish
Roza
Gianfar
They both look red
Neither looks red
Appears bluish
Roza
Gianfar
They both look blue
Neither looks blue
Has a greater energy output
Roza
Gianfar
They have the same energy output
Roza
Sortho
Energy Output
Peaks at a longer wavelength
V B G Y O R
Characteristic
Analysis
Peaks at a longer wavelength
Roza
Sortho
They peak at the same wavelength
Has a lower temperature
Roza
Sortho
They have the same surface temperature
Appears reddish
Roza
Sortho
They both look red
Neither looks red
Appears bluish
Roza
Sortho
They both look blue
Neither looks blue
Has a greater energy output
Roza
Sortho
They have the same energy output
48
Tool #12: Finding Temperature
Practice Problems:
Use the continuous spectra below to calculate the temperature of the stars and the type of light (radio,
infrared, visible, ultraviolet, etc.) at the peak wavelength.
Temperature: __________________
Type of Light: ________________________
Temperature: __________________
Type of Light: ________________________
49
50
Tool #13: Finding Size
There are two properties of a star that determines how much energy it emits: its temperature
and its size. Wien’s Law will let you find the temperature, but there’s no way to directly
observe the size of a star. Even the closest stars just look like a tiny dot through our most
powerful telescopes. Fortunately, if we know the luminosity of the star and the temperature
(using Wien’s Law) we can find the size of the star using a new tool, the Stefan-Boltzmann Law.
The Stefan-Boltzmann Law says that luminosity depends on the size (surface area) and the
temperature of the object. That makes sense: If two objects are the same size, then the hotter
one must be putting out more energy; if the two objects are the same temperature, then the
larger one must be putting out more energy. Look at the examples below:
Each of the pictures compares the continuous spectrum of Star A to the continuous spectrum of
three other stars. Circle the appropriate response based on these spectra.
Characteristic
Analysis
Has a higher temperature
Star A
Star B
They have the same surface temperature
Has a greater energy output
Star A
Star B
They have the same energy output
Is larger in size (surface area)
Star A
Star B
They are the same size
Has a higher temperature
Star A
Star C
They have the same surface temperature
Has a greater energy output
Star A
Star C
They have the same energy output
Is larger in size (surface area)
Star A
Star C
They are the same size
Has a higher temperature
Star A
Star D
They have the same surface temperature
Has a greater energy output
Star A
Star D
They have the same energy output
Is larger in size (surface area)
Star A
Star D
They are the same size
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Take a look at the continuous spectra again. One thing you may have noticed is that the only
important part of these spectra is the location of the peak of the curve. That means that we
could represent each spectrum by just a dot showing the position of the peak on the
luminosity-temperature graph. For example, we could show all four stars like this:
This is usually more convenient when we want to show lots of stars on the same graph. Try to visualize
the continuous spectrum curve rising up to the dot and then falling off.
Indicate the following on this graph:
•
•
•
Label where the largest stars are found
Label where the smallest stars are found
Circle the only two stars which could be the same size
Finally, based on your answers above, draw an arrow on the graph pointing from smallest stars to
largest stars.
In summary, remember that in order to use this tool, you have to first find the luminosity and the find
the temperature. Plot these two values on the graph, and you’ll have an idea of the size of the star.
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Tool #13: Finding Size
Practice Problems:
Use the Luminosity vs. Temperature graph to determine the correct ranking for the sizes
of the 3 stars H, J, and K from LARGEST to SMALLEST.
Largest
Middle
Smallest
Suppose Star A gives off the same amount of energy as Star B, but Star A is hotter than star B. What can
you say about the sizes of Stars A and B?
•
•
•
•
Star A has a larger surface area than Star B
Star B has a larger surface area than Star A
Star A has the same surface area as Star B
We need more information to decide
Examine the continuous spectra of the four objects shown below:
Which two objects could be the same size?
__________________________
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54
Tool #14: Finding Composition
The only information we get from space is contained in the light we receive. That means that we have
to squeeze every last bit of information out of all of that light. Most things in space put out more than
one wavelength (color) of light. In addition to the brightness, color, and energy of light, we can also
measure how much of each wavelength of light we receive. This will tell us what the object is made of,
which is critical for exploring space.
Atoms are made of a nucleus of protons and neutrons surrounded by electrons in shells. Each electron
shell has a specific energy level; in order for the electron to be in that shell, it has to have exactly the
correct energy. The arrangement of those shells is unique to every atom – it makes a “fingerprint” of
the atom. An example atom with its electron shells is shown below:
 EMITS PHOTON 
When an electron moves from an outer shell (higher energy) to an inner shell (lower energy), it has to
get rid of some energy to make the transition. The electron gets rid of that energy by emitting a photon
– light – of a specific energy. As you have learned, each energy of light corresponds to a specific
wavelength (or color) of light. If we can detect the wavelengths of each photon given off by the
electrons in the atom, we will have a map of the atomic structure of that atom – and we can identify it.
In order to assemble this fingerprint, we need to look at all of the wavelengths of light received from the
object and see at what wavelengths we get “spikes” or “peaks” of brightness. This map of wavelengths
versus brightness is called a spectrum; if the spectrum is from photons emitted from an atom, it’s called
an emission spectrum. It’s produced whenever you have a hot, diffuse (low density) glowing gas.
Both of these pictures are showing the same thing. The picture on the left is how the emission spectrum
would appear to your eye. The graph on the right is a bit more useful because we can easily get the
specific wavelengths of the photons emitted. Both emission spectra are fingerprints for the atom.
This is not the only kind of spectrum, however. Earlier you learned about the continuous spectrum,
which looks like a rainbow. When a continuous spectrum from a hot object shines through a cooler
atom, the electrons in the atom will jump from their lowest energy shells into higher energy shells. This
causes the electrons to “eat” the photons they need to have exactly the right energy to jump to a higher
energy shell. Because the electrons absorb these specific wavelengths, we are left with gaps in the
continuous spectrum. Because the arrangement of these energy shells is always the same for each
55
atom, the gaps in the continuous spectrum will always be at exactly the same wavelengths as the
electrons would emit when they drop down from a higher energy shell.
 ABSORBS PHOTON 

The black lines that we see in the spectrum are the wavelengths that were absorbed by the electrons in
the atom. We call this type of spectrum an absorption spectrum. It’s produced whenever a hot, dense
glowing object (which produces a continuous spectrum) shines through a cooler, diffuse gas. Stars
have a hot, dense core surrounded by a cooler, diffuse atmosphere, so produce an absorption spectrum.
We can also plot the wavelengths versus brightness to produce the other type of spectrum map
Also, as you can see below, the bright lines of the emission spectrum and the dark lines of the
absorption spectrum are in exactly the same place, so we can use either one to identify the atom:
VERY IMPORTANT: Emission or absorption lines tell us nothing about the color or temperature of the
object! Just because we have lots of blue lines does not mean the object is hot! Only the continuous
spectrum can give us that information, and that almost always means we need the wavelength versus
brightness form of the spectrum map.
By comparing the emission or absorption lines that you observe from the object to a set of “reference
spectra” (maps of the spectra of atoms you already know), you can determine what elements are
present in the object. In most cases, the object is made of many different kinds of atoms, so they will all
be superimposed on top of each other. Puzzling out which lines go with which atom can be a challenge!
56
Tool #14: Finding Composition
Practice Problems:
Circle the atom producing an emission line:
Circle the atom producing an absorption line:
What type of spectrum is produced by a hot, dense object?
EMISSION ABSORPTION CONTINUOUS
What type of spectrum is produced by a hot, diffuse object?
EMISSION ABSORPTION CONTINUOUS
What type of spectrum is produced by a hot, dense object
shining through a cooler, diffuse gas?
EMISSION ABSORPTION CONTINUOUS
What type of spectrum is produced by a star?
EMISSION ABSORPTION CONTINUOUS
Using the spectra below, identify all of the substances in the object (compare the lines to the reference
spectrum library):
57
58
Reference Spectrum Library
Calcium
Carbon
Helium
Hydrogen
Iron
Oxygen
Sodium
Titanium
Water
59
60
Tool #15: Finding Velocity
When you hear the sound of a car coming towards you, the pitch is higher than when the car is moving
away from you. The car emits sound waves, but because it’s moving, it is “catching up” to its sound
waves. The result is that the sound waves will be compressed as shown below:
YOU
Notice that the wavelength – the distance between the wave peaks – is shorter than it would be if the
car were standing still and emitting sound waves normally. The faster the car goes, the more it will
catch up to its sound waves and the shorter the wavelength will be. As you’ve learned, a shorter
wavelength is a higher frequency, so the pitch of the car’s sound is higher than normal.
When the car is moving away from you, it is leaving it sound waves behind. The faster the car is going,
the greater the distance between peaks. The result in this case looks like the picture below:
YOU
Because the distance between peaks is longer, the wavelength is longer. Since the wavelength is longer,
the frequency of the sound is lower – so the pitch of the car’s sound is lower than normal. The change
in wavelength in either direction is called the Doppler effect.
Light is a wave as well, so if the light source is moving towards us, the light waves will have a shorter
wavelength than they would if the source were standing still. (Because shorter wavelengths are at the
“blue” end of the spectrum, we say the light is “blue-shifted”.) If the light source is moving away from
us, the light waves will have a longer wavelength than normal (and so we say the light is “red-shifted”).
The more the wavelength is shifted from its normal (or “rest” or “laboratory”) wavelength, the faster
the light source is moving. If we measure the actual, observed wavelength of the object (which will be
shifted from normal) and subtract from that the normal, rest wavelength, we can calculate the velocity
of the light source using the Doppler formula:
𝑣𝑣 =
(𝜆𝜆𝑜𝑜𝑜𝑜𝑜𝑜 − 𝜆𝜆𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 )
𝜆𝜆𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
or
𝑐𝑐
𝑣𝑣 =
∆λ
𝜆𝜆𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝑐𝑐
v is the velocity of the light source in km/sec, λobs is the observed wavelength of the light, λrest is the
normal wavelength of the light when at rest, and c is the speed of light (3 x 108 m/sec). We can also
61
write the formula in terms of the “shift” in wavelength, which we write as ∆λ. This is just shorthand for
λobs - λrest. The “shift” is how many nm from the rest wavelength the line has moved.
With this tool, you can calculate the velocity of an object just by taking its spectrum. If you can identify
some lines in the spectrum that have the right pattern but are all shifted in one direction or the other
(and they all have to be shifted in the same direction!), then you can figure out the velocity of the
object. If the observed spectral line is at a longer wavelength than it “should be”, then the object is redshifted and so is moving away from you. If the observed spectral line is at a shorter wavelength than it
should be, then the object is blue-shifted and so is moving towards you.
Suppose you took the spectrum of an object, as shown below:
Compare the spectrum to your Spectrum Reference Library. Can you identify the substance producing
the spectrum? Notice that none of the wavelengths match your reference library, but there is one
substance that matches the pattern of the spectral lines. Which atom is it?
Unknown substance: _______________________________
Now that you have identified the substance, look at the line that has a wavelength of 656 nm. What
wavelength “should” it have been (what is its “rest” wavelength)? Always measure to the nearest 1 nm!
Don’t just round off to the nearest mark, since that is 5 nm.
Observed wavelength: ____________________
Rest wavelength:
____________________
Now use the Doppler formula to figure out the velocity of the object. Show your work below:
Velocity of object:
______________________
Things can get more complicated if have lots of substances mixed together in the same spectrum. Just
remember that the patterns of lines (the spacing between them) will stay the same, but all the lines will
all be shifted. For the most accurate results, use the longest wavelength (reddest) line available, since
the shift will be largest and easiest to see. Identify the substances first so that you can get the observed
wavelength, then compare that to the “rest” wavelength from the Spectrum Reference Library and
figure out the velocity.
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Tool #15: Finding Velocity
Practice Problems:
Examine the spectra shown below and answer the questions. Remember to examine the longest
wavelength line because it will have the largest shift.
Substance(s): ___________________________________________________________________
Line Observed Wavelength: _______________ Line Rest Wavelength: ____________________
(Compare the longest wavelength line)
Velocity of Object: ______________________
Substance(s): ___________________________________________________________________
Line Observed Wavelength: _______________ Line Rest Wavelength: ____________________
(Compare the longest wavelength line)
Velocity of Object: ______________________
Substance(s): ___________________________________________________________________
Line Observed Wavelength: _______________ Line Rest Wavelength: ____________________
(Compare the longest wavelength line)
Velocity of Object: ______________________
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Tool #16: Star Classes
When we learned about finding the size of a star, we saw that we could plots the continuous spectra of
a lot of stars at once by just plotting a dot that shows the peak of each curve. Since all we really care
about from the continuous spectrum is the luminosity and wavelength of the peak, that’s really all we
need to see. If we plot the peaks of all of the stars that we see in the sky, we get a luminositytemperature graph that looks something like this:
As you know from using the Stefan-Boltzmann Law (Tool #13), the smallest objects are the in the
bottom left corner, while the largest objects are in the top right corner. The colors on the diagram
remind you that the hottest stars are blue and the coolest stars are red.
Interestingly, 90% of all of the stars in the universe lie on the line labeled “Main Sequence” in the
diagram. A main sequence star is a star that is stably fusing hydrogen into helium. Remember that
each dot is just the peak of that star’s continuous spectrum. Astronomers have assigned “spectral
classes” to group the stars on the Main Sequence that have similar properties. It is critical that you
memorize the order of the spectral classes from hottest to coolest! The spectral classes are:
Hottest  O B
A
F
G
K
M Coolest
A good way to remember the order is by learning a mnemonic that uses the first letter of each word to
represent a spectral class. A popular mnemonic is “Oh Boy Another Final’s Gonna Kill Me” – but feel
free to come up with your own!
65
If we add spectral classes to our diagram we get something like this:
If we know the star is a Main Sequence star, then the spectral class tells us a lot of information about
the star:
O B A
Hottest
Blue
Bright
High Mass
Short Life
F
G
K M
Coolest
Red
Dim
Low Mass
Long Life
It’s important that you memorize these relationships! O-type stars burn hot, burn fast, and burn out.
They typically only live about 10 million years. G-type stars like our Sun will live about 10 billion years.
M-type stars are the laid back stars of the universe: they burn cool and slow, and can live for as much as
10 trillion years.
If we know the luminosity (from Tool #9) and the temperature (from Tool #12), and if we know the star
is a Main Sequence star (from this tool #13), then we can calculate many other properties of the star
using the following equations:
Stellar Radius (solar radii): 𝑅𝑅𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
1
𝐿𝐿0.5
𝑇𝑇
(5800)2
Mass (solar masses): 𝑀𝑀𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝐿𝐿(3.5)
Lifetime (years): Lifetime = 1𝐸𝐸10 × 𝑀𝑀𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 −2.5
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Tool #16: Star Classes
Practice Problems:
Given the luminosity and temperature of the stars, determine the type of object (Main Sequence star,
white dwarf, giant, or supergiant) and if it is a Main Sequence star, determine its spectral class.
Luminosity (Lsun)
Temperature (K)
0.167
4480
510
3500
17.9
8800
Spectral Class
(if Main Sequence)
Object Type
Each of the following is a Main Sequence star. Determine each star’s radius, mass, and lifetime.
Spectral Type
Luminosity
(Lsun)
Temperature
(K)
B
3160
18600
G
1
5800
M
0.015
3150
Star Radius
(Rsun)
Star Mass
(Msun)
Circle the appropriate response:
Characteristic
Analysis
Hottest
O-type
M-type
They are the same temperature
Smallest
O-type
M-type
They are the same size
Lowest mass
O-type
M-type
They are the same mass
Blue star
O-type
M-type
They are the same mass
Brightest
O-type
M-type
They are the same brightness
Longest lifetime
O-type
M-type
They have the same lifetime
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Star Lifetime
(years)
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Tool #17: Star Structure
Stars generate huge amounts of energy every second. The Sun produces more energy in one second
than has been used by all the humans who have ever lived on planet Earth. How it generates that
energy was a mystery for a long time. If the Sun were “burning” something (like wood burns), then it
would run out of fuel in just a few years. Since the Sun is four and a half billion years old, there’s no way
it is generating energy through burning. In the early part of the twentieth century, we learned about
nuclear reactions and realized that was the only way the Sun could be producing energy. We now know
that the Sun is essentially a continuously exploding hydrogen bomb. It takes four hydrogen atoms and
fuses them together to form two helium atoms. The helium atoms that come out have 0.7% less mass
than the hydrogen atoms that went in. This 0.7% of the mass is converted to energy, and since E = mc2
(as you might have heard once or twice), this is actually a lot of energy converted per reaction. The Sun
is converting 600 million tons of hydrogen per second, so this is a lot of energy total.
In order to fuse, the nuclei of the hydrogen atoms have to be forced together. A hydrogen nucleus is just
a proton, which has a positive charge. Like charges repel each other, so it takes a lot of pressure to force
the hydrogen atoms together.
In fact, it takes the weight nearly the entire star just to provide enough pressure to cause fusion to
happen in the very center. For this reason, we think of the star has having two parts:
•
•
The hot (15 million degrees for the Sun), dense core where energy is being produced
The cooler (5800 degrees for the Sun) and less dense atmosphere (also called the envelope),
which is squeezing the core to keep fusion running.
If a star is a hydrogen bomb, why doesn’t it blow itself apart? The reason is the mass of the envelope is
holding everything together. The star is in hydrostatic equilibrium – it is perfectly balanced between
the escaping energy keeping the envelope from collapsing and the mass of the envelope keeping the
energy from blowing up the star. In many ways it’s like the thermostat in your house.
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Suppose, for some reason, the energy was too strong and started pushing the envelope outward, so the
envelope isn’t squeezing the core quite as hard.
What happens to the rate of fusion reactions in the core?
INCREASES DECREASES STAYS THE SAME
Based on your choice in the previous question, what happens to the amount of energy generated?
INCREASES DECREASES STAYS THE SAME
Based on your choice in the previous question, what happens to the envelope?
EXPANDS COLLAPSES
STAYS THE SAME
Suppose, for some reason, the energy wasn’t strong enough and the envelope started to collapse,
squeezing the core harder.
What happens to the rate of fusion reactions in the core?
INCREASES DECREASES STAYS THE SAME
Based on your choice in the previous question, what happens to the amount of energy generated?
INCREASES DECREASES STAYS THE SAME
Based on your choice in the previous question, what happens to the envelope?
EXPANDS COLLAPSES
STAYS THE SAME
So, hydrostatic equilibrium keeps the star stably fusing hydrogen into helium for billions of years. A star
that is stably fusing hydrogen into helium is called a Main Sequence star. A star spends 90% of its
lifetime as a Main Sequence star, so 90% of all of the stars in the galaxy are main sequence stars.
In the space below draw a diagram of the two parts of a star. Indicate where fusion is taking place and
the approximate temperatures in each part. Draw arrows in the star to show where the squeezing is
taking place that is forcing the hydrogen to fuse.
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Tool #17: Star Structure
Practice Problems:
In the space below draw a diagram of the two parts of a star, labeling each with its name. Indicate
where fusion is taking place and the approximate temperatures in each part. Draw arrows in the star to
show where the squeezing is taking place that is forcing the hydrogen to fuse.
Suppose a star were producing too much energy in its core. Describe what happens to return the star to
balance:
Suppose a star were producing too little energy in its core. Describe what happens to return the star to
balance:
What is hydrostatic equilibrium?
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Tool #18: Star Birth
Stars are like everything else in nature: They are born, they age, they grow old, and they die. When
stars die, they produce the materials for the next generation of stars – and for everything that orbits
that next star, including creatures like us. This leads to an important fact to remember:
The elements that are present in the star are the elements that present on the bodies that orbit it.
So, for example, if you’re looking for iron in an asteroid belt, you will have to have iron present in the
star’s spectrum for there to be a chance of it being there. If the star has only hydrogen and helium in its
spectrum, then it can’t have terrestrial planets like Earth.
Stars are born from giant clouds of gas and dust (but mostly hydrogen gas) called, appropriately
enough, giant molecular clouds. Each cloud has thousands of solar masses of material, and so will form
thousands of stars. The stars that form from a single giant molecular cloud are called a star cluster.
Eventually, the stars in the cluster will drift apart and blend in with the rest of the galaxy.
Giant Molecular Cloud
Left to itself, the cloud will never collapse and condense to form a star – it needs some sort of a “kick” to
get the process started. That kick will usually come from the death of a previous-generation star. When
very large stars die, they explode in a supernova, the largest explosion in the universe. The force of this
explosion can start a nearby cloud collapsing. Once the collapse starts, the gravity of the cloud becomes
the dominant force and causes it to continue to collapse. Because the cloud is clumpy, the clumps’
gravity attract the nearby material, so the cloud fragments into many stars instead of just one massive
stars. In fact, giant molecular clouds are so clumpy that it’s rare for large stars to form at all! Tiny Mtype stars are the most common stars in the galaxy.


Each of these fragments will eventually become a star. When the fragment collapses, it converts
gravitational potential energy (the “stored” energy any object has before it falls downward due to
gravity) into increasing its temperature. The fragment isn’t a star yet, because it isn’t fusing in its core
yet. We call this object a protostar because it will eventually collapse enough to start nuclear fusion and
become an actual star.
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Protostars produce strong stellar winds as they collapse. Initially, these winds have difficulty getting
through the disk of leftover gas and dust that will eventually form planets, so the wind gets channeled
above and below the plane of the forming solar system. These jets are a rich source of hydrogen, but
they are extremely violent!
Eventually, the collapsing protostar squeezes its core hard enough to force temperatures to reach 10
million degrees Kelvin – the temperature at which hydrogen can begin to fuse into helium. The star is
born! The energy from the newborn star will eventually sweep away the gas and dust that hasn’t been
used to form planets, and the star system will take on the form that we see throughout the galaxy.
When the heat generated by nuclear fusion balances the collapse of the star’s atmosphere, the star is in
stable hydrostatic equilibrium and is officially a main sequence star.
Test your understanding by circling the correct response:
Each giant molecular cloud produces a single star.
TRUE
FALSE
Large stars only form from extremely smooth fragments.
TRUE
FALSE
A protostar is fusing hydrogen into helium.
TRUE
FALSE
A protostar is a star.
TRUE
FALSE
When fusion first begins in a star, the star is not yet a main sequence star.
TRUE
FALSE
Planets begin to form before the star begins nuclear fusion.
TRUE
FALSE
Protostars produce a strong stellar wind.
TRUE
FALSE
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TOOL #18: Star Birth
Practice Problems:
Fill in the flow chart below with the process of star birth:
Begin with a:
_____________________
Which is composed mostly of:
_____________________
Somewhere nearby a “kick” occurs in the form of a:
_____________________
After the kick, the force which dominates and causes the cloud to continue to collapse is:
_____________________
Because the cloud is clumpy, it begins to fragment into smaller pieces. The pieces will form a group of
stars called a:
_____________________
As each fragment collapses, it decreases its:
_____________________
And increases its:
_____________________
At this point the fragment is called a:
_____________________
These objects have very strong:
_____________________
Composed mostly of:
_____________________
When the temperature in the core reaches 10 million degrees, fusion begins and the object becomes a:
_____________________
When the energy from the star’s fusion balances the collapse, the object becomes a:
_____________________
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Tool #19: Star Death
All things die, and stars are no exception. Remember that if a star has a little more mass, it burns
through its fuel much faster. When a star begins to die, there are just two rules to remember:
•
•
When non-fusing “ash” builds up in the core, the star expands.
When a star expands, its atmosphere cools down (and when it shrinks, its atmosphere heats up)
The star has been stably fusing hydrogen into helium for its whole life. The helium that is created in the
star sinks to the center of the core as “ash” since the temperatures and pressures in the core of a main
sequence star aren’t high enough to fuse the helium into carbon. When enough ash builds up in the
core, convection processes begin to dominate in the star. Convection is heated bubbles of plasma that
rise and sink in the star’s atmosphere like a lava lamp. Convection cells rise up and push the surface of
the star outward, almost like blowing up a balloon. Convection causes the star to produce an
extremely strong stellar wind that is a good source of hydrogen and helium.
Helium ash has built up in the center of the core. What happens?
Characteristic
Analysis
Star size
Giant
Dwarf
Stays the same
Star temperature
Gets hotter
Gets cooler
Stays the same
Based on what you decided, what color does the star become?
RED
BLUE
UNCHANGED
Put the color with the size you decided upon and name this phase: _______________________
The helium ash is collapsing due to its gravity, so it’s squeezing harder and harder. Eventually, it
squeezes hard enough so that the core temperature reaches 100 million Kelvin and begins to fuse
helium into carbon. The core of the star now has two parts: in the very center it is fusing helium to
carbon, and in a shell surrounding the center it is still fusing hydrogen into helium. Now that the star is
stably fusing helium into carbon, convection no longer dominates, so the star shrinks and the
atmosphere gets hotter. At this point, the star is called a horizontal branch star.
What is the key characteristic of a horizontal branch star?
Stably fusing _____________________ into _________________
Convection no longer dominates in the star. What happens?
Characteristic
Analysis
Star size
Gets larger
Gets smaller Stays the same
Star temperature
Gets hotter
Gets cooler
Stays the same
After a few million years, carbon ash builds up in the center of the core, and convection dominates
again. This time, though, the star is generating much more energy due to the helium -> carbon fusion,
so convection is much stronger.
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Carbon ash is triggering extremely strong convection. What happens?
Characteristic
Analysis
Star size
Bigger giant!
Sub-dwarf
Stays the same
Star temperature
Gets hotter
Gets cooler
Stays the same
The star can be as much as 200 solar radii across. Even though this is technically a supergiant, this phase
is usually called an asymptotic red giant. “Asymptotic” just means that it is “approaching” a supergiant,
but it doesn’t live long enough to get there.
For a star with less than 8 solar masses, this is nearly the end of the story. A low-mass star doesn’t have
enough mass to squeeze the core hard enough to fuse carbon into oxygen. As the carbon ash collapses,
the energy generation in the core becomes unstable. The atmosphere shrinks and expands then shrinks
again, pulsing in an irregular manner. Eventually, the atmosphere pulses far enough that it molecules
can form in the cooler outer layers. The energy released from these molecules gives the star’s
atmosphere a final kick, launching the atmosphere into space as a planetary nebula. Fusion stops
immediately. The star is dead, leaving behind only its rapidly cooling core of mostly carbon, called a
white dwarf. Even though the core is no longer at 100 million degrees, like a giant piece of charcoal, it
still glows hot for a long time – starting at around 30,000 degrees Kelvin.
Low-mass stars end their lives by puffing their atmospheres into space as a _________ _________
leaving behind their core as a _____________ ____________.
A star with greater than 8 solar masses follows a similar path, but it has so much mass that convection
takes it straight to the supergiant phase. It has enough mass to fuse elements all the way to iron. A
high-mass supergiant has a core like an onion, each layer producing a different element from helium to
iron. Iron, however, can’t be fused into anything. The iron ash collapses, heating up as falls. In the very
center of the iron core, the pressure is so high that the electrons of its atoms are forced into its protons,
creating neutrons. The outer layers falling onto this super-hot neutron ball bounce back, creating the
largest explosion in the universe – a supernova.
The object left behind is very small (roughly the size of a city) but contains around one and a half times
the mass of the Sun (remember that the rest of the star blew out into space). This object is composed
entirely of neutrons, so is called a neutron star (even though it isn’t really a star). A rapidly-rotating
neutron star has an ultra-strong magnetic field that can generate a power beam of electromagnetic
energy, usually in the radio band. These rapidly-rotating neutron stars are called pulsars.
If the neutron star is larger than about three times the mass of the Sun, it will continue to collapse
under its own gravity, but now no force in the universe is strong enough to stop the collapse. The
neutron star collapses to literally zero size – it becomes a black hole, an object with so much gravity that
not even light can escape and where all kinds of strange warping of time and space can occur.
High-mass stars end their lives in a powerful ____________ leaving behind their core as either a
_____________ ____________ or a ______________ _____________.
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Tool #19: Star Death
PRACTICE PROBLEMS:
Starting with a G-type (low-mass) Main Sequence star, plot how the star moves on the HR Diagram
below as it begins to die. Plot the following phases:
A.
B.
C.
D.
E.
Main Sequence
Red Giant
Horizontal Branch
Asymptotic Red Giant
White Dwarf
In what event does a low-mass star die? ___________________________
Starting with an A-type (high-mass) Main Sequence star, plot how the star moves on the HR Diagram
below as it begins to die. Indicate where the star is when the supernova event occurs and plot the
following phases:
A. Main Sequence
B. Supergiant
C. Neutron Star/
Black Hole
(HINT: Think about
the size!)
In what event does a high-mass star die? ___________________________
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80
Tool #20: Black Holes
Black holes are some of the most famous – and strangest – objects in the universe. When a massive star
explodes, the remnant core forms a neutron star which is kept from collapsing more by neutron
degeneracy. If the neutron star has more than around 3 Msun of material, then even neutron degeneracy
can’t stop the collapse. The core collapses to size zero (at least ignoring quantum mechanics) -- all of
the mass of the object is contained in an infinitesimal point called the singularity. The gravity is
extremely high (because we can get very close to the singularity) that even light can’t travel fast enough
to escape. As we move further from the singularity, the gravity gets weaker; eventually it is weak
enough that light can escape. This distance marks the Schwarzschild radius or more commonly, the
event horizon (because you can witness any events inside of this distance).
𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝒉𝒉𝒉𝒉𝒉𝒉𝒉𝒉𝒉𝒉𝒉𝒉𝒉𝒉 (𝒌𝒌𝒌𝒌) = 𝟑𝟑 𝒙𝒙 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 𝒐𝒐𝒐𝒐 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 (𝑴𝑴𝒔𝒔𝒔𝒔𝒔𝒔 )
The event horizon isn’t a “thing” – it’s nothing physical. If you fall towards the singularity, you won’t feel
anything especially different as you cross the event horizon. The outside world will never see you again,
though!
The first black hole discovered was an x-ray companion to a blue supergiant in the constellation Cygnus.
Called Cygnus X-1, astronomers soon realized that the only way to produce these x-rays was by material
falling from the companion star towards the event horizon.
Only the largest stars will have more than 3 solar masses of material left after the supernova blows it
apart!
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In addition to these “stellar” black holes, most galaxies seem to have a supermassive black hole with a
mass of millions of solar masses in their centers.
The Milky Way’s supermassive black hole lies in the constellation Sagittarius, so it is called “Sag A*”. Sag
A* is over two million solar masses.
We can measure the mass of a black hole by measuring the speeds of stars that are orbiting the hole.
The image below shows the two closest stars to our galaxy’s central black hole.
SO-102 is the closest star, with S0-2 being the second closest. Because a thick dust cloud lies between
Earth and the galactic center, we can only observe these stars in infrared or x-rays – visible light is
blocked. The black hole is located in the center of the small area where SO-102 and SO-2’s orbits
overlap.
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Tool #20: Black Holes
Practice Problems:
In the black hole image below, label the event horizon and singularity (draw a line to each).
What is the minimum mass of a stellar black hole? _______________________________ Msun
What is the size of the event horizon for this smallest black hole? ____________________ km
If the mass of the Milky Way’s supermassive black hole is 2,000,000 Msun, what is the size of its event
horizon?
________________________________________________ km
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84
Tool #21: Interstellar Medium
The space between the stars isn’t empty. Gas and dust fill the galaxy, but in most places its too diffuse to
detect. This thin material is referred to as the interstellar medium or the ISM. A giant cloud of gas and
dust called a nebula is thick enough to affect the surrounding starlight and so can be seen from Earth. A
nebula can be a star-forming region or the remnant of a supernova. How the nebula appears to us
depends on whether and how its being lit by the surrounding stars.
Nebulae have two effects on starlight: extinction and reddening. Extinction occurs when the cloud is
simply blocking the light of the stars behind it. You can see this in the center of the cloud shown below.
In the center, the cloud is thick enough to block
light from the background stars completely. At
the edges, the brightest stars can shine through
the cloud. What color are the brightest stars in
the galaxy?
RED
BLUE
What color are the stars just inside the edges
of the cloud n the image?
RED
BLUE
The light is reddened the same process that makes the sky blue on Earth. The particles in the cloud (or
Earth’s atmosphere) scatter the blue light away from the line of sight. Only the red light makes it
through. This is the same reason why sunsets can be red: the blue light is scattered away by the
atmosphere.
There are three types of nebula; one dark, two bright. The cloud above Is a dark nebula. A dark nebula
can’t be detected directly. We can “see” it when it blocks the light of the background stars. In radio we
can see more light, as hydrogen emits radio with a wavelength of 21 cm. Other trace substances in the
cloud give off radio emission as well:
•
•
•
•
•
Carbon monoxide (CO)
Ammonia (NH3)
Methyl alcohol (CH3OH)
Ethyl alcohol (CH3CH2OH)
Formaldehyde (H2CO)
85
Don’t forget that there are just tiny amounts of these molecules present! The cloud is made up almost
entirely of hydrogen and helium, so don’t be fooled!
The other two types of nebulae are bright nebulae. These clouds are being lit up by stars, and so we can
see them in visible light. Many of the “pretty pictures” you’ve seen in astronomy are bright nebulae.
All nebula are made of very diffuse (low-density) gases. If a star within the nebula is heating the diffuse
gas to make it very hot, what kind of spectrum is produced?
CONTINUOUS
EMISSION
ABSORPTION
Since hydrogen is the primary gas, emission nebulae look mostly pinkish-red. There are other elements
present, though, so m emission nebula often has many beautiful colors due to the spectra of the various
gases.
If the star is behind the nebula relative to Earth, we already know that will make the nebula seem to be
a dark nebula. The blue light is being scattered away from the center of the cloud. If we happen to be on
this side of the cloud, we can see that blue light as it is being scattered towards us. A reflection nebula
therefore looks mostly blue.
Identify the types of nebulae below:
_____________________________________
__________________________________
Nebula can be produced by the remnants of a supernova explosion, or they can be areas where new
stars are being born – in fact, nebula are how “star stuff” is recycled in the galaxy so that the next
generation of stars can be born. The nebula fragments into hundreds of stars that make up a new star
cluster. These stars eventualy drift away from each other until they are mixed in with the rest of the
galaxy. When these stars die, they will produce new nebulae that can create new stars. Even in space,
the stars must live out their lives and die so that the next generation can take their places.
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Tool #21: Interstellar Medium
Practice Problems:
Identify the types of nebulae shown below.
_____________________________________
__________________________________
Examine the image below:
What type of nebula is in the
background of this image?
_______________________________
What type of nebula is in the
foreground (the horse’s head) of this
image?
_______________________________
What two elements is this nebula mostly composed of?
__________________________________
_____________________________________
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88
Tool #22: Kepler’s Laws
In 1609 a shy geek named Johannes Kepler was a student of the great astronomical observer, Tycho
Brahe. Newton’s theory of gravity hadn’t been discovered, but just by looking at the data Tycho
gathered, Kepler saw three patterns. He didn’t know why they worked, he just knew that they did.
These three laws serve as the basis of modern astronomy.
Kepler’s First Law
“Each planet’s orbit around the Sun is an ellipse with the Sun at one focus.” The keys are that the orbit
is an ellipse, not a circle, and the Sun is at one focus and not at the center of the orbit. A typical orbit is
shown below. Label the orbit with the following:
major axis
semimajor axis
center of orbit
1st Law Key Fact:
perihelion
aphelion
Sun
Planets are closest to the Sun at perihelion
Kepler’s Second Law
“The planet’s speed changes so that a line from the planet to the Sun sweeps out equal areas in equal
times.” Kepler’s 2nd Law says that if the areas swept out are equal, then the times between the points
are equal. In the diagram below, assume the areas marked out between points A and B and between
points C and D are equal.
Characteristic
Planet travels farther
Time between points
Planet travels faster
Analysis
CD
CD
CD
AB
AB
AB
89
Same
Same
Same
Because the areas from AB and from CD are equal, Kepler’s 2nd Law says so is the time to travel
between them. Kepler didn’t know why this was true, but he just knew it was true from Tycho’s data.
We now understand why it’s true. Imagine you throw a ball straight up as shown below:
The ball goes slower and slower as it gets farther from Earth. At its
maximum distance from the Earth, it starts to fall back towards the
Earth – and starts going faster and faster.
In this case, the ball eventually hits the Earth. But the planets – fortunately! – miss the Sun and go
around again. Just like the ball, the planet is moving at its fastest when it is closest to the Sun.
2nd Law Key Fact:
Planets move fastest at perihelion
Kepler’s 3rd Law
“The cube of the semi-major axis of the planet’s orbit is directly proportional to the square of its orbital
period.”
If two things are directly proportional, then we can write an equation for them that looks like this:
𝑎𝑎3 = (𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) 𝑝𝑝2
a is the semi-major axis of the orbit; p is the period of the orbit. If we chose a to be in astronomical
units (AU, the average distance between the Earth and the Sun), and if we chose p to be in Earth years,
then the constant is the mass of the central object (the Sun in this case) in solar masses.
𝑎𝑎3 = 𝑀𝑀 𝑝𝑝2
This relationship holds for all orbits – planets going around the Sun, moons going around Jupiter, and
spacecraft going around the Earth. You will often use Kepler’s 3rd Law to find the semi-major axis of the
orbit (also called the orbital distance or the orbital radius). If so, you’ll need to take the cube root of
both sides:
3
𝑎𝑎 = �𝑀𝑀𝑝𝑝2 = (𝑀𝑀𝑝𝑝2 ).3333333333
If we know the semi-major axis, we can solve Kepler’s 3rd Law for the period:
𝑎𝑎3
𝑀𝑀
𝑝𝑝 = �
If we know the semi-major axis and the period of the planet, we can also find the mass of the star:
Fill in the missing information:
Mass (Solar Masses)
1.00
1.75
𝑀𝑀 =
𝑎𝑎3
𝑝𝑝2
Semi-Major Axis (AU)
1.52
0.75
90
Period (Years)
3.56
2.14
Tool #22: Kepler’s Laws
Practice Problems:
In the orbit diagram below, the time between each planet position shown is exactly 1 month.
1. Label perihelion on the diagram.
2. Label aphelion on the diagram.
3. Between which two dots is the planet moving the fastest?
____________________
4. At which dot is the planet moving the slowest?
____________________
5. What is happening to the planet’s speed between B and C?
INCREASE DECREASE CONSTANT
6. What is happening to the planet’s speed between H and I?
INCREASE DECREASE CONSTANT
Fill in the missing values:
Mass (Solar Masses)
1.00
11.4
2.00
Semi-Major Axis (AU)
5.45
0.426
5.45
91
Period (Years)
2.87
0.341
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Tool #23: Binary Stars
As you know from Kepler’s 3rd Law, in order to measure the mass of an object we need to know the
orbital period and semi-major axis of something that is orbiting the object. In most cases, we can’t see
planets that are orbiting stars, so we have no direct way to measure the mass of a single star. Most of
what we know about the masses of stars comes from studying binary stars, two stars that orbit each
other. Binary stars are actually fairly common in the galaxy – nearly 85% of the bright stars in the galaxy
are multiple star systems. If we know the masses of the stars in the binary, then we know those masses
are approximately the same for all stars of the same spectral type – including the single stars. Finding
the masses of the members of a binary pair is therefore critical to the study of astronomy.
We can find the total mass of a binary system using Kepler’s 3rd Law:
𝑎𝑎3 = 𝑀𝑀 𝑝𝑝2
Solving this for the mass – the total mass of the binary system – we get:
𝑀𝑀𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =
𝑎𝑎3
𝑝𝑝2
The equation tells us that we need to find two things: a) the semi-major axis (in this case, the distance
between the stars) and the b) period (in this case, the time it takes the two stars to orbit each other).
We can get the semi-major axis just by observing the stars in our telescope. The only problem is that our
telescope measures separations in arcseconds (1/3600 of a degree), but we need the separation in AU
for Kepler’s 3rd Law. We can follow a four-step process to do the conversion.
Let’s look at the Alpha Centauri system, the closest star system to Earth.
Name: Alpha Centauri A (on left) and B (on right)
Parallax: 0.75481 arcsecs (for both stars, obviously)
Separation of Stars in Telescope: 17.6639 arcsecs
Find the semi-major axis:
Step 1: Find the Distance:
Step 2: Convert Separation to Radians:
Step 3: Find the Semi-major Axis:
Step 4: Convert Semi-major Axis to AU:
distance (D) = 1 / parallax (p)
1 radian = 206265 arcsecs
semi-major axis (a) = distance (D) * separation (s)
1 pc = 206265 AU
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Now you do it for Alpha Centauri:
Step 1: Find the Distance (D):
_________________________ parsecs
Step 2: Convert Separation to Radians (s): _________________________ radians
Step 3: Find the Semi-major Axis (a):
__________________________ pc
Step 4: Convert the Semi-major Axis to AU: _________________________ AU
Kepler’s 3rd Law needs the period in years. The light from the binary changes in a regular way whenever
one star passes in front of the other. Look at the light curve below and determine the period:
____________________________ days
Period of Alpha Centauri’s Orbit:
Period of Alpha Centauri’s Orbit (p): ____________________________ years
Now you have everything you need to find the total mass of the system. Plug into Kepler’s 3rd Law the
values you found for the semi-major axis in AU and the period of the orbit in years:
𝑎𝑎3
𝑀𝑀𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 2
𝑝𝑝
𝑴𝑴𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 = ______________________ solar masses
Congratulations! You just found the total mass of the two stars! We now know that:
𝑀𝑀𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑀𝑀𝐴𝐴 + 𝑀𝑀𝐵𝐵
You may recall from algebra that if we have two unknowns (MA and MB in this case), then we need two
equations to find out the values. The total mass above gives us one equation. If we know the ratio of the
masses – for example, star A might be 1.5 times as massive as star B – then we can find the individual
masses. The ratio of the masses can be found by finding the velocities of the two stars: the lighter star
will be “slung around” faster than the heavier star. In the case of Alpha Centauri, we have found that:
𝑀𝑀𝐴𝐴 = 1.21 𝑀𝑀𝐵𝐵
Put this relationship in the total mass above and you can find the mass of Alpha Centauri B:
𝑀𝑀𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 1.21 𝑀𝑀𝐵𝐵 + 𝑀𝑀𝐵𝐵
𝑴𝑴𝑩𝑩 = ______________________ solar masses
Now that you know 𝑀𝑀𝐵𝐵 , subtract it from the total mass to find the mass of Alpha Centauri A:
𝑴𝑴𝑨𝑨 = ______________________ solar masses
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Tool #23: Binary Stars
Practice Problems:
Using the data below, calculate the masses of the stars in the Capella star system.
Name: Capella A (on left) and B (on right)
Parallax: 0.07620 arcsecs
Separation of Stars in Telescope: 0.05662 arcsecs
Mass Ratio: MA = 1.04 MB
Light Curve:
_________________________ parsecs
Step 1: Find the Distance (D):
Step 2: Convert Separation to Radians (s): _________________________ radians
Step 3: Find the Semi-major Axis (a):
_________________________pc
Step 4: Convert the Semi-major Axis to AU: _________________________AU
Period of Capella’s Orbit:
Period of Capella’s Orbit (p):
_________________________ days
_________________________ years
Total Binary Mass:
_________________________ solar masses
Mass of Capella A: ____________________________ solar masses
Mass of Capella B: ____________________________ solar masses
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Tool #24: Magnitudes
You know from Tool #1 that we can find the distance to a star by measuring its parallax. You also know
that the further away the star is, the smaller the parallax. For stars that are very far away, the parallax is
too small for us to measure. Our very best space-based telescopes can measure parallaxes of stars out
to about 200 pc. That’s a tiny bubble compared to the size of the galaxy.
You know that as a star gets further away, its brightness drops. If we know the actual brightness of the
star, all we need to do is to measure how bright it seems from Earth, and we can determine its distance.
As you learned, the actual brightness is the luminosity, and how bright it seems (the “apparent”
brightness) is the flux. These are numbers that are a bit tedious to work with when we are talking about
distant objects. Instead, we use something called the magnitude of the star. The actual brightness
(similar to the luminosity) is called the absolute magnitude, while the apparent brightness (similar to
the flux) is called the apparent magnitude. The only thing that might confuse you is that magnitudes
are like “importance”. The brightest star is the “most important”, so it is a 1st magnitude star. The next
dimmer star is “less important” and so is a 2nd magnitude star. The faintest star you can see with your
naked eye (assuming you have perfectly dark skies and perfect vision) is 6th magnitude. Anything “less
important” (in other words, dimmer) than that needs a telescope!
This system was developed by the Greek astronomer Hipparcos in 150 BCE. 1st magnitude stars were
the brightest stars he could see. But there are stars brighter than his 1st magnitude star that he just
couldn’t see from Greece. What do we do? Smaller magnitude numbers are brighter (“more
important”), so stars even brighter that 1st magnitude must have an even smaller number. So, a “0
magnitude” star is brighter than a 1st magnitude star, a “-1 magnitude” star is brighter than a 0
magnitude star, and so on.
Consider the stars below and complete the analysis:
ACANA
BADE
CERIERO
DARNAN
APPARENT
MAGNITUDE
5
4
0
0
Characteristic
ABSOLUTE
MAGNITUDE
4
4
2
0
Analysis
Brighter as seen from Earth
Acana
Bade
They are the same brightness (from Earth)
More luminous
Acana
Bade
They are the same luminosity
Closer to Earth
Acana
Bade
They are the same distance from Earth
Brighter as seen from Earth
Ceriero Darnan
They are the same brightness (from Earth)
More luminous
Ceriero Darnan
They are the same luminosity
Closer to Earth
Ceriero Darnan
They are the same distance from Earth
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If we know both the absolute and the apparent magnitude, then we can calculate the distance. You just
figured out which star was closer than another, but sometimes we need to know the exact distance. To
do that we use the “distance modulus” equation:
(𝑚𝑚 − 𝑀𝑀)
𝐷𝐷
= log( )
5
10
m = apparent magnitude
M = absolute magnitude
D = distance in pc
The easiest way to use this equation to find the distance to a star is to follow these steps:
1. Work out the fraction
(𝑚𝑚−𝑀𝑀)
5
in your calculator. Hit the “=” key to get a number.
2. Take the “anti-log” of this number to get rid of the log. This will be a single button on your
calculator. It might be “INV LOG”, “2nd LOG”, “10x”, or “LOG-1”. Depending on your calculator,
you will either hit this button and then type in the number from step 1, or type in the number
from step 1 and hit that button. It sounds more complicated than it is – once you figure out
how your particular calculator is set up, it’s literally press a single button.
3. The number from step 2 is actually
𝐷𝐷
10
so just multiply your answer from step 2 by 10.
You will be following these same three steps over an over for most of the rest of the course, so we need
to be sure to learn it now! Let’s do a sample:
Star: Acana
Apparent Magnitude (m) = 5
Absolute Magnitude (M) = 4
Step 1: Work out the fraction
(𝑚𝑚−𝑀𝑀)
5
=
(
−
5
)
= _________
Step 2: Take the “anti-log” (we’ll use “10𝑋𝑋 ” to represent your anti-log button – USE THE BUTTON ON
YOUR CALCULATOR!)
10𝑋𝑋 (step 1 number) = 10𝑋𝑋 (_________) = ______________
Step 3: Multiply by 10
D = 10 ∗ (step 2 number) = 10 ∗ (__________) = __________ pc
Distance to Acana = _____________________ pc
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Tool #24: Magnitudes
Practice Problems:
Find the distance in parsecs to these three stars.
Star: Bade
Apparent Magnitude (m) = 4
Absolute Magnitude (M) = 4
Step 1: Work out the fraction
Step 2: Take the “anti-log”
(𝑚𝑚−𝑀𝑀)
5
=
(
−
5
)
= _________
10𝑋𝑋 (step 1 number) = 10𝑋𝑋 (_________) = ______________
Step 3: Multiply by 10
D = 10 ∗ (step 2 number) = 10 ∗ (__________) = __________ pc
Distance to Bade = _____________________ pc
Star: Ceriero
Apparent Magnitude (m) = 0
Absolute Magnitude (M) = 2
Distance to Ceriero = _____________________ pc
Star: Darnan
Apparent Magnitude (m) = 0
Absolute Magnitude (M) = 0
Distance to Darnan = _____________________ pc
Based on the above, what is the distance to any star whose apparent magnitude is the same as its
absolute magnitude? _____________________ pc
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100
Tool #25: Finding Large Distances
In Tool #24 (Magnitudes), you saw that if we know the apparent and absolute magnitudes of a star,
then we can find the distance to the star. We always know the apparent magnitude of the star – just
look at it! Finding the absolute magnitude is usually a lot harder. To help with this, astronomers have a
number of “standard candles” (objects that we know the absolute magnitude) that we can use to
measure distances. We will learn two of the standard candles in this tool: Cepheid variables and type 1
supernovae.
Red giants are huge because their envelopes are dominated by convection. Huge stars are very bright
(even though they may not be hot). Some red giants, however, are right on the border between
convection dominating and convection not dominating. When convection dominates, the star gets
bigger and so gets brighter. When these “borderline” stars get bigger, their envelopes expand enough
that the energy from the core can escape more easily and so convection no longer dominates. The star
shrinks – and so gets dimmer. The shrinking makes the envelope trap energy again, so convection
dominates, the star expands, and the whole cycle starts over. These “pulsing” stars are called Cepheid
variables. Their lights curves (a plot of brightness versus time) have a very unique, easily recognized
shape:
In 1908 Henrietta Levitt discovered that there is a relationship between the period of Cepheid variables
and their absolute magnitudes. All we need to do is observe the brightness of the star over several days
so that we can get the period of the pulsing. We can then use Henrietta Levitt’s “period-luminosity”
graph to look up the absolute magnitude of the Cepheid variable:
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Notice that there is also an equation for this line. If you want, you can simply put in the period in days
for P into this equation and you’ll also get the absolute magnitude. This might be more precise that
reading it from the graph itself!
To find the distance to the Cepheid, we need the absolute and the apparent magnitudes. We just found
the absolute magnitude, but because it’s a variable star the apparent brightness (by definition!)
changes. For Cepheid variables then (and only Cepheid variables), use the average apparent magnitude.
To find the average apparent magnitude, just add up the maximum and minimum apparent magnitudes
and divide by 2.
Let’s find the distance to the Cepheid variable shown on the previous page.
Period: _____________________ days
[Find the number of days between two peaks on the graph]
Absolute Magnitude: __________
[ Use the graph or the equation]
Maximum Apparent Magnitude: ________ [Remember this is the SMALLEST number for the curve!]
Minimum Apparent Magnitude: ________ [This is the LARGEST number for the curve]
Average Apparent Magnitude: ________ [Add the max and the min, hit equals, and divide by 2]
Distance to Cepheid Variable: ________ pc [Use the distance modulus from Tool #24]
Another “standard candle” is a Type I supernova. High-mass stars die in a Type II supernova – that’s
NOT what we are talking about here. A Type I supernova is caused by two white dwarfs colliding into
each other. The force of the collision compresses their carbon enough to fuse it, but this fusion takes
place in the presence of degeneracy (since that is all that keeping each white dwarf from collapsing in on
itself). Fusion goes so crazy as a result, that the two stars blow themselves apart – a supernova. These
are standard candles because all Type I supernovae are caused by the same thing: two white dwarfs
colliding. Therefore, all Type I supernovae have approximately the same absolute magnitude at their
peak brightness:
Type 1 Supernova Peak Absolute Magnitude = -19.3
Remember that negative magnitudes are even brighter than 1st magnitude stars, so (as you would
expect) Type I supernovae are very, very bright! A Type I supernova light curve looks like this:
All you need to do is read the peak apparent magnitude from the graph and use that with the -19.3 peak
absolute magnitude. Put those two numbers in the distance modulus equation, and you can find the
distance. For supernovae, we can see them a VERY long way away – even millions of parsecs!
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Tool #25: Finding Large Distances
Practice Problems:
Find the distance to the star whose light curve is shown below:
Period:
Absolute Magnitude:
_________________ days
_________________
Maximum Apparent Magnitude:
Minimum Apparent Magnitude:
Average Apparent Magnitude:
_________________
_________________
_________________
Distance to Star:
_________________ pc
Find the distance to the object whose light curve is shown below:
Peak Apparent Magnitude:
Peak Absolute Magnitude:
__________________
__________________
Distance to Object:
__________________ pc
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Tool #26: The Milky Way
Some people are surprised that the galaxy we know the least about is our own. The problem is that it’s
hard to study something when you’re inside it! As we saw in the previous tool, the stars you see in the
sky are actually only the tiniest fraction of the stars in our galaxy. If you go outside on a very dark, very
clear night – far away from cities – you can a band of faint light stretching across the sky. This Is the
Milky Way, but it’s actully you looking out into the plane of our galaxy. The light from billions of stars
combines to give the faint band of light you see. The picture below shows the Milky Way taken by a
camera that left its shutter open for a long period of time, enabling it to gather lots of the faint light.
As you can see, Earth (and our solar system, actually) is tilted relative to the disk of the galaxy. In the
picture above you can see the central bulge of the galaxy in the lower left corner. The disk extends up
and to the right. The Milky Way actually wraps all the way around the sky – which isn’t surprising since
we are inside the disk and surrounded by it. Looking away from the bulge means we are looking away
from the center of the galaxy and towards the outer edge.
Finding our position in the galaxy isn’t easy. The best way is to look at stars that are in the halo, which
orbit around the center and see where we are relative to them. Globular clusters are ancient
collections of over 100,000 stars that orbit the galaxy in the halo. Most of the globular clusters contain
Cepheid variables, so we can measure the distance and direction to them fairly accurately. When we
plot the location of these globulars from Earth, we find that they are centered on a point about 8000 pc
from the Sun. This map also tells us that the outer edge of the galaxy is about 7000 pc away from us, so
we are almost exactly halfway between the center and the edge. Since the radius of the disk is 15,000
pc, the size of the entire galaxy must be around 30,000 pc.
105
Looking at the picture of the Milky Way from Earth, we can imagine that our galaxy is a spiral – there’s a
bulge, a disk, and a halo. Finding the spiral arms themselves is challenging. The molecules of gas and
dust in the spiral arms have transitions that produce emission spectra in the radio band. By measuring
where the material is blue shifted or redshifted, we can see where the arms must be.
We have one major problem, though: If the spiral arms were rotating as fast as Kepler’s laws say they
should be, then they shouldn’t exist! Very soon after the formation of the galaxy, the arms would have
wound so tightly that they would disappear.
The answer to this dilemma is that the arms are actually “density waves,” areas of increased density that
triggers star formation. You can think of it like a traffic jam: Individual cars go in and out (eventually) of
the traffic jam, but the location of the jam – the area of increased density – stays in the same place.
Both blue and red stars are made in these locations, but the blue stars outshine the red stars, so they
are all we see. Both the blue and red stars move out of the density wave area as they orbit the center of
the galaxy, but blue stars don’t make it very far before they die. In between the density wave arms,
then we see only red stars – which are much fainter than the blue stars in the arms. As the blue stars
die, the shock wave from their supernovae cause the next wave to form a little beyond the original arm
– and the arm appears to rotate.
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Tool #26: The Milky Way
Practice Problems:
Pictured below is our best guess for the shape of the Milky Way. The Sun’s position is marked in its
approximate location in the Orion Spiral Arm. The white bar at the bottom of the image is 1000 pc long.
The locations of Polaris (the North Star), Pleiades (a young star cluster), and Cygnus X-1 (the first black
hole ever discovered) are also marked on the image.
Here are four of the brightest stars in the night sky and their distances from Earth. Which letter are they
closest to?
Sirius (2.6 pc): _____ Arcturus (10.4 pc): _____ Vega (7.7 pc): _____ Betelgeuse (430 pc): _____
Betelgeuse is one of the most distant of the bright stars. Is it actually that far away from the Sun? _____
The Crab Nebula is a remnant from a supernova that occurred in 1054 and was bright enough to read by
at night. The nebula is approximately 1900 pc away from Earth. Which letter is it closest to? _____
Cygnus X-1, near position C, is the first black hole ever discovered. It is not visible to the naked eye, but
shows up brightly in X-rays. Use the scale bar to determine the approximate distance to Cygnus X-1:
Check One: 1000 pc _____ 5000 pc _____ 8,000 pc _____ 10,000 pc _____ 20,000 pc _____
What is the approximate distance to the galactic center?
Check One: 1000 pc _____ 5000 pc_____ 8,000 pc _____ 10,000 pc _____ 20,000 pc _____
The Andromeda Galaxy is the closest galaxy to the Milky Way and is approximately 750,000 pc from
away. If the 1000 pc scale bar is 1 cm long, how many cm away would it be on this image? _________
How many meters away is that (1 meter = 100 cm)? _______ Would it fit on this page? ______
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Tool #27: Galaxies
As we look further out into the universe we begin to realize that stars are organized into giant
“islands” that all orbit a common center – as you probably know, these are galaxies. Each galaxy can
contain as many as 100 billion stars and are often more than 30,000 pc across. To give you a sense of
scale, practically every star you see in the sky is within 500 pc of Earth. To give you an idea of the
scale, the image below shows our location in the Milky Way; the white circle shows where nearly all
the stars you see in the sky are located.
Until the early part of the 20th century, we thought our galaxy was the entire universe. We now know
that when we look at other galaxies in space, we are looking out of our galaxy and across regions of
space that are nearly empty – less than one hydrogen atom per cubic meter, on average. Galaxies seem
to have three general types: spirals, ellipticals, and irregulars. As we discuss these types, it’s important
to keep in mind that a galaxy is not a “thing,” it’s a collection of stars. It has an apparent shape only
because of the orbits of the stars in it. Because there are billions of stars, basically there is a star at any
given point along an orbit, so it seems like the galaxy is a solid thing, but really it’s not. That will be
important later on.
Our galaxy, the Milky Way, is a spiral galaxy. Spiral galaxies have three basic parts: a bulge in the center,
a disk (which is made up of the spiral arms), and a halo of stars orbiting more or less in a sphere around
the disk. That looks something like this:
109
Notice that the spiral arms are blue. This is a clue that active star formation is taking place. There are
new red stars in the arms, too, but they are being outshone by the incredibly bright blue stars. Since
blue stars don’t live very long, we know that the blue stars in the arms must have formed fairly recently.
In between the arms, it’s not empty space – there are lots of old red stars, but most are too dim to see.
In an elliptical galaxy the stars are orbiting from all angles instead of in a disc. This gives the galaxy a
“football” shape. Remember, the galaxy isn’t a “thing” – it’s shape just shows you where the stars orbit.
Notice that the galaxy seems reddish (yes, it’s really yellowish, but remember there are only two colors
in astronomy – red and blue!). This tells us that there is not active star formation taking place, because
the only stars that are left in the galaxy are the red stars. The blue stars must have already died off and
weren’t replaced by new stars. The red stars live a very long time, and now that the blue stars aren’t
outshining them, we can see them – there’s nothing but red stars left and no gas and dust remaining to
make new red and blue stars.
Irregular galaxies are galaxies that have been ripped apart by other galaxies. This is often because two
galaxies have had a close pass of each other, or even collided with each other. When galaxies collide,
the stars never actually touch each other – remember the galaxy isn’t a “thing”. The gravity of the stars
interact, however, causing a “slingshot effect” that scatters the stars. This is why irregular galaxies have
no defined shape. The galaxy below is the Large Magellanic Cloud, visible in the southern hemisphere.
The Large and Small Magellanic Clouds are satellite galaxies of the Milky Way (they actually orbit the
Milky Way) that were ripped apart by the Milky Way’s gravity.
As you can see, like most irregular galaxies, the LMC is bluish. That means that there are lots of blue
stars (in addition to the red stars that are being out-shown by the blue stars). Since blue stars don’t live
very long, we know that there must be active star formation taking place (so there are new red and blue
stars in addition to the older red stars), likely due to compression from the collision.
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Tool #27: Galaxies
Practice Problems:
Examine the galaxies shown below and circle the appropriate responses in the analysis.
Galaxy A
Galaxy B
Galaxy C
Characteristic
Analysis (NOTE: You might circle more than one!)
Spiral
Galaxy A
Galaxy B
Galaxy C
Elliptical
Galaxy A
Galaxy B
Galaxy C
Irregular
Galaxy A
Galaxy B
Galaxy C
Active Star Formation
Galaxy A
Galaxy B
Galaxy C
Blue Stars
Galaxy A
Galaxy B
Galaxy C
Red Stars
Galaxy A
Galaxy B
Galaxy C
Old Stars
Galaxy A
Galaxy B
Galaxy C
Young Stars
Galaxy A
Galaxy B
Galaxy C
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Tool #28: Active Galaxies
Most of the light coming from “normal” galaxies is primarily in the visible portion of the electromagnetic
spectrum. This makes sense: Most of the light from a galaxy is being produced by stars. Some galaxies,
however, are much brighter than they should be. More imprtantly, these “active” galaxies are
producing light in all parts of the electromagnetic spectrum, not just the visible. This light can’t be
comiing from stars, so we call it “non-stellar radiation”. Active galaxies are galaxies that have an
extreme amount of non-stellar radiation. The spectrum below compares the light received from a
normal and an active galaxy.
The two most common types of active galaxies are Seyfert galaxies and radio galaxies:
Seyfert Galaxy
Radio Galaxy
Both of these types of galaxies are actually the same thing. In the case of a Seyfert galaxy, the emission
is pointed directly at us, so we see a “core-dominated” active galaxy. In the case of a radio galaxy, we
are looking at the galaxy edge-on, so we see a “lobe-dominated” active galaxy. The picture below
illustrates these situations:
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Quasars – “quasi-stellar objects” – were first discovered in the late 1950’s. These objects appear to be
star-like dots (“quasi-stellar” literally means “star-like” – sometimes astronomers aren’t terribly
creative), but their spectra are not star-like at all.
In addition to showing emission lines (rather than absorption lines), the lines are extremely redshifted.
Using Hubble’s Law, we can see that these objects must be very far away. Since they are far away yet
still bright, they must be emitting a truly huge amount of energy.
The power source for these active galaxies was a mystery until the mid-1970s. Observations of the
central area of active galaxies showed that the energy emitted changes on a very short timescale.
The flare is wide, because the energy from the near side of the core reaches us before the energy of the
far side. The number of days between tells us the width of the core in light-days.
In the table below, circle the characteristics of a radio galaxy:
Characteristic
Analysis
Luminosity
Bright
Dim
Energy Emission
Visible only
All types of light
Size of Emitting Region
Large
Small
Flares and Jets
Many
Few
The only object we know of that meets these requirements is a supermassive black hole. As the gas
from the accretion disk falls into the hole, it gives off x-ray and gamma rays. This energy is absorbed by
the disk of the galaxy and re-emitted as radio and infrared, thus producing the full-spectrum light that
we observe. Quasars are only found far away, so are galaxies when they are young (since we look back
in time when look farther out in space). As the gas in the core began to be used up, the emission went
down, and these galaxies became radio galaxies. When most of the gas is used up, we have a normal
galaxy like our own.
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Tool #28: Active Galaxies
PRACTICE PROBLEMS:
Examine the light received from the quasar RXJ 0911.4+0551, shown below:
The x-axis shows the number of days the quasar was observed. In this light curve, we can see a flare that
was produced by a sudden energy burst from the quasar.
One what day did the flare begin?
___________
One what day did the flare end?
___________
What is the size of the region emitting the flare?
___________ light-days
What is the size of the emitting region in kilometers?
___________ km
What is the size of the emitting region in AU?
___________ AU
Given the above calculation, what is most likely responsible for the energy being emitted by the flare?
___________________________________________________________________________________
Conversions you may need:
1 light-day = 2.59 x 1010 km
1 AU = 1.50 x 108 km
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Tool #29: Dark Matter
Kepler’s 3rd Law is the tool we use to find the mass of objects in space:
a3 = M p2
a is the semimajor axis (the distance from the center of the orbit) in astronomical units
p is the orbital period (how long it takes to orbit the center one time) in years
M is the total mass inside the orbit in solar masses
In the case of our solar system, the Sun makes up 99% of the mass. As the semimajor axis (distance
from the Sun) gets bigger, we technically need to add in the mass of the planets, but they don’t
contribute enough to make a difference.
The only reason we know the mass of the Sun is because we know the semimajor axis and periods of the
planets. No matter which planet you use, you get the same answer for the mass of the Sun – and you’d
better! The mass of the Sun doesn’t change just because you use a different planet to measure it!
Solving Kepler’s 3rd Law for M gives:
Use Jupiter to measure the mass of the Sun:
Semimajor axis = 5.203 AU
𝑀𝑀 =
𝑎𝑎3
𝑝𝑝2
Orbital Period = 11.866 years
Mass of Sun = _____________ solar masses
Your answer shouldn’t surprise you much…
Now let’s think about how the period changes if the mass is constant:
𝑝𝑝2 =
𝑎𝑎3
𝑀𝑀
If mass inside the orbit is constant, what happens to the period as the distance increases? ___________
If the orbital period gets longer, what happens to the orbital speed? _______________
Think about that last question: If it takes you longer to travel a distance, are you moving faster or
slower? _____________
Because the mass inside the orbit doesn’t change, as we move outward (as the semimajor axis gets
bigger), the orbital period gets longer and so the orbital speed gets slower. In our solar system, this
looks like:
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In the Milky Way, however, the mass inside the orbits does change. Remember that the galaxy isn’t a
solid object, it’s collection of stars. In the central bulge of the galaxy, as we move out from the center,
the number of stars contained within a certain distance increases rapidly. In fact, it increases so fast that
stars orbit faster as we move out rather than slower as in our solar system. Look at Kepler’s 3rd Law
again and see if you can see why this is true. Plotting the speed versus distance for our galaxy looks like
this:
Once we are outside the bulge, but before we get to the first spiral arm, we aren’t adding much mass
(because there aren’t as many stars). If we aren’t adding any mass, then just like our Solar System, the
orbital speed decreases. When we reach the first spiral arm, we start adding mass very quickly once
again, and the orbital speed goes up like in the bulge. Between the first and second arms there aren’t
many stars, so the orbital speed goes down. In the second spiral arm (which contains the Sun), the
orbital speed goes back up.
Here’s were things get odd. Once we leave the second spiral arm, there aren’t any more areas in the
galaxy where we add large numbers of stars. The orbital speed should then just continually drop, just as
it does in the Solar System. But when we measure the actual orbital speeds, we find that they are nearly
constant! We don’t see any stars or other matter here, but something must be adding mass to keep the
orbital velocities from decreasing.
We currently have no idea what this mysterious matter might be. It doesn’t give off energy like stars do
(not even infrared or x rays or anything), so astronomers call it “dark matter”. Based on the rotation
curve above, this “dark matter” must make up 95% of the galaxy! Think about that: The parts of the
galaxy we can see (the stars) that we’ve been studying for centuries makes up almost none of the actual
galaxy.
There are four possible explanations for dark matter than are currently under consideration:
•
•
•
•
Stellar-mass black holes (there’s no way they make up 95% of the galaxy)
Faint stars (Hubble Space Telescope observations have pretty much discounted this)
“Weakly Interacting Massive Particles” -- W.I.M.P.s (some strange new sub-atomic particle that
we haven’t seen on Earth yet)
New physics (maybe our understanding of physics is incomplete)
For now, we really don’t know what dark matter is. If it truly makes up 95% of the galaxy, then to a
certain extent we are wasting our time studying stars – they aren’t telling us much about what the
galaxy really is. Astronomers are working diligently to narrow down what dark matter might be, but
currently all we’ve been able to do is prove what it isn’t. Perhaps in the future we’ll be able to solve the
mystery.
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Tool #29: Dark Matter
Practice Problems:
Given the information below, calculate how much mass (in solar masses) must be inside the orbit of the
Sun. (HINT: You will need to solve Kepler’s 3rd Law for mass!)
a3 = M p 2
•
•
•
Orbital Period of Sun: 230,000,000 years
Distance of Sun from Galactic Center: 8,000 pc
1 pc = 206,265 AU
Distance from Sun to Galactic Center in AU:
Amount of mass inside of Sun’s orbit:
________________________ AU
_________________________ solar masses
Label each portion of the rotation curve shown below with the corresponding part of the galaxy.
Choices:
Dark Matter Dominant Region
Central Bulge
Spiral Arms (first and second)
Galactic Center
A: ________________________
B: ________________________
C: ________________________
D: ________________________
E: ________________________
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Tool #30: Finding Very Large Distances
Cepheid variables can be identified as far as 25 million parsecs (25 Mpc) away. If we want to observe
galaxies at the edge of the universe, we’ll have to do better than that. Type I supernova can be seen out
a billion parsecs away, but they are rare – we have to just get lucky. Fortunately, there is another way.
In 1929, Edwin Hubble (for whom the Hubble Space Telescope was named) published his observations
of the spectra of several galaxies. These galaxies were fairly nearby, so he was also able to observe
Cepheid variables in those galaxies. He noticed that all of the galaxies he observed had a redshifted
spectrum. He used that redshift to calculate the galaxies’ velocities and plotted that versus the
distances he measured from their Cepheid variables. His resulting plot looked soething like this:
All the galaxies fell onto a straight line! The slope of this line was a constant, which he (naturally) named
after himself. The line goes through the origin of the graph, so the equation of the line is:
v = H0 D
where v is the velocity in km/s, D is the distance in Mpc, and H0 is the slope of the line. The current
value for Hubble’s constant, H0, is 67.8 km/s/Mpc. To use Hubble’s Law, we just need to measure the
spectrum of a galaxy (which we can almost always do), find the velocity from the spectrum’s redshift,
and solve for D. Solving Hubble’s Law for D gives:
𝐷𝐷 =
𝑣𝑣
𝐻𝐻0
Let’s try a sample. Suppose you measure the wavelength of a hydrogen absorption line to be 665.1 nm.
The actual wavelength of this line is 656.3 nm. What is the distance to the galaxy?
First, find the redshift: ∆λ = λobs – λrest =>
∆λ = ____________ - __________ = __________nm
Next, use the Doppler formula to get the velocity: v = c * (∆λ / λrest) =>
v = 3.00E5 * ( ____________ / ____________ ) = ______________km/s
Finally, use Hubble’s Law to get the distance: D = v / H0 =>
D = _______________ / 67.8 = ________________ Mpc
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But wait! Scientists spent centuries trying to convince people that we aren’t at the center of the
universe. If Hubble’s observations are correct and everything is moving away from us, doesn’t that
mean we’re at the center of the universe? Actually, no. Imagine that universe is the surface of the
balloon (ignore the interior of the balloon – our analogy doesn’t work there). Tape some coins to the
surface of the balloon – these are our galaxies.
If you inflate the balloon, the coins all move away from each other. Each coin, however, thinks all of the
other coins are moving away from it because the distance to each other coin is getting greater. So,
there really isn’t a center to the universe, we’re just all expanding on the same “balloon”. The light from
galaxies is getting redshift because space itself is expanding, and so the light’s wavelength gets longer.
Hubble’s observation also led to what was then a dramatic idea: If the universe is expanding, then in the
distant past the universe must have been compressed to a single point. This would have immense
amounts of energy, so would have exploded outward: the Big Bang. One question many people ask is
“Where did the Big Bang happen?” The answer is: everywhere! Remember that there is nothing outside
of the universe. You aren’t on the outside of the balloon looking at it, you’re actually on the surface. As
we move back in time, the balloon shrinks so everything in the universe gets closer to you. Eventually,
we are at the very beginning – the “explosion” is the entire balloon suddenly beginning to expand.
One other piece of information that Hubble’s Law gives us is the age of the universe. Take any galaxy in
the universe. Take its spectrum and determine its velocity. Using Hubble’s Law, figure out the distance
to that galaxy. If we know how fast the galaxy is travelling, and we know where it is now, we can figure
out how long ago it started. For example, if you have driven 120 miles, and we know that you drove at
60 mph, then we can figure out that you left 2 hours ago. Mathematically, that’s just:
t=D/v
All you need to do is convert the distance to the galaxy from Mpc to km and divide by the velocity in
km/s. That will be how many seconds ago the universe began. Divide those seconds by 3.15 x 107 will
convert seconds to years – the age of the universe.
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Tool #30: Finding Very Large Distances
PRACTICE PROBLEMS:
Given the following information, calculate the distance to each galaxy. In each case, use 656.3 nm as the
rest wavelength for the hydrogen absorption line and 67.8 km/sec/Mpc for Hubble’s constant
Galaxy A:
•
•
•
Observed hydrogen line: 658.4 nm
Velocity of galaxy:
______________ km/sec
Distance to galaxy:
______________ Mpc
Galaxy B:
•
•
•
Observed hydrogen line: 661.2 nm
Velocity of galaxy:
______________ km/sec
Distance to galaxy:
______________ Mpc
Galaxy C:
•
•
•
Observed hydrogen line: 672.6 nm
Velocity of galaxy:
______________ km/sec
Distance to galaxy:
______________ Mpc
Suppose we observe two different galaxies from Earth:
Characteristic
Galaxy Distance
Longer hydrogen wavelength
Closer
Farther
They are the same distance from Earth
Faster Doppler velocity
Closer
Farther
They are the same distance from Earth
Suppose we measure a galaxy to be moving away from us at 3000 km/sec. How fast will astronomers in
that galaxy measure us to be moving? Mark your response below.
____ 3000 km/sec towards them
____ 3000 km/sec away from them
____ We won’t seem to be moving
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Tool #31: Cosmology
Cosmology Is literally the study of the universe: how it was born, how it changes over time, and how it
will die. Edwin Hubble’s discovery that the universe is expanding led to the inescapable conclusion that
the universe has a beginning – it started as a singularity of size zero with an unimaginable amount of
energy. This instant of creation was a collosal explosion called (derisviely, at first) the Big Bang. It’s
important to remember that the Big Bang was the universe – it happened everywhere at once. There
was no “outside” from which to observe the birth of the universe.
The birth of the universe can be divided into two general phases: before and after the first 20 minutes of
its existence. Initially, the temperature and density of the universe was so high that matter literally
couldn’t exist – the entire universe was nothing but energy. As the universe cooled, matter began to be
more and more important until a point 380,000 years after the Big Bang, matter dominated the
universe. This point marks the farthest back in time we can see with our telescopes. We see this leftover
energy as the Cosmic Microwave Background.
Here are the key events in the birth of the universe:
The First Twenty Minutes of Creation
•
•
•
•
t=10-43 sec (The Plank Era)
o T = 1032 K, Density= 1095 g/cm3
o Universe is radiation and subatomic particles from pair production
t=10-10 sec (The Inflationary Period)
o T = 1029 K, Density= 1075 g/cm3
o Universe expands by a factor of 1050 in 10-32 seconds
t=3 minutes (The Destruction of Antimatter)
o T = 1012 K, Density= 106 g/cm3
o Hydrogen into helium fusion of the universe begins
o Protons and antiprotons annihilate each other
t=20 minutes (End of the Fusion Era)
o T = 109 K, Density= 104 g/cm3
o Hydrogen into helium fusion of the universe ceases
o Universe is 75% hydrogen, 25% helium
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The first stars did not begin to form until around 100 million years to a billion years after the Big Bang,
but we can now see that they were not composed entirely of hydrogen. The universe fused hydrogen
into helium itself before the first star was ever created.
In 1965, Arno Penzias and Robert Wilson discovered that there is microwave radiation coming from
space itself – in all directions. This cosmic microwave background has a near-perfect spectrum of 2.7
degrees Kelvin. This is exactly the temperature that universe should have cooled to after 14 billion years.
Penzias and Wilson were observing the radiation from the Big Bang. When we look at the CMB, we are
seeing back in time to about 380,000 years after the birth of the universe. This is the farthest back in
time that we can directly observe the universe with any technology we can think of.
One particularly puzzling observation is that the universe seems to have expanded in size by a factor of
1050 in less than 10-32 seconds. This is much faster than the speed of light! It does, however, explain why
the universe is basically the same in all directions: Small fluctuations in the “quantum foam” got inflated
to huge sizes, so each part of the universe is the same temperature as every other part. This “foam” has
tiny variations – these variations become the “seeds” of today’s galaxies. This explains why the universe
is not perfectly smooth with stars and galaxies evenly distributed throughout the universe. In stead of
stars everywhere, we see clumps of galaxies and voids in between them.
There are three possibilities for the death of the universe:
• It could expand forever.
• It could slow down so that it reaches zero velocity after an infinite amount of time.
• It could reach zero velocity and begin to collapse back down to a point (the Big Crunch).
Gravity is the only force that can act at distances on the scale of the universe. If there is enough mass in
the universe so that gravity can halt the expansion, then the universe will die (and be reborn) in the Big
Crunch. If there is not enough mass, then the universe will die a cold, dark and lonely death as it
expands forever until all of the stars go out forever with no more matter left to form new stars.
We can calculate the amount of mass needed to halt the expansion. If the average density of the
universe is more than 9 x 10 -27 kg/m3 – about 5 hydrogen atoms per cubic meter – then the universe will
eventually collapse back to a point. This doesn’t seem like much, but the universe is very, very empty.
Even including dark matter (or the effects of dark matter), the universe has only about 30% of the mass
needed to reach the critical density. Given that the rate of expansion is (mysteriously) increasing, we can
say with some certainty that the universe will not be reborn in the Big Crunch. Ultimately, the universe
we have is the only one we will ever have. It will die in the dark with no one left to mourn its passing.
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Tool #31: Cosmology
Practice Problems:
Sort the following into the groups based on whether they occurred before or after the first twenty
minutes of the universe’s life.
1. First stars are born
BEFORE
AFTER
2. First galaxies are created
BEFORE
AFTER
3. Universe expands faster than light
BEFORE
AFTER
4. Universe (not stars) fuses hydrogen into helium
BEFORE
AFTER
5. Almost all of matter and antimatter annihilate each other
BEFORE
AFTER
6. The Cosmic Microwave Background forms
BEFORE
AFTER
Describe the conditions that will produce the listed death of the universe.
1. The universe collapses back to a point.
DENSITY < CRITICAL
DENSITY = CRITICAL
DENSITY > CRITICAL
2. The universe expands forever, never slowing down to near zero velocity.
DENSITY < CRITICAL
DENSITY = CRITICAL
DENSITY > CRITICAL
3. The universe expands forever, reaching zero velocity after an infinite amount of time.
DENSITY < CRITICAL
DENSITY = CRITICAL
DENSITY > CRITICAL
How will our actual universe die?
Big Crunch
Cold death of everything
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Universe is constant and unchanging
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Tool #32: Dark Energy
In a previous tool (Finding Very Large Distances) you learned that Hubble plotted the velocity of the
galaxies he observed versus their distances and found that they lie on a straight line.
The slope of this line is the Hubble constant, and we have been using 67.8 km/s/Mpc for its value. The
Hubble constant is also a measure of how fast the universe is expanding. The higher the slope of the
line, the faster the universe is expanding. Take a look at the hypothetical universes shown below:
In our universe (the first plot), for example, a galaxy (and therefore spacetime) at 500 Mpc is moving at
around 25,000 km/sec. In the hypothetical universe on the left, that galaxy is moving at almost 75,000
km/sec – much faster! In the hypothetical universe on the right, that galaxy is only moving at around
10,000 km/sec – much slower. The slope of the line is the Hubble constant, so the size of the Hubble
constant is a measure of how fast the universe is expanding – the higher the slope the faster the
universe is expanding.
To this point, we’ve assumed that the rate of the expansion of the universe has been constant. Hubble’s
data certainly shows that the slope of the line doesn’t change. What would the plot look like if the
universe were slowing down over time? This would make sense if the gravity of everything in the
universe were all pulling together to stop the expansion from the Big Bang. The key to making these
plots is to realize that when we look out in distant space, we are looking back in time. The light from
distant galaxies can take billions of years to reach us, so we are seeing the galaxy as it was billions of
years ago. The farther out into space we look, the farther back in time we are seeing.
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Walk through the logic:
•
•
•
If the universe were slowing down, then it would have been expanding faster in the past than it
is today
If the universe were expanding faster in the past, then the slope of the line (Hubble’s constant)
would be greater than it is today.
Since the past is what we see at greater distances, then the slope at greater distances (the early
universe) must be greater than the slope close to us (the universe today).
The plot on the left shows this logic. The plot on the right takes the two regions and connects them with
a smooth curve (because the universe in this example was slowing down gradually and not instantly).
What if the universe were accelerating? In other words, what if the universe is expanding faster today
than it was in the past? On the left plot below, draw two line segments that show the slope of the line
today and in the past. On the right plot below, connect those two segments with a smooth curve.
Hubble used Cepheid variables to measure the distances to the galaxies he observed, but we can’t use
Cepheids for galaxies at the edge of the observable universe (which would we see as they were near just
after the universe was created). As you know, a Type I supernova allows us to make measurements at
extremely large distances. Over time, we have collected enough data to determine if the expansion of
the universe is constant (a straight line, as Hubble thought), slowing down, or speeding up. In the
practice problems, you will determine what our real universe is doing and what could be causing the
effects that we see.
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Tool #32: Dark Energy
Practice Problems:
Pictured below is the Hubble plot for our actual universe. Answer the questions below.
1. On the diagram, label the location of galaxies seen from the early universe in the distant past.
2. On the diagram, label the location of galaxies seen from the universe today.
Characteristic
Analysis
Slope of the line in the past
High
Low
Zero
Slope of the line today
High
Low
Zero
Expansion rate in the past
Fast
Slow
Zero
Expansion rate today
Fast
Slow
Zero
Type of universe we live in
Accelerating
Decelerating
Constant
Gravity could cause the universe to decelerate (slow down) by pulling all the galaxies together. We
don’t know of anything that could actually cause the universe to accelerate (speed up). It requires
energy to cause something to accelerate. We can’t see any of this mysterious universal energy in space,
so it is given the name “dark energy”.
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Tool #33: Cluster Ages
Star clusters are simply groups of stars. There are 2 types of star clusters: open clusters and globular
clusters. Globular clusters are tight groups of ancient stars, and are less well-understood. Open clusters,
however, are an incredibly useful tool for astronomers. It is thought that all of the stars in an open
cluster formed from the same collapsing cloud of interstellar gas, and are roughly the same age. Open
clusters tend to be much smaller and younger than globular clusters. The stars within a cluster have a
range of stellar masses, and so when the cluster is young (and its stars have just begun to fuse hydrogen
into helium in their cores), the stars will all lie along the main sequence. This forms the zero-age main
sequence (ZAMS).
A star will remain on the main sequence as long as it has
hydrogen fuel in its core. Recall that a star will live a mainsequence lifetime of
𝑙𝑖𝑓𝑒𝑡𝑖𝑚𝑒 = 𝑀!"#$ %&.( ∗ 10)* 𝑦𝑒𝑎𝑟𝑠
where Mstar is expressed in solar masses. A star’s mass and
lifetime are inversely proportional: the higher its mass,
the shorter its lifetime.
As a star ages, it begins to run out of hydrogen fuel in its core. Hydrogen then begins to burn in a shell
around the core, and the star expands and cools, becoming a supergiant or red giant. The first stars to
become supergiants are the most massive stars in the cluster, followed by the less massive stars
becoming red giants. Therefore, as a cluster ages, the stars turn off the main sequence and evolve onto
the red giant branch. The main sequence gets shorter and the red giant region becomes increasingly
populated (as shown in the diagram at right).
The main sequence turnoff point is indicated in the
diagrams. The turnoff point is the highest-mass star still
left on the main sequence: all higher-mass stars have
already evolved off.
As the cluster ages further, the turnoff point moves further down the main sequence to less luminous
stars. The stars now at the turnoff point are less massive and therefore they live a longer lifetime, which
is why they are still on the main sequence while all higher-mass stars are gone already. All stars with
higher masses than the stars at the turnoff point have run out of
hydrogen fuel, evolved off the main sequence, and are in the
process of dying (if they aren’t already dead).
The stars at the turnoff point are next. They are right at the end
of their lifetime. They are about to run out of fuel and begin to
die. This is how the turnoff point tells us the age of the cluster.
What is the luminosity of the stars at the turnoff point in the HR
diagram above? __________________
What is the mass of a star with that luminosity?
__________________
What is the lifetime of a star with this mass?
__________________
You have just calculated how long these stars will live until they run out of hydrogen fuel. However,
since these stars at the turnoff point are about to run out of fuel, they have reached the end of their
lifetime and will soon begin dying. Hence, the lifetime of this star is equal to its age.
What is the age of the star at the turnoff point in the HR diagram above? __________________
Finally, remember that all of these stars in the cluster were born together at the same time. Therefore
all of the stars in the cluster are the same age. By finding the age of the star at the turnoff point, you
have found the age of the whole cluster and everything in it.
What is the age of the star cluster in the HR diagram above? __________________
Lastly, a warning. When binary stars orbit each other very closely, they can transfer mass between
them. Stars gaining or losing mass can affect their evolution, in some cases causing them to live a longer
time than they otherwise would. These stars are called “blue stragglers.” Don’t confuse the blue
stragglers for the turnoff point, or you will think a cluster
is younger than it really is. Refer to the HR diagram for the
cluster M55 at right to see what these look like.
Tool #33: Cluster Ages
Practice Problems:
1. What is the luminosity of the turnoff point for the cluster below? ____________________
What is the turnoff mass? ____________________
What is the cluster age? ____________________
2. What is the luminosity of the turnoff point for this cluster? ____________________
What is the turnoff mass? ____________________
What is the cluster age? ____________________