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Name:
1
KEY
Math Activity
Intoduction
This activity is meant to give you a review of math that will be both useful and important in
Astronomy 101. If this is not a review and you find it difficult, please stop by my office hours, or
study Appendix C of your text.
1.1
Scientific notation
The idea behind scientific notation is that it’s a shorter way of expressing very large or very small
numbers. This is something that often is very useful in astronomy.
1. What are some large objects or distances that you saw in the powers of 10 video?
There are many answers. Some include the speed of light, the size of the sun, the age of the
sun, the age of the galaxy, the distance between the Earth and the Sun, and the distances
between galaxies.
2. What are some small objects that you saw in the powers of 10 video? Which ones are relevant
to astronomy?
Some answers include the size of atoms and molecules, the wavelengths of light, and the sizes
of protons and electrons.
Basics
Write these numbers in non-scientific notation:
3. 5 ×106 = 5,000,000
4. 6.3 ×1010 = 63,000,000,000
5. 4.2 ×10−3 = 0.0042
6. 3 ×10−6 = 0.000003
Write these numbers in scientific notation
7. 100,000 = 1 ×105
8. 8,230 = 8.23 ×103
9. 0.0023 = 2.3 ×10−3
10. 0.0000000734 = 7.34 ×10−8
1
Multiplication and Division
There is a slightly different method we might use for multiplying and dividing numbers written
in scientific notation. I’ll work out an example or two on the board, and then ask you to finish
the rest of these. Please first try them in your head, then check with a calculator:
11. 2.1 ×104 × 3 ×10−2 =
2.3 × 3 = 6.3
104 × 10−2 = 102
6.3 ×102
12. 5 ×105 × 4.2 ×106 =
5 × 4.2 = 21
105 × 106 = 1011
21 ×1011 = 2.1 ×1012
13. 6 ×10−9 × 1.2 ×104 =
6 × 1.2 = 7.2
10−9 × 104 = 10−5
7.2 ×10−5
14. 1.5 ×109 / 5 ×104 =
1.5 / 5 = 0.3
109 / 104 = 105
0.3 ×105 = 3 ×104
15. 4.8 ×10−2 / 2 ×106 =
4.8 / 2 = 2.4
10−2 × 106 = 10−8
1.2
2.4 ×10−8
Units
Units are very important in any calculation. If you are doing homework for this class and don’t
have a unit after a number, you should check your work and find the appropriate units.
Typical Units
16. What are some typical units of length, or distance?
centimeters, meters, astronomical units, parsecs, Angstroms
17. What are some typical units of time?
seconds, minutes, days, years
18. What are some typical units of speed?
miles per hour, meters per second
Conversion Factors
When you are looking at a number that are given in one unit, you will often need it in another
set of units entirely. Sometimes you need to do these in order to move on to the next step of a
calculation, other times its necessary in order to make the number make more sense to you.
19. Convert 2 ×107 cm to km
1km
1m
2
2 ×107 cm × 1×10
2 cm × 1×103 m = 2 ×10 km = 200 km
20. Convert 789 days to years
1year
798 days × 365days
= 2.16 years
21. Convert 1.5 ×10−3 days to seconds
60s
1.5 ×10−3 days × 24hours
× 60min
× 1min
= 130 s
1day
1hour
2
1.3
Common Sense and Math
An essential part of math is applying some reasoning to the problems you are answering to
make sure your answer makes sense. Is it expressed in the appropriate units, with a reasonable
number of digits? Is it around the right number?
1.3.1
Digits
If you’ve taken a math or science class where you discussed significant figures, those rules would
certainly give you an answer with the appropriate amount of digits. We don’t need to be that
exact in Astronomy 101, but need to apply common sense. Fill in the columns below with the
answers to the math problem in the left column.
math
calculator output
answer
1000 / 3
333.3333333
333
π×2
6.283185307
6.28
1.414213562
1.41
6×1024 / 7 ×1017
8571428.571
8.6 ×106
log(99)
1.995635195
2
√
1.3.2
2
Checking The Size of Things
Match each distance or size to its approximate length in centimeters (on the right) and a more
appropriate unit to express it in.
3.9 ×108 cm PPP
PP
PP
PP
angstrom (Å)
height of a human Q
Pdistance
Q
Q
Q kiloparsec (kpc)
Q
Q
Q
Q
Q #
S
Q
Q
#Q
S
Q
#
Q
1.54 ×1023 cm Q
radius
of
an
atom
parsec (pc)
Q
S
#
Q
Q
Q
S# QQ
Q
Q
Q
Q
#
S
Q
Q
−8
#
S
1 ×10 cm
Q
distance
to
a
nearby
star
meter (m)
#
Q
S
#
Q
S
Q
#
Q
#
S
13
X
Qdiameter of galaxy
1.5 ×10 cm X
astronomical
unit (AU)
XX
S
XXX
S
XXX
X
SS
2
18
9.3 ×10
cm Q
2 ×10 cm
from here to NYC S
distance to the sun
3
Q
kilometer (km)
1.4
Example Problems
22. How long does it take light to travel from the Sun to the Earth?
The speed of light is c = 3.00 ×108 m/s
The distance from the Sun to the Earth is 1 AU =1.50 ×1011 m.
time =
distance
speed
=
1.50×1011 m
3.00×108 m/s
= 500s
If you wanted to know this in minutes, you could use a conversion factor:
500s × 1min
= 8 minutes
60s
23. How long does it take light to travel from the Moon to the Earth?
The distance from the Moon to the Earth is 3.844 ×105 km.
First convert the distance to meters because the speed of light is in m/s:
3m
= 3.844 ×105 km
3.844 ×105 km × 1×10
1km
Now use the same method as the previous problem:
3.844×105 m
= 3.00×10
time = distance
8 m/s = 1.28s
speed
24. How fast are we traveling around the galaxy?
We are located 28,000 Ly from the center of the Galaxy
We orbit the galactic center once every 230 million years
One Ly is 9.46 ×1015 m
The distance we travel is related to the radius: d = 2πr
First, we’ll convert light-years into meters:
15 m
= 2.6488 ×1020 m
2.8 ×104 Ly × 9.46×10
1Ly
Then convert the galactic radius into the circumference of our orbit:
d = 2 ×π× 2.6488 ×1020 m = 1.664 ×1021 m
Note that 230 million years = 2.3 ×108 yr. We need to convert that into seconds:
24hr
60s
2.3 ×108 yr = 365days
× 1day
× 60min
× 1min
= 7.2533 ×1015 s
1yr
1hr
speed =
distance
time
=
1.664×1021 m
7.2533×1015 s
= 229412.819
Rounded to two significant digits, the answer is: 2.3 ×105 m/s
This is the same as 510,000 miles per hour - fast!
4
Name:
2
KEY
Kepler Activity
Intoduction
This activity will discuss the properties of ellipses and Kepler’s laws of orbital motion.
2.1
Ellipses
First, please name the parts of the ellipse that are
labeled in the picture to the right:
• a semi-major axis
• b semi-minor axis
• c foci (focuses)
• d center
Eccentricity
The eccentricity of an ellipse is defined as the “distance from the center to the focus of the ellipse
divided by the semi-major axis.” Another way of saying that is that it’s a measure of how squashed
an ellipse is. An ellipse with a higher eccentricity would be more squashed.
1. In the space below, draw and label an ellipse with a higher eccentricity than the one above,
and an ellipse with lower eccentricity than the one above.
2. What is an ellipse with an eccentricity of zero (note: the distance between the focus and the
center would be zero)?
A circle!
5
2.2
Kepler’s the Laws of Planetary Motion
Johannes Kepler was the first to explain the motions of the planets in the sky. We study these
laws because they are similar to the laws the govern the orbits of all objects around each other.
We’ll discuss these laws in class - under each of them, write 2-3 sentences to explain them in
your own words.
Kepler’s Laws
• First Law: The orbit of each planet around the Sun is an ellipse with the Sun at one
focus.
This one is in pretty plain language. The important thing to remember is that the Sun is
not at the center of the ellipse, but at the focus. That means a planet on a highly eccentric
orbit can be close to the Sun sometimes, and very far at other times.
• Second Law: As a planet moves around its orbit, it sweeps out equal areas in equal
times.
This means that when the planet is closer to the Sun, it moves faster than it does when
it is far away. This also means that a planet with a highly eccentric orbit will move at a
variety of speeds, while a planet with a nearly-circular orbit will only vary a little in its
speed.
• Third Law: The square of the orbital period of a planet is directly proportional to the
cube of the semi-major axis of its orbit.
The semi-major axis is the planet’s distance form the sun, and the period is the amount
of time it takes for the planet to complete an orbit. So this law is saying that the farther
a planet is from the Sun, the longer it will take to orbit. Planets follow this with the
relation P 2 = a2 , where P is the period, and a is the semi-major axis.
Kepler’s First Law
3. In the space below, draw the Sun and the highly eccentric orbit of a planet around it. (Don’t
worry about drawing it to scale.)
Kepler’s Second Law
4. On your picture above, label the point where the planet would be moving fastest, and where it
would be moving the slowest.
6
Kepler’s Third Law
5. In the table below, “P” stands for period, or number of years it takes the planets to go around
the sun, and “a” stands for semi-major axis, which measures how far the planet is from the
sun. The middle column is the calculated square of the period and the cube of the semi-major
axis. Fill in the blank spaces; you can round off any decimal places.
P (years)
1
8
27
64
90
125
P 2 = a3
1
64
729
4096
8100
15625
a (AU)
1
4
9
16
20
25
6. On the axes below, graph P and a from the table that you filled in. Draw a line through the
points that you graph. Note that the axes are on different scales.
120
Period (years)
100
80
60
40
20
0
0
5
10
15
semi−major axis (AU)
20
25
7. Draw and label the line that would show P = a. It’s in red above
8. Is it surprising how sharply the period of the planet’s orbit rises? I think it’s pretty drastic.
The planets both have farther to travel, and are moving slower, so it’s a large effect.
7
Name:
3
KEY
Light Activity
Intoduction
This activity goes through some of the important properties of light that we need to know about in
order to understand astronomy.
3.1
The Electromagnetic Spectrum
Wavelength, Frequency, and Energy
1. What is wavelength?
Wavelength is defined as the distance between two successive peaks or two successive troughs
of a wave.
2. What is frequency?
Frequency is the number of peaks (or troughs) that are passing a point in a unit of time (like
a second).
Wavelength and frequency are inversely proportional, which means that if one increases, the
other decreases. We write their relationship like this:
c
c
or
f=
(1)
λ=
f
λ
Where f is the frequency, λ is the wavelength, and c is the speed of light (c = 3 × 108 m
s)
3. Below, draw a wave with a wavelength of 3 cm.
0
1
2
3
4
5
6
7
8
9
10
Centimeters
4. Calculate the frequency of the wave you drew:
First, convert the units of either λ or c so that they match, then calculate:
c = 3.0 × 108 m
s ×
f=
c
λ
=
100 cm
1 m
3.0×1010 cm
s
3 cm
11
12
= 3.0 × 1010 cm
s
= 1010 s1 = 1010 Hz
Energy is proportional to frequency, which we write as:
E =h×f
(2)
2
where E is energy, f is frequency, and h is Plank’s Constant (h = 6.626 × 10−34 m skg ). One
way to think about this is that a wave with a shorter wavelength will peak more frequently,
which will carry more energy.
8
5. Calculate the Energy of the wave you drew:
2
2
E = h × f = h = 6.626 × 10−34 m skg × 1010 s1 = 6.626 × 10−24 ms2kg = 6.626 × 10−24 Joule
Naming the Different Types of Light
Each region of the spectrum has a name that we use to refer to it. We’ll go over the information
below as part of a lecture on the electromagnetic spectrum. Please fill out the columns to
characterize the types of light.
Type of Light
Wavelength Size
λ (m)
Energy
(eV)
Source on Earth
Frequency
(Hz)
gamma ray
smaller than an
atom
10
106
radioactive
ments
1020
X-ray
thickness of
DNA strand
10−9
103
X-ray machine
1018
ultraviolet
groove in a CD
10−7
10
1016
Visible Light
red blood cell or
bacterium
10−6
1
The sun - it
burns!
fire, light bulbs
Infrared
width of a human
hair
10−4
10−1
stovetop, people
1013
microwave
coffee bean
10−2
10−4
radar, microwave
oven
1010
radio
city bus
10
10−7
Radio
station
transmitter
108
3.2
a
ele-
1014
Thermal Emission
This activity goes through two main types of light emission. The first is thermal emission,
which comes from very dense material. When I say dense, I’m talking about the sun or other
stars, planets, or even human beings. On the next page is the thermal emission spectrum from
an object with a temperature of T = 6000K, like our sun.
3.5
Relative Intensity
3.0
2.5
2.0
1.5
1.0
0.5
0.0
200
400
600
Wavelength (nm)
9
800
1000
The temperature affects the emission in two ways: the amount of light emitted, and its peak
wavelength.
Peak Wavelength
The peak wavelength of a spectrum of a given temperature can be calculated using Wein’s Law:
λmax =
2.9 × 106 nm K
T
(3)
where λmax is the peak wavelength of the spectrum and T is the temperature of the object.
The peak wavelength is inversely proportional to temperature, meaning that hotter objects
peak at shorter wavelengths and cooler objects peak at longer wavelengths. This means that
hot objects are bluer, and cold objects are red.
Amount of Light
Additionally, cooler objects emit less light than hotter objects at every wavelength. For example, this is how the thermal radiation spectrum of a T = 5000K (in grey) object would compare
to the T = 6000K object (in black):
3.5
Relative Intensity
3.0
2.5
2.0
1.5
1.0
0.5
0.0
200
400
600
Wavelength (nm)
10
800
1000
In the graph below, a thermal emission spectrum with a temperature of T = 6000K is shown.
6. Calculate the peak wavelength and draw in the thermal emission spectra a T = 8000K object:
λmax =
2.9 × 106 nm K
2.9 × 106 nm K
=
= 362.5 nm (shown in blue below)
T
8000 K
(4)
7. Calculate the peak wavelength and draw in the thermal emission spectra a T = 4000K object:
λmax
2.9 × 106 nm K
2.9 × 106 nm K
=
=
= 725.0 nm (shown in red below)
T
4000 K
(5)
14
Relative Intensity
12
10
8
6
4
2
0
200
3.3
300
400
500
600
Wavelength (nm)
700
800
900
Atomic Spectra
At this point, we’ll go back into lecture mode to learn about atoms and their emission/absorbtion
spectra.
8. In the space below, draw an energy level diagram. Show what happens when an atom absorbs
light, and what happens when an atom emits light.
Next, we’ll turn on four atomic lamps, as well as a white light lightbulb. For each of the lamps
and the lightbulb, first look at the color of the light without a diffraction grating. Then hold
the grating close to your eye and look to the side of the light. You should see spectra to the left
and the right - the one on the right should line up with the colors listed on the spaces provided
for drawing.
11
White light: The normal lightbulb is on the top of the tower.
9. What color is the light without the diffraction grating? white
violet
400
blue
450
green
500
550
600
Wavelength (nm)
yellow
650
red
700
750
Hydrogen lamp: The atomic hydrogen lamp is the top lamp, just under the light bulb.
10. What color is the lamp without the diffraction grating? purple
violet
400
blue
450
green
500
550
600
Wavelength (nm)
yellow
650
red
700
750
Helium lamp: The atomic helium lamp is second from the top lamp.
11. What color is the light without the diffraction grating? white
violet
400
blue
450
green
500
550
600
Wavelength (nm)
yellow
650
red
700
750
Mercury lamp: The Mercury (Hg) lamp is just above the bottom lamp.
12. What color is the light without the diffraction grating? blue
violet
400
blue
450
green
500
550
600
Wavelength (nm)
yellow
650
red
700
750
Neon lamp: The Neon lamp is on the bottom of the tower.
13. What color is the light without the diffraction grating? orange
violet
400
blue
450
green
500
550
600
Wavelength (nm)
12
yellow
650
red
700
750
3.4
Emission and Absorption
The last step in understanding light in astronomy is figuring out how the thermal radiation
spectrum, from hot dense objects, and the atomic emission spectrum, from diffuse matter, go
together.
14. In the space below, draw a hot dense object and a gas cloud. Draw and label three observers
who see:
• Only a thermal emission spectrum
• Only an atomic emission spectrum
• A thermal emission spectrum with atomic absorbtion
Each spectrum represents one of the observers, which is looking in the direction of the nearby
black arrow. The red and blue2 arrows show that each of the objects emit light in all directions.
13
Name:
4
KEY
Solar Activity
Intoduction
Solar activity is a blanket term for all the different kinds of variations in the Sun that we can observe
from Earth. Most solar activity is caused by magnetic fields on the Sun. There is a large magnetic
field that runs through the entire Sun and also many smaller magnetic fields that are tangled with
the hot, convective plasma at the Sun’s surface.
In the first part of this activity, you’ll measure the solar activity cycle using sunspot data. In the
second part of the activity, you’ll use pictures of sunspots to calculate the rotation period of the Sun.
Note that the rotation isn’t caused by solar activity, but is due to the angular momentum that the
Sun has conserved since formation. In the third part of the activity, you’ll measure the speed of a
coronal mass ejection and figure out how long it takes for the energetic particles to reach the Earth.
4.1
The Solar Cycle
The Sun goes through a cycle of more and less activity. Records as far back as the 1600s show a
continuous cycle in the number of sunspots, which are the most visible tracer of the magnetic fields
that cause all forms of solar activity.
The solar cycle seems to be linked to changes in the Sun’s global magnetic field. At one solar activity
maximum, the magnetic field is aligned so that a compass needle on the Sun would point north, and
at the next maximum the needle would point south. The reversal of the poles of the magentic field
is the basic cause of the cycle.
The graph below shows the number of Sun spots per year over a period of almost 50 years.
14
Our task is to measure and calculate the length of the solar cycle using the data in the chart at
the bottom of the previous page. Because the x-axis is in years, measuring the period is similar to
measuring wavelength - you calculate the distance between each of the the solar maximums and the
solar minimums on the graph.
Because the data doesn’t seem to resemble a perfect wave, we need to measure the distances between
several solar maximums and several solar minimums in order to determine the best possible period.
1. Fill in the table below with the year of each of the solar maximums and solar minimums along
the cycle. It doesn’t matter if every year is exactly correct; focus more on being consistent from
maximum to maximum and minimum to minimum.
2. Subtract the year of the previous maximum from each solar maximum year; do the same for
each solar minimum year.
Solar
Maximum
Year
Solar
Minimum
Year
1958
Difference
Between
Maxima
————
————
1954
Difference
Between
Minima
————
————
1969
11
1965
11
1980
11
1976
10
1990
10
1986
10
2000
10
1996
10
3. The period of the solar cycle should be the average of all the distances between the maxima
and minima. What is the average period?
The answers above give an average of 10.4 years.
4. More detailed analysis has shown that the Solar activity cycle is about 10.7 years long. How
does your period compare?
The period I got is only a little shorter. The solar cycle isn’t actually constant - it varies a
little bit from cycle to cycle. Astronomers haven’t yet figured out why.
5. Based on your period and the years of the last maximum and minimum, should this year be
closer to solar maximum or solar minimum?
My answers give a minimum around 2006 and a maximum around 2010. Based on those numbers, we should be in a solar maximum. However, this cycle seems to be longer than usual, so
we are still in a period of increasing solar activity.
15
4.2
Sunspots and Rotation
The two images below were taken two days apart. Due to the distinctive pattern of the sunspots, we
can use them to measure and calculate the Sun’s rotation period.
Image Scale
The first thing we need to do is figure out the scale of the image by measuring the size of the Sun in
the image and comparing it to the actual size of the Sun. That will allow us figure out how far the
sunspots are moving by how far they look like they are moving.
6. What is the diameter of the Sun in the image?
I measured about 6.7 cm
7. What is the radius of the Sun in the image?
Half of 6.7 cm is 3.35 cm
The actual radius of the sun is R = 6.96 × 108 m. What we are looking for is a scale factor,
which is:
actual radius
scale factor =
(6)
measured radius
Don’t worry too much about the units of your scale factor. In actuality, they are something like
meters in reality over centimeters in image. If this helps you, write it with your scale factor.
Otherwise, use it as a dimensionless number.
8. What is your scale factor?
6.96×108
3.35
real m
= 2.08 × 108 image
cm
16
Sunspot Motion
Choose one of the active regions near the Sun’s equator. Make sure that you can see it and recognize
it in both of the images.
9. Measure the distance from the left side of the Sun to the sunspot in both of the images and
write those measurements below.
In the first image, 2.5 cm. In the second image, 4 cm.
10. What is the difference between the two measurements (the amount the sunspot has moved)?
4 cm - 2.5 cm = 1.5 cm
11. Using the scale factor you found above, calculate the actual distance that the sunspot has
moved.
real m = 3.12 x 108 m
1.5 cm x 2.08 x 108 image
cm
The Sun’s Period
We can find the period of the Sun from the distance the sunspots have moved. We need to use the
ratio of:
total solar period
distance around the Sun
=
(7)
two days
distance traveled by a spot in two days
The distance around the Sun is its circumference: D = 2πR.
12. Calculate the Sun’s rotation period:
period =
2π6.96×108 m
3.12×108 m
× 2 days = 28 days
13. The Sun’s rotation period is roughly 27 days, as viewed from Earth. How does your period
compare?
The period I measured is just about equal to the 27 day period. If you have a different answer,
it could be due to the sunspot you measured not being near the equator, or due to a projection
effect - things appear to move slower when they are closer to the sides of the circle. That’s
because they are also moving away.
17
4.3
Coronal Mass Ejections
A Coronal Mass Ejection (CME) is an event where a mass of energetic particles are ejected from
the surface of the Sun due to interactions of magnetic fields with the surface. In this section of
the activity, we’ll figure out the velocity of a CME using images from the Solar And Heliospheric
Observatory (SOHO). We’ll also calculate how long the particles would take to reach the Earth.
Image Scale
The four panels of the image above show the CME at different points in time. The Sun is blocked in
the image - the white circle represents both the size and the position of the Sun. The first thing we
need to do is figure out the scale of the image, like we did in the previous section.
18
14. What is the diameter of the Sun in the image? What is the radius?
The diameter is about 1.3 cm, so the radius is about 0.65 cm.
15. Calculate a scale factor using the same equation as the previous section.
6.96×108
0.65
real m
= 1.07 × 109 image
cm
Measuring the Image
The first column gives the time the image was taken at (given in the lower left corner of each image)
and the second column gives the time between each image and the one before it, in seconds. You’ll
need to fill in the distance and velocity columns.
16. Select a clump of material that seems to be moving away from the star in each image. Measure
the distance (on the image) between that clump and the surface of the star. It’s ok if you can
only see the clump in three of the four images.
17. The number we need is how much the clump has moved between each image. Subtract the
distance of the clump in each image from the distance in the image before.
18. Convert your image distances to real distances using the scale factor your found above.
19. Divide your change in distance by the elapsed time to calculate velocity.
Image
Time
5:53
Time
Elapsed
(seconds)
————
————
1360
6:05
780
6:53
2820
5:30
Measured
Distance
0.9 cm
Measured
Change
Distance
————
————
1.7 cm
0.8 cm
In
Actual
Change
Distance
————
————
Velocity
In
8.56 x 108 m
8
2.4 cm
0.7 cm
7.49 x 10 m
————
————
————
————
————
————
————
————
6.29 x 105 m/s
9.60 x 105 m/s
————
————
You will likely have two velocities filled into the last column - the differences between the first, second
and third images. The CME has mostly moved out of the frame in the fourth image.
20. What is the average of the velocities?
7.95 x 105 m/s
21. If this CME was headed in the direction of Earth, how long would it take to reach Earth?
Convert your answer into days.
1 AU = 1.5 x 1011 m
1.5×1011 m
5
time = distance
speed = 7.95×105 m/s = 1.89 x 10 s
1 day
1hour
1.89 x 105 s x 1 60min
s × 60min × 24 hours = 2.18 days
19
Name:
5
KEY
H-R Diagram
Intoduction
Around the year 1910, both Ejnar Hertzsrung and Henry Norris Russel came up with the idea of arranging stars on a graph according to their color (temperature) and absolute brightness (luminosity).
The resulting diagram organizes stars in a logical way, and has proved so useful that astronomers
still use it today. We still call a graph that shows star’s luminosity as a function of temperature a
Hertzsprung-Russel (or H-R) Diagram.
One of the most important features of the H-R Diagram is the main sequence, which organizes
hydrogen burning stars by their mass. Stars that don’t fall along the main sequence are either still
forming or evolved, which means that they’ve burnt all the hydrogen in their cores and are now
radiating light from some other process. This activity goes through the reasons why main sequence
and evolved stars fall into certain regions of the H-R Diagram.
5.1
Mass and the Main Sequence
In class, we discussed the concept of gravitational equilibrium, which is the reason that the core of
the Sun is hot and dense. Gravity pulls each part of the Sun towards the center. To keep the Sun
from collapsing in on itself, the inside of the Sun pushes back with thermal pressure. This pressure
allows the Sun to maintain a very hot (T∼ 106 K) temperature in the core. The high temperature,
pressure, and density in the Sun’s core are necessary for fusion to occur.
Stars which are burning hydrogen into helium are arranged on the main sequence according to their
mass. The mass of a star determines its temperature and luminosity. In order to understand why
and how, we will estimate the properties of a star with 0.1 times the mass of the Sun and one with
10 times the mass of the Sun.
1. The star with 10 times the mass of the Sun has more matter for gravity to pull towards core.
Will the pressure pushing back against that gravity need to higher or lower than the pressure
in the center of the Sun? Why?
The pressure will be higher due to the increased amount of gravity pushing down onto the core.
2. What about the star with 0.1 times the mass of the Sun? With less material for gravity to pull
inward, will its core pressure and be lower or higher?
The core pressure will be lower due to the decreased amount of gravity pushing down.
20
An increase in the core pressure will result in an increase in the core temperature. Remember
that temperature is a measure of the average kinetic energy of particles in a substance. If
you put more pressure on that substance, the particles will move faster, increasing the kinetic
energy and temperature.
3. Of the three stars we’re thinking about (0.1, 1, and 10 times the mass of the Sun), which one
would have the highest core temperature? How about the lowest? Why?
The star with 0.1 times the mass of the Sun should have the lowest core temperature, while
the star with 10 times the mass of the Sun should have the highest core temperature.
Fusion is very sensitive to the conditions in the core of the star. The process of hydrogen
turning into helium is based on a series of collisions - each collision where the particles stick
together forms a new nucleus. The faster particles are moving, the more collisions will happen.
The more collisions that happen, the more energy fusion generates per time.
4. Based on the core temperatures that you figured out above, which star has the fastest (most
efficient) fusion? Which star has the slowest?
The star with the higher core temperature and larger mass should have faster fusion, while the
star with the lower core temperature and smaller mass should have slower fusion.
The rate of energy produced in the core is the same as the rate of energy released from the
surface of the star. The energy is released in the form of light, so the rate of energy production
is equal to the star’s luminosity. As we discussed, luminosity is energy per time, in units of
joules per second.
5. Which of the three stars has the highest luminosity? Which has the lowest?
The largest star has the highest luminosity, while the smallest star has the lowest.
Stars emit absorption spectra - which are thermal spectra that have some atomic absorption
over them. Right now we are only concerned with the thermal spectrum of the stars, which
depends on the temperature in the photosphere.That temperature is related to the amount of
energy that’s released by that layer.
6. Which star should be the bluest? Which star will be the reddest?
The higher luminosity star should have bluer light, while the lower luminosity star should emit
redder light.
21
7. To sum up your answers to questions 1-6, fill in the the blanks below:
The star with 10 times the mass of the Sun should have a
higher
luminosity than the
Sun. The temperature at its surface should be hotter and its color should be bluer .
The star with 0.1 times the mass of the Sun should have a
lower luminosity than the
Sun. The temperature at its surface should be cooler and its color should be redder .
5.2
The Main Sequence on an H-R Diagram
The table on the next page has the properties of 19 stars. Each of the stars are hydrogen burning
stars that should fall along the main sequence.
8. The stars are listed in the order of their distances. The apparent brightness of each star is not
given; which star should look the brightest when viewed form Earth?
The Sun. Without using the formula for apparent vs. absolute brightness (or flux vs. luminosity), we can see that the Sun is MUCH (106 times) closer so it should be MUCH brighter.
9. Fill in the last column of the table by converting the Luminosity given into scientific notation.
The graph on the page after the table has the correct axes to be an H-R diagram. The axes
on the graph are on what we call a “log scale,” which makes it easier to compare numbers that
span many powers of ten. On the y-axis, each evenly spaced tick mark represents a power of
10 in luminosity. On the x-axis, the tick marks are in increments of either 1000 or 10000. Note
that the temperature is backwards - low temperatures are on the right, and high temperatures
are on the left.
10. How many times larger is the largest luminosity on the graph than the smallest luminosity?
The largest luminosity is 106 and the smallest is 10−5 . So the largest luminosity is 1011 times
larger than the smallest - that’s 100 billion times larger.
11. Plot each data point in the table on the H-R diagram according to its temperature and luminosity.
12. Now that the data is on the graph, take a look at the spectral type column. Label each point
or group of points with the letter in the spectral type column.
13. The spectral types should form a sequence. From hottest temperature and highest luminosity
to the coolest, faintest stars, what order are the spectral types in?
OBAFGKM
22
Star Name
Distance
(Lyr)
Spectral
Type
Surface
Temp (K)
Sun
2.0 x 10−5
G
5800
Luminosity
(compared to
Sun)
1.0
Alpha Centauri A
4.3
G
5800
1.5
Alpha Centauri B
4.3
K
4200
0.33
Lalande 21185
8.1
M
3200
0.0055
Sirius A
8.7
B
10400
23.0
Luyten726-8 B
8.7
M
2700
0.00002
Ross 154
10.3
M
2800
0.00041
Ross 248
10.3
M
2700
0.00011
Epsilon Eridani
10.8
K
4500
0.3
61 Cygni B
11.1
K
3900
0.039
Procyon A
11.3
F
6500
7.3
Altair
16.5
A
8000
11.0
70 Ophiuchi A
17
G
5100
0.6
Vega
26
B
10700
55.0
Achernar
65
B
14000
200
Beta Centauri
300
B
21000
5000
Zeta Persei A
465
B
24000
16000
Delta Persei
590
B
17000
1300
Zeta Pupis
1090
O
39000
790000
23
Luminosity
(scientific
notation)
100
1.5 x100
3.3 x 10−1
5.5 x 10−3
2.3 x 101
2 x 10−5
4.1x 10−4
1.1 x 10−4
3 x 10−1
3.9 x 10−2
7.3 x 100
1.1 x 101
6 x 10−1
5.5 x 101
2 x 102
5 x 103
1.6 x 104
1.3 x 103
7.9 x 105
106
104
Luminosity
102
100
10−2
10−4
40000 30000
5.3
20000
10000
6000
Temerature (K)
4000
2000
The Lifetimes of Main Sequence Stars
Using the luminosity of stars, we can figure out their comparative lifetimes. Again, we will think
about the stars that are 0.1 times the mass of the Sun and 10 times the mass of the Sun.
14. The star with 10 times the mass of the Sun is a B star, and it should have a temperature
around T = 25000 K. According to the H-R diagram, how many times greater is its luminosity
compared to the Sun’s?
It should be about 104 times the luminosity of the Sun.
15. The luminosity of a star is the same as its rate of energy generation, and the rate of energy
generation is proportional to the mass lost to fusion. With 10 times as much mass as the Sun,
and the luminosity from the H-R Diagram, how many times longer or shorter than the Sun will
the massive B star be burning hydrogen?
It it 104 times as luminous, but only 10 times as massive. It should burn through its mass in
about 10−3 times as long as the Sun, or burn its mass 103 times faster.
24
16. The lifetime of the sun is approximately 1010 years. What is the approximate lifetime of the
massive B star?
Since the Sun has a lifetime to 1010 years and the more massive star will live 10−3 times as
long, it will have a lifetime of 107 years.
17. The star with 0.1 times the mass of the Sun is an M star, and should have a temperature
around T = 3000K. How does the luminosity of that star compare to the Sun’s luminosity?
It has a luminosity of 10−3 times the luminosity of the Sun.
18. Using a similar logic, how much faster or slower will the small M star burn its mass? What is
its approximate lifetime?
It will take a longer time. It will burn through its mass 102 times slower, and this will lead to
a lifetime of 1012 years.
19. To sum up, fill in the blanks with the answers from questions 14 to 19:
5.4
A star with 0.1 times the mass of the Sun is an
than the sun.
M
A star with 10 times the mass of the Sun is a
than the sun.
B
star. It will burn hydrogen for
longer
star. It will burn hydrogen for
shorter
Other Stars on the H-R Diagram
In addition to the stars on the main sequence, there are two other main types of stars whose temperatures and luminosities are not directly related to mass; these stars fall in different places on the
diagram.
Red Giants
Red giants are stars that have used up all the hydrogen in their cores. This has caused their cores
to collapse, heat up, and begin fusing helium into carbon. We will talk more aout this in the next
class, but for now all you need to know is that the cores of red giants produce more energy per time
than the fusion of hydrogen into helium.
20. When a star turns into a red giant, how should its luminosity compare to the luminosity it had
while burning hydrogen?
It should have a larger luminosity than it did on the main sequence.
25
For stars along the main sequence, their temperatures and luminosities follow a set relation.
That’s not true for red giants because of their size; red giants are enormous. Because they are
so big, the material at the surface is very cool. The energy coming out of the star is spread
over a larger surface doesn’t heat up the outer layer as much.
21. Where (compared to the main sequence) should red giants fall if they were shown on the H-R
diagram?
They should fall in the upper right portion, because they are both cooler and more luminous.
Below is data for 8 red giant stars.
22. Rewrite the luminosities of the red giants in scientific notation.
23. On the next page is a blank H-R diagram. Plot each red giant on it according to its temperature
and luminosity.
24. Draw a line on your new H-R diagram that goes through the space where the main sequence
was in the previous H-R diagram.
Star Name
Distance
(Lyr)
Surface
Temp (K)
Arcturus
36
4500
Luminosity
(compared to
Sun)
110
Capella
47
5900
170
Alderbaran
53
4200
100
Canopus
100
7400
1500
Antares
400
3400
5000
Betelgeuse
500
3200
17000
Rigel
800
11800
40000
Delta Aquarii B
1030
6000
4300
26
Luminosity
(scientific
notation)
1.1 x 102
1.7 x 102
1 x 102
1.5 x 103
5 x 103
1.7 x 104
4 x 104
4.3 x 103
106
104
Luminosity
102
100
10−2
10−4
40000 30000
20000
10000
6000
Temerature (K)
4000
2000
White Dwarfs
White dwarfs are the last type of star we’ll go through today. They are stars that are no longer
burning anything; they are small and dense and the energy that they radiate comes from the slow
process of cooling over the course of billions of years.
25. The cooling process that causes white dwarfs to radiate releases very little energy per unit time.
How should their luminosities compare to those of main sequence stars?
They should be less luminous than the main sequence.
26. While white dwarfs are radiating very little energy, they are very small so their outer layers
are hot. What color should they be, compared to larger stars generating the same amount of
energy?
They should be bluer, because their surfaces are hotter.
27
In the table below are 5 white dwarf stars.
27. Convert their luminosities into scientific notation.
28. Plot each white dwarf on the H-R diagram according to their temperature and luminosity.
Star Name
Distance
(Lyr)
Surface
Temp (K)
Sirius B
8.7
10700
Luminosity
(compared to
Sun)
0.0024
Procyon B
11.3
7400
0.00055
Van Maanen’s Star
14.0
7500
0.00016
GJ 440
15.0
8500
0.0005
HU Aquarii B
587
12500
0.0022
Luminosity
(scientific
notation)
2.4 x 10−3
5.5 x 10−4
1.6 x 10−4
5 x 10−4
2.2 x 10−3
You’ve now created a complete H-R diagram! We will learn more about the processes that happen
when stars leave the main sequence to become red giants and white dwarfs during the next class.
28
Name:
6
6.1
KEY
Post Main Sequence Activity
Red Giant Evolution
The flow chart below is designed to show the evolution of stars after the main sequence until they
become white dwarfs, neutron stars, or black holes. Some of the arrows point directly at the next
box, while others split the stars into a lower mass and a higher mass group. Be careful to follow the
correct arrow when you fill in each box.
29
Two H-R diagrams are below, each with the main sequence drawn in for reference. On the top
diagram, draw and label the post main sequence evolutionary path of the Sun. On the bottom
diagram, draw and label the post main sequence evolutionary path of a 10 solar mass star.
106
104
Luminosity
102
100
10−2
10−4
40000 30000
20000
10000
6000
Temerature (K)
4000
2000
20000
10000
6000
Temerature (K)
4000
2000
106
104
Luminosity
102
100
10−2
10−4
40000 30000
30
6.2
Planetary Nebulae
Below is a Hubble Space Telescope image of the Helix Nebula. We will use the image, and a few
other facts, to determine the age of the nebula.
Note that the scale bar shows both actual (light-years and parsecs) and angular (arcminutes) size.
In order to show both, they had to figure out a distance to the Nebula, which is about 200 parsecs.
The first thing we want to use the scale along the bottom of the image to determine the size of the
planetary nebula.
1. How many cm long is the scale bar?
5.9 cm
2. How many cm long is the Helix Nebula, along its widest part?
8.4 cm
31
To find the physical size of the nebula, you divide its size on the page by the length of the scale
bar on the page, then multiply by the actual size of the scale bar in parsecs.
3. What is the diameter of the nebula? What is its radius?
8.4 cm ×
0.6 parsecs
5.9 cm
= 0.85 parsecs
r = 0.43 parsecs
4. What is the radius of the nebula in km?
1 parsec = 3.09 ×1013 km
r = 0.43 parsecs ×
3.09×1013 km
1 parsec
= 1.32 × 1013 km
You take a spectrum of the center of the Helix Nebula and notice that the hydrogen line you
expected to be at 656.3 nm is actually at 656.2 nm. You remember the formula for doppler
shift is:
v =c×
λrest − λshif t
λrest
(8)
and that c = 3 ×105 km/s.
5. That shift comes from material that is moving away from the center of the Nebula. What is
that material’s velocity?
v=
656.3
nm−656.2
656.3 nm
nm × 3 × 105 km/s = 46 km/s
We can calculate the age of the nebula by figuring out how long it has been expanding - the
radius of the nebula divided by the current velocity of the material that’s moving out.
6. What is the age of the Helix Nebula, in years?
age =
1.32×1013 km
46 km/s
= 2.87 × 1011 s ×
1
min ×
s
60
32
1 hr
60 min
×
1 day
24 hr
×
1 year
365 days
= 9157 years
6.3
Density and Escape Velocity of Stellar Remnants
For the following questions, you will need the equation for density:
density =
mass
M
= 4 3
volume
πR
3
(9)
And the equation for escape velocity:
s
v=
2GM
R
(10)
And the following numbers and constants:
mass of Sun = 2 ×1030 kg
radius of Sun = 6.95 ×105 km = 6.95 ×108 m
radius of Earth = 6.4 ×103 km = 6.4 ×106 m
m3
G = 6.67 ×10−11 kgs
2
9. What are the density and escape velocity of the Sun?
density =
s
kg = 1422 kg
m3
m)3
2×1030
4
π(6.95×108
3
m3 ×2×1030 kg
kg s2
= 6.2 × 105 m/s
6.95×108 m
2×6.67×10−11
v=
10. What are the density and escape velocity of a one solar mass white dwarf?
density =
s
2×1030
kg = 1 × 1010 kg
m3
m)3
4
π(6.4×106
3
m3 ×2×1030 kg
kg s2
= 6.5 × 106 m/s
6.4×106 m
2×6.67×10−11
v=
11. What are the density and escape velocity of a one solar mass Neutron star?
density =
s
v=
kg
18 kg
m)3 = 3 × 10 m3
2×1030
4
π(5000
3
m3 ×2×1030 kg
kg s2
= 2.3 × 108 m/s
5000 m
2×6.67×10−11
Take home point: The densities of white dwarfs and neutron stars are huge, and because
they are small and massive, it is very hard for things to leave their surfaces. Not as hard as it
would be to leave a black hole, though!
33
Name:
7
7.1
KEY
Milky Way Activity
Structure of the Galaxy
1. Dwarf a basic diagram of the Milky Way Galaxy face-on. Label the disk, bulge, bar, and spiral
arms.
2. Draw a basic diagram of the Milky Way Galaxy edge-on. Label the disk, bulge, halo, and
globular clusters.
34
7.2
The Star-Gas Cycle
The diagram below is a version of the star gas cycle presented in class. For each of the arrows, fill in
the process that moves the cycle from one step to the next.
5. atomic hydrogen to molecular hydrogen
Warm atomic hydrogen gas gradually cools and becomes more dense to form molecular hydrogen
clouds.
6. molecular clouds to star formation
Pockets of the molecular clouds collapse due to their own gravity to form protostellar clouds.
These protostellar clouds starts to form stars.
7. star formation to stellar fusion
The stars ignite fusion in their cores. After their main sequence lifetime, they start to burn
heavier elements in their cores.
8. stellar fusion to returning gas
During the late stages of stars’ lifetimes, their outer layers expand, carrying with them enriched materials. Higher mass stars can also have dramatic supernova, which return gas to the
interstellar medium.
9. returning gas to hot bubbles
When many supernovae happen in a young cluster, this can blow a bubble of hot gas that
breaks out of the disk and releases hot gas into the halo of the galaxy.
10. hot bubbles to atomic hydrogen
The hot gas high above the plane of the galaxy gradually cools until it is the warm gas, closer
to the plane.
35
7.3
Unseen Parts of the Milky Way
For this part of the activity, you’ll need the equation for the mass enclosed within an orbit with the
speed v and radius r, which we will derive on the board. Write the equation here:
M=
r×v 2
G
You will need the following constants:
3
G = 6.67 × 10−11 m 2
kg s
Msun = 2 × 1030 kg
Supermassive Black Hole
Suppose you observe a star orbiting the galactic center at a speed of 106 m
s in a circular orbit with a
radius of 20 light-days (5.18 × 1014 m).
11. Calculate the mass of the object in kilograms, based on the equation given in class.
M=
r×v 2
G
=
2
m×(106 m
s ) = 7.77 × 1036 kg
3
6.67×10−11 m 2
kg s
5.18×1014
12. Convert the mass into units of solar masses.
7.77 × 1036 kg ×
Msun
6
kg = 3.89 × 10 Msun
2×1030
13. Compare the mass and radius of the object in the center of the galaxy to a globular cluster.
Globular clusters have around 105 stars in a region that’s about 50 light-years (4.7×1017 m) in
radius.
The radius of the object has to be less than 20 light-days, because a star is orbiting around it
at that distance. It also must be more massive than the globular cluster, because its mass is
10 to one hundred times larger. Because it is more massive and much smaller than a globular
cluster, the object in the center of the galaxy is unlikely to be made of stars.
36
Dark Matter Halo
The Large Magallenic Clouds is a small galaxy that orbits the Milky Way. It is currently orbitting
at roughly 160000 light-years (1.5 × 1021 m) from the galactic center at a speed of 3 × 105 m
s.
14. Calculate the mass within the radius of the LMC’s orbit in kilograms.
M=
r×v 2
G
=
2
m×(3×105 m
s ) = 2.04 × 1042 kg
3
6.67×10−11 m 2
kg s
1.51×1021
15. Convert the mass into units of solar masses.
2.04 × 1042 kg ×
Msun
12
kg = 1.02 × 10 Msun
2×1030
16. The number of stars in the galaxy should be something more like 2 × 1011 . Does this match
the mass you calculated? What are some reasons the two answers might not match?
We calculated a mass of about 1012 Msun , which is about 5 times more than the number of
stars in the galaxy, if we assume that each star is the mass of the Sun. Because most stars are
less massive than the Sun, our answer is more like 10 times as much mass as there are stars.
One source of error in our answer is that the equation assumes that the orbit of the LMC is
circular, and it shouldn’t be exactly circular. But that is likely to be a small effect - our answer
actually indicates the presence of matter that’s not in the form of stars. That matter could be
dust, gas, or dark matter.
37
Name:
8
8.1
KEY
Dark Matter Activity
Measuring a Rotation Curve
The graph below shows the observed rotation curve of a galaxy. The x-axis shows the distance from
the center of the galaxy, and the y-axis shows the velocity of the material at each of those distances.
1. What measurements did we need in order to calculate the velocities? How does the redshift of
the galaxy affect those measurements?
In order to calculate velocities, we need to take a spectrum of the galaxy at each radius, and
compare the wavelength of the lines to the rest wavelength. This is complicated by the redshift
because a galaxy will also have shifted lines due to its cosmological redshift.
2. Can you measure the velocities of material in a face-on galaxy? Why or why not?
In order to measure the shifts of lines, we need to measure a galaxy that is somewhat edge on.
In a face-on galaxy, the stars will orbit on the plane of the sky, and not have a redshift.
3. How were the distances from the center of the galaxy measured and calculated?
In order to calculate the distance from the center of the galaxy, we need to measure the angular
distance from the center of the galaxy in an image, and then multiply by the galaxy’s distance
to get a real distance from the center of the galaxy.
Velocity (km/s)
200
100
0
−100
−200
−10
−8
−6
−4
−2
0
2
Radius (kpc)
38
4
6
8
10
150
Velocity (km/s)
100
50
0
−50
−100
−150
−12 −10
−8
−6
−4
−2
0
2
Radius (kpc)
4
6
8
10
12
6•1010
4•1010
3•1010
2•1010
1•1010
0
0
1
2
3
4
5
6
Radius (kpc)
7
8
9
10
11
1
2
3
4
5
6
Radius (kpc)
7
8
9
10
11
0.8
Fraction of "Normal" Mass
Mass (Msun)
5•1010
0.6
0.4
0.2
0.0
0
39
8.2
Calculating Mass From a Rotation Curve
The table below shows data for a rotation curve similar to the one shown on the first page. The only
difference is that we are working with one half of the curve, which should give us similar results for
the other side.
5. Plot the velocities onto the rotation curve graph (velocity vs. radius) on the previous page.
6. Convert the radii from kpc to meters: 1 kpc = 3.08 ×1019 m
7. Convert the velocities from km/s to m/s: 1 km/s = 103 m/s
8. Calculate the mass enclosed in each radius. The equation below is the one we’ve used before to
determine how much mass is inside an orbit. You’ll need the universal gravitational constant:
3
G = 6.67 × 10−11 m 2
kg s
r × v2
M=
(11)
G
9. Convert the masses from kg to solar masses: Msun = 2 × 1030 kg
10. Plot the masses onto the mass curve graph (mass vs. radius) on the previous page.
r (kpc)
radius (m)
V (km/s)
Velocity (m/s)
Mass (kg)
Mass (Msun )
0
0
0
0
0
0
1
3.08 × 1019
50
5 × 104
1.15 × 1039
5.75 × 108
2
6.16 × 1019
80
8 × 104
5.91 × 1039
2.96 × 109
3
9.24 × 1019
110
1.1 × 105
1.68 × 1040
8.38 × 109
4.5
1.39 × 1020
130
1.3 × 105
3.52 × 1040
1.76 × 1010
6
1.85 × 1020
140
1.4 × 105
5.44 × 1040
2.72 × 1010
7.5
2.31 × 1020
150
1.5 × 105
7.79 × 1040
3.90 × 1010
9
2.77 × 1020
150
1.5 × 105
9.35 × 1040
4.68 × 1010
10.5
3.23 × 1020
150
1.5 × 105
1.09 × 1041
5.45 × 1010
Each of the masses that you calculate is the mass inside the radius that you are using. So the mass
inside the radius of 3 kpc includes all the mass inside 2 kpc, plus the mass between 2 kpc and 3 kpc.
So it’s not surprising that the mass increases at every radius, what’s important is how it increases
compared to the observed mass in stars.
40
8.3
Luminous Mass
The visible mass in a galaxy includes the gas, dust, and stars in that galaxy. The stars are the main
contributor to the light that we see. Sometimes we call this “normal” matter because it is made up
of particles that we understand. The dark matter that we infer exists is made up of different stuff
entirely.
The graph below shows the light emitted from the galaxy, in terms of the luminosity of the Sun. It’s
similar to the mass plots, in that the light shown at each radius is all of the light inside that radius
added up. We can use the numbers on the graph to estimate the amount of luminous, or normal,
matter in the galaxy. You’ll need those numbers for the questions on the next page.
5•109
Light emitted (Lsun)
4•109
3•109
2•109
1•109
0
0
1
2
3
4
5
6
Radius (kpc)
7
8
9
10
11
We’ll also need to estimate the mass in stars that is generating the amount of luminosity that we
detect. It’s a pretty safe bet to estimate that for every unit of solar luminosity, there are two units
of solar mass.
11. Why is that a better estimate than assuming one solar mass for each solar luminosity?
This is a really rough estimate, but there’s some logic to it. Most stars are smaller than the
Sun, and those stars are a lot less luminous than the Sun is. We estimate that those stars are
producing less luminosity, which makes us estimate more mass for each unit of luminosity.
41
12. Fill in the table below with the total Mass you calculated based on the rotation curve.
13. Use the graph on the previous page to estimate the light that we observe within each radius.
14. Calculate the amount of luminous mass from the light emitted.
15. Add that amount of luminous mass to the graph of mass as a function of radius. Use different
symbols, or a different color, to distinguish it from the total mass.
16. Calculate the fraction of normal matter in the galaxy - divide luminous mass by total mass.
17. Plot the fraction of normal matter on the last graph.
D (kpc)
Mass (Msun )
Light (Lsun )
luminous M (Msun )
Fraction of “Normal” M
0
0
0
0
0
1
5.75 × 108
2.5 × 108
5.0 × 108
0.87
2
2.96 × 109
6.0 × 108
1.2 × 109
0.42
3
8.38 × 109
1.2 × 109
2.4 × 109
0.29
4.5
1.76 × 1010
2.1 × 109
4.2 × 109
0.24
6
2.72 × 1010
2.9 × 109
5.8 × 109
0.21
7.5
3.90 × 1010
3.9 × 109
7.8 × 109
0.2
9
4.68 × 1010
4.1 × 109
8.2 × 109
0.17
10.5
5.45 × 1010
4.6 × 109
9.2 × 109
0.17
18. The outside few points on your rotation curve show a constant velocity. Why does the mass
plot show mass increasing even through the rotation curve is flat?
If the mass stayed the same, the rotation curve wouldn’t be flat. The rotation curve would
actually dip, decreasing as the radius increases. Because the curve is flat, we know that there’s
more and more mass at farther radii.
19. Where is the largest difference between the mass in stars and the mass in dark matter?
Near the outside of the galaxy. There are very few stars at outer radii, but there is still a lot
of dark matter.
20. Where is most of the dark matter located? How can you tell?
The dark matter is mostly located outside the center of the galaxy. While there is dark matter
at the inner parts of the galaxy, it is much less than the amount of luminous matter. We can
tell there is dark matter at the edges of the galaxy because there is more mass that we can’t
see.
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8.4
Mass in a Galaxy Cluster
Now we will determine the mass of a galaxy cluster through its temperature. In lecture, we discussed
that the temperature of hot gas in a cluster was related to the average velocities of the particles in
the gas. The specific equation is:
√
m
vparticle = 140 √ × T
(12)
s K
A cluster of galaxies has a radius of about 5.1 million light-years and a temperature of 6×107 K. The
observed light of all the stars in the cluster is 8 ×1012 Lsun .
one light-year = 9.46 × 1015 m
21. What is the average particle speed of gas in the cluster?
vparticle = 140 m
s ×
√
T = 140 m
s ×
√
6 × 107 K = 1.08 × 106 m
s
22. Using the outer radius of the cluster and the particle speed for the velocity of material, what
is the mass of the cluster in terms of the mass of the Sun?
M=
r×v 2
G
5.1×106
m
2
×(1.08×106 m
s)
light-year
= 8.44 × 1044 kg
3
6.67×10−11 m 2
kg s
light-years×9.46×1015
=
M = 8.44 × 1044 kg ×
1Msun
2×1030 kg
= 4.22 × 1014 Msun
23. We often talk about the mass to light ratio, which is the number of solar masses over the
number of solar luminosities. What is the mass to light ratio of the cluster?
mass-to-light =
M
L
=
4.22×1014 Msun
8×1012 Lsun
= 53
24. Calculate the ratio of the mass of luminous matter to dark matter. How does the amount of
dark matter in the cluster compare to the amount of dark matter in the galaxy we worked with
before?
luminous mass = L × 2 = 1.6 × 1013 Msun
fraction of normal mass =
1.6×1013 Msun
4.22×1014 Msun
= 0.0379
This is about 20 times less than the ratio we found for the galaxy we were working with. There’s
some dark matter in galaxies, but there’s even more in clusters of galaxies.
43
Name:
9
9.1
KEY
Cosmology
Accelerating Expansion
Below is a graph similar to the one that you constructed in the Hubble Law Lab. There are no units
on it, but it extends to farther distances than can be measured using the sizes of galaxies as standard
rods.
1. What standard candle allows us to find distances to the farthest galaxies?
The white dwarf supernovae are the best standard candle. They are both very bright and have
a consistent luminosity.
2. What measurements would you need in order to construct the graph below?
Velocity
We need to measure the brightness of supernovae in order to calculate a distance, and we also
need to measure a redshift of spectral lines in order to get a velocity.
Distance
44
The solid line on the graph on the previous page shows how velocity and distance are related through
the Hubble constant. In the local universe, the points for each galaxy fall neatly along the line.
3. At a given distance, where do the farthest galaxies fall compared to the Hubble constant line?
The most distant galaxies fall under the Hubble constant line, at a smaller velocity for a given
distance.
4. The more distant galaxies emitted the light that we see now longer ago, so they trace the way
the universe was expanding at that time. Was the expansion in the past larger or smaller than
it is today?
The expansion in the past was smaller, because the velocities are each smaller.
5. Has the rate of expansion increased or decreased over time?
If the expansion was smaller in the past than it is today, then it must have increased over time.
Velocity
6. We use graphs similar to this one to infer that the universe is accelerating in its expansion. In
the graph below, draw and label three lines that represent constant expansion, accelerating
expansion, and deccelerating expansion.
Distance
45
9.2
Eras in the Early Universe
As lecture progresses through each of these eras, fill in the duration, average temperature, and a
description of each one.
Era Name
Planck Era
GUT Era
Electroweak Era
Particle Era
Era of Nucleosynthesis
Era of Nuclei
Era of Atoms
Era of Galaxies
Duration
Temperature
Description
10−43 s
1032 K
Physics was incomprehensible; the strong,
weak, electromagnetic, and gravitational
forces acted as one.
10−38 s
1029 K
The electromagnetic and weak forces separate from the strong force. This releases
the energy that causes inflation.
10−10 s
1015 K
Energy to randomly created particles and
anti-particles. Ends when electromagnetic
and weak forces separate.
0.001 s
1012 K
Energy continued to spontaneously generate particles and anti-particles through the
end of the particle era.
5 min
109 K
The fusion of hydrogen was balanced by
the destruction of helium. This era determined the abundance of helium today.
3.8 × 105 yrs
3000 K
The universe was a hot, ionized plasma
where photons bounced between particles,
never traveling very far.
109 year
18 K
The universe consists mostly of atoms and
photons. Protogalactic clouds form.
still going
3K
Galaxies and clusters of galaxies dominate
the universe on large scales.
7. What is the basic argument for the universe having been smaller, denser, and hotter in the
past?
We can see the universe expanding at the current time - it is currently getting larger, less dense
and cooler. If we look backward, it must have been smaller, dense, and hotter becuase we have
expanded since that time.
8. Inflation happened during which era?
At the end of the GUT era.
9. Between which two eras was the radiation that now forms the cosmic microwave background
emitted?
Between the Era of Nuclei and the Era of Atoms. That’s when the universe became diffuse
enough for photons.
46
10. When were the last particles and anti-particles created from energy?
At the end of the Particle Era. After that, the universe was too cold for energy to spontaneously
create particles.
9.3
The Cosmic Microwave Background
The Cosmic Microwave Background (CMB) is now observed to have a thermal spectrum that corresponds to a temperature of T = 2.73 K. The law that relates peak wavelength of a thermal radiation
spectrum (λpeak ) to the temperature of the object emitting that spectrum (T) is known as Wein’s
Law:
2.9 × 106 nm K
T
Which we’ve used before to determine the peak wavelength of stars.
λpeak =
(13)
11. What is the peak wavelength of the CMB today?
λpeak =
2.9×106 nm
2.73 K
K = 1.06 × 106 nm ×
1 mm
106 nm
= 1.06 mm
12. What was the peak wavelength of the CMB when it was emitted (refer to the table you filled
in on the previous page)?
λpeak =
2.9×106 nm
3000 K
K = 967 nm
13. The stretching of the wavelengths is proportional to the stretching of the universe over the
time between the emission of the light and our observation of the light today. How many times
bigger is the universe today than when the CMB was emitted?
λ
stretch of universe = λ now =
past
1.06×106 nm
967 nm
= 1097 times larger today
14. What do the over densities in the map of the CMB have to do with the galaxies and clusters
that we see today? How do those relate to the quantum fluctuations in the first 10−36 seconds
of the universe?
The over densities in the CMB have their origin in the pre-inflation universe. On the small
scales that existed then, the fluctuations that cause the CMB to be different temperatures were
quantum fluctuations on the scale of tiny particles. Those same fluctuations in the density of
matter and energy then are the same over densities that are the size of galaxies and clusters of
galaxies today.
47
Name:
10
KEY
Drake Equation Activity
How common is life in the Milky Way Galaxy?
This activity focuses on the Drake Equation, which was put together to calculate the chance of
detecting other civilizations in our Galaxy. As we talk about each factor, put an estimate of the
number in the box next to it. At the end, multiply all the numbers for your own estimate.
R
rate of star formation
5 per year
fplanets
nhabitible
flif e
fintellegence
ftechnology
L
N
fraction of stars that have
planets
0.2
average number habitable
planets per star with planets
1
fraction of habitable planets
that develop life
1
fraction of planets with life
that develop intelligent life
0.002
fraction of civilizations that
develop detectable technology
0.1
the length of time for which
civilizations continue to be
detectable
1,000 years
the number of civilizations
that we could detect
0.2
My answer isn’t necessarily “right,” there are many possible answers, and none yet confirmed.
48