Name: KEY

1 Math Activity

Intoduction

This activity is meant to give you a review of math that will be both useful and important inAstronomy 101. If this is not a review and you find it difficult, please stop by my office hours, orstudy 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 smallnumbers. 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 thesun, the age of the galaxy, the distance between the Earth and the Sun, and the distancesbetween galaxies.

2. What are some small objects that you saw in the powers of 10 video? Which ones are relevantto astronomy?Some answers include the size of atoms and molecules, the wavelengths of light, and the sizesof 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

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Multiplication and Division

There is a slightly different method we might use for multiplying and dividing numbers writtenin scientific notation. I’ll work out an example or two on the board, and then ask you to finishthe 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 2.4 ×10−8

1.2 Units

Units are very important in any calculation. If you are doing homework for this class and don’thave 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 anotherset of units entirely. Sometimes you need to do these in order to move on to the next step of acalculation, other times its necessary in order to make the number make more sense to you.

19. Convert 2 ×107 cm to km2 ×107 cm × 1m

1×102cm× 1km

1×103m= 2 ×102 km = 200 km

20. Convert 789 days to years798 days × 1year

365days= 2.16 years

21. Convert 1.5 ×10−3 days to seconds1.5 ×10−3 days ×24hours

1day×60min

1hour× 60s

1min= 130 s

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1.3 Common Sense and Math

An essential part of math is applying some reasoning to the problems you are answering tomake sure your answer makes sense. Is it expressed in the appropriate units, with a reasonablenumber 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 wouldcertainly give you an answer with the appropriate amount of digits. We don’t need to be thatexact in Astronomy 101, but need to apply common sense. Fill in the columns below with theanswers to the math problem in the left column.

math calculator output answer

1000 / 3 333.3333333 333

π × 2 6.283185307 6.28

√2 1.414213562 1.41

6×1024 / 7 ×10178571428.571 8.6 ×106

log(99) 1.995635195 2

1.3.2 Checking The Size of Things

Match each distance or size to its approximate length in centimeters (on the right) and a moreappropriate unit to express it in.

3.9 ×108 cm PPPPPPPPPP

height of a human QQQQQQQQQQQQQQ

angstrom (A)

9.3 ×1018 cm QQQQQQQQQQ

distance from here to NYCSSSSSSSSSSSSSS

kiloparsec (kpc)

1.54 ×1023 cm QQQQQQQQQQQ

radius of an atom ����

���

���

��

parsec (pc)

1 ×10−8 cm ������

������

distance to a nearby star ����

������

�

meter (m)

1.5 ×1013 cm XXXXXXXXXXX

diameter of galaxy ###########

astronomical unit (AU)

2 ×102 cm ������������������

distance to the sun���

����

�

kilometer (km)

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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/sThe distance from the Sun to the Earth is 1 AU =1.50 ×1011m.

time = distancespeed

= 1.50×1011m3.00×108m/s

= 500s

If you wanted to know this in minutes, you could use a conversion factor:500s ×1min

60s= 8 minutes

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:3.844 ×105 km ×1×103m

1km= 3.844 ×105 km

Now use the same method as the previous problem:time = distance

speed= 3.844×105m

3.00×108m/s= 1.28s

24. How fast are we traveling around the galaxy?We are located 28,000 Ly from the center of the GalaxyWe orbit the galactic center once every 230 million yearsOne Ly is 9.46 ×1015 mThe distance we travel is related to the radius: d = 2πr

First, we’ll convert light-years into meters:2.8 ×104 Ly ×9.46×1015m

1Ly= 2.6488 ×1020 m

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:2.3 ×108 yr = 365days

1yr× 24hr

1day× 60min

1hr× 60s

1min= 7.2533 ×1015 s

speed = distancetime

= 1.664×1021m7.2533×1015s

= 229412.819

Rounded to two significant digits, the answer is: 2.3 ×105 m/sThis is the same as 510,000 miles per hour - fast!

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Name: KEY

2 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 arelabeled 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 ellipsedivided by the semi-major axis.” Another way of saying that is that it’s a measure of how squashedan 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 thecenter would be zero)?A circle!

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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 theselaws 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 inyour own words.

Kepler’s Laws

• First Law: The orbit of each planet around the Sun is an ellipse with the Sun at onefocus.This one is in pretty plain language. The important thing to remember is that the Sun isnot at the center of the ellipse, but at the focus. That means a planet on a highly eccentricorbit 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 equaltimes.This means that when the planet is closer to the Sun, it moves faster than it does whenit is far away. This also means that a planet with a highly eccentric orbit will move at avariety of speeds, while a planet with a nearly-circular orbit will only vary a little in itsspeed.

• Third Law: The square of the orbital period of a planet is directly proportional to thecube 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 amountof time it takes for the planet to complete an orbit. So this law is saying that the farthera planet is from the Sun, the longer it will take to orbit. Planets follow this with therelation 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’tworry 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 itwould be moving the slowest.

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Kepler’s Third Law

5. In the table below, “P” stands for period, or number of years it takes the planets to go aroundthe sun, and “a” stands for semi-major axis, which measures how far the planet is from thesun. The middle column is the calculated square of the period and the cube of the semi-majoraxis. Fill in the blank spaces; you can round off any decimal places.

P (years) P 2 = a3 a (AU)

1 1 1

8 64 4

27 729 9

64 4096 16

90 8100 20

125 15625 25

6. On the axes below, graph P and a from the table that you filled in. Draw a line through thepoints that you graph. Note that the axes are on different scales.

0 5 10 15 20 25semi−major axis (AU)

0

20

40

60

80

100

120

Pe

rio

d (

ye

ars

)

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.

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Name: KEY

3 Light Activity

Intoduction

This activity goes through some of the important properties of light that we need to know about inorder 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 troughsof a wave.

2. What is frequency?Frequency is the number of peaks (or troughs) that are passing a point in a unit of time (likea second).

Wavelength and frequency are inversely proportional, which means that if one increases, theother decreases. We write their relationship like this:

λ =c

for f =

c

λ(1)

Where f is the frequency, λ is the wavelength, and c is the speed of light (c = 3× 108ms )

3. Below, draw a wave with a wavelength of 3 cm.

0 1 2 3 4 5 6 7 8 9 10 11 12

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× 108ms ×

100 cm1 m = 3.0× 1010 cm

s

f = cλ

=3.0×1010 cm

s3 cm = 1010 1

s = 1010Hz

Energy is proportional to frequency, which we write as:

E = h× f (2)

where E is energy, f is frequency, and h is Plank’s Constant (h = 6.626 × 10−34m2kgs ). One

way to think about this is that a wave with a shorter wavelength will peak more frequently,which will carry more energy.

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5. Calculate the Energy of the wave you drew:

E = h× f = h = 6.626× 10−34m2kgs × 1010 1

s = 6.626× 10−24m2kgs2 = 6.626× 10−24Joule

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 informationbelow as part of a lecture on the electromagnetic spectrum. Please fill out the columns tocharacterize the types of light.

Type of Light Wavelength Size λ (m) Energy(eV)

Source on Earth Frequency(Hz)

gamma ray smaller than anatom

10 106 radioactive ele-ments

1020

X-ray thickness of aDNA strand

10−9 103 X-ray machine 1018

ultraviolet groove in a CD 10−7 10 The sun - itburns!

1016

Visible Light red blood cell orbacterium

10−6 1 fire, light bulbs 1014

Infrared width of a humanhair

10−4 10−1 stovetop, people 1013

microwave coffee bean 10−2 10−4 radar, microwaveoven

1010

radio city bus 10 10−7 Radio stationtransmitter

108

3.2 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 otherstars, planets, or even human beings. On the next page is the thermal emission spectrum froman object with a temperature of T = 6000K, like our sun.

200 400 600 800 1000Wavelength (nm)

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

Rela

tive Inte

nsity

9

The temperature affects the emission in two ways: the amount of light emitted, and its peakwavelength.

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 objectspeak at shorter wavelengths and cooler objects peak at longer wavelengths. This means thathot objects are bluer, and cold objects are red.

Amount of Light

Additionally, cooler objects emit less light than hotter objects at every wavelength. For exam-ple, this is how the thermal radiation spectrum of a T = 5000K (in grey) object would compareto the T = 6000K object (in black):

200 400 600 800 1000Wavelength (nm)

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

Re

lative

In

ten

sity

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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

T=

2.9× 106 nm K

8000 K= 362.5 nm (shown in blue below) (4)

7. Calculate the peak wavelength and draw in the thermal emission spectra a T = 4000K object:

λmax =2.9× 106 nm K

T=

2.9× 106 nm K

4000 K= 725.0 nm (shown in red below) (5)

200 300 400 500 600 700 800 900Wavelength (nm)

0

2

4

6

8

10

12

14

Rela

tive Inte

nsity

3.3 Atomic Spectra

At this point, we’ll go back into lecture mode to learn about atoms and their emission/absorbtionspectra.

8. In the space below, draw an energy level diagram. Show what happens when an atom absorbslight, 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 lampsand the lightbulb, first look at the color of the light without a diffraction grating. Then holdthe grating close to your eye and look to the side of the light. You should see spectra to the leftand the right - the one on the right should line up with the colors listed on the spaces providedfor drawing.

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White light: The normal lightbulb is on the top of the tower.

9. What color is the light without the diffraction grating? white

400 450 500 550 600 650 700 750Wavelength (nm)

violet blue green yellow red

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

400 450 500 550 600 650 700 750Wavelength (nm)

violet blue green yellow red

Helium lamp: The atomic helium lamp is second from the top lamp.

11. What color is the light without the diffraction grating? white

400 450 500 550 600 650 700 750Wavelength (nm)

violet blue green yellow red

Mercury lamp: The Mercury (Hg) lamp is just above the bottom lamp.

12. What color is the light without the diffraction grating? blue

400 450 500 550 600 650 700 750Wavelength (nm)

violet blue green yellow red

Neon lamp: The Neon lamp is on the bottom of the tower.

13. What color is the light without the diffraction grating? orange

400 450 500 550 600 650 700 750Wavelength (nm)

violet blue green yellow red

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3.4 Emission and Absorption

The last step in understanding light in astronomy is figuring out how the thermal radiationspectrum, from hot dense objects, and the atomic emission spectrum, from diffuse matter, gotogether.

14. In the space below, draw a hot dense object and a gas cloud. Draw and label three observerswho 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 nearbyblack arrow. The red and blue2 arrows show that each of the objects emit light in all directions.

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Name: KEY

4 Solar Activity

Intoduction

Solar activity is a blanket term for all the different kinds of variations in the Sun that we can observefrom Earth. Most solar activity is caused by magnetic fields on the Sun. There is a large magneticfield that runs through the entire Sun and also many smaller magnetic fields that are tangled withthe 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 thesecond 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 theSun has conserved since formation. In the third part of the activity, you’ll measure the speed of acoronal 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 acontinuous cycle in the number of sunspots, which are the most visible tracer of the magnetic fieldsthat 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 activitymaximum, the magnetic field is aligned so that a compass needle on the Sun would point north, andat the next maximum the needle would point south. The reversal of the poles of the magentic fieldis the basic cause of the cycle.

The graph below shows the number of Sun spots per year over a period of almost 50 years.

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Our task is to measure and calculate the length of the solar cycle using the data in the chart atthe bottom of the previous page. Because the x-axis is in years, measuring the period is similar tomeasuring wavelength - you calculate the distance between each of the the solar maximums and thesolar minimums on the graph.

Because the data doesn’t seem to resemble a perfect wave, we need to measure the distances betweenseveral 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 alongthe cycle. It doesn’t matter if every year is exactly correct; focus more on being consistent frommaximum to maximum and minimum to minimum.

2. Subtract the year of the previous maximum from each solar maximum year; do the same foreach solar minimum year.

SolarMaximumYear

DifferenceBetweenMaxima

SolarMinimumYear

DifferenceBetweenMinima

1958———————— 1954

————————

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 maximaand 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. Howdoes your period compare?

The period I got is only a little shorter. The solar cycle isn’t actually constant - it varies alittle 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 becloser to solar maximum or solar minimum?

My answers give a minimum around 2006 and a maximum around 2010. Based on those num-bers, we should be in a solar maximum. However, this cycle seems to be longer than usual, sowe are still in a period of increasing solar activity.

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4.2 Sunspots and Rotation

The two images below were taken two days apart. Due to the distinctive pattern of the sunspots, wecan 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 inthe image and comparing it to the actual size of the Sun. That will allow us figure out how far thesunspots 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 × 108m. What we are looking for is a scale factor,which is:

scale factor =actual radius

measured radius(6)

Don’t worry too much about the units of your scale factor. In actuality, they are something likemeters 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= 2.08× 108 real m

image cm

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Sunspot Motion

Choose one of the active regions near the Sun’s equator. Make sure that you can see it and recognizeit in both of the images.

9. Measure the distance from the left side of the Sun to the sunspot in both of the images andwrite 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 hasmoved.

1.5 cm x 2.08 x 108 real mimage cm = 3.12 x 108 m

The Sun’s Period

We can find the period of the Sun from the distance the sunspots have moved. We need to use theratio of:

total solar period

two days=

distance around the Sun

distance traveled by a spot in two days(7)

The distance around the Sun is its circumference: D = 2πR.

12. Calculate the Sun’s rotation period:

period = 2π6.96×108 m3.12×108 m × 2 days = 28 days

13. The Sun’s rotation period is roughly 27 days, as viewed from Earth. How does your periodcompare?

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 projectioneffect - things appear to move slower when they are closer to the sides of the circle. That’sbecause they are also moving away.

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4.3 Coronal Mass Ejections

A Coronal Mass Ejection (CME) is an event where a mass of energetic particles are ejected fromthe surface of the Sun due to interactions of magnetic fields with the surface. In this section ofthe activity, we’ll figure out the velocity of a CME using images from the Solar And HeliosphericObservatory (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 inthe image - the white circle represents both the size and the position of the Sun. The first thing weneed to do is figure out the scale of the image, like we did in the previous section.

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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= 1.07× 109 real m

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’llneed 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. Measurethe distance (on the image) between that clump and the surface of the star. It’s ok if you canonly 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 thedistance 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.

ImageTime

TimeElapsed(seconds)

MeasuredDistance

MeasuredChange InDistance

ActualChange InDistance

Velocity

5:30 ———————— 0.9 cm

————————

————————

————————

5:53 13601.7 cm 0.8 cm 8.56 x 108 m

6.29 x 105 m/s

6:05 7802.4 cm 0.7 cm 7.49 x 108 m

9.60 x 105 m/s

6:53 2820————————

————————

————————

————————

You will likely have two velocities filled into the last column - the differences between the first, secondand 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

time = distancespeed = 1.5×1011 m

7.95×105 m/s= 1.89 x 105 s

1.89 x 105 s x 1 min60 s ×

1hour60min ×

1 day24 hours = 2.18 days

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Name: KEY

5 H-R Diagram

Intoduction

Around the year 1910, both Ejnar Hertzsrung and Henry Norris Russel came up with the idea of ar-ranging 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 astronomersstill use it today. We still call a graph that shows star’s luminosity as a function of temperature aHertzsprung-Russel (or H-R) Diagram.

One of the most important features of the H-R Diagram is the main sequence, which organizeshydrogen burning stars by their mass. Stars that don’t fall along the main sequence are either stillforming or evolved, which means that they’ve burnt all the hydrogen in their cores and are nowradiating light from some other process. This activity goes through the reasons why main sequenceand 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 ofthe Sun is hot and dense. Gravity pulls each part of the Sun towards the center. To keep the Sunfrom collapsing in on itself, the inside of the Sun pushes back with thermal pressure. This pressureallows the Sun to maintain a very hot (T∼ 106K) 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 theirmass. The mass of a star determines its temperature and luminosity. In order to understand whyand how, we will estimate the properties of a star with 0.1 times the mass of the Sun and one with10 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 pressurein 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 pullinward, will its core pressure and be lower or higher?

The core pressure will be lower due to the decreased amount of gravity pushing down.

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An increase in the core pressure will result in an increase in the core temperature. Rememberthat temperature is a measure of the average kinetic energy of particles in a substance. Ifyou put more pressure on that substance, the particles will move faster, increasing the kineticenergy and temperature.

3. Of the three stars we’re thinking about (0.1, 1, and 10 times the mass of the Sun), which onewould 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, whilethe 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 hydrogenturning into helium is based on a series of collisions - each collision where the particles sticktogether 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 (mostefficient) fusion? Which star has the slowest?

The star with the higher core temperature and larger mass should have faster fusion, while thestar 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 thesurface of the star. The energy is released in the form of light, so the rate of energy productionis equal to the star’s luminosity. As we discussed, luminosity is energy per time, in units ofjoules 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 absorptionover them. Right now we are only concerned with the thermal spectrum of the stars, whichdepends on the temperature in the photosphere.That temperature is related to the amount ofenergy 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 emitredder light.

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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 theSun. 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 theSun. 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 burningstars 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 notgiven; which star should look the brightest when viewed form Earth?

The Sun. Without using the formula for apparent vs. absolute brightness (or flux vs. lumi-nosity), 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 axeson the graph are on what we call a “log scale,” which makes it easier to compare numbers thatspan many powers of ten. On the y-axis, each evenly spaced tick mark represents a power of10 in luminosity. On the x-axis, the tick marks are in increments of either 1000 or 10000. Notethat the temperature is backwards - low temperatures are on the right, and high temperaturesare 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 timeslarger 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 lumi-nosity.

12. Now that the data is on the graph, take a look at the spectral type column. Label each pointor group of points with the letter in the spectral type column.

13. The spectral types should form a sequence. From hottest temperature and highest luminosityto the coolest, faintest stars, what order are the spectral types in?

OBAFGKM

22

Star Name Distance(Lyr)

SpectralType

SurfaceTemp (K)

Luminosity(compared toSun)

Luminosity(scientificnotation)

Sun 2.0 x 10−5 G 5800 1.0100

Alpha Centauri A 4.3 G 5800 1.51.5 x100

Alpha Centauri B 4.3 K 4200 0.333.3 x 10−1

Lalande 21185 8.1 M 3200 0.00555.5 x 10−3

Sirius A 8.7 B 10400 23.02.3 x 101

Luyten726-8 B 8.7 M 2700 0.000022 x 10−5

Ross 154 10.3 M 2800 0.000414.1x 10−4

Ross 248 10.3 M 2700 0.000111.1 x 10−4

Epsilon Eridani 10.8 K 4500 0.33 x 10−1

61 Cygni B 11.1 K 3900 0.0393.9 x 10−2

Procyon A 11.3 F 6500 7.37.3 x 100

Altair 16.5 A 8000 11.01.1 x 101

70 Ophiuchi A 17 G 5100 0.66 x 10−1

Vega 26 B 10700 55.05.5 x 101

Achernar 65 B 14000 2002 x 102

Beta Centauri 300 B 21000 50005 x 103

Zeta Persei A 465 B 24000 160001.6 x 104

Delta Persei 590 B 17000 13001.3 x 103

Zeta Pupis 1090 O 39000 7900007.9 x 105

23

40000 30000 20000 10000 6000 4000 2000Temerature (K)

10−4

10−2

100

102

104

106

Lum

inosity

5.3 The Lifetimes of Main Sequence Stars

Using the luminosity of stars, we can figure out their comparative lifetimes. Again, we will thinkabout 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 temperaturearound T = 25000 K. According to the H-R diagram, how many times greater is its luminositycompared 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 energygeneration 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 willthe massive B star be burning hydrogen?

It it 104 times as luminous, but only 10 times as massive. It should burn through its mass inabout 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 themassive B star?

Since the Sun has a lifetime to 1010 years and the more massive star will live 10−3 times aslong, 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 temperaturearound 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 isits approximate lifetime?

It will take a longer time. It will burn through its mass 102 times slower, and this will lead toa lifetime of 1012 years.

19. To sum up, fill in the blanks with the answers from questions 14 to 19:

A star with 0.1 times the mass of the Sun is an M star. It will burn hydrogen for longerthan the sun.

A star with 10 times the mass of the Sun is a B star. It will burn hydrogen for shorterthan the sun.

5.4 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 tem-peratures and luminosities are not directly related to mass; these stars fall in different places on thediagram.

Red Giants

Red giants are stars that have used up all the hydrogen in their cores. This has caused their coresto collapse, heat up, and begin fusing helium into carbon. We will talk more aout this in the nextclass, but for now all you need to know is that the cores of red giants produce more energy per timethan the fusion of hydrogen into helium.

20. When a star turns into a red giant, how should its luminosity compare to the luminosity it hadwhile 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 areso big, the material at the surface is very cool. The energy coming out of the star is spreadover 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-Rdiagram?

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 temperatureand luminosity.

24. Draw a line on your new H-R diagram that goes through the space where the main sequencewas in the previous H-R diagram.

Star Name Distance(Lyr)

SurfaceTemp (K)

Luminosity(compared toSun)

Luminosity(scientificnotation)

Arcturus 36 4500 1101.1 x 102

Capella 47 5900 1701.7 x 102

Alderbaran 53 4200 1001 x 102

Canopus 100 7400 15001.5 x 103

Antares 400 3400 50005 x 103

Betelgeuse 500 3200 170001.7 x 104

Rigel 800 11800 400004 x 104

Delta Aquarii B 1030 6000 43004.3 x 103

26

40000 30000 20000 10000 6000 4000 2000Temerature (K)

10−4

10−2

100

102

104

106

Lum

inosity

White Dwarfs

White dwarfs are the last type of star we’ll go through today. They are stars that are no longerburning anything; they are small and dense and the energy that they radiate comes from the slowprocess 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 layersare hot. What color should they be, compared to larger stars generating the same amount ofenergy?

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)

SurfaceTemp (K)

Luminosity(compared toSun)

Luminosity(scientificnotation)

Sirius B 8.7 10700 0.00242.4 x 10−3

Procyon B 11.3 7400 0.000555.5 x 10−4

Van Maanen’s Star 14.0 7500 0.000161.6 x 10−4

GJ 440 15.0 8500 0.00055 x 10−4

HU Aquarii B 587 12500 0.00222.2 x 10−3

You’ve now created a complete H-R diagram! We will learn more about the processes that happenwhen stars leave the main sequence to become red giants and white dwarfs during the next class.

28

Name: KEY

6 Post Main Sequence Activity

6.1 Red Giant Evolution

The flow chart below is designed to show the evolution of stars after the main sequence until theybecome white dwarfs, neutron stars, or black holes. Some of the arrows point directly at the nextbox, while others split the stars into a lower mass and a higher mass group. Be careful to follow thecorrect 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 topdiagram, draw and label the post main sequence evolutionary path of the Sun. On the bottomdiagram, draw and label the post main sequence evolutionary path of a 10 solar mass star.

40000 30000 20000 10000 6000 4000 2000Temerature (K)

10−4

10−2

100

102

104

106

Lum

ino

sity

40000 30000 20000 10000 6000 4000 2000Temerature (K)

10−4

10−2

100

102

104

106

Lu

min

osity

30

6.2 Planetary Nebulae

Below is a Hubble Space Telescope image of the Helix Nebula. We will use the image, and a fewother 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 theplanetary 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 scalebar 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 parsecs5.9 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 km1 parsec = 1.32× 1013 km

You take a spectrum of the center of the Helix Nebula and notice that the hydrogen line youexpected to be at 656.3 nm is actually at 656.2 nm. You remember the formula for dopplershift is:

v = c× λrest − λshiftλrest

(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 isthat material’s velocity?

v = 656.3 nm−656.2 nm656.3 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 - theradius 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 km46 km/s

= 2.87× 1011 s× 1 min60 s ×

1 hr60 min ×

1 day24 hr ×

1 year365 days = 9157 years

32

6.3 Density and Escape Velocity of Stellar Remnants

For the following questions, you will need the equation for density:

density =mass

volume=

M43πR3

(9)

And the equation for escape velocity:

v =

√2GM

R(10)

And the following numbers and constants:mass of Sun = 2 ×1030 kgradius of Sun = 6.95 ×105 km = 6.95 ×108 mradius of Earth = 6.4 ×103 km = 6.4 ×106 mG = 6.67 ×10−11 m3

kgs2

9. What are the density and escape velocity of the Sun?

density = 2×1030 kg43π(6.95×108 m)3

= 1422 kgm3

v =

√2×6.67×10−11 m3

kg s2×2×1030 kg

6.95×108 m = 6.2× 105 m/s

10. What are the density and escape velocity of a one solar mass white dwarf?

density = 2×1030 kg43π(6.4×106 m)3

= 1× 1010 kgm3

v =

√2×6.67×10−11 m3

kg s2×2×1030 kg

6.4×106 m = 6.5× 106 m/s

11. What are the density and escape velocity of a one solar mass Neutron star?

density = 2×1030 kg43π(5000 m)3

= 3× 1018 kgm3

v =

√2×6.67×10−11 m3

kg s2×2×1030 kg

5000 m = 2.3× 108 m/s

Take home point: The densities of white dwarfs and neutron stars are huge, and becausethey are small and massive, it is very hard for things to leave their surfaces. Not as hard as itwould be to leave a black hole, though!

33

Name: KEY

7 Milky Way Activity

7.1 Structure of the Galaxy

1. Dwarf a basic diagram of the Milky Way Galaxy face-on. Label the disk, bulge, bar, and spiralarms.

2. Draw a basic diagram of the Milky Way Galaxy edge-on. Label the disk, bulge, halo, andglobular 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 inthe 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 hydrogenclouds.

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 burnheavier elements in their cores.

8. stellar fusion to returning gas

During the late stages of stars’ lifetimes, their outer layers expand, carrying with them en-riched materials. Higher mass stars can also have dramatic supernova, which return gas to theinterstellar medium.

9. returning gas to hot bubbles

When many supernovae happen in a young cluster, this can blow a bubble of hot gas thatbreaks 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, closerto 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 thespeed v and radius r, which we will derive on the board. Write the equation here:

M = r×v2G

You will need the following constants:G = 6.67× 10−11 m3

kg s2

Msun = 2× 1030kg

Supermassive Black Hole

Suppose you observe a star orbiting the galactic center at a speed of 106ms 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×v2G

=5.18×1014 m×(106 m

s )2

6.67×10−11 m3

kg s2

= 7.77× 1036 kg

12. Convert the mass into units of solar masses.

7.77× 1036 kg× Msun2×1030 kg = 3.89× 106 Msun

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) inradius.

The radius of the object has to be less than 20 light-days, because a star is orbiting around itat that distance. It also must be more massive than the globular cluster, because its mass is10 to one hundred times larger. Because it is more massive and much smaller than a globularcluster, 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 orbittingat roughly 160000 light-years (1.5× 1021 m) from the galactic center at a speed of 3× 105m

s .

14. Calculate the mass within the radius of the LMC’s orbit in kilograms.

M = r×v2G

=1.51×1021 m×(3×105 m

s )2

6.67×10−11 m3

kg s2

= 2.04× 1042 kg

15. Convert the mass into units of solar masses.

2.04× 1042 kg× Msun2×1030 kg = 1.02× 1012 Msun

16. The number of stars in the galaxy should be something more like 2 × 1011. Does this matchthe 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 ofstars in the galaxy, if we assume that each star is the mass of the Sun. Because most stars areless 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 iscircular, and it shouldn’t be exactly circular. But that is likely to be a small effect - our answeractually indicates the presence of matter that’s not in the form of stars. That matter could bedust, gas, or dark matter.

37

Name: KEY

8 Dark Matter Activity

8.1 Measuring a Rotation Curve

The graph below shows the observed rotation curve of a galaxy. The x-axis shows the distance fromthe 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 ofthe galaxy affect those measurements?

In order to calculate velocities, we need to take a spectrum of the galaxy at each radius, andcompare the wavelength of the lines to the rest wavelength. This is complicated by the redshiftbecause 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 angulardistance from the center of the galaxy in an image, and then multiply by the galaxy’s distanceto get a real distance from the center of the galaxy.

−10 −8 −6 −4 −2 0 2 4 6 8 10Radius (kpc)

−200

−100

0

100

200

Velo

city (

km

/s)

38

−12 −10 −8 −6 −4 −2 0 2 4 6 8 10 12Radius (kpc)

−150

−100

−50

0

50

100

150

Ve

locity (

km

/s)

0 1 2 3 4 5 6 7 8 9 10 11Radius (kpc)

0

1•1010

2•1010

3•1010

4•1010

5•1010

6•1010

Ma

ss (

Msu

n)

0 1 2 3 4 5 6 7 8 9 10 11Radius (kpc)

0.0

0.2

0.4

0.6

0.8

Fra

ctio

n o

f "N

orm

al"

Ma

ss

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 onlydifference is that we are working with one half of the curve, which should give us similar results forthe 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 todetermine how much mass is inside an orbit. You’ll need the universal gravitational constant:G = 6.67× 10−11 m3

kg s2

M =r × v2

G(11)

9. Convert the masses from kg to solar masses: Msun = 2× 1030kg

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 massinside 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 increasescompared 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 maincontributor to the light that we see. Sometimes we call this “normal” matter because it is made upof particles that we understand. The dark matter that we infer exists is made up of different stuffentirely.

The graph below shows the light emitted from the galaxy, in terms of the luminosity of the Sun. It’ssimilar to the mass plots, in that the light shown at each radius is all of the light inside that radiusadded 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.

0 1 2 3 4 5 6 7 8 9 10 11Radius (kpc)

0

1•109

2•109

3•109

4•109

5•109

Lig

ht

em

itte

d (

Lsu

n)

We’ll also need to estimate the mass in stars that is generating the amount of luminosity that wedetect. It’s a pretty safe bet to estimate that for every unit of solar luminosity, there are two unitsof 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 theSun, and those stars are a lot less luminous than the Sun is. We estimate that those stars areproducing 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 differentsymbols, 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 massplot 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 wouldactually dip, decreasing as the radius increases. Because the curve is flat, we know that there’smore 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 lotof 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 matterat the inner parts of the galaxy, it is much less than the amount of luminous matter. We cantell there is dark matter at the edges of the galaxy because there is more mass that we can’tsee.

42

8.4 Mass in a Galaxy Cluster

Now we will determine the mass of a galaxy cluster through its temperature. In lecture, we discussedthat the temperature of hot gas in a cluster was related to the average velocities of the particles inthe gas. The specific equation is:

vparticle = 140m

s√K×√T (12)

A cluster of galaxies has a radius of about 5.1 million light-years and a temperature of 6×107 K. Theobserved light of all the stars in the cluster is 8 ×1012Lsun.one light-year = 9.46× 1015 m

21. What is the average particle speed of gas in the cluster?

vparticle = 140ms ×√T = 140m

s ×√

6× 107K = 1.08× 106ms

22. Using the outer radius of the cluster and the particle speed for the velocity of material, whatis the mass of the cluster in terms of the mass of the Sun?

M = r×v2G

=5.1×106 light-years×9.46×1015 m

light-year×(1.08×106 m

s )2

6.67×10−11 m3

kg s2

= 8.44× 1044 kg

M = 8.44× 1044 kg× 1Msun2×1030 kg = 4.22× 1014Msun

23. We often talk about the mass to light ratio, which is the number of solar masses over thenumber of solar luminosities. What is the mass to light ratio of the cluster?

mass-to-light = ML

=4.22×1014Msun8×1012 Lsun

= 53

24. Calculate the ratio of the mass of luminous matter to dark matter. How does the amount ofdark matter in the cluster compare to the amount of dark matter in the galaxy we worked withbefore?

luminous mass = L× 2 = 1.6× 1013 Msun

fraction of normal mass =1.6×1013 Msun4.22×1014 Msun

= 0.0379

This is about 20 times less than the ratio we found for the galaxy we were working with. There’ssome dark matter in galaxies, but there’s even more in clusters of galaxies.

43

Name: KEY

9 Cosmology

9.1 Accelerating Expansion

Below is a graph similar to the one that you constructed in the Hubble Law Lab. There are no unitson it, but it extends to farther distances than can be measured using the sizes of galaxies as standardrods.

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 havea consistent luminosity.

2. What measurements would you need in order to construct the graph below?

We need to measure the brightness of supernovae in order to calculate a distance, and we alsoneed to measure a redshift of spectral lines in order to get a velocity.

Distance

Ve

locity

44

The solid line on the graph on the previous page shows how velocity and distance are related throughthe 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 givendistance.

4. The more distant galaxies emitted the light that we see now longer ago, so they trace the waythe universe was expanding at that time. Was the expansion in the past larger or smaller thanit 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.

6. We use graphs similar to this one to infer that the universe is accelerating in its expansion. Inthe graph below, draw and label three lines that represent constant expansion, acceleratingexpansion, and deccelerating expansion.

Distance

Ve

locity

45

9.2 Eras in the Early UniverseAs lecture progresses through each of these eras, fill in the duration, average temperature, and adescription of each one.

Era Name Duration Temperature DescriptionPlanck Era

10−43 s 1032 K Physics was incomprehensible; the strong,weak, electromagnetic, and gravitationalforces acted as one.

GUT Era10−38 s 1029 K The electromagnetic and weak forces sep-

arate from the strong force. This releasesthe energy that causes inflation.

Electroweak Era10−10 s 1015 K Energy to randomly created particles and

anti-particles. Ends when electromagneticand weak forces separate.

Particle Era0.001 s 1012 K Energy continued to spontaneously gener-

ate particles and anti-particles through theend of the particle era.

Era of Nucleosynthesis5 min 109 K The fusion of hydrogen was balanced by

the destruction of helium. This era deter-mined the abundance of helium today.

Era of Nuclei3.8× 105 yrs 3000 K The universe was a hot, ionized plasma

where photons bounced between particles,never traveling very far.

Era of Atoms109 year 18 K The universe consists mostly of atoms and

photons. Protogalactic clouds form.Era of Galaxies

still going 3 K Galaxies and clusters of galaxies dominatethe universe on large scales.

7. What is the basic argument for the universe having been smaller, denser, and hotter in thepast?

We can see the universe expanding at the current time - it is currently getting larger, less denseand cooler. If we look backward, it must have been smaller, dense, and hotter becuase we haveexpanded 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 backgroundemitted?

Between the Era of Nuclei and the Era of Atoms. That’s when the universe became diffuseenough for photons.

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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 spontaneouslycreate particles.

9.3 The Cosmic Microwave Background

The Cosmic Microwave Background (CMB) is now observed to have a thermal spectrum that corre-sponds to a temperature of T = 2.73 K. The law that relates peak wavelength of a thermal radiationspectrum (λpeak) to the temperature of the object emitting that spectrum (T) is known as Wein’s

Law:

λpeak =2.9× 106 nm K

T(13)

Which we’ve used before to determine the peak wavelength of stars.

11. What is the peak wavelength of the CMB today?

λpeak = 2.9×106 nm K2.73 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 filledin on the previous page)?

λpeak = 2.9×106 nm K3000 K = 967 nm

13. The stretching of the wavelengths is proportional to the stretching of the universe over thetime between the emission of the light and our observation of the light today. How many timesbigger is the universe today than when the CMB was emitted?

stretch of universe =λnowλpast

= 1.06×106 nm967 nm = 1097 times larger today

14. What do the over densities in the map of the CMB have to do with the galaxies and clustersthat we see today? How do those relate to the quantum fluctuations in the first 10−36 secondsof the universe?

The over densities in the CMB have their origin in the pre-inflation universe. On the smallscales that existed then, the fluctuations that cause the CMB to be different temperatures werequantum fluctuations on the scale of tiny particles. Those same fluctuations in the density ofmatter and energy then are the same over densities that are the size of galaxies and clusters ofgalaxies today.

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Name: KEY

10 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 ofdetecting other civilizations in our Galaxy. As we talk about each factor, put an estimate of thenumber 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 fraction of stars that haveplanets 0.2

nhabitible average number habitableplanets per star with planets 1

flife fraction of habitable planetsthat develop life 1

fintellegence fraction of planets with lifethat develop intelligent life 0.002

ftechnology fraction of civilizations thatdevelop detectable technol-ogy

0.1

L the length of time for whichcivilizations continue to bedetectable

1,000 years

N the number of civilizationsthat we could detect 0.2

My answer isn’t necessarily “right,” there are many possible answers, and none yet confirmed.

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