1 Math Activity. Intoduction. 1.1 Scientific notation

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1 Name: KEY Intoduction 1 Math Activity 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: = 5,000, = 63,000,000, = = Write these numbers in scientific notation ,000 = ,230 = = =

2 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: = = = = = = = = = = / = 1.5 / 5 = / 10 4 = = / = 4.8 / 2 = = 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 cm to km cm 1m 1km = cm m 102 km = 200 km 20. Convert 789 days to years 798 days 1year = 2.16 years 365days 21. Convert days to seconds days 24hours 60min 60s = 130 s 1day 1hour 1min 2

3 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? 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 / π / log(99) 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 cm height of a human angstrom (Å) cm distance from here to NYC kiloparsec (kpc) cm radius of an atom parsec (pc) cm distance to a nearby star meter (m) cm diameter of galaxy astronomical unit (AU) cm distance to the sun kilometer (km) 3

4 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 = m/s The distance from the Sun to the Earth is 1 AU = m. time = distance speed = m 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 km. First convert the distance to meters because the speed of light is in m/s: km m = km 105 km Now use the same method as the previous problem: time = distance = m = 1.28s speed m/s 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 m The distance we travel is related to the radius: d = 2πr First, we ll convert light-years into meters: Ly m = m 1Ly Then convert the galactic radius into the circumference of our orbit: d = 2 π m = m Note that 230 million years = yr. We need to convert that into seconds: yr = 365days 24hr 60min 60s = yr 1day 1hr 1min 1015 s speed = distance time = m s = Rounded to two significant digits, the answer is: m/s This is the same as 510,000 miles per hour - fast! 4

5 Name: KEY Intoduction 2 Kepler Activity 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

6 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 = a 2, 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

7 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) P 2 = a 3 a (AU) 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 Period (years) semi major axis (AU) 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

8 Name: KEY Intoduction 3 Light Activity 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: λ = c f or f = c λ (1) Where f is the frequency, λ is the wavelength, and c is the speed of light (c = m s ) 3. Below, draw a wave with a wavelength of 3 cm 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 = m s 100 cm 1 m = cm s f = c λ = cm s 3 cm = s = 1010 Hz 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 = m 2 kg s ). One way to think about this is that a wave with a shorter wavelength will peak more frequently, which will carry more energy. 8

9 5. Calculate the Energy of the wave you drew: E = h f = h = m 2 kg s s 1 = m 2 kg s 2 = 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 radioactive elements X-ray thickness of a DNA strand X-ray machine ultraviolet groove in a CD The sun - it burns! Visible Light red blood cell or bacterium fire, light bulbs Infrared width of a human hair stovetop, people microwave coffee bean radar, microwave oven radio city bus Radio station transmitter 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. Relative Intensity Wavelength (nm) 9

10 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: λ max = nm K T where λ max 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. (3) 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): Relative Intensity Wavelength (nm) 10

11 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: λ max = nm K T = nm K 8000 K = nm (shown in blue below) (4) 7. Calculate the peak wavelength and draw in the thermal emission spectra a T = 4000K object: λ max = nm K T = nm K 4000 K = nm (shown in red below) (5) Relative Intensity Wavelength (nm) 3.3 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

12 White light: The normal lightbulb is on the top of the tower. 9. What color is the light without the diffraction grating? white violet blue green yellow red Wavelength (nm) 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 blue green yellow red Wavelength (nm) Helium lamp: The atomic helium lamp is second from the top lamp. 11. What color is the light without the diffraction grating? white violet blue green yellow red Wavelength (nm) Mercury lamp: The Mercury (Hg) lamp is just above the bottom lamp. 12. What color is the light without the diffraction grating? blue violet blue green yellow red Wavelength (nm) Neon lamp: The Neon lamp is on the bottom of the tower. 13. What color is the light without the diffraction grating? orange violet blue green yellow red Wavelength (nm) 12

13 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

14 Name: KEY Intoduction 4 Solar Activity 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

15 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 1958 Difference Between Maxima Solar Minimum Year Difference Between Minima 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 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

16 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 = 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? = real m image cm 16

17 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 cm = 1.5 cm 11. Using the scale factor you found above, calculate the actual distance that the sunspot has moved. 1.5 cm x 2.08 x 10 8 real m image 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 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 = 2π m 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

18 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

19 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 = 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 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 Measured Measured Actual Velocity Time Elapsed Distance Change In Change In (seconds) Distance Distance 5: cm 5: cm 0.8 cm 8.56 x 10 8 m 6.29 x 105 m/s 6: : cm 0.7 cm 7.49 x 10 8 m 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 10 5 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 m time = distance speed = m m/s = 1.89 x 105 s 1.89 x 10 5 s x 1 min 60 s 60min 1hour 1 day 24 hours = 2.18 days 19

20 Name: KEY Intoduction 5 H-R Diagram 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 10 6 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

21 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

22 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 (10 6 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 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 10 6 and the smallest is So the largest luminosity is 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

23 Star Name Distance (Lyr) Spectral Type Surface Temp (K) Sun 2.0 x 10 5 G Alpha Centauri A 4.3 G Alpha Centauri B 4.3 K Lalande M Sirius A 8.7 B Luminosity (compared to Sun) Luyten726-8 B 8.7 M Ross M Ross M Epsilon Eridani 10.8 K Cygni B 11.1 K Procyon A 11.3 F Altair 16.5 A Ophiuchi A 17 G Vega 26 B Achernar 65 B Beta Centauri 300 B Zeta Persei A 465 B Delta Persei 590 B Zeta Pupis 1090 O Luminosity (scientific notation) x x x x x x x x x x x x x x x x x x

24 Luminosity Temerature (K) 5.3 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 = K. According to the H-R diagram, how many times greater is its luminosity compared to the Sun s? It should be about 10 4 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 10 4 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 10 3 times faster. 24

25 16. The lifetime of the sun is approximately years. What is the approximate lifetime of the massive B star? Since the Sun has a lifetime to years and the more massive star will live 10 3 times as long, it will have a lifetime of 10 7 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 10 2 times slower, and this will lead to a lifetime of 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 longer than the sun. A star with 10 times the mass of the Sun is a B star. It will burn hydrogen for shorter than 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 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

26 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 Capella Alderbaran Canopus Antares Betelgeuse Rigel Delta Aquarii B Luminosity (compared to Sun) Luminosity (scientific notation) 1.1 x x x x x x x x

27 Luminosity Temerature (K) 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

28 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 Luminosity (compared to Sun) Procyon B Van Maanen s Star GJ HU Aquarii B Luminosity (scientific notation) 2.4 x x x x 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

29 Name: KEY 6.1 Red Giant Evolution 6 Post Main Sequence Activity 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

30 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 Luminosity Temerature (K) Luminosity Temerature (K) 30

31 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

32 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 = km r = 0.43 parsecs km 1 parsec = km You take a spectrum of the center of the Helix Nebula and notice that the hydrogen line you expected to be at nm is actually at nm. You remember the formula for doppler shift is: and that c = km/s. v = c λ rest λ shift λ rest (8) 5. That shift comes from material that is moving away from the center of the Nebula. What is that material s velocity? v = nm nm nm 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 = km 46 km/s = s 1 min 60 s 1 hr 60 min 1 day 24 hr 1 year 365 days = 9157 years 32

33 6.3 Density and Escape Velocity of Stellar Remnants For the following questions, you will need the equation for density: And the equation for escape velocity: density = v = And the following numbers and constants: mass of Sun = kg radius of Sun = km = m radius of Earth = km = m 11 m3 G = kgs 2 mass volume = 2GM 9. What are the density and escape velocity of the Sun? R M 4 3 πr3 (9) (10) density = v = kg 4 = 1422 kg 3 π( m) 3 m m3 kg s kg m = m/s 10. What are the density and escape velocity of a one solar mass white dwarf? density = v = kg 4 = kg 3 π( m) 3 m m3 kg s kg m = m/s 11. What are the density and escape velocity of a one solar mass Neutron star? density = kg 4 = kg 3 π(5000 m m) m3 v = kg s kg 5000 m = m/s 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

34 Name: KEY 7.1 Structure of the Galaxy 7 Milky Way Activity 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

35 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

36 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 v2 G You will need the following constants: G = m3 kg s 2 M sun = kg Supermassive Black Hole Suppose you observe a star orbiting the galactic center at a speed of 10 6 m s radius of 20 light-days ( m). in a circular orbit with a 11. Calculate the mass of the object in kilograms, based on the equation given in class. M = r v2 = m (10 6 m s ) 2 G m3 kg s 2 = kg 12. Convert the mass into units of solar masses kg M sun kg = Msun 13. Compare the mass and radius of the object in the center of the galaxy to a globular cluster. Globular clusters have around 10 5 stars in a region that s about 50 light-years ( 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

37 Dark Matter Halo The Large Magallenic Clouds is a small galaxy that orbits the Milky Way. It is currently orbitting at roughly light-years ( m) from the galactic center at a speed of m s. 14. Calculate the mass within the radius of the LMC s orbit in kilograms. M = r v2 = m ( m s ) 2 G m3 kg s 2 = kg 15. Convert the mass into units of solar masses kg M sun kg = Msun 16. The number of stars in the galaxy should be something more like Does this match the mass you calculated? What are some reasons the two answers might not match? We calculated a mass of about 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

38 Name: KEY 8.1 Measuring a Rotation Curve 8 Dark Matter Activity 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. 200 Velocity (km/s) Radius (kpc) 38

39 Velocity (km/s) Radius (kpc) Mass (M sun ) Radius (kpc) 0.8 Fraction of "Normal" Mass Radius (kpc) 39

40 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 = m 7. Convert the velocities from km/s to m/s: 1 km/s = 10 3 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: G = m3 kg s 2 M = r v2 G 9. Convert the masses from kg to solar masses: M sun = 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) 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. (11) 40

41 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 Light emitted (L sun ) Radius (kpc) 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