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PREVIEW: THE COSMIC LANDSCAPE The Earth, Our Home The Moon The Planets The Sun The Solar System A Sense of Scale The Astronomical Unit The Milky Way Galaxy The Light Year Galaxy Clusters and the Universe Gravity Atoms and Other Forces The Scientific Method Goals This Chapter summarizes our knowledge of the overall structure of the Universe. Students should come away from it with a sense of the hierarchical structure of the Universe and how the Earth fits into it. Basic inventory of Universe (planets, stars, galaxies, etc... general appearance and features and relative sizes). The chapter also introduces powers-of-ten notation, vocabulary and astronomical distance units (AU and light year) that are used throughout the text. The video "Powers of Ten" is an excellent visual summary of the material. Some narration by the instructor may help students better understand the numbers that flash on the screen during the video. Key Terms Planet Satellite Star Solar System Astronomical Unit Milky Way Galaxy Gravity Universe Atoms Nucleus Electron Proton Electric Charge Light year Local Group Galaxy Cluster The Local Supercluster Electric Force Strong or Nuclear Force Weak Force Scientific Method Answers to Problems 1. RSun = 7x105 km. REarth = 6.4x103 km. We can determine how much bigger the Sun is than the earth by dividing the Earth's radius into the Sun's. Therefore RSun/ REarth= 7x105 km/ 6.4x103 km = 7/6.4 x(105/103) = 1.09x(105-3) = 1.09x(102) = 109. 2. The average distance between the Earth and the Sun is called the astronomical unit. 1 AU = 1.5x108 km. The distance traveled by a light ray in 1 year is called a light year (ly). 1 ly = 1x1013 km. We can find the number of AU in a light year by dividing the number of km in a light year (1013) by number of km in an AU (1.5x108). Thus, Number of AU in a light year = 1013/(1.5x108) = 1013-8/1.5 = 105/1.5 = 6.7x104 = about 68,000. 3. If the Earth had a radius of 0.1 cm (1 mm), the Sun would have a radius of 100 x 0.1 = 10 cm. The Solar System is 8600 x R(Sun) and so, R(solar system) = 8600 x 10 cm = 8.6x104 cm = 8.6x102 m, since there are 100 cm in 1 m. R(Milky Way)= 1x108x R(solar system) = 8.6x1010 m = 8.6x107 km, since there are 1000 m in 1 km. 4. Using the scale in Q.3, the radius of the Solar System (to Pluto) is 40 AU = 860 m. In our model, 1 AU therefore = 860 / 40 = 21.5 m Planet a (AU) a (m) R (mm) -----------------------------------------------------------------Mercury 0.39 8.4 0.38 Venus 0.72 15.5 0.95 Earth 1.00 21.5 1.00 Mars 1.52 32.7 0.53 Jupiter 5.2 111.8 11.21 Saturn 9.55 205.3 9.45 Uranus 19.18 412.4 4.01 Neptune 30.06 646.3 3.88 Pluto 39.44 848.0 0.18 5. The Milky Way has a radius of 50,000 ly = 5x104 ly. We scale this to 2 cm, and let the Local Group have a scaled radius of x. The Local Group's true radius is about 1.5x106 ly. Thus we can form the proportion that x/1.5x106 ly as 2 cm/5x104 ly. Thus, x = 1.5x106x2 cm/5x104 = 300/5 cm or 60 cm, or about the size of a standard bed pillow. If we let the scaled size of the local supercluster be y, then y/40x106 ly = 2 cm/5x104 ly. Thus, y = 2 cmx40x106 ly/5x104 ly = 16x102 cm = 16 m. The visible Universe would be 2 cmx15x109 ly/5x104 ly = 6x105 cm = 6 km, or about 4 miles. 6. (4x108)3/(5x10-6)2 = 43x(108)3/(52x(10-6)2)=64x1024/(25x10-12)=2.56x1036. 7. t=D/v. For light, v=c, so t=D/c. Pluto is about 40 AU from the Sun and 1 AU =1.5x108 km. Thus, Pluto is 40x1.5x108 = 6x109 km from the Sun. The speed of light is 3x105 km/s. Thus, t = 6x109/3x105 = 2x104 seconds. Divide by 60 to get the number of minutes = about 330 minutes. 8. To find the number of times one thing is smaller than another, divide the larger by the smaller. Thus the hydrogen atom is 10-6 m/10-10 m = 10-6-(-10) = 104 times smaller than the bacterium. 9. (3x104)2/(4x10-6)1/2 = 32x104x2/41/2x10-6x1/2 = 9x108/2x10-3 = 4.5x108-(-3) = 4.5x1011. 10. Given an object's speed and distance from another, we can find how long it took them to move apart by using the relation D=vt, where D is the distance, v is the speed, and t is the time. Solving for t gives, t=D/v. Now inserting values for v and D, we find t = 3x108 ly/6,000 km-seconds-1. Notice that are units are mixed here. We have ly on top and km below. Thus, we need to express ly in km. From the appendix we can find that 1 ly = 9.5x1012 km. Thus, t= 3x108 lyx9.5x1012 km/ly/6,000 km-seconds-1 = (3x9.5/6)x108+12-3 seconds = 4.8x1017 seconds. To express this in years, we need to divide by the number of seconds in a year = about 3.2x107. Thus, t = 4.8x1017 seconds/3.2x107 seconds/yr = 1.5x1010 yrs. This is one way that astronomers deduce the age of the Universe. Answers to Self-Test 1 (b) The light year is a unit of distance. 2 (c) 3 (e) 4 (c) 5 (e) Your cosmic address is Earth, Solar System, Milky Way, Local Group. All of them are. Jupiter's diameter is about 10 times larger than the Earth's diameter. All except (d). Choice (d) is a matter of opinion. CHAPTER 1 - HISTORY OF ASTRONOMY 1-1 Prehistoric Astronomy The Celestial Sphere 1-2 1-3 1-4 1-5 Constellations Motions of the Sun and Stars Daily or Diurnal Motion Annual Motion The Ecliptic The Seasons The Ecliptic's Tilt Solstices and Equinoxes The Planets and the Zodiac The Moon Eclipses Early Ideas of the Heavens: Classical Astronomy The Shape of the Earth The Size of the Earth Distance and Size of Sun and Moon Extending Our Reach: Measuring the Diameter of Astronomical Objects The Motion of the Planets Ptolemy Islamic Contributions Asian Contributions Astronomy in the Renaissance Copernicus Tycho and Kepler Galileo Isaac Newton and the Birth of Astrophysics The Growth of Astrophysics New Discoveries New Technologies The Nature of Matter and Heat The Kelvin Temperature Scale Goals Basic ideas of the heavens as seen by the naked eye: Features of the celestial sphere. Definition of celestial equator, poles. Definition of angular diameter. Ecliptic, Zodiac. Cause of seasons. Cause and appearance of eclipses and phases of Moon. A sense of the date and contributions of Aristarchus, Eratosthenes, Aristotle, Ptolemy; the methods they used, and approximately when they lived. Contribution of Kepler, Copernicus, Galileo and Newton and about when they lived. Kepler's laws and their use. The Kelvin Temperature scale. This chapter introduces some important ideas that will be needed later, such as parallax (also described in Chapter 12), angular size (also discussed in Chapter 6), and Kepler's third law. Nevertheless, it may be omitted if these ideas are dealt with as needed. If pressed for time, omit details of methods used by ancient astronomers, angular diameter, or naked eye astronomy. The material on the celestial sphere, zodiac, and so forth, appear nowhere else apart from Essay 1. Lecture Suggestions The simple cardboard overhead projector planetarium will be helpful in explaining rising and setting motions as well as retrograde motion. An orrery (mechanical model of the Sun and Earth) will help illustrate that the constellations that are visible change with the seasons. I set such a gadget on a table in the front of the room and then mark off constellations on the wall with chalk. Moving the model Earth around the model Sun then allows students to see how the stars visible change and how the Sun "moves" through the Zodiac. You can also make decorative constellations on paper and tape them to the wall. In effect, this turns the room into a simple planetarium. A flashlight with a fat beam shows the importance of angle to seasonal heating. When directed directly at the wall the energy is concentrated in a small area. When shined obliquely at the wall, the beam covers a larger area, implying less concentration of heat and therefore a lower temperature. A tensor lamp or other bright light source and a tennis and ping-pong ball serve to demonstrate features of eclipses. A volley ball or basket ball and a bright lamp can illustrate lunar phases. Key Terms Celestial sphere Model Constellation Celestial pole Solstice Zodiac Retrograde motion Phases Parallax Geocentric Epicycle Heliocentric Celestial equator Ecliptic Equinox Solar eclipse Lunar eclipse Angular size Focus Semi-major axis Kepler's Laws Answers to Thought Questions 1. If you were standing on the Earth's equator, you would see the north celestial pole on the horizon (and the south celestial pole too). You cannot see the north celestial pole from Australia, only the south celestial pole. 2. SKETCH FOR STUDENTS 3. The main astronomical reason why there are 12 zodiacal signs is that the Sun appears to travel about 30 degrees per month across the sky. 4. The basic proofs are as follows: - The Earth casts a circular shadow on the Moon during a lunar eclipse. Only a sphere will cast a circular shadow from any illumination angle. - As a traveler moves south, stars previously hidden below the southern horizon come into view. - Ships disappear over the horizon gradually, the mast being the last part to disappear from view. The newer proofs come from modern technology (I think the last flat Earth society existed until about 1930). Some modern proofs are: - Supersonic aircraft fly high enough to see the curvature of the Earth. - Spacecraft send back pictures of the spherical Earth. 5. The position of sunrise along the eastern horizon changes during the year because the Earth's axis is tilted at 23.5 degrees to the plane of its orbit (the ecliptic) and the Earth maintains this same tilt throughout the year. At the equinoxes (March 21 and Sept. 23), the Sun lies on the celestial equator. Because the celestial equator cuts the horizon at the east and west points, the Sun will rise and set due east and due west, respectively. At the winter solstice (Dec. 21), the Sun lies 23.5 degrees south of the celestial equator on the sky. It will therefore rise south of the east point on the horizon and set south of the west point. At the summer solstice (June 21), the Sun lies 23.5 degrees north of the celestial equator. It will therefore rise north of the east point and set north of the west point. 6. If the stars were very much closer than they really are, Aristarchus would have been able to demonstrate the stellar parallax caused by the Earth's orbital motion around the Sun. 7. The phases of Venus are caused by Venus orbiting the Sun, so keeping all distances and periods constant, it wouldn't matter whether Earth was orbiting the Sun or the Sun (and Venus) were orbiting the Earth. Answers to Problems 1. The Sun is 36 degrees from straight overhead, so by the parallel line theorem, the angle between you, the center of the planet and the missile silo is also 36 degrees. 36 / 360 = 1/10 of a complete circle. You are 1/10th of the circumference of Myrmidon from the missile silo. If the distance to the silo is 1,000 miles, then the circumference of the planet is 10,000 miles. The circumference = 2 x radius, therefore Radius = 10,000 miles / (2) = 1,592 miles. 2. Historical question. Student must come up with own answers. 3. If P = 64 years, you can use Kepler's 3rd Law to estimate its distance from the Sun. P2 = a3, where P=period in years and a=average orbital radius in AU. We can find a by taking the cube root of both sides to get P2/3 = a, or a = P2/3 = (64)2/3 = 16 AU 4. Aliens have a 1Mo star and are 4 AU from it. According to Kepler's 3rd law P2 = a3 P=square root of (43) = (43)1/2 = 8 years. 5. The Andromeda galaxy has an angular diameter of 5 degrees at a distance of 2.2x106 ly. We can find its true size by using the angular diameter formula L = 2DA/360, where D = distance, A = angle subtended, and L = linear diameter. Thus, L = 2 x 2.2x106 ly x 5 degrees / 360 degrees = 1.92x105 ly 6. A shell of gas has an angular diameter A = 0.1 degrees and a linear diameter L = 1 ly. We can find its distance D by reversing the procedure in the previous problem. We begin by writing out the angular diameter formula L = 2DA/360. We next solve it for D by dividing both sides by 2A and multiplying both sides by 360 to get D = 360L/(2A). Next inserting values for L and D we find D = 360 x 1 ly/( 2 x 0.1) = 573 ly. 7. This problem is a modern version of the method Eratosthenes used to measure the size of the Earth. Given that the shadow length is 15 degrees, the distance in latitude between the two points on the asteroid must be 15 degrees, or 15/360 = 1/24 th the circumference of the asteroid. If the 15 degrees corresponds to 10 km, then the total distance around the asteroid must be 10x24 = 240 km. The radius, R, of the asteroid is related to its circumference, C, by C=2R. Thus R = C/2240/2= 38 km 8. This is an application of Kepler's third law, P2 = a3, where a is in AU and P is in years. If P = 125 yrs, then a3 = 1252. Solving for a, we take the cube root of both sides to get a = (1252)1/3, where we have used the fact that the cube root of a number is the number to the 1/3 power. Using your pocket calculator, you will find that a = 25 AU. If the planet's orbit is circular, then that is also the planet's orbital radius. 9. This problem is another application of Kepler's third law, P2 = a3, where a is in AU and P is in years. In this case, we are given a, and are asked to find P. Thus P2 = a3 = 163. Solving for P by taking the square root and recalling that the square root is the number to the 1/2 power, we find that P = (163)1/2 = 64 yrs. (Note: in solving this problem, you can simplify the math by reversing the order of the power and the square root. That is take the square root of 16 (=4) and then cube it to get 64. Answers to Self-Test 1. (d) The north celestial pole is directly overhead. 2. (b) Retrograde motion causes planets to stop their regular eastward motion with respect to the stars and move westwards for a time. 3. (a) The Earth is spherical. 4. (e) Venus orbits the Sun. 5. (a) Kepler's 3rd Law relates a planet's orbital period to the size of its orbit around the Sun. Additional Readings: Aveni, Anthony F. "Emissaries to the Stars: The Astronomers of Ancient Maya." Mercury 24 (January/February 1995): 15. Gingerich, Owen. The Eye of Heaven: Ptolemy, Copernicus, Kepler. New York: American Institute of Physics, 1993. Gurshtein, Alexander. "When the Zodiac Climbed the Sky." Sky and Telescope 90 (October 1995): 28. Hetherington, Barry. A Chronicle of Pre-Telescopic Astronomy. New York: John Wiley and Sons, 1996 ESSAY 1 - Backyard Astronomy Learning the Constellations Star Lore Amateur Astronomy Small Telescopes Celestial Coordinates Star Charts Planetary Configurations Your Eyes at Night Goal Introduce students to star lore and some terms used in naked eye and backyard astronomy. Key Terms Asterism Declination Right Ascension Altitude Azimuth Conjunction Inferior Conjunction Superior Conjunction Transit Opposition Morning/evening star Greatest Elongation Synodic period Dark Adaption Averted Vision Answers to Thought Questions 1. You would look for Mercury in the west, towards the sunset. 2. A "Morning Star", either Mercury or Venus, is seen close to the eastern horizon just before sunrise. 3. If a planet is at opposition and you see it overhead it must be midnight because the planet is 180 degrees from the Sun and rises at sunset. 4. You cannot see Mercury in the western sky at dawn because the Sun rises in the east and Mercury can never get more than 28 degrees from the Sun. 5. Dark adaptation refers to the state of the human eye after 20 minutes or so in very low light levels, when the pupil has widened and the retina has become more than a million times more sensitive to low light than it is in daylight. The eye also becomes more sensitive to blue light (which is why many astronomers use dim flashlights with red cellophane over them). 6. The pupil of your eye gets bigger in dim light to let more light into the eye, in the same way that a photographer adjusts the aperture of a camera to match the light levels. 7. Averted vision refers to the practice of looking at faint objects out of the corner of your eye where there are more receptors for low light levels than at the center of the retina. 8. Right ascension (R.A.) is an astronomical coordinate that locates objects east-west on the sky, much like longitude locates objects east-west on the Earth. Technically, right ascension is measured eastwards along the celestial equator from the vernal equinox (point where the Sun crosses the celestial equator going north) to the hour circle (a great circle in the equatorial system that passes through the celestial poles perpendicular to the celestial equator) of the object. 9. Declination (Dec.) is an astronomical coordinate that locates objects north-south on the sky, much as latitude locates objects north-south on the Earth. Technically, declination measures the angular distance of an object from the celestial poles. The celestial equator is 0 degrees, the north celestial pole is +90 degrees and the south celestial pole is -90 degrees. 10. Azimuth can be defined in a few ways, but most of the time it refers to the angular distance of an object from due north, measured eastwards up to 360 degrees. Answers to Self-Test 1. (c) As a star travels across the sky, its azimuth changes but the R.A. and Dec. remain the same. 2. (b) If a planet is at inferior conjunction, it is seen against the disk of the Sun, so it must rise and set with the Sun. 3. (a) If Mercury is at greatest western elongation it is close to 28 degrees west of the Sun, so that it rises just before the sun and is a morning star. 4. (d) Venus is an inferior planet (orbiting the Sun closer than the Earth) and so can be at inferior conjunction, i.e., it can pass between the Earth and the Sun. 5. (e) When your eye is dark adapted, its pupil is dilated and it is much more sensitive to light than during the day.