Introductory Lab 1 ASTRO

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Astronomy

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Oct 30, 2023

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Introductory Lab: Scales ( 75pts.) It has been suggested that man is about halfway in size between an atom and a star, a somewhat deceptive description (Sir Arthur Eddington). However, if we use mass as the measurement then man is about halfway between a proton and the Earth. 1.0 Scales The study of Planetary Science involves having some basic understanding of a wide variety of scientific fields. Some are well known (astronomy and physics) and we have been collecting information and data about them for hundreds, if not thousands of years. Others, such as remote sensing, atmospherics and atomic chemistry, are younger and we are just really beginning to discover what they are and how they work. Because planets are fundamental objects in our Universe (after all, we live on one) it is important for us to understand how the Earth came to be, how it works and how we can use knowledge of these things to our best advantage. It turns out, that the formation of a star and the planets and moons around it, involves the chemistry of matter, the atomic interactions of matter and energy, geologic processes, atmospheric processes, and the physics of large interacting bodies. In order to study these things, we need a fundamental understanding of such things as optics, the Electromagnetic Spectrum, chemistry, geologic processes and products, remote sensing, orbital mechanics…the list goes on and on. Before we get into these things though, we must understand the scales involved in the construction and study of atoms to galaxies. 1.1 Standard and metric (4pts) The meter is the basic distance measure used worldwide in science today. Many general web sites and books use both measurements systems but almost every serious science publication lists measurement values in metric only. This is a short exercise to help review or familiarize you with that system. It is base 10 and so easy to learn, particularly if you pay attention to the prefix as it relates to the Latin word determining size (see Table 1 below). Centi- is the metric prefix denoting division of a meter by 100 (or multiplication by 0.01). On the answer sheet under 1.1 fill in the appropriate distance equivalents based on a meter. 1 kilometer = _______1,000 _______ 1 hectometer = ______100_______ 1 decameter = _____10________ 1 decimeter = ______10_______ 1 centimeter = _____100________ 1 millimeter = _____1,000________ 1 nanometer = _______1.0x10^9 ____________ Fill in the appropriate distance equivalents based on feet : 1 yard = _____3__________

1 mile = _____5,280 __________ Converting standard to metric if there are 2.54cm in an inch : 1 inch = _____25.4________mm 1 millimeter = ___0.0393701______in 1 foot = _______30.48______cm 1 centimeter = ____0.393701 _____in 1 meter = ___1.09361_______yd 1 kilometer = ____0.621371____mi 1 mile = ______1.60934 _______km 1.2 Scientific Notation (6pts) S.N. is base 10. As a result, 500 = 5.0 X 10 2 . The number “5” is called the coefficient. To write a number in scientific notation: Put the decimal after the first digit and drop the zeroes. In the number 123,000,000,000 the coefficient will be 1.23 and to find the exponent count the number of places from the decimal to the end of the number. In 123,000,000,000 there are 11 places. Therefore, we write 123,000,000,000 as: 1.2.1 On the answer sheet determine and fill in the S.N. values for the numbers below. a. 10 = _____1.0x10^1___________ b. 1,000,000 = _____1.0x10^6__________ c. 52,300,000 = ____5.23x10^7__________ d. .00127 = _____1.27x10^-3__________ e. .00000000755 = ______7.55x10^-9___________ f. -32,956,888 = -3.2956888 * 10^7

Adding and multiplying S.N. (12) 1.2.2 Determine the values for the problems below (show your work & answers in S.N.) a) (1.2 x 10 3 ) + (2.4 x 10 3 ) 3.6x10^3 b) (4.5 x 10 4 ) + (2.1 x 10 6 ) 45000 + 2100000 2.145*10^6 c) (4.9 × 10 2 ) + (7.9 × 10 3 ) 490 + 7900 8.39x10^3 d) (4.9 × 10 -6 ) – (7.9 × 10 -5 ) .0000049 - .000079 0.0000741 -7.41 x 10^-5 e) (2.6 × 10 5 ) (9.2 × 10 –13 ) 23.92 = 2.6 * 9.2 10^5*10^-13 = 10^-8 2.392 * 10^-7 f) (7.942 x 10 -3 ) / (3.897 x 10 -7) 2.03797793174 = 7.942 / 3.897 (-3) - (-7) = 4 2.03797793174 x 10^4

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1.3 Orders of Magnitude Planetary Science is an integrated field that involves the physical characteristics of matter ranging from the minute to the cosmic. As a result, it requires a grasp of the concept “Orders Of Magnitude” (OOM). In its most common form this scale is logarithmic, meaning that each unit increase (or decrease) is a power of 10 greater (or smaller) than the preceding unit (see Table 1). In words (long scale) In words (short scale) Prefi x S y m b o l Decimal Powe r of te n Order of magnitude quadrillionth septillionth yocto - y 0.0000000000000000000000 01 10 −24 −24 trilliardth sextillionth zepto- z 0.000000000000000000001 10 −21 −21 trillionth quintillionth atto- a 0.000000000000000001 10 −18 −18 billiardth quadrillionth femto - f 0.000000000000001 10 −15 −15 billionth trillionth pico- p 0.000000000001 10 −12 −12 milliardth billionth nano- n 0.000000001 10 −9 −9 millionth millionth micro - µ 0.000001 10 −6 −6 thousandth thousandth milli- m 0.001 10 −3 −3 hundredth hundredth centi- c 0.01 10 −2 −2 Tenth tenth deci- d 0.1 10 −1 −1 One one - - 1 10 0 0 Ten ten deca- d 10 10 1 1 hundred hundred hecto- h 100 10 2 2 thousand thousand kilo- k 1,000 10 3 3 million million mega- M 1,000,000 10 6 6 milliard billion giga- G 1,000,000,000 10 9 9 billion trillion tera- T 1,000,000,000,000 10 12 12 billiard quadrillion peta- P 1,000,000,000,000,000 10 15 15 trillion quintillion exa- E 1,000,000,000,000,000,000 10 18 18 trilliard sextillion zetta- Z 1,000,000,000,000,000,000,0 00 10 21 21 quadrillion septillion yotta- Y 1,000,000,000,000,000,000,0 00,000 10 24 24 Table 1. Metric prefixes, symbols, powers and Orders Of Magnitude. This table shows orders of magnitude in a base 10 system. Other systems can be used and have been used over time, but most

science work is done in base 10. Familiarity with this system helps when dealing with integrated sciences as they can run the gamut of scales from sub-atomic (the size of a proton or other sub-atomic particle) to cosmologic (the numbers involved when measuring the distances to other galaxies and stars). Although it is difficult for us to truly understand some of these numbers, reducing large, unwieldy numbers to simpler concepts like “orders of magnitude” begin to provide a baseline for us to comprehend the scale of the various physical matter we will deal with in studying Planetary Science. 2.0 Size of the Atom The nucleus of an atom has a diameter of about 10 -15 m, whereas the atomic diameter is about 10 -11 m. This means that the nucleus of an atom has a diameter 10,000 times smaller than the entire atom when the orbital cloud of its electrons is taken into account. As a result, we find that atoms are dominated by empty space. The great amount of empty space in an atom can be illustrated by the following analogy. Imagine the nucleus to be the size of a golf ball (~4cm). Then on this scale the first electron shell would be about one kilometer from the golf ball, the second shell about four kilometers, the third nine kilometers and so on. If you find that hard to visualize then try this: the period at the end of this sentence, (depending on your monitor and the font you are using), is probably about 1/2 a millimeter in diameter. If that period represents the nucleus then the electrons in the first shell would be orbiting with a diameter about 50 meters around you. In fact, the actual diameter of an atom is very small and it would require some two hundred million of them, side by side, to form a line a centimeter long. 3.0 A light-year and the Astronomical Unit (AU) A light-year is not a unit of time but a unit of length, equal to just less than 10 trillion kilometers (10 16 m or 6 trillion miles). The International Astronomical Union (IAU) defines a light-year as the distance light travels in a vacuum in one Julian year. Because of the great distance that light can travel in a year, using this distance as a unit of length can make it easier to comprehend the vast distances between objects in our Universe. Another common unit used in measuring distances, mostly in our Solar System, is the Astronomical Unit (A.U.). An A.U. is the average distance from our Sun to the Earth, 149,597,870.7km (92,955,807.27mi) or about 93million miles. For comparison purposes, it takes light from the Sun about 8mins to reach Earth. One light-year is equal to: ● exactly 9,460,730,472,580.8 km ● about 5,878,625,373,183.608 miles (about 6 trillion miles) ● about 63,241.1 Astronomical Units ● about 0.306601 parsecs 4.0 Size of the Solar System

The Sun is the center of our Solar System (SS). Until recently, our SS was the only one we knew of and so completely biased our understanding of planetary formation. Today, we know of about 450 other solar systems (most have a single planet that we know of but that too, is a function of bias due to the limitations of our technology) so we can extrapolate from that to theorize most stars will have some type or evolutionary phase of solar system associated with them. Our SS can be thought of having evolved to a mature age. In this stage, the similarity of our SS, albeit at a much greater scale, is very similar in construction to an atom; i.e., a central dense nucleus (the Sun) with much smaller particles orbiting around it at mathematically predictable distances. One interesting aspect of this comparison is that both entities are composed of mostly the same thing…….space. The size of our SS depends upon what objects we assume are an integral part of such a celestial community (Figure 1). If we just consider a star and planets then our SS is about 4.5 x 10 9 or 30 AU If we include the Sun’s influence on space then it extends out to the heliopause, the sphere around the Sun where its solar wind is stopped by the interstellar wind within the Milky Way, ~90 AU from the Sun. If we use the definition including the furthest object ever detected (Sedna), its orbit varies out to a distance of 975 A.U. If we include objects in the Oort Cloud, a theoretical cloud of cometary material surrounding the Sun in all directions, then the diameter of the SS grows to between 50 – 100,000 AU, a distance covering ~1/3 the distance to the closest star, Proxima Centauri. Finally, if we used the gravitational influence of the Sun, some astronomers believe it extends roughly 2 light-years out, not quite half the distance to Proxima Centauri. Take a moment to examine the table above and determine the OOM difference between the sizes of these various Solar Systems. If we consider our SS based on the total amount of mass of material contained within it, we discover it is mostly just a star. 99.87% of all matter in the SS is contained in the Sun while most of the other 0.13% is contained in Jupiter, much like the atom where the preponderance of mass is in the nucleus, not the orbiting electron cloud. It doesn’t take long at these scales to realize how insignificant our tiny blue planet and its anthropocentric population really are. Figure 1. Our Solar System has many parts in many locations relative to the Sun. As a result, the size of the SS is dependent on what components you include in the definition of the SS. 5.0 Size of the Universe The Universe is vast place. Even the concept of it is difficult to comprehend at a fundamental level. The better our technology gets the more of our Universe we seem able to see. However, recent discoveries suggest there is a limit to how far we can see and it’s not restricted by the complexity and sophistication of our technology. It is limited by time and the physical nature and make-up of the Universe itself. How did it start? Where did it come from? How old is it? These are all questions we are beginning to answer

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and understand. Much of our understanding is based on electromagnetic energy, its physical properties and how it enables us to “see” the Universe. It turns out, that the physical properties of matter during a period shortly after the “Big Bang” may be the wall beyond which we will never be able to see. The age of the Universe is about 13.75 billion years. This age is based on the speed of light and the distance light has traveled from very distant but visible objects in our Universe. But, due to the expansion of space we are now observing objects that are now considerably farther away than they should be given that age. The diameter of the observable universe is estimated to be about 28 billion parsecs (93 billion light-years) putting the edge of the observable universe at about 46–47 billion light-years away. To give you an idea of the magnitude of this distance, the most distant space probe humans have ever launched into space is Voyager 1. It is about 16 light-hours away from Earth. This translates to ~21,750,000,021,025km and ~145 A.U. from the Earth. It took that space probe 41 years to cover that distance and will take over 18,000 years to reach one light-year at the same speed. Voyager 1 reached the heliopause on Aug. 25 th , 2012, at a distance of >133A.U entering interstellar space; the first man-made object to exit our Solar System. Questions (53pts) (9) Find and record the values for the three celestial objects below (provide both metric and standard/imperial values . Radius Diameter Circumference a. Sun 432,690mi or 696,000km 865,000mi or 1,400,000km 2,715,396mi or 4,370,006km b. Earth 3,958.8mi or 6,371km 7,917.5mi or 12,742km 24,901mi or 40,075km c. Moon 1,080mi or 1,740km 2,159mi or 3,475km 6,783.5mi or 10,917km 2. (1) Write out the values for the speed of light in both miles/second and kilometers/sec using SN. The rounded speed of light is 1.86282397 x 10^5 miles per second or 2.99792458 x 10^5 kilometers per second 3. (1) Write out the value for an A.U. in both km & mi in SN. 149,597,870.7km (92,955,807.27mi) The value of an astronomical unit is 9.295580727 x 10^7 miles or 1.495978707 x 10^8 kilometers 4. (4) The “Light Year” is the distance light travels in a vacuum in one year. Determine that distance in km, miles, A.U.’s and parsec’s, again using SN. ● exactly 9,460,730,472,580.8 km ● about 5,878,625,373,183.608 miles (about 6 trillion miles)

● about 63,241.1 Astronomical Units ● about 0.306601 parsecs The value of a light year is 9.4607304725828 x 10^12 kilometers, 5.878625373183608 x 10^12 miles, 6.32411 x 10^4 Astronomical Units or about 3.06601 x 10^-1 parsecs 5. (6) The red-giant star Betelgeuse is ~1,180x the Sun’s diameter. Use the Sun’s diameter that you researched in Question 1, above to calculate the diameter of Betelgeuse. What is its diameter in km & mi? What is the diameter of this giant in A.U.? How many Orders Of Magnitude (OOM) larger is Betelgeuse than our Sun? If you were to replace our Sun with Betelgeuse as the center of our SS, provide a brief description of how its size would change the appearance of our Solar System. 1.18 * 10^3 * 8.65*10^5mi = 1.0207*10^9mi and 1.18*10^3 * 1.4*10^6km = 1.652*10^9km The Diameter of Betelgeuse is 1.0207*10^9mi or 1,020,700,000mi and 1.652*10^9km or 1,652,000,000km 1 AU=149,600,000km or 1.496*10^8, so Betelgeuse is (1,652,000,000km/149,600,000km) = 11.0427807487 AU Betelgeuse is 1 order of magnitude larger than our sun. Our Solar System is 1,921.56 AU, based on the influence that the sun has on Sedna, and with Betelgeuse at 11.04 AU, the planets Mercury, Venus, Earth, Mars, Jupiter and Saturn would be completely engulfed by the replacement of the sun by this new star. Our solar system would then be centered by Betelguese orbited by Uranus, Neptune, Pluto and Sedna. 6. (17) Measure and record the diameter of the blue bead you’ve been given in cm and meters (put the value in “a.” below). If the Earth is scaled down to the size of the blue bead, find the values for b - f at the same scale as the blue bead : a. (1) Diameter of blue bead The diameter of the blue is about 1 cm. b. (2) Diameter of the Sun (in meters) 1,400,000km or 1,400,000,000m. The sun’s diameter is 109 times larger than earth, so the sun is 109cm. The diameter of the sun is 1.09 meters. c. (2) Diameter of Betelgeuse (in meters) 1,652,000,000,000m. The diameter of Betelgeuse is 1286.2 meters. d. (2) Distance from the Sun to the Earth (in meters)

149,600,000/1400000 = about 107 117.175m is the distance from the sun to the earth. 149,600,000km/12742 = 11,740,700 is how many times larger the AU is to the earth. The distance from the sun to the earth is 11,740,700 meters. e. (6) Take the bead and the meter stick into the hallway or outside. In pairs, have one of you step off the distance you calculated for A.U. in “d” above. Determine the length of each stride at this scale between the Sun and Earth. Have whoever is the “Earth” look back at the “Sun”. Holding your arm out straight, does your finger cover the “Sun” or not? Tell me what this means regarding measurement of objects in the sky (i.e., how does the size of your thumb relate to apparent size or distance in the sky?). The length of each stride at thcme scale of the blue bead is 1 meter. When holding my arm straight out, my thumb did cover the student pretending to be the sun. At this scale the sun would be about 1 arc degree in diameter. Many other objects in the sky, such as the moon, would be about 1 arc degree in diameter. f. (4) What is the name of the nearest star to our Sun? How far away is it? At the scale determined by your measurement of the blue bead, if you were going to pace off the distance to this star, how far would you have to walk? The nearest star to our sun is Proxima Centauri 4.2 light years. A lightyear is 9.46 * 10^12km. 3.973*10^13km is the distance from our sun to Proxima Centauri in KM. (3.973*10^13)/12742 = 318034845.39. This makes the distance from the sun to Proxima Centauri about 318034845.39 earths, or cm. This means that I would have to walk about 3180.3484539 km when the distance is scaled down to the measurement of the blue bead. 7. (4) Take the atom example in the text above (Section 2.0) and convert it to a scale so that the nucleus is the same size as your blue bead. At this new scale, calculate the distance from the nucleus to the first electron shell. How does this distance compare to your example of the Sun- Earth system in question 6? Show your work. Nucleus = 1cm. If a golf ball is 4cm and the first electron shell is 1 km or 100,000cm away. Divide both the golf ball and the distance by 4. 4/4 = 1cm and 100,000/4 = 25,000cm. This means, therefore, that the distance that the first Electron Shell from the 1cm Atomic Nucleus is 25,000cm. In comparison to the distance I would have to walk to reach Proxima Centauri when in scale of the blue bead, the distance from the Atomic Nucleus to the first electron shell is extremely small. This distance is about .25km. That is just over .15mi. Overall, in comparison the distance to walk from the Atomic Nucleus to the first electron shell is much smaller than the distance to walk from the sun to Proxima Centauri. 8. (11) Using Table 1 from above, answer the following questions in Orders of Magnitude (OOM) : a. (1) Assume the average human is 2 meters tall. Approximately how much larger is the Earth (using its diameter)?

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12,742km is the diameter of the earth. The earth is 63,781,000 times larger than the average human. b. (1) How much larger is the Sun than the average human (using its diameter) ? 1,400,000km is the diameter of the Sun. The sun is 7.0x10^8 times larger than the average human. c. (1) How much longer is a light-year than the average human? 9,460,730,472,580.8km is a light year. A lightyear is 4.73037*10^15 times larger than the average human. d. (2) The Moon is approximately 384,403km from the Earth. How long does it take for light reflected off the surface of the Moon to reach your eye on the Earth? Show your work. 1.28 seconds. The speed of light is 3.0*10^5kps. 384,403km/300,000kps = 1.28 seconds. e. (2) How long would take you to drive around the Earth if you were traveling at 80mph ? Give your answer in hrs., and days. 24,901mi is the circumference of the earth. 24,901/80 = 311.27 hours or 12.97 days. It would take me 311.27 hours or 12.97 days to travel around the earth if I was traveling at a constant pace of 80mph. f. (2) To drive to the Moon at 80mph (in days)? 384,403km = 238,856.95mi. 238,856.95mi/80mph = 2985.71 hours or 124.4 days. g. (2) To drive to the Sun at 80mph (in yrs.)? 149,600,000km = The distance from the sun to the earth. 149,600,000km = 92957130.3587mi. 92957130.3587hr/80mph = 1161964.12948 hours, or 48415.17 days or 132.64 years.