Complete Lab Manual for The Planets | ASTR 105G, Lab Reports of Astronomy

Download Complete Lab Manual for The Planets | ASTR 105G and more Lab Reports Astronomy in PDF only on Docsity! ASTR 105G Lab Manual Spring 2009 Astronomy Department New Mexico State University 3. If one meter equals 40 inches, how many meters are there in 400 inches? 4. How many centimeters are there in 400 inches? 5. In August 2003, Mars made its closest approach to Earth for the next 50,000 years. At that time, it was only about .373 AU away from Earth. How many km is this? 1.4.1 Map Exercises One technique that you will use this semester involves measuring a photograph or image with a ruler, and converting the measured number into a real unit of size (or distance). One example of this technique is reading a road map. Figure 1.1 shows a map of the state of New Mexico. Down at the bottom left hand corner of the map is a scale in both miles and kilometers. Use a ruler to determine (2 points each): 6. How many kilometers is it from Las Cruces to Albuquerque? 7. What is the distance in miles from the border with Arizona to the border with Texas if you were to drive along I-40? 8. If you were to drive 100 km/hr (kph), how long would it take you to go from Las Cruces to Albuquerque? 9. If one mile = 1.6 km, how many miles per hour (mph) is 100 kph? 3 ere ea Figure 1.1: Map of New Mexico. 1.5 Squares, Square Roots, and Exponents In several of the labs this semester you will encounter squares, cubes, and square roots. Let us briefly review what is meant by such terms as squares, cubes, square roots and exponents. The square of a number is simply that number times itself: 3 × 3 = 32 = 9. The exponent is the little number “2” above the three. 52 = 5 × 5 = 25. The exponent tells you how many times to multiply that number by itself: 84 = 8 × 8 × 8 × 8 = 4096. The square of a number simply means the exponent is 2 (three squared = 32), and the cube of a number means the exponent is three (four cubed = 43). Here are some examples: • 72 = 7 × 7 = 49 • 75 = 7 × 7 × 7 × 7 × 7 = 16,807 • The cube of 9 (or “9 cubed”) = 93 = 9 × 9 × 9 = 729 • The exponent of 1216 is 16 • 2.563 = 2.56 × 2.56 × 2.56 = 16.777 Your turn (2 points each): 10. 63 = 11. 44 = 12. 3.12 = The concept of a square root is fairly easy to understand, but is much harder to calculate (we usually have to use a calculator). The square root of a number is that number whose square is the number: the square root of 4 = 2 because 2 × 2 = 4. The square root of 9 is 3 (9 = 3 × 3). The mathematical operation of a square root is usually represented by the symbol “ √ ”, as in √ 9 = 3. But mathematicians also represent square roots using a fractional exponent of one half: 91/2 = 3. Likewise, the cube root of a number is represented as 271/3 = 3 (3 × 3 × 3 = 27). The fourth root is written as 161/4 (= 2), and so on. Here are some example problems: 5 19. −0.121 = There is one issue we haven’t dealt with, and that is when to write numbers in scientific notation. It is kind of silly to write the number 23.7 as 2.37 × 101, or 0.5 as 5.0 × 10−1. You use scientific notation when it is a more compact way to write a number to insure that its value is quickly and easily communicated to someone else. For example, if you tell someone the answer for some measurement is 0.0033 meter, the person receiving that information has to count over the zeros to figure out what that means. It is better to say that the measurement was 3.3 × 10−3 meter. But telling someone the answer is 215 kg, is much easier than saying 2.15 × 102 kg. It is common practice that numbers bigger than 10,000 or smaller than 0.01 are best written in scientific notation. 1.7 Calculator Issues Since you will be using calculators in nearly all of the labs this semester, you should become familiar with how to use them for functions beyond simple arithmetic. 1.7.1 Scientific Notation on a Calculator Scientific notation on a calculator is usually designated with an “E.” For example, if you see the number 8.778046E11 on your calculator, this is the same as the number 8.778046 ×1011. Similarly, 1.4672E-05 is equivalent to 1.4672 ×10−5. Entering numbers in scientific notation into your calculator depends on layout of your cal- culator; we cannot tell you which buttons to push without seeing your specific calculator. However, the “E” button described above is often used, so to enter 6.589 ×107, you may need to type 6.589 “E” 7. Verify that you can enter the following numbers into your calculator: • 7.99921 ×1021 • 2.2951324 ×10−6 1.7.2 Order of Operations When performing a complex calculation, the order of operations is extremely important. There are several rules that need to be followed: i. Calculations must be done from left to right. ii. Calculations in brackets (parenthesis) are done first. When you have more than one set of brackets, do the inner brackets first. 8 iii. Exponents (or radicals) must be done next. iv. Multiply and divide in the order the operations occur. v. Add and subtract in the order the operations occur. If you are using a calculator to enter a long equation, when in doubt as to whether the calculator will perform operations in the correct order, apply parentheses. Use your calculator to perform the following calculations (2 points each): 20. 7+34 2+23 = 21. (42 + 5) - 3 = 22. 20 ÷ (12 - 2) × 32 - 2 = 1.8 Graphing and/or Plotting Now we want to discuss graphing data. You probably learned about making graphs in high school. Astronomers frequently use graphs to plot data. You have probably seen all sorts of graphs, such as the plot of the performance of the stock market shown in Fig. 1.2. A plot like this shows the history of the stock market versus time. The “x” (horizontal) axis represents time, and the “y” (vertical) axis represents the value of the stock market. Each place on the curve that shows the performance of the stock market is represented by two numbers, the date (x axis), and the value of the index (y axis). For example, on May 10 of 2004, the Dow Jones index stood at 10,000. Plots like this require two data points to represent each point on the curve or in the plot. For comparing the stock market you need to plot the value of the stocks versus the date. We call data of this type an “ordered pair.” Each data point requires a value for x (the date) and y (the value of the Dow Jones index). Table 1.2 contains data showing how the temperature changes with altitude near the Earth’s surface. As you climb in altitude, the temperature goes down (this is why high mountains can have snow on them year round, even though they are located in warm areas). The data points in this table are plotted in Figure 1.3. 1.8.1 The Mechanics of Plotting When you are asked to plot some data, there are several things to keep in mind. 9 Figure 1.2: The change in the Dow Jones stock index over one year (from April 2003 to July 2004). Altitude Temperature (feet) oF 0 59.0 2,000 51.9 4,000 44.7 6,000 37.6 8,000 30.5 10,000 23.3 12,000 16.2 14,000 9.1 16,000 1.9 Table 1.2: Temperature vs. Altitude First of all, the plot axes must be labeled. This will be emphasized throughout the semester. In order to quickly look at a graph and determine what information is being con- veyed, it is imperative that both the x-axis and y-axis have labels. Secondly, if you are creating a plot, choose the numerical range for your axes such that the data fit nicely on the plot. For example, if you were to plot the data shown in Table 1.2, with altitude on the y-axis, you would want to choose your range of y-values to be something like 0 to 18,000. If, for example, you drew your y-axis going from 0 to 100,000, then all of the data would be compressed towards the lower portion of the page. It is important to choose your ranges for the x and y axes so they bracket the data points. 10 Figure 1.4: Graph paper for plotting the hourly temperatures in Tucson and Honolulu. One of our primary goals this semester is to help you develop intuition about our solar sys- tem. This includes recognizing if an answer that you get “makes sense.” For example, you may be told (or you may eventually know) that Mars is 1.5 AU from Earth. You also know that the Moon is a lot closer to the Earth than Mars is. So if you are asked to calculate the Earth-Moon distance and you get an answer of 4.5 AU, this should alarm you! That would imply that the Moon is three times farther away from Earth than Mars is! And you know that’s not right. Use your intuition to answer the following questions. In addition to just giving your answer, state why you gave the answer you did. (4 points each) 13 27. Earth’s diameter is 12,756 km. Jupiter’s diameter is about 11 times this amount. Which makes more sense: Jupiter’s diameter being 19,084 km or 139,822 km? 28. Sound travels through air at roughly 0.331 kilometers per second. If BX 102 suddenly exploded, which would make more sense for when people in Mesilla (almost 5 km away) would hear the blast? About 14.5 seconds later, or about 6.2 minutes later? 29. Water boils at 100 ◦C. Without knowing anything about the planet Pluto other than the fact that is roughly 40 times farther from the Sun than the Earth is, would you expect the surface temperature of Pluto to be closer to -100◦ or 50◦? 1.10 Putting it All Together We have covered a lot of tools that you will need to become familiar with in order to complete the labs this semester. Now let’s see how these concepts can be used to answer real questions about our solar system. Remember, ask yourself does this make sense? for each answer that you get! 30. To travel from Las Cruces to New York City by car, you would drive 3585 km. What is this distance in AU? (4 points) 14 31. The Earth is 4.5 billion years old. The dinosaurs were killed 65 million years ago due to a giant impact by a comet or asteroid that hit the Earth. If we were to compress the history of the Earth from 4.5 billion years into one 24-hour day, at what time would the dinosaurs have been killed? (4 points) 32. The New Horizons spacecraft is traveling at approximately 20 kilometers per second. How long will it take to reach Jupiter, which is roughly 4 AU from Earth? [Hint: see the definition of an AU in Section 1.3 of this lab.] (4 points) 15 look at the planets in our Solar System, we can see that the planet Mercury, which orbits nearest to the Sun, has an average distance of 0.4 AU and Pluto, the planet almost always farthest from the Sun, has an average distance of 40 AU. Thus, Earth’s distance from the Sun, which we think of as large (75 million times greater than the average distance you commute to campus!) is only 2.5 percent of the distance between the Sun and planet Pluto!! This is a tremendous distance indeed. Now, how can we put all these distances into perspective on the field? For our Scale Model, the Sun will be located at the goal line of the North end zone. Below you will proceed through a number of steps that will allow for the development of a scale-model of the solar system. 2.2 Distance From the Sun Fill in the first and second columns of Table 2.1. In other words, list, in order of increasing distance from the Sun, the planets in our solar system and their average distances from the Sun in units of Astronomical Units (AU). You can find these numbers in Appendix E of your textbook. (21 points) Average Distance From Sun Planet AU Yards Earth 1 2.5 Pluto 40 100 Table 2.1: Planets’ average distances from Sun. Determine the SCALED orbital semi-major axes of the planets, based upon the assumption that the Sun-to-Pluto average distance in Astronomical Units (which you have written down above) is represented by 100 yards, or goal-line to goal-line, on the football field. To deter- mine similar scalings for each of the planets, use the following equation: Distance of the planet from the Sun’s goal line (in units of yards) = (Planet’s average distance in AU) × (100 yards / 40 AU ) 18 Write these values on the lines provided in Table 2.1 (fill in the third column). 2.3 Sizes of Planets You have just determined where on the football field the planets are located (how far from the Sun they are located in our scale model). Now it is time to determine how large (or small) the planets themselves are on the same scale. We have already indicated above that the diameter of the Earth is 12,756 kilometers, while the distance from the Sun to Earth (1 AU) is equal to 150,000,000 km. We already deter- mined above that in our scale model, 1 AU is represented by 2.5 yards (90 inches). We will start here by using the largest object in the solar system as our ‘tool’ for deter- mining how large the objects will be in our scale model. The Sun is the largest single object in the Solar System. It has a diameter of ∼ 1,400,000 (1.4 million) kilometers. [Note that the Sun’s diameter is ∼100 times greater than the Earth’s di- ameter!!] Since in our scaled model 150,000,000 kilometers (1 AU) is equivalent to 90 inches, how many inches will correspond to 1,4000,000 kilometers (the Sun’s actual diameter)? This can be determined by the following calculation: Sun’s scaled diameter (inches) = Sun’s true diameter (km) × (90 in.) (150,000,000 km) Sun’s scaled diameter = 0.84 inches So, on the scale of our football field solar system, the scaled Sun has a diameter of only 0.84 inches!! Now that we have established the scaled Sun’s size (0.84 inch diameter), let’s proceed through a similar exercise for each of the nine planets, and the Moon (diameter = 3476 kilometers), using the following formula: Scaled diameter (inches) = actual diameter (km) × (90 in.) (150,000,000 km) Fill in the values in Table 2.2 (8 points). 19 Object Actual Diameter (km) Scaled Diameter (inches) Sun ∼ 1,400,000 0.84 Mercury 4,878 Venus 12,104 Earth 12,756 0.0075 Moon 3,476 Mars 6,794 Jupiter 142,800 Saturn 120,540 Uranus 51,200 Neptune 49,500 Pluto 2,200 0.0013 Table 2.2: Planets’ diameters in a football field scale model. Now you have all the information required to create your scale model of the Solar System. Use any of the items listed in Table 2.3 (spheres of different diameter) to select your scaled planets, Sun and Moon. NOTE: Some of the items are not the appropriate size for this scale model!! Object Diameter (inches) Basketball 15 Tennis ball 2.5 Golf ball 1.625 Marble 0.5 Sesame seed 0.07 Poppy seed 0.04 Ground flour 0.001 Table 2.3: Everyday objects that could represent the planets. Designate one person for each planet, one person for the Sun, and one person for the Earth’s Moon. Each person should choose the model object which represents their solar system object, and then walk (or run) to that object’s scaled orbital semi-major axis on the football field. The Sun will be on the goal line of the North end zone (towards the Pan Am Center) and Pluto will be on the south goal line. 20 1. List the planets in our solar system and their average distances from the Sun in units of Astronomical Units (AU). Then, using a scale of 40 AU = 455 miles (1 AU = 11.375 miles), determine the scaled planet-Sun distances and the city near the location of this planet’s scaled average distance from the Sun. Insert these values into Table 2.4, and draw on your map of New Mexico (on the next page) the locations of the solar system objects. (22 points) Average Distance from Sun Planet in AU in Miles City Earth 1 11.375 Pluto 40 455 3 miles north of Raton Table 2.4: Planets’ average distances from Sun. 2. Determine the scaled size (diameter) of objects in the Solar System for a scale in which 40 AU = 455 miles, or 1 AU = 11.375 miles). Insert these values into Table 2.5. (19 points) Scaled diameter (feet) = actual diameter (km) × (11.4 mi. × 5280 ft/mile) 150,000,000 km Object Actual Diameter (km) Scaled Diameter (feet) Object Sun ∼ 1,400,000 561.7 Mercury 4,878 Venus 12,104 Earth 12,756 5.1 height of 12 year old Moon 3,476 Mars 6,794 Jupiter 142,800 Saturn 120,540 Uranus 51,200 Neptune 49,500 Pluto 2,200 Table 2.5: Planets’ diameters in a New Mexico scale model. 23 et cine ae yl os eee AAS San Poo, mae BE "oi eh Name(s): Date: 3 Phases of the Moon 3.1 Introduction You will need the following materials for this lab: • small spheres (representing the Moon), with two different colored hemispheres. The dark hemisphere represents the portion of the Moon not illuminated by the Sun. • flashlight (representing the Sun) • yourself (representing the Earth, with your nose representing the location of Las Cruces) Work in Groups of Three People! The objective of this lab is to improve your understanding of the Moon phases [a topic that you WILL see on future exams!]. This concept, the phases of the Moon, involves 1. the position of the Moon in its orbit around the Earth, 2. the illuminated portion of the Moon that is visible from here in Las Cruces, and 3. the time of day that a given Moon phase is at the highest point in the sky as seen from Las Cruces. For this lab, you will finish by demonstrating to your instructor that you do clearly under- stand the concept of Moon phases, including an understanding of • which direction the Moon travels around the Earth • how the Moon phases progress from day-to-day • at what time of the day the Moon is highest in the sky at each phase You will use the colored sphere and flashlight as props for this demonstration. Carefully read and thoroughly answer the questions associated with each of the five Exercises on the following pages. [Don’t be concerned about eclipses as you answer the questions in these Exercises]. Using the dual-colored sphere to represent the Moon, the flashlight to represent the Sun, and a member of the group to represent the Earth (with that person’s nose repre- senting Las Cruces’ location), ‘walk through’ and ‘rotate through’ the positions indicated in the Exercise figures to fully understand the situation presented. Each Exercise is worth 10 points. There are additional questions at the end. 25 3.4 Exercise 3 Shown below are different phases of the Moon as seen by an observer in the Northern Hemisphere. A B C D E Ranking Instructions: Beginning with the waning gibbous phase of the Moon, rank the moon phases shown below in the order that the observer would see them over the next four weeks. Ranking Order: First phase following waning gibbous phase 1 2 3 4 5 Last phase seen Or, all of these phases would be visible at the same time. (indicate with a check mark). 28 3.5 Exercise 4 In the set of figures below, the Moon is shown in the first quarter phase at different times of the day (or night). Assume that sunset occurs at 6 p.m. and that sunrise occurs at 6 a.m. Instructions: Determine the time at which each view of the Moon would have been seen, and write it on each panel of the figure. 29 3.6 Exercise 5 In the set of figures below, the Moon is shown overhead, at its highest point in the sky, but in different phases. Assume that sunset occurs at 6 p.m. and that sunrise occurs at 6 a.m. Instructions: Determine the time at which each view of the Moon would have been seen, and write it on each panel of the figure. 30 Name: Date: 3.9 Take-Home Questions The following questions are worth 11 points each. 1. If the Earth was one-half as massive as it actually is, how would the time interval (number of days) from one Full Moon to the next in this ‘small Earth mass’ situa- tion compare to the actual time interval of 29.5 days between successive Full Moons? Assume that all other aspects of the Earth and Moon system, including the Moon’s orbital semi-major axis, the Earth’s rotation rate, etc. do not change from their cur- rent values. 2. If you were on Earth looking up at a Full Moon at midnight, and you saw an astronaut at the center of the Moon’s disk, what phase would the astronaut be seeing the Earth in? Draw a diagram to support your answer. 33 34 Name(s): Date: 4 The Origin of the Seasons 4.1 Introduction The origin of the science of Astronomy owes much to the need of ancient peoples to have a practical system that allowed them to predict the seasons. It is critical to plant your crops at the right time of the year—too early and the seeds may not germinate because it is too cold, or there is insufficient moisture. Plant too late and it may become too hot and dry for a sensitive seedling to survive. In ancient Egypt, they needed to wait for the Nile to flood. The Nile river would flood every July, once the rains began to fall in Central Africa. Thus, the need to keep track of the annual cycle arose with the development of agriculture, and this required an understanding of the motion of objects in the sky. The first devices used to keep track of the seasons were large stone structures (such as Stonehenge) that used the positions of the rising Sun or Moon to forecast the coming seasons. The first recognizable calendars that we know about were developed in Egypt, and appear to date from about 4,200 BC. Of course, all a calendar does is let you know what time of year it was, it does not provide you with an understanding of why the seasons occur! The ancient people had a variety of models for why seasons occurred, but thought that everything, including the Sun and stars, orbited around the Earth. Today, you will learn the real reason why there are seasons. • Goals: To learn why the Earth has seasons. • Materials: a meter stick, a mounted styrofoam globe, an elevation angle apparatus, string, a halogen lamp, and a few other items 4.2 The Seasons Before we begin today’s lab, let us first talk about the seasons. In New Mexico we have rather mild Winters, and hot Summers. In the northern parts of the United States, however, the winters are much colder. In Hawaii, there is very little difference between Winter and Summer. As you are also aware, during the Winter there are fewer hours of daylight than in the Summer. In Table 4.1 we have listed seasonal data for various locations around the world. Included in this table are the average January and July maximum temperatures, the latitude of each city, and the length of the daylight hours in January and July. We will use this table in Exercise #2. In Table 4.1, the “N” following the latitude means the city is in the northern hemisphere of the Earth (as is all of the United States and Europe) and thus North of the equator. An “S” following the latitude means that it is in the southern hemisphere, South of the Earth’s equator. What do you think the latitude of Quito, Ecuador (0.0o) means? Yes, it is right on the equator. Remember, latitude runs from 0.0o at the equator to ±90o at the poles. If north of the equator, we say the latitude is XX degrees north (or sometimes “+XX degrees”), and 35 6) At that time of year, what season is it in Las Cruces? What do you conclude about the statement “the seasons are caused by the changing distance between the Earth and the Sun”? (4pts) Exercise #2. Characterizing the nature of the seasons at different locations. For this exercise, we are going to be exclusively using the data contained in Table 4.1. First, let’s look at Las Cruces. Note that here in Las Cruces, our latitude is +32.5o. That is we are about one third of the way from the equator to the pole. In January our average high temperature is 57oF, and in July it is 96oF. It is hotter in Summer than Winter (duh!). Note that there are about 10 hours of daylight in January, and about 14 hours of daylight in July. Thus, for Las Cruces, the Sun is “up” longer in July than in January. 7) Is the same thing true for all cities with northern latitudes: Yes or No ? (1pt) Ok, let’s compare Las Cruces with Fairbanks, Alaska. Answer these questions by filling in the blanks: 8) Fairbanks is the North Pole than Las Cruces. (1pt) 9) In January, there are more daylight hours in . (1pt) 10) In July, there are more daylight hours in . (1pt) Now let’s compare Las Cruces with Sydney, Australia. Answer these questions by filling in the blanks: 12) While the latitudes of Las Cruces and Sydney are similar, Las Cruces is of the Equator, and Sydney is of the Equator. (2pts) 13) In January, there are more daylight hours in . (1pt) 14) In July, there are more daylight hours in . (1pt) 38 15) Summarizing: During the Wintertime (January) in both Las Cruces and Fairbanks there are fewer daylight hours, and it is colder. During July, it is warmer in both Fair- banks and Las Cruces, and there are more daylight hours. Is this also true for Sydney?: . (1pt) 16) In fact, it is Wintertime in Sydney during , and Summertime during . (2pts) 17) From Table 4.1, I conclude that the times of the seasons in the Northern hemisphere are exactly to those in the Southern hemisphere. (1 pt) From Exercise #2 we learned a few simple truths, but ones that maybe you have never thought about. As you move away from the equator (either to the north or to the south) there are several general trends. The first is that as you go closer to the poles it is generally cooler at all times during the year. The second is that as you get closer to the poles, the amount of daylight during the Winter decreases, but the reverse is true in the Summer. The first of these is not always true because the local climate can be moderated by the proximity to a large body of water, or depend on the elevation. For example, Sydney is milder than Las Cruces, even though they have similar latitudes: Sydney is on the eastern coast of Australia (South Pacific ocean), and has a climate like that of San Diego, California (which has a similar latitude and is on the coast of the North Pacific). Quito, Ecuador has a mild climate even though it sits right on the equator due to its high elevation–it is more than 9,000 feet above sea level, similar to the elevation of Cloudcroft, New Mexico. The second conclusion (amount of daylight) is always true—as you get closer and closer to the poles, the amount of daylight during the Winter decreases, while the amount of daylight during the Summer increases. In fact, for all latitudes north of 66.5o, the Summer Sun is up all day (24 hrs of daylight, the so called “land of the midnight Sun”) for at least one day each year, while in the Winter there are times when the Sun never rises! 66.5o is a special latitude, and is given the name “Arctic Circle”. Note that Fairbanks is very close to the Arctic Circle, and the Sun is up for just a few hours during the Winter, but is up for nearly 22 hours during the Summer! The same is true for the southern hemisphere: all latitudes south of −66.5o experience days with 24 hours of daylight in the Summer, and 24 hours of darkness in the Winter. −66.5o is called the “Antarctic Circle”. But note that the seasons in the Southern Hemisphere are exactly opposite to those in the North. During Northern Winter, the North Pole experiences 24 hours of darkness, but the South Pole has 24 hours of daylight. 4.3 The Spinning, Revolving Earth It is clear from the preceeding that your latitude determines both the annual variation in the amount of daylight, and the time of the year when you experience Spring, Summer, Autumn and Winter. To truly understand why this occurs requires us to construct a model. One of the key insights to the nature of the motion of the Earth is shown in the long exposure photographs of the nighttime sky on the next two pages. 39 Figure 4.2: Pointing a camera to the North Star (Polaris, the bright dot near the center) and exposing for about one hour, the stars appear to move in little arcs. The center of rotation is called the “North Celestial Pole”, and Polaris is very close to this position. The dotted/dashed trails in this photograph are the blinking lights of airplanes that passed through the sky during the exposure. What is going on in these photos? The easiest explanation is that the Earth is spinning, and as you keep your camera shutter open, the stars appear to move in “orbits” around the North Pole. You can duplicate this motion by sitting in a chair that is spinning—the objects in the room appear to move in circles around you. The further they are from the “axis of rotation”, the bigger arcs they make, and the faster they move. An object straight above you, exactly on the axis of rotation of the chair, does not move. As apparent in Figure 4.3, the “North Star” Polaris is not perfectly on the axis of rotation at the North Celestial Pole, but it is very close (the fact that there is a bright star near the pole is just random chance). Polaris has been used as a navigational aid for centuries, as it allows you to determine the direction of North. As the second photograph shows, the direction of the spin axis of the Earth does not change during the year—it stays pointed in the same direction all of the time! If the Earth’s spin axis moved, the stars would not make perfect circular arcs, but would wander around in whatever pattern was being executed by the Earth’s axis. 40 Table 4.3: Position #1: Length of Night and Day Latitude Daylight Hours Nighttime Hours Equator 45oN Arctic Circle Antarctic Circle 18) The caption for Table 4.2 was “Equinox data”. The word Equinox means “equal nights”, as the length of the nighttime is the same as the daytime. While your numbers in Table 4.3 may not be exactly perfect, what do you conclude about the length of the nights and days for all latitudes on Earth in this experiment? Is this result consistent with the term Equinox? (3pts) Experiment #2: Now we are going to re-orient the globe so that the (top) polar axis points exactly away from the Sun and repeat the process of Experiment #1. Near the North Pole, there is a black line on the globe that allows you to precisely orient the globe: just make sure the shadow of the wooden axis falls on this line segment. With this alignment, the Earth’s axis should point exactly away from the Sun (you can spin the globe slightly for better alignment). Fill in the following two tables (4 pts): Table 4.4: Position #2: Solstice Data Table Latitude Length of Daylight Arc Length of Nightime Arc Equator 45oN Arctic Circle Antarctic Circle Table 4.5: Position #2: Length of Night and Day Latitude Daylight Hours Nighttime Hours Equator 45oN Arctic Circle Antarctic Circle 19) Compare your results in Table ?? for +45o latitude with those for Minneapolis in Table 4.1. Since Minneapolis is at a latitude of +45o, what season does this orientation of 43 the globe correspond to? (2 pts) 20) What about near the poles? In this orientation what is the length of the nighttime at the North pole, and what is the length of the daytime at the South pole? Is this consistent with the trends in Table 4.1, such as what is happening at Fairbanks or in Ushuaia? (4 pts) Experiment #3: Now we are going to approximate the Earth-Sun orientation six months after that in Experiment #2. To do this correctly, the globe and the lamp should now switch locations. Go ahead and do this if this lab is confusing you—or you can simply rotate the globe apparatus by 180o so that the North polar axis is tilted exactly towards the Sun. Try to get a good alignment by looking at the shadow of the wooden axis on the globe. Since this is six months later, it easy to guess what season this is, but let’s prove it! Complete the following two tables (4 pts): Table 4.6: Position #3: Solstice Data Table Latitude Length of Daylight Arc Length of Nightime Arc Equator 45oN Arctic Circle Antarctic Circle Table 4.7: Position #3: Length of Night and Day Latitude Daylight Hours Nighttime Hours Equator 45oN Arctic Circle Antarctic Circle 21) As in question #19, compare the results found here for the length of daytime and nighttime for the +45o degree latitude with that for Minneapolis. What season does this appear to be? (2 pts) 44 22) What about near the poles? In this orientation, how long is the daylight at the North pole, and what is the length of the nighttime at the South pole? Is this consistent with the trends in Table 4.1, such as what is happening at Fairbanks or in Ushuaia? (2 pts) 23) Using your results for all three positions (Experiments #1, #2, and #3) can you ex- plain what is happening at the Equator? Does the data for Quito in Table 4.1 make sense? Why? Explain. (3 pts) We now have discovered the driver for the seasons: the Earth spins on an axis that is inclined to the plane of its orbit (as shown in Figure 4.4). But the spin axis always points to the same place in the sky (towards Polaris). Thus, as the Earth orbits the Sun, the amount of sunlight seen at a particular latitude varies: the amount of daylight and nighttime hours change with the seasons. In Northern Hemisphere Summer (approximately June 21st) there are more daylight hours, at the start of the Autumn (∼ Sept. 20th) and Spring (∼ Mar. 21st) the days are equal to the nights. In the Winter (approximately Dec. 21st) the nights are long, and the days are short. We have also discovered that the seasons in the Northern and Southern hemispheres are exactly opposite. If it is Winter in Las Cruces, it is Summer in Sydney (and vice versa). This was clearly demonstrated in our experiments, and is shown in Figure 4.4. The length of the daylight hours is one reason why it is hotter in Summer than in Winter: the longer the Sun is above the horizon the more it can heat the air, the land and the seas. But this is not the whole story. At the North Pole, where there is constant daylight during the Summer, the temperature barely rises above freezing! Why? We will discover the reason for this now. 45 Figure 4.6: An ellipse with the major and minor axes defined. So, why are we making you measure these areas? Note that the black tube restricts the amount of light coming from the flashlight into a cylinder. Thus, there is only a certain amount of light allowed to come out and hit the paper. Let’s say there are “one hundred units of light” emitted by the flashlight. Now let’s convert this to how many units of light hit each square centimeter at angles of 90◦ and 45◦. At 90◦, the amount of light per centimeter is 100 divided by the Area of circle = units of light per cm2 (1 pt). At 45◦, the amount of light per centimeter is 100 divided by the Area of the ellipse = units of light per cm2 (1 pt). Since light is a form of energy, at which elevation angle is there more energy per square centimeter? Since the Sun is our source of light, what happens when the Sun is higher in the sky? Is its energy more concentrated, or less concentrated? How about when it is low in the sky? Can you tell this by looking at how bright the ellipse appears versus the circle? (4 pts) As we have noted, the Sun never is very high in the arctic regions of the Earth. In fact, at the poles, the highest elevation angle the Sun can have is 23.5◦. Thus, the light from the 48 Sun is spread out, and cannot heat the ground as much as it can at a point closer to the equator. That’s why it is always colder at the Earth’s poles than elsewhere on the planet. You are now finished with the in-class portion of this lab. To understand why the Sun appears at different heights at different times of the year takes a little explanation (and the following can be read at home unless you want to discuss it with your TA). Let’s go back and take a look at Fig. 4.3. Note that Polaris, the North Star, barely moves over the course of a night or over the year—it is always visible. If you had a telescope and could point it accurately, you could see Polaris during the daytime too. Polaris never sets for people in the Northern Hemisphere since it is located very close to the spin axis of the Earth. Note that as we move away from Polaris the circles traced by other stars get bigger and bigger. But all of the stars shown in this photo are always visible—they never set. We call these stars “circumpolar”. For every latitude on Earth, there is a set of circumpolar stars (the number decreases as you head towards the equator). Now let us add a new term to our vocabulary: the “Celestial Equator”. The Celestial Equator is the projection of the Earth’s Equator onto the sky. It is a great circle that spans the night sky that is directly overhead for people who live on the Equator. As you have now learned, the lengths of the days and nights at the equator are nearly always the same: 12 hours. But we have also learned that during the Equinoxes, the lengths of the days and the nights everywhere on Earth are also twelve hours. Why? Because during the equinoxes, the Sun is on the Celestial Equator. That means it is straight overhead (at noon) for people who live in Quito, Ecuador (and everywhere else on the equator). Any object that is on the Celestial Equator is visible for 12 hours per night from everywhere on Earth. To try to understand this, take a look at Fig. 4.7. In this figure is shown the celestial geometry explicitly showing that the Celestial Equator is simply the Earth’s equator projected onto the sky (left hand diagram). But the Earth is large, and to us, it appears flat. Since the objects in the sky are very far away, we get a view like that shown in the right hand diagram: we see one hemisphere of the sky, and the stars, planets, Sun and Moon rise in the east, and set in the west. But note that the Celestial Equator exactly intersects East and West. Only objects located on the Celestial Equator rise exactly due East, and set exactly due West. All other objects rise in the northeast or southeast and set in the northwest or the southwest. Note that in this diagram (for a latitude of 40◦) all stars that have latitudes (astronomers call them “Declinations”, or “dec”) above 50◦ never set–they are circumpolar. What happens is that during the year, the Sun appears to move above and below the Celestial Equator. On, or about, March 21st the Sun is on the Celestial Equator, and each day after this it gets higher in the sky (for locations in the Northern Hemisphere) until June 21st. After which it retraces its steps until it reaches the Autumnal Equinox (September 20th), after which it is South of the Celestial Equator. It is lowest in the sky on December 21st. This is simply due to the fact that the Earth’s axis is tilted with respect to its orbit, and this tilt does not change. You can see this geometry by going back to the illuminated globe model used in Exercise #3. If you stick a pin at some location on the globe away from the equator, turn on the halogen lamp, and slowly rotate the entire apparatus around (while keeping the pin facing the Sun) you will notice that the shadow of the pin will increase and decrease in size. This is due to the apparent change in the elevation angle of the “Sun”. 49 Figure 4.7: The Celestial Equator is the circle in the sky that is straight overhead (“the zenith”) of the Earth’s equator. In addition, there is a “North Celestial” pole that is the projection of the Earth’s North Pole into space (that almost points to Polaris). But the Earth’s spin axis is tilted by 23.5◦ to its orbit, and the Sun appears to move above and below the Celestial Equator over the course of a year. 4.5 Summary (35 points) Summarize the important concepts covered in this lab. Questions you should answer include: • Why does the Earth have seasons? • What is the orgin of the term “Equinox”? • What is the orgin of the term “Solstice”? • Most people in the United States think the seasons are caused by the changing distance between the Earth and the Sun. Why do you think this is? • What type of seasons would the Earth have if its spin axis was exactly perpendicular to its orbital plane? Make a diagram like Fig. 4.4. • What type of seasons would the Earth have if its spin axis was in the plane of its orbit? (Note that this is similar to the situation for the planet Uranus.) • What do you think would happen if the Earth’s spin axis wobbled randomly around on a monthly basis? Describe how we might detect this. 4.6 Extra Credit We have stated that the Earth’s spin axis constantly points to a single spot in the sky. This is actually not true. Look up the phrase “precession” of the Earth’s spin axis. Describe what is happening and the time scale of this motion. Describe what happens to the timing of the seasons due to this motion. Some scientists believe that precession might help cause ice ages. Describe why they believe this. (5 pts) 50 Many credit Kepler with the origin of modern physics, as his discoveries were what led Isaac Newton (1643 − 1727) to formulate the law of gravity. Today we will investigate Kepler’s laws and the law of gravity. 5.2 Gravity Gravity is the fundamental force governing the motions of astronomical objects. No other force is as strong over as great a distance. Gravity influences your everyday life (ever drop a glass?), and keeps the planets, moons, and satellites orbiting smoothly. Gravity affects everything in the Universe including the largest structures like super clusters of galaxies down to the smallest atoms and molecules. Experimenting with gravity is difficult to do. You can’t just go around in space making extremely massive objects and throwing them together from great distances. But you can model a variety of interesting systems very easily using a computer. By using a computer to model the interactions of massive objects like planets, stars and galaxies, we can study what would happen in just about any situation. All we have to know are the equations which predict the gravitational interactions of the objects. The orbits of the planets are governed by a single equation formulated by Newton: Fgravity = GM1M2 R2 (1) A diagram detailing the quantities in this equation is shown in Fig. 5.1. Here Fgravity is the gravitational attractive force between two objects whose masses are M1 and M2. The distance between the two objects is “R”. The gravitational constant G is just a small number that scales the size of the force. The most important thing about gravity is that the force depends only on the masses of the two objects and the distance between them. This law is called an Inverse Square Law because the distance between the objects is squared, and is in the denominator of the fraction. There are several laws like this in physics and astronomy. Figure 5.1: The force of gravity depends on the masses of the two objects (M1, M2), and the distance between them (R). Today you will be using a computer program called “Planets and Satellites” by Eugene Butikov to explore Kepler’s laws, and how planets, double stars, and planets in double star 53 systems move. This program uses the law of gravity to simulate how celestial objects move. • Goals: to understand Kepler’s three laws and use them in conjunction with the com- puter program “Planets and Satellites” to explain the orbits of objects in our solar system and beyond • Materials: Planets and Satellites program, a ruler, and a calculator 5.3 Kepler’s Laws Before you begin the lab, it is important to recall Kepler’s three laws, the basic description of how the planets in our Solar System move. Kepler formulated his three laws in the early 1600’s, when he finally solved the mystery of how planets moved in our Solar System. These three (empirical) laws are: I. “The orbits of the planets are ellipses with the Sun at one focus.” II. “A line from the planet to the Sun sweeps out equal areas in equal intervals of time.” III. “A planet’s orbital period squared is proportional to its average distance from the Sun cubed: P2 ∝ a3” Let’s look at the first law, and talk about the nature of an ellipse. What is an ellipse? An ellipse is one of the special curves called a “conic section”. If we slice a plane through a cone, four different types of curves can be made: circles, ellipses, parabolas, and hyperbolas. This process, and how these curves are created is shown in Fig. 5.2. Before we describe an ellipse, let’s examine a circle, as it is a simple form of an ellipse. As you are aware, the circumference of a circle is simply 2πR. The radius, R, is the distance between the center of the circle and any point on the circle itself. In mathematical terms, the center of the circle is called the “focus”. An ellipse, as shown in Fig. 5.3, is like a flattened circle, with one large diameter (the “major” axis) and one small diameter (the “minor” axis). A circle is simply an ellipse that has identical major and minor axes. Inside of an ellipse, there are two special locations, called “foci” (foci is the plural of focus, it is pronounced “fo-sigh”). The foci are special in that the sum of the distances between the foci and any points on the ellipse are always equal. Fig. 5.4 is an ellipse with the two foci indentified, “F1” and “F2”. Exercise #1: On the ellipse in Fig. 5.4 are two X’s. Confirm that that sum of the distances between the two foci to any point on the ellipse is always the same by measuring the distances between the foci, and the two spots identified with X’s. Show your work. (2 points) 54 Figure 5.2: Four types of curves can be generated by slicing a cone with a plane: a circle, an ellipse, a parabola, and a hyperbola. Strangely, these four curves are also the allowed shapes of the orbits of planets, asteroids, comets and satellites! Figure 5.3: An ellipse with the major and minor axes identified. 55 Now let’s put the Initial Velocity down to a value of 1.0. Run the simulation. What is happening here? The orbit is now a circle. Where are the two foci located? In this case, what is the distance between the focus and the orbit equivalent to? (4 points) The point in the orbit where the planet is closest to the Sun is called “perihelion”, and that point where the planet is furthest from the Sun is called “aphelion”. For a circular orbit, the aphelion is the same as the perihelion, and can be defined to be anywhere! Exit this simulation (click on “File” and “Exit”). Exercise #4: Kepler’s Second Law: “A line from a planet to the Sun sweeps out equal areas in equal intervals of time.” From the simulation window, click on the “Second Law” after entering the Kepler’s Law window. Move the Initial Velocity slide bar to a value of 1.2. Hit Go. Describe what is happening here. Does this confirm Kepler’s second law? How? When the planet is at perihelion, is it moving slowly or quickly? Why do you think this happens? (4 points) 58 Look back to the equation for the force of gravity. You know from personal experience that the harder you hit a ball, the faster it moves. The act of hitting a ball is the act of applying a force to the ball. The larger the force, the faster the ball moves (and, generally, the farther it travels). In the equation for the force of gravity, the amount of force generated depends on the masses of the two objects, and the distance between them. But note that it depends on one over the square of the distance: 1/R2. Let’s explore this “inverse square law” with some calculations. • If R = 1, what does 1/R2 = ? • If R = 2, what does 1/R2 = ? • If R = 4, what does 1/R2 = ? What is happening here? As R gets bigger, what happens to 1/R2? Does 1/R2 de- crease/increase quickly or slowly? (2 points) The equation for the force of gravity has a 1/R2 in it, so as R increases (that is, the two objects get further apart), does the force of gravity felt by the body get larger, or smaller? Is the force of gravity stronger at perihelion, or aphelion? Newton showed that the speed of a planet in its orbit depends on the force of gravity through this equation: V = √ (G(Msun + Mplanet)(2/r − 1/a)) (2) where “r” is the radial distance of the planet from the Sun, and “a” is the mean orbital radius (the semi-major axis). Do you think the planet will move faster, or slower when it is closest to the Sun? Test this by assuming that r = 0.5a at perihelion, and r = 1.5a at aphelion, and that a=1! [Hint, simply set G(Msun + Mplanet) = 1 to make this comparison very easy!] Does this explain Kepler’s second law? (4 points) 59 What do you think the motion of a planet in a circular orbit looks like? Is there a definable perihelion and aphelion? Make a prediction for what the motion is going to look like–how are the triangular areas seen for elliptical orbits going to change as the planet orbits the Sun in a circular orbit? Why? (3 points) Now let’s run a simulation for a circular orbit by setting the Initial Velocity to 1.0. What happened? Were your predictions correct? (3 points) Exit out of the Second Law, and start-up the Third Law simulation. Exercise 4: Kepler’s Third Law: “A planet’s orbital period squared is proportional to its average distance from the Sun cubed: P2 ∝ a3”. As we have just learned, the law of gravity states that the further away an object is, the weaker the force. We have already found that at aphelion, when the planet is far from the Sun, it moves more slowly than at perihelion. Kepler’s third law is merely a reflection of this fact–the further a planet is from the Sun (“a”), the more slowly it will move. The more slowly it moves, the longer it takes to go around the Sun (“P”). The relation is P2 ∝ a3, where P is the orbital period in years, while a is the average distance of the planet from the Sun, and the mathematical symbol for proportional is represented by “∝”. To turn the proportion sign into an equal sign requires the multiplication of the a3 side of the equation by a constant: P2 = C × a3. But we can get rid of this constant, “C”, by making a ratio. We will do this below. In the next simulation, there will be two planets: one in a smaller orbit, which will represent the Earth (and has a = 1), and a planet in a larger orbit (where a is adjustable). 60 many orbits (or what fraction of an orbit) have Neptune and Pluto completed since their discovery? (3 points) 5.4 Going Beyond the Solar System One of the basic tenets of physics is that all natural laws, such as gravity, are the same everywhere in the Universe. Thus, when Newton used Kepler’s laws to figure out how gravity worked in the solar system, we suddenly had the tools to understand how stars interact, and how galaxies, which are large groups of billions of stars, behave: the law of gravity works the same way for a planet orbiting a star that is billions of light years from Earth, as it does for the planets in our solar system. Therefore, we can use the law of gravity to construct simulations for all types of situations—even how the Universe itself evolves with time! For the remainder of the lab we will investigate binary stars, and planets in binary star systems. First, what is a binary star? Astronomers believe that about one half of all stars that form, end up in binary star systems. That is, instead of a single star like the Sun, being orbited by planets, a pair of stars are formed that orbit around each other. Binary stars come in a great variety of sizes and shapes. Some stars orbit around each other very slowly, with periods exceeding a million years, while there is one binary system containing two white dwarfs (a white dwarf is the end product of the life of a star like the Sun) that has an orbital period of 5 minutes! To get to the simulations for this part of the lab, exit the Third Law simulation (if you haven’t already done so), and click on button “7”, the “Two-Body and Many-Body” simulations. We will start with the “Double Star” simulation. Click Go. In this simulation there are two stars in orbit around each other, a massive one (the blue one) and a less massive one (the red one). Note how the two stars move. Notice that the line connecting them at each point in the orbit passes through one spot–this is the location of something called the “center of mass”. In Fig. 5.6 is a diagram explaining the center of mass. If you think of a teeter-totter, or a simple balance, the center of mass is the point where the balance between both sides occurs. If both objects have the same mass, this point is halfway between them. If one is more massive than the other, the center of mass/balance point is closer to the more massive object. Most binary star systems have stars with similar masses (M1 ≈ M2), but this is not always the case. In the first (default) binary star simulation, M1 = 2M2. The “mass ratio” (“q”) in this case is 0.5, where mass ratio is defined to be q = M2/M1. Here, M2 = 1, and M1 = 2, so q = M2/M1 = 1/2 = 0.5. This is the number that appears in the “Mass Ratio” 63 Figure 5.6: A diagram of the definition of the center of mass. Here, object one (M1) is twice as massive as object two (M2). Therefore, M1 is closer to the center of mass than is M2. In the case shown here, X2 = 2X1. window of the simulation. Exercise 5: Binary Star systems. We now want to set-up some special binary star orbits to help you visualize how gravity works. This requires us to access the “Input” window on the control bar of the simulation window to enter in data for each simulation. Clicking on Input brings up a menu with the following parameters: Mass Ratio, “Transverse Velocity”, “Velocity (magnitude)”, and “Direction”. Use the slide bars (or type in the numbers) to set Transverse Velocity = 1.0, Velocity (magnitude) = 0.0, and Direction = 0.0. For now, we simply want to play with the mass ratio. Use the slide bar so that Mass Ratio = 1.0. Click “Ok”. This now sets up your new simulation. Click Run. Describe the simulation. What are the shapes of the two orbits? Where is the center of mass located relative to the orbits? What does q = 1.0 mean? Describe what is going on here. (4 points) 64 Ok, now we want to run a simulation where only the mass ratio is going to be changed. Go back to Input and enter in the correct mass ratio for a binary star system with M1 = 4.0, and M2 = 1.0. Run the simulation. Describe what is happening in this simulation. How are the stars located with respect to the center of mass? Why? [Hint: see Fig. 5.6.] (4 points) Finally, we want to move away from circular orbits, and make the orbit as elliptical as possible. You may have noticed from the Kepler’s law simulations that the Transverse Velocity affected whether the orbit was round or elliptical. When the Transverse Velocity = 1.0, the orbit is a circle. Transverse Velocity is simply how fast the planet in an elliptical orbit is moving at perihelion relative to a planet in a circular orbit of the same orbital period. The maximum this number can be is about 1.3. If it goes much faster, the ellipse then extends to infinity and the orbit becomes a parabola. Go back to Input and now set the Transverse Velocity = 1.3. Run the simulation. Describe what is happening. When do the stars move the fastest? The slowest? Does this make sense? Why/why not? (4 points) The final exercise explores what it would be like to live on a planet in a binary star system–not so fun! In the “Two-Body and Many-Body” simulations window, click on the 65 cause it to be ejected completely out of the solar system. (You can see an example of the latter process by changing the Planet–Star Distance = 0.4 in the current simulation.) 5.5 Summary (35 points) Please summarize the important concepts of this lab. Your summary should include: • Describe the Law of Gravity and what happens to the gravitational force as a) as the masses increase, and b) the distance between the two objects increases • Describe Kepler’s three laws in your own words, and describe how you tested each one of them. • Mention some of the things which you have learned from this lab • Astronomers think that finding life on planets in binary systems is unlikely. Why do they think that? Use your simulation results to strengthen your argument. Use complete sentences, and proofread your summary before handing in the lab. 5.6 Extra Credit Derive Kepler’s third law (P2 = C × a3) for a circular orbit. First, what is the circumference of a circle of radius a? If a planet moves at a constant speed “v” in its orbit, how long does it take to go once around the circumference of a circular orbit of radius a? [This is simply the orbital period “P”.] Write down the relationship that exists between the orbital period “P”, and “a” and “v”. Now, if we only knew what the velocity (v) for an orbiting planet was, we would have Kepler’s third law. In fact, deriving the velocity of a planet in an orbit is quite simple with just a tiny bit of physics (go to this page to see how it is done: http://www.go.ednet.ns.ca/∼larry/orbits/kepler.html). Here we will simply tell you that the speed of a planet in its orbit is v = (GM/a)1/2, where “G” is the gravitational constant mentioned earlier, “M” is the mass of the Sun, and a is the radius of the orbit. Rewrite your orbital period equation, substituting for v. Now, one side of this equation has a square root in it–get rid of this by squaring both sides of the equation and then simplifying the result. Did you get P2 = C × a3? What does the constant “C” have to equal to get Kepler’s third law? (5 points) 5.7 Possible Quiz Questions 1) Briefly describe the contributions of the following people to understanding planetary motion: Tycho Brahe, Johannes Kepler, Isaac Newton. 2) What is an ellipse? 3) What is a “focus”? 4) What is a binary star? 5) Describe what is meant by an “inverse square law”. 6) What is the definition of “semi-major axis”? 68 Name: Date: 6 The Orbit of Mercury Of the five planets known since ancient times (Mercury, Venus, Mars, Jupiter, and Saturn), Mercury is the most difficult to see. In fact, of the 6 billion people on the planet Earth it is likely that fewer than 1,000,000 (0.0002%) have knowingly seen the planet Mercury. The reason for this is that Mercury orbits very close to the Sun, about one third of the Earth’s average distance. Therefore it is always located very near the Sun, and can only be seen for short intervals soon after sunset, or just before sunrise. It is a testament to how carefully the ancient peoples watched the sky that Mercury was known at least as far back as 3,000 BC. In Roman mythology Mercury was a son of Jupiter, and was the god of trade and commerce. He was also the messenger of the gods, being “fleet of foot”, and commonly dipicted as having winged sandals. Why this god was associated with the planet Mercury is obvious: Mercury moves very quickly in its orbit around the Sun, and is only visible for a very short time during each orbit. In fact, Mercury has the shortest orbital period (“year”) of any of the planets. You will determine Mercury’s orbital period in this lab. [Note: it is very helpful for this lab exercise to review Lab #1, section 1.5.] • Goals: to learn about planetary orbits • Materials: a protractor, a straight edge, a pencil and calculator Mercury and Venus are called “inferior” planets because their orbits are interior to that of the Earth. While the planets Mars, Jupiter, Saturn, Uranus, Neptune, and Pluto are called “superior” planets, as their orbits lie outside that of the Earth. Because the orbits of Mercury and Venus are smaller than the Earth’s, these planets can never be located very far from the Sun as seen from the Earth. As discovered by Galileo in 1610 (see Fig. 6.1), the planet Venus shows phases that look just like those of the Moon. Mercury also shows these same phases. As can be envisioned from Figure 6.1, when Mercury or Venus are on the far side of the Sun from Earth (a configuration called “superior” conjunction), these two planets are seen as “full”. Note, however, that it is almost impossible to see a “full” Mercury or Venus because at this time the planet is very close to, or behind the Sun. When Mercury or Venus are closest to the Earth, a time when they pass between the Earth and the Sun (a configuration termed “inferior” conjunction), we would see a “new” phase. During their new phases, it is also very difficult to see Mercury or Venus because their illuminated hemispheres are pointed away from us, and they are again located very close to the Sun in the sky. The best time to see Mercury is near the time of “greatest elongation”. At the time of greatest elongation, the planet Mercury has its largest angular separation from the Sun as seen from the Earth. There are six or seven greatest elongations of Mercury each year. At the time of greatest elongation, Mercury can be located up to 28◦ from the Sun, and sets (or rises) about two hours after (or before) the Sun. In Figure 6.2, we show a diagram for the greatest elongation of Mercury that occurred on August 14, 2003. In this diagram, we 69 Figure 6.1: A diagram of the phases of Venus as it orbits around the Sun. The planet Mercury exhibits the same set of phases as it too is an “inferior” planet like Venus. plot the positions of Mercury and the Sun at the time of sunset (actually just a few minutes before sunset!). As this diagram shows, if we started our observations on July 24th, Mercury would be located close to the Sun at sunset. But as the weeks passed, the angle between Mercury and the Sun would increase until it reached its maximum value on August 14th. After this date, the separation between the Sun and Mercury quickly decreases as it heads towards inferior conjunction on September 11th. Figure 6.2: The eastern elongation of August, 2003. Mercury was at superior conjunction on July 5th, and quickly moved around its orbit increasing the angular separation between it and the Sun. By July 24th, Mercury could be seen just above the Sun shortly after sunset. As time passed, the angular separation between the Sun and Mercury increased, reaching its maximum value on August 14th, the time of greatest Eastern elongation. As Mercury continued in its orbit it comes closer to the Earth, but the angular separation between it and the Sun shrinks. Eventually, on September 11th, the time of inferior conjunction, Mercury passed directly between the Earth and the Sun. You can see from Figure 6.2 that Mercury is following an orbit around the Sun: it was “behind” the Sun (superior conjunction) on July 5th, and quickly races around its orbit until the time of greatest elongation, and then passes between the Earth and the Sun on September 11th. If we used a telescope and made careful drawings of Mercury throughout this time, we would see the phases shown in Figure 6.3. On the first date in Figure 6.2 (July 70 left of the Sun. As described earlier, if Mercury is to the left of the Sun, it is an eastern elongation. Figure 6.5: A diagram showing the orbital geometry of the Earth and Mercury during a greatest Eastern elongation. The orbits of the Earth and Mercury are the two large circles. The line of sight to Mercury at the time of greatest elongation is indicated. Note that at this time the angle between the Earth, Mercury, and the Sun is a right angle. The direction of motion of the two planets is shown by the arrows on the orbits. 6.3 The Orbits of Earth and Mercury As shown in the previous diagram, both the Earth and Mercury are orbiting the Sun. That means that every single day they are at a different position in their orbits. Before we can begin this lab, we must talk about how we can account for this motion! A year on Earth, the time it takes the Earth to complete one orbit around the Sun, is 365 days. If we assume that the Earth’s orbit is a perfect circle, then the Earth moves on that circle by about 1 degree per day. Remember that a circle contains 360 degrees (360◦). If it takes 365 days to go 360◦, the Earth moves 360◦/365 = 0.986 degrees per day (◦/day). For this lab, we will assume that the Earth moves exactly one degree per day which, you can see, is very close to the truth. How far does the Earth move in 90 days? 90 degrees! How about 165 days? 6.4 The Data In Table 6.1, we have listed thirteen dates for greatest elongations of Mercury, as well as the angle of each greatest elongation. Note that the elongations are either East or West! In the third column, we have listed something called the Julian date. Over long time intervals, our common calendar is very hard to use to figure out how much time has elapsed. For example, how many days are their between March 13th, 2001 and December 17th 2004? Remember that 2004 is a leap year! This is difficult to do in your head. To avoid such calculations, astronomers have used a calendar that simply counts the days that have passed since some distant day #1. The system used by astronomers sets Julian date 1 to January 1st, 4713 BC (an arbitrary date chosen in the sixteenth century). Using this calendar, March 13th, 2001 has a Julian date of 2451981. While December 17th 2004 has a 73 Table 6.1: Elongation Data For Mercury # Date Elongation Angle Julian Date Days Degrees #1 Sep. 1, 2002 27.2 degrees east 2452518 − − #2 Oct. 13, 2002 18.1 degrees west 2452560 42 42 #3 Dec. 26, 2002 19.9 degrees east 2452634 #4 Feb. 4, 2003 25.4 degrees west 2452674 #5 Apr. 16, 2003 19.8 degrees east 2452745 #6 Jun. 3, 2003 24.4 degrees west 2452793 #7 Aug. 14, 2003 27.4 degrees east 2452865 #8 Sep. 27, 2003 17.9 degrees west 2452909 #9 Dec. 09, 2003 20.9 degrees east 2452982 #10 Jan. 17, 2004 23.9 degrees west 2453021 #11 Mar. 29, 2004 18.8 degrees east 2453093 #12 May 14, 2004 26.0 degrees west 2453139 #13 Jul. 27, 2004 27.1 degrees east 2453213 Julian date of 2453356. Taking the difference of these two numbers (2453356 − 2451981) we find that there are 1,375 days between March 13th, 2001 and December 17th 2004. Exercise #1: Fill-in the Data Table. The fourth and fifth columns of the table are blank, and must be filled-in by you. The fourth column is the number of days that have elapsed between elongations in this table (that is, simply subtract the Julian date of the previous elongation from the following elongation). We have worked the first one of these for you as an example. The last column lists how far the Earth has moved in degrees. This is simply the number of days times the number 1.0! As the Earth moves one degree per day. (If you wish, instead of using 1.0, you could multiply this number by 0.986 to be more accurate. You will get better results doing it that way.) So, if there are 42 days between elongations, the Earth moves 42 degrees in its orbit (or 41.4 degrees using the correct value of 0.986 ◦/day). (10 points) Exercise #2: Plotting your data. Before we describe the plotting process, review Figure 6.5. Unlike that diagram, you do not know what the orbit of Mercury looks like–this is what you are going to figure out during this lab! But we do know two things: the first is that the Earth’s orbit is nearly a perfect circle, and two, that the Sun sits at the exact center of this circle. On the next page is a figure containing a large circle with a dot drawn at the center to represent the Sun. At one position on the large circle we have put a little “X” as a reference point. The large circle here is meant to represent the Earth’s orbit, and the “X” is simply a good starting point. To plot the first elongation of Mercury from our data table, using a pencil, draw a line 74 connecting the X, and the Sun using a straight edge (ruler or protractor). The first elongation in the table (September 1, 2002) is 27.2 degrees East. Using your protractor, put the “X” that marks the Earth’s location at the center hole/dot on your protractor. Looking back to Figure 6.5, that elongation was also an eastern elongation. So, using that diagram as a guide, measure an angle of 27.2 degrees on your protractor and put a small mark on the worksheet. Now, draw a line from the Earth’s location through this mark just like the “line of sight” arrow in Figure 6.5. Now, rotate your protractor so that the 90 degree mark is on this line and towards the position of the Earth, while the reference hole/dot is on the same line. Slide the protractor along the line until the 0◦ (or 180◦) reference line intersects the center of the Sun. Mark this spot with a dark circle. This is the position of Mercury! This is the procedure that you will use for all of the elongations, so if this is confusing to you, have your TA come over and clarify the technique for you so that you don’t get lost and waste time doing this incorrectly. Ok, now things become slightly more difficult—the Earth moves! Looking back to Figure 6.5, note the arrows on the orbits of Earth and Mercury. This is the direction that both planets are moving in their orbits. For the second elongation, the Earth has moved 42 degrees. We have to locate where the Earth is in its orbit before we can plot the next elongation. So, now put the center hole/dot of your protractor on the Sun. Line up the 0/180 degree mark with the first line that connected the Earth and Sun. Measure an angle of 42 degrees (in the correct direction) and put a small mark. Draw a line through this mark that intersects the postion of the Sun, the mark, and the orbit of the Earth. Put an X where this line intersects the Earth’s orbit. This is the spot from where you will plot the next elongation of Mercury. Now, repeat the process for plotting this next elongation angle. Note, however, that this elongation is a western elongation, so that the line of sight arrow this time will be to the right of the Sun. It is extremely important to remember that on eastern elongations the line of sight arrow to Mercury goes to the left of the Sun, while during western elongations it goes to the right of the Sun. [Hints: It is helpful to label each one of the X’s you place on the Earth’s orbit with the elongation number from Table 6.1. This will allow you to go back and fix any problems you might find. Note that you will have a large number of lines drawn in this plot by the time you get finished. Use a sharp pencil so that you can erase some/all/pieces of these lines to help clean-up the plot and reduce congestion. You might also find it helpful to simply put a “left” or a “right” each time you encounter East and West in Table 6.1 to insure you plot your data correctly.] Plot all of your data (28 points). 75 Perihelion (p) = mm. Aphelion (q) = mm. Astronomers use the term eccentricity (“e”) to measure how out-of-round a planet’s orbit is, and the eccentricity is defined by the equation: e = (q − p)/(p + q) = Plug your values into this equation and calculate the eccentricity of Mercury’s orbit. (4 points) 4) The eccentricity for the Earth’s orbit is e = 0.017. How does your value of the eccentricity for Mercury compare to that of the Earth? Does the fact that we used a circle for the Earth’s orbit now seem justifiable? (5 points) Exercise #5: The orbital period of Mercury. Looking at the positions of Mercury at elongation #1, and its position at elongation #2, approximately how far around the orbit did Mercury move in these 42 days? Estimate how long you think it would take Mercury to complete one orbit around the Sun: days. (2 points) Using Kepler’s laws, we can estimate the orbital period of a planet (for a review of Kepler’s laws, see lab #4). Kepler’s third law says that the orbital period squared (P2) is proportional to the cube of the semi-major axis (a3): P2 ∝ a3. This is a type of equation you might not remember how to solve (if you have not done so already, review Lab #1 section 1.5). But let’s take it in pieces: 78 a3 = a× a× a = Plug-in your value of a for Mercury and find its cube. To find the period of Mercury’s orbit, we now need to take the square root of the number you just calculated (see your TA if you do not know whether your calculator can perform this operation). (4 points) P = √ a3 = . (8) Now, the number you just calculated probably means nothing to you. But what you have done is calculate Mercury’s orbital period as a fraction of the Earth’s orbital period (that is because we have been working in AU, a unit that is defined by the Earth-Sun distance). Since the Earth’s orbital period is exactly 365.25 days, find Mercury’s orbital period by multiplying the number you just calculated for Mercury by 365.25: Porb(Mercury) = days. 5) How does the orbital period you just calculated using Kepler’s laws compare to the one you estimated from your plot at the beginning of this exercise? (4 points) 6.5 Summary (35 points) Before you leave lab, your TA will give you the real orbital period of Mercury, as well as its true semi-major axis (in AU) and its orbital eccentricity. • Compare the precisely known values for Mercury’s orbit with the ones you derived. How well did you do? • What would be required to enable you to do a better job? 79 • Describe how you might go about making the observations on your own so that you could create a data table like the one in this lab. Do you think this could be done with just the naked eye and some sort of instrument that measured angular separation? What else might be necessary? 6.6 Extra Credit In this lab you have measured three of the five quantities that completely define a planet’s orbit. The other two quantities are the orbital inclination, and the longitude of perihelion. Determining the orbital inclination, the tilt of the plane of Mercury’s orbit with respect to the Earth’s orbit, is not possible using the data in this lab. But it is possible to determine the longitude of perihelion. Astronomers define the zero point of solar system longitude as the point in the Earth’s orbit at the time of the Vernal Equinox (the beginning of Spring in the northern hemisphere). In 2004, the Vernal Equinox occurs on March 20. If you notice, one of the elongations in the table (#11) occurs close to this date. Thus, you can figure out the true location of the Vernal Equinox by moving back from position #11 by the right number of degrees. The longitude of Mercury’s perihelion is just the angle measured counterclockwise from the Earth’s vernal equinox. You should find that your angle is larger than 180 degrees. Subtract off 180 degrees. How does your value compare with the precise value of 77◦? (3 points) 6.7 Possible Quiz Questions • What does the term “inferior planet” mean? • What is meant by elongation angle? • Why do Mercury and Venus show phases like the Moon? 80 Material Density (g/cm3) Gold 19.3 Lead 11.4 Iron 7.9 Aluminum 2.7 Rock (typical) 2.5 Liquid Water 1.0 Wood (typical) 0.8 Insulating foam 0.1 Silica Gel 0.02 Air (dry, at sea level) 0.0012 Helium 0.0001785 Table 7.3: Densities of different materials. 2. Compare the density of the shiny weight with the densities listed in Table 7.3. What do you suspect this object is made of? Why do you think this? (10 points) 3. Compare the densities of 3, 4, and 5 and consider Table 7.3. Do you agree that the Piñon meteorite could be rich in iron? Why? (10 points) 4. Referring only to Table 7.3, what material (or combination of materials) would you guess the golf ball to be composed of? Is this a reasonable composition for a golf ball? What does this suggest about the use of average density information only when we make ‘guestimates’ about what materials solar system bodies might be composed of? (10 points) 83 Name: Date: 7.3 Take-Home Questions 1. Some scientists have measured the density of a substance to be 3.1 g/cm3. Based on what you have learned so far, is the substance more likely to be a solid, a liquid, or a gas? (5 points) 2. A solid cylinder of plastic has a density of 1.6 g/cm3. It is then cut exactly in half. What is the density of each of the pieces now? Explain your answer. (5 points) 3. Liquid A has a density of 0.90 g/cm3, liquid B has a density of 1.15 g/cm3, and liquid C has a density of 0.65 g/cm3. They are poured into a graduated cylinder and allowed to sit overnight. Assuming that the liquids do not mix into one another, which liquid will be on the bottom, in the middle, and at the top of the graduated cylinder? Draw a picture to illustrate this scenario. (5 points) 4. Use your textbook (Appendix E) or information on-line to fill in the following table (12 points): 84 Object Mass (kg) Average Density (g/cm3) Sun Mercury Venus Earth Moon Mars Ceres (largest asteroid) 8.7 × 1020 1.98 Jupiter Saturn Titan (Saturn’s largest moon) Uranus Neptune Pluto Comet Halley (nucleus) 4 × 1015 0.1 Table 7.4: Densities of different solar system objects. 5. Based on the density information in Table 7.4, can you group the objects in our solar system into different classes or categories? How would you divide them up? (5 points) 6. How does the density of the Sun compare to the densities of other objects in our solar system? Can you say anything about the chemical composition of the Sun as compared to the composition of the planets based on Table 7.4 alone?(5 points) 7. Which planet is the most massive? (2 points) Which planet is the most dense? (2 points) 85 and the Earth’s volume is a function of its radius: VOLUME = (4/3) × π × RADIUS3 We will implement Eratosthenes’ circumference measurement method and end up with an estimate of the Earth’s radius. Now, what measurements did Eratosthenes use to estimate Earth’s circumference? Er- atosthenes, knowing that Earth is spherical in shape, realized that the length of an object’s shadow would depend upon how far in latitude (north-or-south) the object was from being directly beneath the Sun. He measured the length of a shadow cast by a vertical post in Egypt at local noon on the day of the northern hemisphere summer solstice (June 20 or so). He made a measurement at the point directly beneath the Sun (23.5 degrees North, at the Egyptian city Syene), and at a second location further north (Alexandria, Egypt). The two shadow lengths were not identical, and it is that difference in shadow length plus the knowl- edge of how far apart the the two posts were from each other (a few hundred kilometers), that permitted Eratosthenes to calculate his estimate of Earth’s circumference. As we conduct this lab exercise we are not in Egypt, nor is today the seasonal date of the northern hemisphere summer solstice (which occurs in June), nor is it locally Noon (since our lab times do not overlap with Noon). But, nonetheless, we will forge ahead and estimate the Earth’s circumference, and from this we will estimate the Earth’s radius. TASKS: • Take a post outside, into the sunlight, and measure the length of the post with the tape measure. • Place one end of the post on the ground, and hold the post as vertical as possible. • Using the tape measure provided, measure to the nearest 1/2 centimeter the length of the shadow cast by the post; this shadow length should be measured three times, by three separate individuals; record these shadow lengths in Table 8.1. • You will be provided with the length of a post and its shadow measured simultaneously today in Boulder, Colorado. • Proceed through the calculations described after Table 8.1, and write your answers in the appropriate locations in Table 8.1. 8.3 ANGLE DETERMINATION: With a bit of trigonometry we can transform the height and shadow length you measured into an angle. As shown in Figure 8.1 there is a relationship between the length (of your shadow in this situation) and the height (of the shadow-casting pole in this situation), where: 88 Table 8.1: Angle Data Location Post Height Shadow Length Angle (cm) (cm) (Degrees) Las Cruces Shadow #1 Las Cruces Shadow #2 Las Cruces Shadow #3 Average Las Cruces Angle: Boulder, Colorado TANGENT of the ANGLE = far-side length/ near-side length Since you know the length of the post (the near side length, which you have measured) and the length of the shadow (the far side length, which you have also measured, three separate times), you can determine the shadow angle from your measurements, using the ATAN, or TAN−1 capability on your calculator (these functions will give you an angle if you provide the ratio of the height to length): ANGLE = ATAN (shadow length / post length) or ANGLE = TAN−1(shadow length / post length) Figure 8.1: The geometry of a vertical post sitting in sunlight. Calculate the shadow angle for each of your three shadow-length measure- ments, and also for the Boulder, Colorado shadow-length measurement. Write these angle values in the appropriate locations in Table 8.1. Then calculate the average of the three Las Cruces shadow angles, and write the value on the “Average Las Cruces Angle” line. The angles you have determined are: 1) an estimate of the angle (latitude) difference between Las Cruces and the latitude at which the Sun appears to be directly overhead (which is currently ∼ 12 degrees south of the equator since we are experiencing early northern autumn), and 2) the angle (latitude) difference between Boulder, Colorado and the latitude 89 at which the Sun appears to be directly overhead. The difference (Boulder angle minus Las Cruces angle) between these two angles is the angular (latitude) separation between Las Cruces and Boulder, Colorado. We will now use this information and our knowledge of the actual distance (in kilometers) between Las Cruces’ latitude and Boulder’s latitude. This distance is: 857 kilometers north-south distance between Las Cruces and Boulder, Colorado In the same way that Eratosthenes used his measurements (just like those you have made today), we can now determine an estimate of the Earth’s circumference from: EARTH CIRCUMFERENCE (kilometers) = 857 kilometers × (360o)/(Boulder angle —Avg LC Angle) = 857 × [360o/( − )] = km Using your calculated Boulder Shadow Angle and your Average Las Cruces Shadow Angle values, calculate the corresponding EARTH CIRCUMFERENCE value, and write it below: AVERAGE EARTH CIRCUMFERENCE = kilometers The CIRCUMFERENCE value you have just calculated is related to the RADIUS via the equation: EARTH CIRCUMFERENCE = 2 × π × EARTH RADIUS which can be converted to RADIUS using: EARTH RADIUS = RE = EARTH CIRCUMFERENCE / (2 × π) For your calculated CIRCUMFERENCE, calculate that value of the Radius (in units of kilometers) in the appropriate location below: AVERAGE EARTH RADIUS VALUE = RE = kilometers Convert this radius (RE) from kilometers to meters, and enter that value in Table 8.3. You have now obtained one important piece of information (the radius of the Earth) needed for determining the density of Earth. We will, in a bit, use this radius value to calculate the Earth’s volume. Next, we will determine Earth’s mass, since we need to know both the Earth’s volume and its mass in order to be able to calculate the Earth’s density. 90 By rearranging the Gravity equation to solve for ME, we can now make an estimate of the Earth’s mass: ME = Average Acceleration × [1000 × RE]2 / G = [The factor of 1000 here converts your Average Radius determined in the previous section in units of kilometers to a radius in units of meters.] Write the value of ME (in kilograms) in Table 8.3 below. 8.5 Determining the Earth’s Density Now that we have estimates for the mass (ME) and radius (RE) of the Earth, we can easily calculate the density: Density = Mass/Volume. You will do this below. Tasks: • Calculate the volume (VE) of the Earth given your determination of its radius (in meters!): VE = (4/3) × π × RE3 and write this value in the appropriate location in Table 8.3 below. • Divide your value of ME (that you entered in Table 8.3) by your estimate of VE that you just calculated (also written in Table 8.3): the result will be your estimate of the Average Earth Density in units of kilograms per cubic meter. Write this value in the appropriate location in Table 8.3. • Divide your AVERAGE ESTIMATE OF EARTH’S DENSITY value that you just calculated by the number 1000.0; the result will be your estimated Earth density value in units of grams per cubic centimeter (the unit in which most densities are tabulated). Write this value in the appropriate location in Table 8.3. Table 8.3: Data for the Earth Estimate of Earth’s Radius: m Estimate of Earth’s Mass: kg Estimate of Earth’s Volume: m3 Estimate of Earth’s Density: kg/m3 Density of the Earth: gm/cm3 93 8.6 IN-LAB QUESTIONS: 1. Is your calculated value of the Earth’s density GREATER THAN, or LESS THAN, or EQUAL TO the actual value (see the Introduction) of the Earth’s density? If your calculated density value is not identical to the known Earth density value, calculate the “percent error” of your calculated density value compared to the actual density value: PERCENT ERROR = 100%× (CALCULATED DENSITY− ACTUAL DENSITY) ACTUAL DENSITY = (9) 2. You used the AVERAGE Las Cruces shadow angle in calculating your estimate of the Earth’s density (which you wrote down in Table 8.3). If you had used the LARGEST of the three measured Las Cruces shadow angles shown in Table 8.1, would the Earth density value that you would calculate with the LARGEST Las Cruces shadow angle be larger than or smaller than the Earth density value you wrote in Table 8.3? Think before writing your answer! Explain your answer. 3. If the Las Cruces to Boulder, Colorado distance was actually 2000 km in length, but your measured fall times did not change from what you measured, would you have calculated a larger or smaller Earth density value? Explain the reasoning for your answer. 94 Name: Earth Density Lab Ast 105 Take-Home Questions 1. Using the post-height (length) and post and shadow lengths your Lab TA will provide to you at the end of the lab meeting, fill in the Table 8.4 below. Table 8.4: Angle Data Location Post Height Shadow Length Angle (cm) (cm) (Degrees) Las Cruces Angle: Boulder, Colorado Angle Now re-calculate the following numbers: Earth Circumferance = kilometers Earth Radius = kilometers Answer Questions #2, #3, and #4 below in typed form on a separate sheet of paper, and then type your response to item #5. These will be handed in with the remainder of your lab materials. 2. List in an appropriate order the approximately eight (8) steps involved in determining the Earth’s density in this lab. 3. Using the “Earth Radius” value you calculated using the provided-by-your-TA Las Cruces and Boulder post and shadow lengths (see Table 8.4 above): Calculate a new estimate of Earth’s density using these values as your starting point. How does your answer compare to your AVERAGE DENSITY OF EARTH value in Table 8.3 of the lab? The shadow angles you wrote down were recorded at 1pm MDT on October 12th, 2008. Do you believe this new density value is a worse, or better estimate than you calculated using the values measured during your lab? Why? 4. If we had conducted this experiment on the Moon rather than here on the Earth, would your measured values (fall time, angles and angle difference between two locations separated north-south by 857 kilometers) be the same as here on Earth, or different? Clearly explain your reasoning. [It might help if you draw a circle representing Earth and then draw a circle with 1/4th of the radius of the Earth’s circle to represent the Moon.] 95