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Star Charts – Instructor’s Guide 25
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
INTRODUCTION
Prerequisite Knowledge
• Basic familiarity with overhead-view and horizon-view star maps
Goals
• Understand how to use overhead-view and horizon-view star maps to locate objects in the observer’s night sky
• Be able to make the translation between where star groups are located on an overhead star map to their location on a horizon star map.
Pre-activity Question
1) The star map provided below shows the sky at midnight on July 1. What is the name of the star group that appears highest in the sky at this time? a) Scorpius b) Corona Borealis c) Draco d) Ursa Minor e) Auriga
26 Star Charts – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy Third Edition
TUTORIAL GUIDE
1) [ Any of the star groups found in the center of the overhead view star map, such as Hercules, Draco, or Bootes, would be acceptable. ]
Many students will incorrectly respond that a star group found at the top of the overhead star map (such as Scorpious) corresponds with the highest position in the night sky. Some students will resist committing themselves to an answer to this question. Encourage these students to provide an answer, and assure them that this answer is just an initial prediction that they will be able to change later. There is no need to directly challenge incorrect responses at this point, as the students will be asked to revisit this issue again in Question 5. It can, however, be helpful to ask students with incorrect responses to articulate their reasoning.
2) [ Hercules or Bootes ]
Note: This horizon view shows only the tip of Draco at the very top of the drawing.
This question requires students to make the translation between how star groups are represented on an overhead star map with how these same star groups are represented on a horizon-view star map. It is particularly important that students recognize that Figure 2 shows a south-facing horizon view star map. It may help to ask students to orient Figure 1 so it matches Figure 2.
3) The overhead star map would need to be folded in half along the east-west line, and then held with the east-west fold pointing upward while looking at the side of the map containing Scorpius near the bottom (top half of the map turned upside down).
This portion of the Lecture-Tutorial can be very difficult for some students. They may not understand exactly what is being asked or may provide an incomplete answer. It can be worthwhile to do this part of the Lecture-Tutorial together as a class after students have had sufficient time to attempt an answer on their own.
4) The map would be folded in same way as in Question 3, but this time, it would be held with the east-west fold pointing upward while looking at the side of the map containing Auriga near the bottom (bottom half of the map).
5) Many students will have already gone back and changed their answer to Question 1 if they were initially incorrect. This question offers an excellent place to engage students that are still struggling with how an overhead view star map is used to find objects in the night sky.
6) a) [ At the center of the overhead view star map. Draco ]
b) [ Along the top of the overhead view star map. Scorpius]
c) [ Along the right side of the overhead view star map. Capricornus or Equuleus ]
Star Charts – Instructor’s Guide 27
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
ADDITIONAL QUESTIONS 1) The star map provided below shows the sky at midnight on July 1. What is the name of the
star group that will be directly overhead when looking to the east at this time? a) Crater b) Ursa Major c) Draco d) Cygnus e) Equuleus
2) The star map provided below shows the sky at midnight on July 1. Which of the following star groups do you see when looking east at this time? a) Auriga b) Scorpius c) Draco d) Equuleus e) Crater
28 Star Charts – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy Third Edition
Kepler’s Second Law – Instructor’s Guide 29
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
INTRODUCTION
Prerequisite Knowledge
• A basic familiarity of how Kepler’s second law describes the motion of an orbiting object in terms of equal areas in equal times
• A fundamental understanding of how the motion of an orbiting object changes as it orbits a central star based on Kepler’s second law
Goals
• Be able to reason about and describe the entire motion of a planet’s orbit based on the area swept out by the planets and the time it takes for that motion to occur
• Practice estimation and analytical reasoning skills Pre-activity Question 1) Kepler’s second law says “a line joining a planet and the Sun sweeps out equal areas in
equal amounts of time.” Which of the following statements means nearly the same thing? a) Planets move farther in each unit of time when they are closer to the Sun. b) Planets move equal distances throughout their orbit of the Sun. c) Planets move the same speed at all points during their orbit of the Sun. d) Planets move slowest when they are moving away from the Sun. e) Planets move fastest when they are moving toward the Sun.
TUTORIAL GUIDE 1) [ Yes ] Since the planet is in a perfectly circular orbit, and the distances between the
lettered positions look the same, the area swept out by the planet between each of the lettered positions would be the same. So, since the time intervals between each of the lettered positions are also the same—one month—it obeys Kepler’s second law: equal areas swept out in equal amount of time.
2) [ Staying the same ] Since the planet is moving the same distance in the same time it must be traveling at the same speed the entire time.
3) Students should have drawn and shaded in an area connecting A, B, and the Sun.
4) [ possible correct answers include C-H or D-I ]
Note: There are many choices that could be offered which should be considered correct provided they identify a distance traveled which is greater than the distance between locations A and B. Students may struggle with deciding which letters to choose since there are several options. For these students, tell them there is no one correct answer. In addition, some students
30 Kepler’s Second Law – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
struggle with understanding what is meant by “swept-out-area.” If students are stuck here, it can be helpful to ask them if they were going to carpet (or paint) a new area with the same size as the shaded in area they made between Positions A and B, which two of the points between C and I would use the same amount of carpet (or paint).
5) [ Equal to ] The lettered positions identified in Answer 4 were chosen because the area swept out between them would be about the same as the area swept out between A and B. And since we are told that the planet obeys Kepler’s second law—equal areas in equal times—the time interval must be equal.
6) [ Question 4 ]
7) [ Question 4 ] Since the planet traveled a greater distance for the motion identified in Question 4 than it did in Question 3, and both motions took place in the same amount of time, the planet had to be moving faster during the interval identified in Question 4. Some students struggle with making the connection that if the planet travels a greater distance in the same amount of time it must be traveling with a greater speed. For these students ask “If we both drive for one hour, but you drive 50 miles and I drive 100 miles, who’s driving faster?”
8) [ No ]
9) [ Fastest: G; Slowest: A ] Since the time intervals between the lettered positions are the same, and the distance traveled in that time interval is greatest at Position G and smallest at Position A, it must be moving fastest at Position G and slowest at Position A.
10) [ Increasing ] Since the time intervals between each of the lettered positions are the same, and the distance between Positions D and E is greater than the distance between Positions D and C, the speed of the planet must be increasing.
11) The closer a planet is to its companion star the faster it moves, and the farther away it is the slower it moves.
12) [ A ] Orbit A is a circle, and Earth’s eccentricity is nearly zero, and since orbits with an eccentricity of zero are “perfectly circular,” Orbit A must be the one that most closely matches Earth’s orbits. Orbits B and C are not circles. Many students have the misconception that Earth’s orbit is highly elliptical.
13) [ Largest: Pluto; Smallest: Venus ]
14) Earth’s orbital speed wouldn’t change very much throughout the year because its orbit is nearly a perfect circle, so it wouldn’t change its distance from the Sun very much. And, since we found in Question 11 that the closer a planet is to its star the faster it moves, and the farther away it is the slower it moves, Earth would have to change its distance from the Sun to change its speed.
Kepler’s Second Law – Instructor’s Guide 31
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
ADDITIONAL QUESTIONS The planet in the orbit shown in the drawing at right obeys Kepler’s Laws. Use this drawing to answer the next four questions. 1) According to Kepler’s Second Law, during
which one of the portion of the planet’s orbit (B, C, or D), would the planet take the same amount of time as it took for the portion of the orbit identified with letter “A”? If you think all the portions of the orbit take the same amount of time, answer “E”. [B is the correct answer]
2) During which part of the planet’s orbit (A, B, C, or D) would the planet move with the
greatest speed? [C is the correct answer] 3) During how many portions of the planet’s orbit (A, B, C and D) would the planet be speeding
up the entire time? a) Only during one of the portions shown. b) During two of the portions shown. c) During three of the portions shown. d) During four of the portions shown. e) None of the above.
4) During which of the portions of the planet’s orbit (A, B, C, or D) would the planet experience
an increase in speed for at least a moment? a) Only during one of the portions shown. b) During two of the portions shown. c) During three of the portions shown. d) During four of the portions shown. e) None of the above.
32 Kepler’s Second Law – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
Kepler’s Third Law – Instructor’s Guide 33
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
INTRODUCTION Prerequisite Knowledge
• Basic familiarity with the motion of planets around the Sun, as presented in the heliocentric model of the solar system
• Basic familiarity with Kepler’s Laws Goals
• Understand that the orbital period of a planet in the solar system is related to the planet’s orbital distance but not the planet’s mass (Note: At this level of instruction, assuming the mass of the planet is negligible is appropriate.)
Pre-activity Question 1) If a small weather satellite and the large International Space Station are orbiting Earth at the
same altitude above Earth’s surface, which object takes longer to orbit once around Earth? a) the large space station b) the small weather satellite c) They would take the same amount of time.
TUTORIAL GUIDE 1) [ The closer (Jupiter-sized) planet called Esus will have the shorter orbital period. ]
Note: At this level of instruction, assuming the mass of the planet is negligible is appropriate. The mass of the central object being orbited is important. In a sense, the planets are falling towards the much more massive central star, and the mass of a falling object is not relevant. Questions 1 through 4 are designed to have students commit to a line of reasoning. However, the answers they provide are not important at this point. Some students will reason that smaller objects move more quickly than do large objects and therefore predict that the Earth-like planet called Sulis will have the shorter orbital period. Again, it is not necessary to engage students answering incorrectly for the first few questions of this Lecture-Tutorial, as their reasoning difficulties will be confronted later.
2) [ Yes, the Earth-like planet called Sulis would now have the shorter orbital period. ]
Kepler’s third law says that the period of a planet’s orbit is related to its distance. Specifically, if the planet is further away, its orbital period will be longer. This means it will also take more time to move around its central star. So, the closer the planet is to its central star, the shorter its orbital period will be, and the faster it will mover around its orbital star
3) [ Stay the same.] Since the orbital period depends only on how far away the planet is from
its central star, changing the mass of the planet won’t effect how long it takes for the planet to go around the star.
34 Kepler’s Third Law – Instructor’s Guide
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4) [ They would both orbit around their central star in the same amount of time. ] Since the orbital period depends only on how far away the planet is from its central star, and not on the mass of the planet, they will move around their star in the same amount of time because they are at the same distance away from the star.
5) [ Increase ]
For students who have answered the previous questions incorrectly, Question 5 is the first point where their ideas are challenged. Having students interpret the graph and describe the general relationship in words may help students better understand that the orbital period increases with distance.
6) [ 1 AU ]
7) [ 2.8 years ]
Using Kepler’s third law gives 2.83 years. However, any answer in this range is acceptable since it is important for students to be able to estimate values using the graph.
8) [ d ]
The specific examples in Questions 6 and 7 help students answer this more easily than if they were asked to interpret the relationship directly from the graph.
9) [ Earth ]
10) [ d ] Based on the information in the table, Mars is less massive than Earth but takes longer to orbit the Sun. Meanwhile, Jupiter is more massive than Earth, but also takes longer to orbit the Sun. In addition, the graph that shows the orbital periods of planets shows that this distance is related to how far the planet is from the star, not how massive it is.
11) [ Mars ] The Student incorrectly thinks that the planets keep getting more massive the farther away they are from the Sun. This is true until Mars, which is less massive than Earth, the next closest planet to the Sun than Mars. The Student is correct, though, that the farther the planets are away from the Sun, the longer they take to go around the Sun.
12) It is important for students to describe how they would correct any of their initial answers to Questions 1-4 as a last step in coming to understand that mass does not affect a planet’s orbital period.
ADDITIONAL QUESTIONS
1) Consider a planet orbiting the Sun. If the mass of the planet doubled but the planet stayed at the same orbital distance, then the planet would take a) more than twice as long to orbit the Sun. b) exactly twice as long to orbit the Sun. c) the same amount of time to orbit the Sun. d) exactly half as long to orbit the Sun. e) less than half as long to orbit the Sun.
Kepler’s Third Law – Instructor’s Guide 35
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
2) Which of the following best describes what would happen if Mercury and Jupiter were to switch places in their orbits about the Sun? a) Jupiter, the larger planet, would have a shorter orbital period than before. b) Mercury, the smaller planet, would have a shorter orbital period than before. c) Neither of the two planets would have any change in their orbital periods.
3) Imagine a new planet in our solar system located 3 AU from the Sun. Which of the following best approximates the orbital period of this planet? a) 1 year b) 3 years c) 5 years d) 9 years
36 Kepler’s Third Law – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy Third Edition
Newton’s Law and Gravity– Instructor’s Guide 37
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
INTRODUCTION
Prerequisite Knowledge
• Basic familiarity with Newton’s law of gravity • Basic familiarity with Newton’s second law • Basic familiarity with Newton’s third law
Goals
• Be able to reason about magnitude and direction of the forces acting on bodies due to the force of gravity
• Be able to apply and predict the how objects will move according to Newton’s second law of motion when being attracted by gravity
• Practice estimation and analytical reasoning skills Pre-activity Question Use the picture below to answer the next three questions. In this picture the Earth-Moon system is shown (not to scale) along with three possible positions (A-C) for a spaceprobe traveling from Earth to the Moon. Note that Position B is exactly halfway between Earth and the Moon.
1) In what direction would the net (total) force point if the spaceprobe were moving very quickly
toward the Moon when at Position B? a) toward Earth b) toward the Moon c) since the force on the spaceprobe by Earth is equal to the force on the spaceprobe by
the Moon the net (total) force would be zero and not point in either direction. 2) At which position (A, B or C) would the spaceprobe feel the greatest acceleration?
a) at Position A b) at Position B c) at Position C d) The acceleration would be the same at all the positions. e) none of the above
38 Newton’s Law and Gravity – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
3) What would the spacecraft do next if it were moving toward the moon when at Position A? a) speed up b) slow down c) travel with a constant acceleration d) travel with a constant speed
TUTORIAL GUIDE
1) [ Equal ] The equation above shows that there is just one force between two objects. To
calculate the force, both masses go into the equation, and no matter which one was M or m (Earth or the Moon), the answer would be the same. This is consistent with the statement of Newton’s third law, which identifies (in this case) that the Moon must be pulling on Earth with a force that is the same type (gravitational), and is of an equal strength, but in the opposite direction to the force that Earth exerts on the Moon. Many students strongly believe that the more massive object, in a pair of objects that are gravitationally attracted, will exert a stronger force on the less massive object. Don’t worry if students answer this question incorrectly. The following Student Debate is designed to confront this known difficulty and help students to clarify their pre-instructional thinking.
2) [ Student 1 is correct ] Student 1 correctly uses Newton’s third law as applied to the problem of two objects gravitationally attracting one another to illustrate that the strength of the force between the Moon and Earth is the same. Student 2 incorrectly thinks that it is possible for one of the objects that are gravitationally attracted to pull harder than the other, and is typically the case, has attributed the greater force to the more massive objects.
For students that struggle to use Newton’s third law to understand this problem, it can be helpful to show them that if one wanted to calculate the force of gravity on Earth by the Moon or the force of gravity on the Moon by Earth, in either case one would use both objects’ masses in the equation, and no matter which one was M or m (Earth or the Moon), the answer would be the same.
3) If the mass of the Moon were made two times as big, the gravitational force between the Moon and Earth would be two times larger, too.
4) [ At the surface of Earth ] Since the gravitational force between two objects is proportional to their masses and inversely proportional to the distance they are apart, then reducing the distance between the objects makes the gravitational force larger. If we want to make the force strongest between Earth and the spaceprobe, the spaceprobe has to be the closest to Earth it can be which is at the surface of Earth.
5) [ At the surface of Mars ] For the same reason as in Question 4, to make the gravitational force between Mars and the spaceprobe the greatest, the spaceprobe needs to be as close to Mars as it can be.
6) [ At the surface of Earth ] The spaceprobe will experience the greatest net (or total) force when it is closest to the more massive planet, which is at the surface of Earth. This will make the gravitational force between Earth and the spaceprobe as great as it possibly can be. It will also make the gravitational force between Mars and the spaceprobe as weak as it
Newton’s Law and Gravity– Instructor’s Guide 39
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
can be since it is at the greatest distance away from Mars. At the surface, Earth will be pulling on the probe with the greatest force it can, and Mars will be pulling in the other direction on the probe with its weakest force. The result is that the “net” force on the probe will be its greatest.
7) At the halfway point the strength and direction of the gravitational force on the spaceprobe by Earth is stronger than, and in the opposite direction of, the gravitational force on the spaceprobe by Mars. The gravitational force is always attractive which will cause these forces to be in opposite directions, and, since the combined mass of Earth and the spaceprobe is greater than the combined mass of Mars and the spaceprobe, while the distance between the spaceprobe and each planet is the same the force on the probe by Earth will be greater. Some students may incorrectly answer Question 7 by stating that the forces would be equal, only taking distance into account instead of also considering that the mass of Earth is greater than the mass of Mars. You do not need to correct them at this point. Question 8 is a Student Debate that will help students confront their reasoning difficulties with this concept.
8) [ Student 2 is correct ] Student 1 is incorrectly only considering that the spaceprobe is the
same distance from both Earth and Mars, and is forgetting to take into account that the gravitational force between the spaceprobe and Earth and the gravitational force between the spaceprobe and Mars are determined by the combined mass of the two objects (either Earth and the spaceprobe or Mars and the spaceprobe) AND their distances those two objects are apart. Since the distances between each of the two objects (either Earth and the spaceprobe or Mars and the spaceprobe) are the same, but the combined mass of Earth and the spaceprobe is much greater than the combined mass of Mars and the spaceprobe, the gravitational force between Earth and the spaceprobe has to be greater than the gravitational force between Mars and the spaceprobe.
9) [ Move ] If the spaceprobe had lost all ability to control its motions and was sitting at rest at
the midpoint between Earth and Mars: (a) The spaceprobe would start moving toward Earth since the gravitational force between Earth and the spaceprobe is greater than the gravitational force between Mars and the spaceprobe at the halfway point based on the answer to Question 7; (b) The spaceprobe would speed up as it moves toward Earth (c) The net (or total) force on the spaceprobe would be increasing as the probe moves closer to the more massive Earth and farther from the less massive Mars; and (d) The spaceprobe would experience the greatest acceleration where it would experience the greatest net force (from Newton’s second law: net force equals mass times acceleration), which is at the surface of Earth (see Question 6.)
10) [ Somewhere closer to Mars than the halfway point ] For the space probe to be stopped and remain at rest, the net (or total) force experienced by the spaceprobe needs to be zero. For the net force on the probe to be zero, it must experience a force toward Earth that is exactly the same strength as the force it feels toward Mars. Since the combined mass of Earth and the spaceprobe is greater than the combined mass of Mars and the spaceprobe, the distance between Mars and the spaceprobe needs to be smaller than the distance between Earth and the spaceprobe to make the gravitational forces felt by the probe from these two planets equal.
40 Newton’s Law and Gravity – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
11) [ Less ] Since my weight on Earth is simply the gravitational force between me and Earth, this means my weight would be less on Mars because Mars is less massive than Earth is, and my mass is the same whether I am on Earth or Mars, so the gravitational force between Mars and me would be less than the gravitation force between Earth and me.
ADDITIONAL QUESTIONS In each figure below two rocky asteroids are shown with masses (m), expressed in arbitrary units, separated by a distance (d), also expressed in arbitrary units. Three of the asteroids are identified with the letters A, B, and C. Use these figures to answer the next two questions.
1) Which of the following correctly describes how the gravitational force exerted BY Asteroid A on its “partner” asteroid compares to the gravitational force exerted BY Asteroid B on its “partner” asteroid. a) The force of A on its partner is greater than the force of B on its partner. b) The force of B on its partner is greater than the force of A on its partner. c) The force of A on its partner is equal to the force of B on its partner.
2) Which of the following is the correct ranking for the acceleration that Asteroids A and C
would experience as a result of the gravitational force exerted on them? a) A=C b) A>C c) C>A
3) Imagine that you throw a ball directly upward. Which of the following statements best
describes how Newton’s second law accounts for the motion of the ball when it reaches its maximum height? a) The ball has a velocity that is zero and an acceleration that is zero. b) The ball has a velocity that is upward and an acceleration that is downward. c) The ball has a net force that is downward and an acceleration that is downward. d) The ball has a net force that is downward and a velocity that is downward. e) The ball has a net force that is downward and an acceleration of zero.
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4) Which of the following would cause the force on the Moon by Earth to increase by the largest amount? a) double the mass of the Moon. b) double the mass of Earth. c) move the moon two times closer to Earth. d) Due to Newton’s third law, the Moon’s force on Earth will always be the same size as
Earth’s force on the Moon so none of the changes listed in choices a-c could cause the force to increase.
42 Newton’s Law and Gravity – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
Apparent and Absolute Magnitudes of Stars – Instructor’s Guide 43
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy Third Edition
INTRODUCTION
Prerequisite Knowledge
• Basic familiarity with the definitions for apparent magnitude and absolute magnitude
Goals
• Distinguish between what apparent and absolute magnitudes tell us about a star’s brightness
• Understand how apparent magnitude depends on distance • Understand how absolute magnitude is related to luminosity • Be able to rank stellar distances using apparent and absolute magnitudes
Pre-activity Question
1) Imagine that you are viewing a star that has an apparent magnitude of 0.2 and is located about 100 parsecs away from us. Which of the following is most likely the star’s absolute magnitude? a) – 4.8 b) 0.1 c) 0.2 d) 0.3 e) 5.2
TUTORIAL GUIDE
1) a) [ apparent magnitude ]
b) [ absolute magnitude ]
This question is provided to check whether students understand the definitions of apparent and absolute magnitudes. The term “actual brightness” is used in place of “luminosity” because it is more meaningful to students at this stage.
2) a) [ Mars ]
Students may incorrectly identify with absolute value of -12.6 being larger than +2.0 and neglect that the negative sign plays an important role is ranking the size of these numbers.
b) [ Full Moon ] The smaller, or more negative, the number, the brighter the object.
c) [ any number greater than 2.0 is acceptable ]
3) [ Student 1 is correct ] Student 2 has the number scale backward. Numbers get bigger for dimmer stars. Magnitude is a ranking—lower numbers are better (brighter).
44 Apparent and Absolute Magnitudes of Stars – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
If Student 1 had made these claims about absolute magnitude he or she would also have been incorrect.
4) [ my < mz and My > Mz ] Since Star Y appears much brighter from Earth than Star Z, the
apparent magnitude number (m) for Star Y has to be smaller than the apparent magnitude number (m) for Star Z. For example: the apparent magnitude number for Star Y could be -1 and the apparent magnitude number for Star Z could be 1 because -1 is less than 1. However, since Star Y actually gives off much less light than Star Z, the absolute magnitude number (M) for Star Y has to be bigger than the absolute magnitude number (M) for Star Z. For example: the absolute magnitude number for Star Y could be 1 and the absolute magnitude number for Star Z could be -1 because 1 is greater than -1.
Some students will find it difficult to assign arbitrary values for these magnitudes. In these cases it is useful to provide students with a single magnitude number for one of the stars to get them started.
5) [ – 6.9 ] Since Rigel is farther than 10 pc, it would appear brighter at 10 parsecs than at its
true location. This means its absolute magnitude number must be smaller than its apparent magnitude number.
6) a) [ Star D ] D has a smaller apparent magnitude number.
It may help to remind students that magnitude is a ranking. It is better for a team to be #1 than #10. b) [ Star A ] A has a smaller absolute magnitude number. c) [ A = 10 pc, B closer than 10 pc, C farther than 10 pc, D = 10 pc ] Apparent and
absolute magnitudes are equal when the star is 10 parsecs away. A star appearing dimmer than its absolute magnitude (M<m) must be farther than 10 parsecs away. A star appearing brighter than its absolute magnitude (m<M) must be closer than 10 parsecs.
d) [ Apparent magnitude number would increase. Absolute magnitude number would
not change. ] By definition, absolute magnitude does not depend upon a star’s actual distance. Apparent magnitude does depend on distance. Since star A has the same apparent and absolute magnitudes it is initially Star A is 10 pc away. If Star A were moved to 40 pc away, it would appear dimmer, and its apparent magnitude number would increase.
7) [ Very near Earth, The Sun ] With an apparent magnitude of –26.7, this star
unquestionably appears the brightest in our sky. In addition, since its absolute magnitude is 4.0, this star must be very close to Earth.
Apparent and Absolute Magnitudes of Stars – Instructor’s Guide 45
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy Third Edition
ADDITIONAL QUESTIONS
1) Star G has an apparent magnitude of 5.0 and an absolute magnitude of 4.0. Star H has an apparent magnitude of 4.0 and an absolute magnitude of 5.0. Which of the following statements is true about viewing these two stars from Earth? a) Star G will appear brighter than Star H. b) The two stars will appear to have the same brightness. c) Star H will appear brighter than Star G.
2) Vega has an apparent magnitude of 0.03 and an absolute magnitude of 0.58. If it were moved twice as far from Earth as it is now, which of the following would occur? a) apparent magnitude number would increase b) apparent magnitude number would decrease c) apparent magnitude number would stay the same d) absolute magnitude number would increase e) absolute magnitude number would decrease
3) Pollux has an apparent magnitude of 1.1 and an absolute magnitude of 1.1. Epsilon Eridani has an apparent magnitude of 3.72 and an absolute magnitude of 6.1. From which of these stars do we receive more light at Earth? a) Pollux b) Epsilon Eridani
4) Star A has an apparent magnitude of 1.0 and an absolute magnitude of 1.0. How would the apparent and absolute magnitudes of this star change if the distance between Earth and the star were decreased? a) The apparent magnitude number would increase, and the absolute magnitude number
would decrease. b) The apparent magnitude number would decrease, and the absolute magnitude number
would increase. c) The apparent magnitude number would increase, and the absolute magnitude number
would not change. d) The apparent magnitude number would not change, and the absolute magnitude number
would increase. e) The apparent magnitude number would decrease, and the absolute magnitude
number would not change.
46 Apparent and Absolute Magnitudes of Stars – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
The Parsec – Instructor’s Guide 47
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
INTRODUCTION
Prerequisite Knowledge
• Know that distances to different stars vary • Basic familiarity with the concept of parallax • Basic familiarity with the “top-down” view of the solar system
Goals
• Describe a star’s parallax by comparing its motion to background stars • Explain the relationship between parallax angle and parsec • Determine relative distances to stars using their parallax angles
Pre-activity question 1) You observe two stars over the course of a year (or more) and find that both stars have
measurable parallax angles. Star X has a parallax angle of 1 arcsecond. Star Y has a parallax angle of ½ an arcsecond. Which star is closer? a) Star X b) Star Y c) not enough information
TUTORIAL GUIDE 1-2) Students should draw a straight line from both Earth positions through Star A and continue
them to the distant stars. These lines will not necessarily land on one of the distant stars. 3-4) When drawing the positions of Star A on the field of background stars, Star A should be
placed next to the same groups of stars to which it was closest in Questions 1 and 2. The background stars in Questions 3-4 are incorrectly shown in the same formation as in Questions 1-2 to avoid confusion. This is not typically a point on which students dwell.
Note that the scale of this figure is exaggerated. Normal parallax angles are very small (less than 1 arcsecond) and cannot be displayed to scale on a diagram.
5) [ The nearby star moves back and forth with respect to the background stars over the
course of a year. ] 6) [ Star C ]
Students often confuse parallax angle and parsec, interpreting both as units of length. It is acceptable if students answer this question incorrectly, as this issue will be revisited in Questions 11 and 13. The purpose of this question is to get students to commit to an idea that can be reevaluated later.
48 The Parsec – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
7-8) As the directions indicate, students should draw a line from Earth, through Star A, to the
top of the page, and label the angel between their line and the dotted line with pa. A parallax angle pB for a second star will be added to this diagram in Question 12.
9) [ 1 parsec ]
Students should be encouraged to use the definition of the parsec provided immediately before Question 9 rather than looking for an equation in their books. Again, no effort has been made to correctly scale the diagram.
10) [ Length ]
This question reinforces that a parsec is a unit of distance. The confusion that many students have is that since a parallax angle is used to define a parsec, a parsec also must be an angle.
11) [ Student 2 is correct ] There is an inverse relationship between a star’s parallax
angle and its distance. The smaller a star’s parallax angle, the farther it is away from Earth.
Many students have serious difficulty with this concept. Student 1 incorrectly uses a “more means more” reasoning model to associate larger parallax angles with greater distances. Student 2 gives the correct reasoning by using a geometric argument.
12-13) [ The closer star (Star A) has the larger parallax angle. ]
By repeating the process of finding a parallax angle for a second star, students reconstruct the reasoning of Student 2 from Question 11. Students should be encouraged to use what they draw in Question 12 to resolve any discrepancies with Questions 6 and 11.
ADDITIONAL QUESTIONS 1) Which of the following stars is closest to us?
a) Procyon (parallax angle = 0.29”) b) Ross 780 (parallax angle = 0.21”) c) Regulus (parallax angle = 0.04”) d) Sirius (parallax angle = 0.38”)
2) On Earth, the parallax angle measured for the star Procyon is 0.29 arcseconds. If you were to measure Procyon’s parallax angle from Venus, what would the parallax angle be? (Note: Earth’s orbital radius is larger than Venus’s orbital radius.) a) more than 0.29 arcseconds b) 0.29 arcseconds c) less than 0.29 arcseconds d) zero arcseconds (no parallax)
The Parsec – Instructor’s Guide 49
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition
3) Consider two stars (X and Y). If Star X is 3 parsecs away and Star Y is 5 parsecs away, which has the greater parallax angle? a) Star X b) Star Y c) Not enough information
4) Below are two star photos taken six months apart and laid atop one another so the
background stars (circles) line up. There are two nearby stars also shown. Which of these nearby stars is closer?
a) Star b) Star
50 The Parsec – Instructor’s Guide
Instructor’s Guide for Lecture-Tutorials for Introductory Astronomy
Third Edition