1. In the Greek geocentric model, the retrograde motion of a planet occurs when A. Earth is about to pass the planet in its orbit around the Sun. B. the planet actually goes backward in its orbit around Earth. C. the planet is aligned with the Moon in our sky. 2. Which of the following was not a major advantage of Copernicus s Sun-centered model over the Ptolemaic model? A. It made significantly better predictions of planetary positions in our sky. B. It offered a more natural explanation for the apparent retrograde motion of planets in our sky. C. It allowed calculation of the orbital periods and distances of the planets. 3. When we say that a planet has a highly eccentric orbit, we mean that A. it is spiraling in toward the Sun. B. its orbit is an ellipse with the Sun at one focus. C. in some parts of its orbit it is much closer to the Sun than in other parts. 4. Earth is closer to the Sun in January than in July. Therefore, in accord with Kepler s second law, A. Earth travels faster in its orbit around the Sun in July than in January. B. Earth travels faster in its orbit around the Sun in January than in July. C. it is summer in January and winter July. 5. According to Kepler s third law, A. Mercury travels fastest in the part of its orbit in which it is closest to the Sun. B. Jupiter orbits the Sun at a faster speed than Saturn. C. all the planets have nearly circular orbits. 6. Tycho Brahe s contribution to astronomy included A. inventing the telescope. B. proving that Earth orbits the Sun. C. collecting data that enabled Kepler to discover the laws of planetary motion. 7. Galileo s contribution to astronomy included A. discovering the laws of planetary motion. B. discovering the law of gravity. C. making observations and conducting experiments that dispelled scientific objections to the Sun-centered model. 8. Which of the following is not true about scientific progress? A. Science progresses through the creation and testing of models of nature. B. Science advances only through the scientific method. C. Science avoids explanations that invoke the supernatural. 9. When Einstein s theory of gravity (general relativity) gained acceptance, it demonstrated that Newton s theory had been 1 / 5
A. wrong. B. incomplete. C. really only a guess. 10. Which of the following is not true about a scientific theory? A. A theory must explain a wide range of observations or experiments. B. Even the strongest theories can never be proved true beyond all doubt. C. A theory is essentially an educated guess. (a) (b) Figure 1: Study the two graphs above, based on Figure 3.21 from the textbook. Use the information in the graphs to answer questions 11 17. 11. Approximately how fast is Jupiter orbiting the Sun? A. cannot be determined from the information provided km/s B. 20 km/s C. 10 km/s D. a little less than 15 km/s 12. An asteroid with an average orbital distance of 2 AU will orbit the Sun at an average speed that is. A. a little slower than the orbital speed of Mars B. a little faster than the orbital speed of Mars C. the same as the orbital speed of Mars 13. On graph 1(a), you can see Kepler s third law ( p 2 a 3) from the fact that. A. the data fall on a straight line B. the axes are labeled with values for p 2 and a 3 C. the planet names are labeled on the graph 14. Uranus, not shown on graph 1(b), orbits about 19 AU from the Sun. Based on Figure 1, its approximate orbital speed is between about A. 20 and 25 km/s 2 / 5
B. 15 and 20 km/s C. 10 and 15 km/s D. 5 and 10 km/s 15. Kepler s third law is often stated as p 2 a 3. The value a 3 for a planet is shown on. A. the horizontal axis of Figure 1(a) B. the vertical axis of Figure 1(a) C. the horizontal axis of Figure 1(b) D. the vertical axis of Figure 1(b) 16. Suppose Figure 1(a) showed a planet on the red line directly above a value of 1000 AU 3 along the horizontal axis. On the vertical axis, this planet would be at. A. 1000 years 2 B. 1000 2 years 2 C. 1000 years 2 D. 100 years 17. How far does the planet in question 16 orbit from the Sun? A. 10 AU B. 100 AU C. 1000 AU D. 1000 AU 18. Method of Eratosthenes I. You are an astronomer on planet Nearth, which orbits a distant star. It has recently been accepted that Nearth is spherical in shape, though no one knows its size. One day, while studying in the library of Alectown, you learn that on the equinox your sun is directly overhead in the city of Nyene, located 1000 kilometers due north of you. On the equinox, you go outside and observe that the altitude of your sun is 80. What is the circumference of Nearth? (Hint: Apply the technique used by Eratosthenes to measure Earth s circumference.) Solution: We will follow Eratosthenes s logic from the Special Topic section on page 65 in the textbook. In the case of Nearth, we learn that the Sun is straight overhead at Nyene at the same time that it is 10 from the zenith at Alectown and that the two cities are 1000 kilometers apart. So we can set up the same type of relationship as Eratosthenes did: 10 We solve for the circumference of Nearth: (circumference of Tirth) = 1000 km 360 circumference of Nearth = 1000 km 360 = 36000 km 10 19. Method of Eratosthenes II. You are an astronomer on planet Tirth, which orbits a distant star. It has recently been accepted that Tirth is spherical in shape, though no one knows its size. One day, you learn that on the equinox your sun is directly overhead in the city of Tyene, located 400 kilometers due north of you. On the equinox, you go outside and observe that the altitude of your sun is 86. What 3 / 5
is the circumference of Tirth? circumference.) (Hint: Apply the technique used by Eratosthenes to measure Earth s Solution: We will follow Eratosthenes s logic from the Special Topic section on page 65 in the textbook. In the case of Tirth, we learn that the Sun is straight overhead at Tyene at the same time that it is 4 from the zenith at Alectown and that the two cities are 1000 kilometers apart. So we can set up the same sort of relationship as Eratosthenes did: 4 We solve for the circumference of Tirth: (circumference of Tirth) = 400 km 360 circumference of Tirth = 400 km 360 4 = 36000 km 20. Mars Orbit. Find the perihelion and aphelion distances of Mars. (Hint: You ll need data from Appendix E.) Solution: We are told in Mathematical Insight 3.2 on page 73 of the textbook that the perihelion and aphelion distances between the Sun and a planet are, respectively: perihelion distance = a (1 e) aphelion distance = a (1 + e) where a is the semimajor axis and e is the eccentricity. From Appendix E, Marss orbital eccentricity is 0.093 and its semimajor axis is 1.524 AU. So we get: perihelion distance = (1.524 AU) (1 0.093) = 1.382 AU = 2.07 10 8 km aphelion distance = (1.524 AU) (1 + 0.093) = 1.666 AU = 2.50 10 8 km Mars s minimum distance from the Sun is 1.382 AU and its maximum distance is 1.666 AU. 21. Most Eccentric Orbit. Which planet in our solar system has the most eccentric orbit? Find the planet s perihelion and aphelion distances. Solution: From Appendix E, Mercury has an eccentricity of 0.206. Given its semimajor axis, 0.387 AU, we can use the formulas from above to calculate its perihelion and aphelion distances from the Sun: perihelion distance = (0.387 AU) (1 0.206) = 0.307 AU = 4.6 10 7 km aphelion distance = (0.387 AU) (1 + 0.206) = 0.467 AU = 7.01 10 7 km Eris and Pluto are incorrect answers because they are not planets. 22. Least Eccentric Orbit. Which planet in our solar system has the least eccentric orbit? Find the planet s perihelion and aphelion distances. Solution: From Appendix E, Venus has the smallest eccentricity of the planets, at 0.007. Given its semimajor axis, 0.723 AU, we can use the formulas from above to calculate its perihelion and aphelion 4 / 5
distances from the Sun: perihelion distance = (0.723 AU) (1 0.007) = 0.718 AU = 1.08 10 8 km aphelion distance = (0.723 AU) (1 + 0.007) = 0.728 AU = 1.09 10 8 km 23. Eris Orbit. The recently discovered Eris, which is slightly larger than Pluto, orbits the Sun every 560 years. What is its average distance (semimajor axis) from the Sun? How does its average distance compare to that of Pluto? Solution: Eris s period is 560 years. a 3 = p 2 a = 3 p 2 = 3 (560) 2 = 67.9 AU Eris has an average distance (semimajor axis) of 67.9 AU. 24. New Planet Orbit. A newly discovered planet orbits a distant star with the same mass as the Sun. The planet s average distance from the star is 112 million kilometers. Its orbital eccentricity is 0.3. Find the planet s orbital period and its nearest and farthest orbital distances fro m its star. Solution: Convert the planet s average distance from the star from kilometers to AU: 1.12 10 8 km 1 AU 1.5 10 8 km = 0.747 AU Using Kepler s 3rd law: p = a 3 = (0.747) 3 = 0.645 yr The new planet has an orbital period of 0.645 year, or about 7.75 months. 25. Halley Orbit. Halley s Comet orbits the Sun every 76.0 years and has an orbital eccentricity of 0.97. (a) Find its average distance from the Sun (semimajor axis). Solution: Since Halley has an orbital period of 76 years, we can calculate the semimajor axis: Comet Halley has a semimajor axis of 17.9 AU. a = 3 p 2 = 3 (76) 2 = 17.9 AU (b) Find its perihelion and aphelion distances. Does Halley s Comet spend most of its time near its perihelion distance, near its aphelion distance, or halfway in between? Explain. Solution: We are told that Halley s comet has an eccentricity of 0.97 and in part (a), we calculated its semimajor axis to be 17.9 AU: perihelion distance = (17.9 AU) (1 0.97) = 0.537 AU aphelion distance = (17.9 AU) (1 + 0.97) = 35.3 AU Halley s comet comes as close as 0.537 AU to the Sun and travels as far away as 35.3 AU from the Sun. 5 / 5