PTYS/ASTR 206 Section 2 Spring 2007 Homework #1 (Page 1/4) NAME: KEY Due Date: start of class 1/25/2007 5 pts extra credit if turned in before 9:00AM (early!) (To get the extra credit, the assignment must be placed the box outside of room 330!) 5 pts taken off if turned in after the start of the class period and before 5PM of the due date (late!) The assignment will not be graded if turned in later than 5PM 1/25/2007 (too late!) Turn in your own work. It is not acceptable to turn in work that is identical to that of another student. You may work together to understand the material, but the work you turn in must be in your own words. There are 3 parts, plus an extra-credit part. Please put your answers to all questions to PARTS 1-2 in the space provided (you may use the back of the page if necessary). PART 1 (25 pts total): Conceptual. Please provide a concise short answer (not more than a few sentences) for each of the following. # 1. (5 pts) (Chapters 2 and 4) Explain how the motion of planets in the night sky differs from that of stars. Which planet (of the 9) moves most like the stars? Explain your answers. The planets move relative to the stars and can exhibit retrograde motion, something that stars to not do! This is due to the fact that the planets are much closer to us than the stars. Pluto moves most like the stars since it is farthest away. Pluto, which is not longer a planet, is so far away that it moves very nearly at the same rate as the stars (but it still exhibits retrograde motion!). # 2. (5 pts) (Chapter 2) Name three things that are astronomically significant things about the vernal equinox (about March 21 of each year). Any of the following (or more if they are correct!): (1) The amount of time for night and day is equal everywhere on Earth (12 hours) (2) The Sun rises due East, and sets due West everywhere on Earth (3) The Sun crosses the celestial equator, going North (in the Northern hemisphere) (4) To observers at Earth s equator at 12:00 noon, the Sun will be directly overhead.
PTYS/ASTR 206 Section 2 Spring 2007 Homework #1 (Page 2/4) # 3. (5 pts) (Chapter4: #26 of the textbook) Why was the discovery of Neptune an important confirmation of Newton s law of universal gravitation? Its existence and position in the sky was predicted using Newton s law of gravitation using the forces from all of the known planets. It was then discovered rather easily, once the position had been determined. This confirmed the Newton s theory. # 4. (5 pts) (Chapter 2: #11 of the textbook) Give two reasons why it s warmer is summer than in winter. Explain how each of these causes it to be hotter. It is warmer in the summer because the Sun s rays strike the ground more perpendicularly (therefore, the Earth s surface absorbs more energy in the summer than in the winter when the suns rays strike the ground at a much more shallow angle) and the days are longer (meaning that there is more time to heat the surface). # 5. (5 pts) (Chapter 4) Describe one set of observations that support the heliocentric model for our Solar System. Observations of the phases of Venus and moons orbiting Jupiter reinforced the heliocentric model. The observations of the moons of Jupiter gave evidence that objects could orbit around something other than the Earth. However, it was Galileo s observation of the gibbous (or fullish) phase of Venus that definitively ruled out the older Ptolemaic, earth-centered model. In the old model, there was no way for Venus to have such a phase, whereas in the heliocentric model, it can. These observations require the use of a telescope and so were impossible before Galileo s time.
PTYS/ASTR 206 Section 2 Spring 2007 Homework #1 (Page 3/4) PART 2 (15 pts total): Quantitative # 1. (5 pts) (Chapter 4) The thick line in the diagram below is a fictitious near-earth asteroid. Answer each of the questions below based on this diagram (you will need to use a ruler). (a) Determine the radius of the Earth s orbit about the Sun in inches (or cm). What is the conversion factor to go from inches (or cm) to AU for this diagram? (Hint: Think of this like a map and you are asked to determine the legend). If this is printed out on regular paper, the radius of Earth s orbit is 2.2cm. Since we know that the Earth is 1 AU from the Sun. The legend for this figure would be: 2.2 cm = 1 AU or 1 cm = 1/(2.2) AU = 0.45 AU (b) What is the semimajor axis of the asteroid s orbit in AU? The long axis of the ellipse is 6.7 cm in this figure. Thus, the semimajor axis (1/2 of this distance) is 3.85cm. Using the answer to part (a) we find that the semimajor axis is a = 3.85 x 0.45 AU = 1.7 AU (c) What is the period of the asteroids orbit (in years)? From Kepler s 3 rd Law, P 2 =a 3 (where P is in years and a is in AU). Using the answer to part (b) we find P = (1.7 3 ) ½ years = 2.2 years # 2. (5 pts) (Chapter 1) If we ever send humans to Mars, it will be difficult to have conversations with them because of the communication time lag due to the finite speed of light. Compute the time (in minutes) it takes for light to travel from Mars to Earth when Mars is 0.5 AU from Earth (near its closest approach to Earth). The distance from Mars to the Earth is 0.5 AU = 0.5 x (1.496 10 8 km) = 7.48x10 7 km. If you express the Mars-Earth distance in kilometers, you must express the speed of light in km/s. The light-travel time is 0.5 AU divided by the speed of light: t = d/v = (7.48x10 7 km)/(3x10 5 km/s)=249.3 seconds=4.15min.
PTYS/ASTR 206 Section 2 Spring 2007 Homework #1 (Page 4/4) # 3. (5 pts) (Chapter 4) Calculate the gravitational force exerted on Earth by the (a) Moon and (b) by the Sun. Some potentially useful information: mass of Earth = 6x10 24 kg, diameter of Earth = 1.2x10 7 m, mass of Moon = 7x10 22 kg, diameter of moon = 3x10 6 m, distance between Earth and Moon = 3.8x10 8 m, mass of Sun = 2x10 30 kg, diameter of Sun = 7x10 8 m, distance between Earth and Sun = 1.5x10 11 m. (a) Force exerted on Earth by the Moon (b) Force exerted on Earth by Sun PART 3 (10 pts total): Starry Night Backyard Observing Problems Please provide your answers on a separate sheet of paper and attach it to the rest of your homework. # 1. (5 pts) Chapter 2, #59 of the textbook The stars move counterclockwise (assuming you set Tucson as the viewing location) when looking towards the northern horizon. The rotation of the stars about the celestial pole is caused by the Earth s rotation (diurnal motion). Yes, there are many circumpolar stars (Polaris for one). When looking to the south (again from Tucson), the stars move clockwise, however, we cannot see the south celestial pole. Intuitively we can understand why the motion is in the opposite sense when looking to the south because we have turned our head around to see it note that the stars always move from East to West (e.g. the Sun rises in the east and sets in the west), thus when looking to the north, the starts appear to be moving counterclockwise, but when looking to the south the east-west motion is in the clockwise sense. There are no circumpolar stars to the South. # 2. (5 pts) Chapter 4, #57 of the textbook a) Normally Mars is moving to the left (remember, the stars are stationary). Retrograde motion begins around end of September or beginning of October, and retrograde motion ends about December 10. b) Normally Earth is moving to the left. Retrograde motion begins again around end of September or beginning of October, and retrograde motion ends again about December 10. c) Earth is closest to Mars around the end of October or the beginning of November. This date is right in the middle of the two dates in part a) and b), and shows that retrograde motion occurs right before and after closest approach.
PART 4: Extra Credit Points available: 2 for one, 5 for both Please provide your answers on a separate sheet of paper and attach it to the rest of your homework. Chapter 4, #36 of the textbook (a) a = 3 2 p = 2 3 64 = 16 AU. (b) The perihelion distance is the major axis minus the aphelion distance: Perihelion distance = 2 16 AU 31.5 AU = 0.5 AU. Chapter 4, #47 of the textbook (a) For the closest side: r = 384,400 km 0.5(12,756 km)= 3.78 10 8 m. Then using 11 2 2 22 mm 1 2 (6.67 10 N m /kg )(7.348 10 kg)(1 kg) -5 F = G = 2 8 2 3.43 10 N. r (3.78 x 10 m) (b) For the farthest side: r = 384,400 km + 0.5(12,756 km) = 3.91 10 8 m. Then, 11 2 2 22 (6.67 10 N m /kg )(7.348 10 kg)(1 kg) -5 F = 8 2 = 3.21 10 N. (3.91 x 10 m) (c) The difference is 0.22 10-5 N. (d) There is only a very small deformation of the Earth s surface due to tidal forces because the tidal force is very small compared to the Earth s gravity, which tends to maintain a spherical surface.