CHAPTER 1 T H E CYCLES OF THE SKY Lecture Suggestions Planetarium software may be helpful for this chapter. An orrery (mechanical model of the Sun and Earth) or even just an approximation of one can also help illustrate various motions and that the constellations change with the seasons. Set it on a table in the front of the room and then mark off constellations on the walls with chalk or paper decorations. Moving the model Earth around the model Sun then allows students to see how the stars visibly change and how the Sun “moves” through the Zodiac. In effect, this turns the room into a simple planetarium. A flashlight with a fat beam shows the importance of angle to seasonal heating. When directed directly at the wall the energy is concentrated in a small area. When shined obliquely at the wall, the beam covers a larger area, implying less concentration of heat and therefore a lower temperature. It’s also important to include the idea that the day is longer in the summer, which is sometimes overlooked. A bright light source and tennis balls, basketballs, volleyballs or even golf balls can demonstrate features of eclipses and phases of the moon. Answers to Thought Questions 1. If you were standing on Earth’s equator, looking due north you would see the north celestial pole on the horizon (and the south celestial pole on the horizon, looking due south).You cannot see the north celestial pole from Australia (it’s below the horizon), only the south celestial pole. 2. SKETCH FOR STUDENTS 3. The main astronomical reason why there are 12 zodiacal signs is that the Sun appears to move about 30 degrees per month across the background stars (360 / 30 = 12). 4. SKETCH FOR STUDENTS. At the horizon, setting or rising stars move perpendicularly to the horizon (so they are useful for East-West navigation). At the north pole, stars more or less do not set, they just circle in the sky. At a mid-latitude, the stars make an angle with the horizon. In the Northern Hemisphere, as they set they also move more toward the north. Extremely schematically: setting stars looking west: at the equator, | | |; at a mid-latitude, Northern Hemisphere: \ \ \; at the North Pole, ---. 5. When it is winter in New York, the Northern Hemisphere is tilted away from the Sun; therefore at that time the Southern Hemisphere is tilted towards the Sun and it’s summer in Australia. Although Paris is partway around the world from New York, it’s at about the same latitude and it is also winter there. The part of the sky that you see at night is the part of the sky away from the Sun, so everywhere on Earth sees the same “half” of the sky at night during a 24 hour period, as Earth rotates viewers into nighttime: all three locations should be able to see Orion, which straddles the celestial equator.
Chapter 1
The Cycles of the Sky
6. If Earth’s orbit had no tilt, there would be no variation in the angle of sunlight over the year, nor would there be variation in the length of day—conditions would be somewhat like the equinox all the time and a 12 hour day everywhere, every day. There would be a slight variation in temperature over the year with higher temperatures in January based on the small change in Earth’s distance from the Sun, but this effect would be very small (clearly it does not affect the seasons induced by the tilt very much). Northern and Southern hemispheres would experience these weak seasons at the same time. 7. The position of sunrise along the eastern horizon changes during the year because Earth’s axis (and correspondingly, the celestial equator) is tilted at 23.5 degrees to the plane of its orbit (the ecliptic) and Earth maintains this same tilt throughout the year. At the equinoxes (March 21 and Sept. 23), the Sun lies on the celestial equator. Because the celestial equator cuts the horizon at the east and west points, the Sun will rise and set due east and due west, respectively. In winter, the tilt of Earth results in the Sun rising north of east and setting north of west, and in winter, the Sun rising south of east and setting south of west. At the winter solstice (Dec. 21), the Sun lies 23.5 degrees south of the celestial equator on the sky. It will therefore rise the most to the south of the East on the horizon and set the most to the south of West that year. At the summer solstice (June 21), the Sun lies 23.5 degrees north of the celestial equator. It will therefore rise the most north of East and set the most North of West. 8. We have time zones to keep our local time in approximate alignment with solar time, and to standardize time between different parts of countries and the world—using exact local solar time everywhere would be just as confusing as using one set of hours for the whole world. The sketch can show how it’s solar noon on one part of earth (the Sun is highest in the sky) and a very different time elsewhere (the Sun would be high or low in the sky), or just show a close-up of the difference between the local time in one time zone (say noon) and an adjacent time zone (say 1 pm). 9. Some possible ideas: (One or two of these or related ideas should be sufficient). -You can see some phases during the day, so the geometry is incorrect for the phase to be a shadow. -You can determine the Sun-Earth-Moon angle is not 180 degrees during most phases. -Lunar eclipses (shadow on the Moon) do occur, and only during the 180 degree/full moon alignment. -The curvature of the terminator during Moon phases is not consistent from phase to phase—if it was Earth’s shadow, it would always be the same shape. The changing terminator shape is consistent with a partially illuminated sphere. -The radius of curvature of the terminator for phases does not match the radius of the shadow of Earth during an eclipse. -Eclipses happen over a period of hours, while the phases change slowly over weeks, suggesting they are not caused by the same thing. 10. In this case, the sidereal month would remain 27.3 days as the periodic alignment with the stars would not change, but the solar month would be shorter because the Moon will reach new moon before re-aligning with the stars instead of after. The redrawn figure
Chapter 1
The Cycles of the Sky
of 1.14 should indicate that this is the case (the right part of the diagram will happen in opposite order). Answers to Problems 1. 360 degrees / 24 hours = 15 degrees/hr. A simple problem but a good number for students to know. 2. The equinox (Sept. 22) latitude of the Sun will be 90° - 55° = 35°. The highest angle = 35° + 23.5° = 58.5°, lowest = 35° - 23.5° = 11.5°. 3. MAKE SKETCH FOR STUDENTS. Use the phases diagram and label the time on different parts of Earth; then draw horizon lines tangent to Earth to the waxing crescent moon to determine the time it is 1st visible rising the east is at about 9 AM and it sets in the west at about 9 PM. 4. The Moon’s sidereal period is 27.3 days. Therefore, it moves… 360° / 27.3 days = 13.2°/day. Earth must rotate an extra 13.2° to “catch up” to the Moon. Earth rotates about 15°/hr (Problem 1), so the Moon rises about 13.2°/(15°/hr) = 0.88 hr ≈ 53 minutes later each day. 5. Moon’s draconic orbital period is 27.2122 days = P. Moon’s synodic period is 29.5306 days = S. 242 P = 6,585.3524 days 223 S = 6,585.3238 days 242 P = 6585.3524 days /(365.24 days/yr)= 18.03 years. The result gives a pattern of solar and lunar eclipses which repeat over about 18 years, but shift in location across Earth because there is not an even number of solar days in a year. 6. Circumference of moon’s orbit = C = 2 384,400 km = 2.42 106 km V = C / Period = 2.42 106 km / (27.322 d 24 h/d) = 3680 km/h Earth’s diameter = 6400 km 2 = 12,800 km Time = D / V = 12,800 km / (3680 km/h) = 3.5 hours (approx) Since Earth only moves a small distance around its orbit in a few hours of time, the time for the Moon to travel through Earth’s shadow is the dominant effect in determining the duration of the lunar eclipse. (See also next question). 7. We would need to decide exactly when to start and when to stop counting—the above estimate considers the leading edge of the moon going from one side of Earth to the other, but we should probably add the Moon’s diameter to the distance traveled so we
Chapter 1
The Cycles of the Sky
move all the way out of Earth’s shadow. This would increase the time (by about diameter of Moon / diameter of Earth ~ 30%). However, Earth’s shadow is not exactly the size of Earth’s diameter—it is a bit smaller, and that would decrease the time (a noticeable correction). As the whole system moves around the Sun, there is also a slight shift in the shadows; since Earth and Moon are moving in the same direction this will lengthen the shadow passage very, very slightly. 8. The approach to the estimation is still fine: the Moon is still moving fast enough that any effect on the shadows from the Earth-Moon system moving around the Sun is negligible for our purposes. We do need to treat the radius as a ~ 200 km shadow passing overhead; 200 km / (3680 km/hr) = 0.05 h = 3.26 minutes. Rmoon
Rsun d Sun
Moon
Answers to Test Yourself 1. (d) The north celestial pole is directly overhead. 2. (b) Altitude = latitude = 30°. 3. (a), (c), (d) Sun, Moon, stars would rise in West, rotate around Polaris clockwise. 4. (d) longer day and more concentrated Sun 5. (d) Above the arctic circle, or below the Antarctic circle, the daylight varies from 0 to 24 hours over the year. 6. (b) Waning gibbous. Must be same moon phase seen all over Earth. 7. (c) Full moon rises at 6pm. 8. (b) between first quarter (rises at noon) and full (rises at 6pm) 9. (c) 2 weeks 10. (c) Moon covering sun 11. (b) bigger moon, more eclipses, but not every month.
Chapter 2
The Rise of Astronomy
CHAPTER 2 THE RISE OF ASTRONOMY Answers to Thought Questions 1. Despite the Moon illusion, you are actually closest to the Moon when it is at its highest point in the sky. As seen from over the North Pole,
2. If the stars were much closer than they really are, Aristarchus would have been able to demonstrate the stellar parallax caused by Earth’s orbital motion around the Sun. 3. The phases of Venus are caused by Venus orbiting the Sun, so keeping all distances and periods constant, it wouldn’t matter whether Earth and Venus were orbiting the Sun or Venus was orbiting the Sun and the Sun (and Venus) were orbiting Earth. 4. The Sun has a slightly larger angular diameter in January than in July because Earth’s orbit is an ellipse. The idea that orbits are ellipses is Kepler’s First Law. 5. The apparent motion of Jupiter is primarily a result of Earth’s motion around the Sun, not Jupiter’s motion. A sketch like the one below can help show this—consider the view from Earth over the year, with Jupiter moving only a little along its orbit. At position A, Jupiter is at opposition, and rises when the Sun sets. By the time Earth is at B, Jupiter has not moved very far (so its motion is neglected in this sketch), but now it is only 90 degrees away from the Sun—already high in the sky at Sunset. By position C, Jupiter would be “up” in the daytime. Moving to position D, Jupiter is moving past the Sun on the opposite side as before.
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The Rise of Astronomy
6. If the same comet is only visible every 50 years or more (or much, much more!) then the comet must have a much longer p than Earth’s orbit. Consequently by Kepler’s third law it must have a corresponding larger a. If the comet needs to be close to Earth and Sun to be seen, then at least some of the time it must be at only 1 or 2 AU from the Sun; for this to fit with a large a and p, it must have an elliptical orbit with high eccentricity. 7. This should be less about technology and more about philosophy; and about getting students to look up contemporary work. One key difference is less personal, government and patron-sponsored religious/mystic motivation—astronomy is no longer in the game of predicting planetary positions as portents of war and peace. Also, astronomy has become increasingly based on fact and less influenced by philosophy (Ptolemy’s and Kepler’s motivations looking for perfection in the heavens; Kepler was conflicted but notably broke with his philosophical position (circular orbits) when confronted with scientific evidence). Increased technology has revealed entire branches of astronomy previously invisible to scientists (radio, x-ray, infrared, gravitational waves, etc.). 8. (Students must research the astronomers in question).
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The Rise of Astronomy
Answers to Problems 1. This problem is a modern version of the method Eratosthenes used to measure the size of Earth. Given that the shadow length is 15 degrees, the distance in latitude between the two points on the asteroid must be 15 degrees, or 15°/360° = 1/24th the circumference of the asteroid. If the 15 degrees corresponds to 10 km, then the total distance around the asteroid must be 10 km × 24 = 240 km. The radius, R, of the asteroid is related to its circumference, C, by C = 2R. Thus R = C/2 = 240/2 = 38 km 2. This can be directly calculated, but it is easier to just look at the proportionalities involved. Since C = 360°/angle × distance, the circumference is directly proportional to the distance between Alexandria and Syene. If the distance were three times as much, and the angle the same, Eratosthenes would have calculated that the circumference of Earth was three times larger: 250,000 stadia × 3 = 750,000 stadia or 25,000 miles × 3 = 75,000 miles. (“Three times larger” should be an acceptable answer.) 3. From Earth, the Sun has an angular diameter of 0.5°. Angular size varies as the inverse of the distance, so if Mercury is 0.387 times as far, the Sun is 0.5°/0.387 = 1.29°. Pluto is 39.53 times Earth’s distance, so the Sun is 0.5°/39.53 = 0.0126°. 4. The Andromeda galaxy has an angular diameter of 5 degrees at a distance of 6 2.2 × 10 ly. We can find its true size by using the angular diameter formula L = 2dA/360°, where d = distance, A = angle subtended, and L = linear diameter. Thus, 6 5 L = 2 × 2.2 × 10 ly × 5°/360° = 1.92 × 10 ly. 5. If P = 64 years, you can use Kepler’s 3rd Law to estimate its distance from the Sun. 2 3 P = a , where P = period in years and a = average orbital radius in AU. We can find a by 2/3 2/3 2/3 1/3 taking the cube root of both sides to get P = a, or a = P = (64) = (64 × 64) 1/3
= (8 × 8 × 8 × 8) circular.
= 2 × 2 × 2 × 2 = 16 AU. This is the radius of the orbit if the orbit is
6. a3 = P2. (526)3 = 1.4553 × 108 = P2, so P = 12,063 years. 7. This is an application of Kepler’s third law, P2 = a3, where a is in AU and P is in years. If P = 125 yrs, then a3 = 1252. Solving for a, we take the cube root of both sides to get a = (1252)1/3, where we have used the fact that the cube root of a number is the number to the 1/3 power. This can be solved with a calculator or by noticing that
Chapter 2
The Rise of Astronomy
(125 x 125)1/3 = (25 x 5 x 5 x 25)1/3 = (25 x 25 x 25)1/3 = 25, so a = 25 AU. If the planet’s orbit is circular, then that is also the planet’s orbital radius. 8. This problem is another application of Kepler’s third law, P2 = a3, where a is in AU and P is in years. In this case, we are given a, and are asked to find P. Thus P2 = a3 = 163. Solving for P by taking the square root and recalling that the square root is the number to the 1/2 power, we find that P = (163)1/2 = 64 yrs. (Note: in solving this problem, you can simplify the math by reversing the order of the power and the square root. That is, (163)1/2 = (161/2)3 = 43 = 64.) Answers to Test Yourself 1. (d) Angular size is inversely proportional to distance so LM/LS= 1/400. 2. (b) Retrograde motion causes planets to stop their regular eastward motion with respect to the stars and move westward for a time. 3. (b) simplicity of models 4. (d) P2 = a3. 43 = 4 × 4 × 4 = 4 × 2 × 2 × 4 = 82 5. (a) Kepler’s 3rd Law relates a planet’s orbital period to the size of its orbit. 6. (e) Venus orbits the Sun. 7. (c) Parallax was conjectured but impossible to measure without high quality telescopes.
CHAPTER 3 GRAVITY AND MOTION Answers to Thought Questions 1. You would not want to kick a cinder block in space with your bare foot (ignoring the problems of having your foot bare in space) because it still has the same mass and inertia as on Earth. Kicking the block would hurt your foot as badly as if you’d kicked it on Earth’s surface. 2. Students should draw a sketch something like the ones below (either style or both). It may feel as if the “imaginary” centrifugal force tries to push you out, but really you are trying to travel out at a tangent and the wall is stopping you. Keep in mind that the direction of acceleration in a circular orbit is always towards the center, and the net force must be in the same direction as net acceleration. The wall is constantly pushing you in to keep you going in a circle.
3. Yes, there is a force of gravity between the ISS and Earth; in fact it is the force keeping it in orbit, so yes, the ISS is affected by it. The astronauts also experience a force of gravity and are accelerated in orbit around Earth. They and the space station are accelerated together, and move together in orbit—all are continuously falling towards Earth together and so the astronauts “feel” weightless. There is no (appreciable) net force between the astronauts and the space station. 4. F = ma, so larger masses require larger forces for the same accelerations. The force to accelerate a car or to keep it moving at constant speed against friction comes ultimately from the power produced by burning gasoline in the engine. All other things being equal, you will need to use more energy in the engine to apply a larger force to move a larger vehicle the same distance as a smaller one. 5. Newton’s 3rd law will still apply. The Moon’s gravitational pull on Earth has exactly the same magnitude (but in the opposite direction of) Earth’s gravitational pull on the Moon.
6. When you walk, you accelerate forwards, so the force on you must be in the forward direction. You can’t push yourself forwards, so the ground must push you forwards. The ground pushing on you is a reaction to your foot pushing on the ground, so you must push backwards against the ground. So you push on the ground and the ground pushes on you, and the ground pushing on you makes you go forwards. Since the ground moves very little when you push on it, you can push hard, and it can push you back hard, and by F = ma you have a big force to move forwards. If you are walking on sand, it moves easily when you push back on it with your foot, and it’s much harder to make forward progress because you are more easily able to move the ground (so you can’t push so hard against it). Likewise, it moves you with less force. 7. Yes, because there is a force of gravity between any pair of objects with mass. The force decreases with the square of the distance between the objects. The book falls to the ground even though it is much further from the center of Earth than the center of you because Earth is much, much, much more massive than you are. 8. F = ma means F must have units of [kg] [m] / [s]2. Therefore F = GMm/d2 must also have those units, so [kg] [m] / [s]2 = [units of G] [kg][kg]/[m]2 simplifying, [units of G] = [kg]-1 [m]3 [s]-2 ; the units of G must be m3/(kg s2). (Alternatively, [Newtons] = [units of G][kg]2[m]-2 and [units of G] = [N][m]2[kg]-2, but the intent is to remove the Newton unit, hence including F = ma in the problem setup). 9. The Sun’s gravitational force on Earth equals Earth’s gravitational force on the Sun, as there is only one value for the force of gravity between two objects, although the directions are opposite. F = GMm/d2 = GmM/d2. Earth pulls the Sun towards Earth just as strongly as the Sun pulls Earth towards the Sun. This is also an expression of Newton’s 3rd law of action and reaction. (How much each object actually accelerates depends on the relative masses, of course). 10. The two stars pull on each other with equal force, so the lower mass star will have a greater acceleration. Therefore, it will move more, and have the wider orbit (with a larger circumference). (See also the reasoning in TQ9 above.) 11. Surface gravity is given by g = GM/R2, so it depends on the ratio of the mass to the radius of the planet squared. A planet with a large radius, but made of low density materials such as ice, or light gases such as hydrogen, could have a lower value of g than a planet made of very dense material (iron) that was quite small. Answers to Problems