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
1. F = ma so a = F/m. (a) 2F on m: a’ = F’/m’ = 2F/m = 2 (F/m) = 2a. (b) 2F on 2m: a’ = F’/m’ = 2F/(2m) = (2/2) (F/m) = (1)(a) = a. (c) 10F to m: a’ = F’/m’ = 10F/m = 10(F/m) = 10a. (d) 10F to 3m: a’ = F’/m’ = 10F/3m = (10/3)(F/m) = 10/3 a = 3.33 a 2. This is just F = ma again, but we have to figure out a. a = change in velocity / time = 2 m/s / 25 s = 0.08 m/s2. F = ma = (2500 kg)(0.08 m/s2) = 200 Newtons 3. STUDENTS SHOULD DO THIS FOR THEMSELVES. The weight of a body is just the gravitational force exerted on it. You can therefore find your weight on Earth or on the Moon from their surface gravities. 2 2 • Earth’s surface gravity is 9.8 m/s while the Moon’s is 1.7 m/s . • The Moon’s is thus 9.8/1.7 = about 5.8 times smaller than Earth’s. • Thus your weight on the Moon is your weight on Earth divided by 5.8. Alternatively, students can calculate the surface gravity of the Moon directly using g = GM/R2 and then their weight with F = mg; but they will have to come up with their mass in kilograms to calculate the force in Newtons and/or convert that back to pounds or compare to their Earth weight in Newtons (depending on instructor preference for the format of the answer). 4. From the problem, Neptune’s distance from the Sun is about 30 AU (more exactly it’s 30.05 AU). The orbital velocity, V, of a small mass around a much larger one can be 1/2 found from the formula in the chapter, namely, V = (GM/d) , where M is the Sun’s mass and d is Jupiter’s distance from the Sun. Substituting: -11
30
11
1/2
3
VJupiter = ( (6.7 × 10 m3 kg-1s-2 × 2 × 10 kg)/ (30 × 1.5 × 10 m) ) = 5.44 × 10 m/s. The orbital period is the time it takes a body to complete an orbit. Thus, orbital velocity = circumference/orbital period, V = C/P. Evaluating this we obtain 11 3 9 P = 2d/V = 2 × 30 × 1.5 × 10 m/ (5.44 × 10 m/s)= 5.2 × 10 seconds. 7
Since there are approximately 3.16 × 10 seconds in 1 year, P = about 165 years. Students should be encouraged to check a result like this against the data in the appendix or Kepler’s 3rd law.
11
20
5. The Milky Way Galaxy has a mass of about 10 MSun, and the Sun is 2.6 × 10 m from the center of the galaxy. 1/2 -11 30 11 20 V = (GM/R) = (6.7 × 10 m3 kg-1s-2 × 2 × 10 × 10 kg /2.6 × 10 m)1/2 5
V = 2.3 × 10 km/s 20 5 15 P = 2R/v = 2 × 2.6 × 10 /2.3 × 10 = 7.1 × 10 s. One year is about 3.16 × 107 sec, so the period of the Sun’s orbit around the Milky Way 15 8 in years is 7.1 × 10 s /(3.16 × 107 s/yr) = 2.25 × 10 years = 225 million years. 6. The modified form of Kepler’s third law states that M = 4d3/GP2, where d is the orbital radius (assuming it is circular) and P is the orbital period. In this problem d = 5 × 1010 meters and P = 124 days. Before we can solve the problem, however, we need to convert P in days to P in seconds. 7 124 d × 24 hr/1d × 60 min/1hr × 60 sec/1min = 1.07 × 10 sec. Inserting the values for a and P, gives M = 4 (5 × 1010 m)3/[6.67 × 10-11 m3 kg-1 s-2 (1.07 × 107 s )2] = 4 × 125 × 1030-(-11)/[6.67 × (1.072 × 107x2 )] kg = 4935/7.6 × 1041-14 kg = 6.5 × 1029 kg Since the Sun’s mass is 2 × 1030 kg, the mass of Gliese 581e is about 1/3 of a solar mass: 0.65 × 1030kg / (2 × 1030 kg/solar mass) = 0.33 solar masses. 7. In Section 3.7, Earth’s surface gravity is compared with the Moon’s. We will need the mass and radius of Jupiter and Pluto, which conveniently are given in terms of Earth’s radius and mass in the Appendix. M Earth GM Earth 1 2 2 M Jupiter g Earth REarth 317.9 2 = 11.19 = 125 = 0.4 2 = = = g Jupiter GM Jupiter R 317.9 317.9 1 11.19 Earth 2 R Jupiter R Jupiter The surface gravity of Earth is about 4 tenths of the surface gravity on Jupiter; or Jupiter’s surface gravity is about 2.5 times Earth’s. For Pluto,
M 1 g M Earth 0.1882 g Earth == Pluto2 = 0.0021 1 2 = 0.0021 = 16.8 R Jupiter Earth 0.188 R Pluto The surface gravity of Earth is about 16.8 times as strong as the surface gravity of Pluto, which is about 1/16.8 = 0.059 that of Earth. 8. The escape velocity for Earth is found from:
1/2
Vesc = (2GM/R)
Inserting the given values in the equation, we find that 1/2 -11 24 6 1/2 4 Vesc = (2GM/R) = (2 × 7 × 10 m3 kg-1 s-2 × 6 × 10 kg / (6 × 10 m)) = 1.18 × 10 m/s or Vesc = approx. 12 km/s 9. The conversion to mph helps put this speed into context for students. 11.8 km/s × (1 mile/1.609 km) × (3600 s/ 1h) = 26,400 miles/h = 26,400 mph 1/2
10. The escape velocity for the Sun may be found using the formula Vesc = (2GM/R) with the Sun’s mass and radius inserted. This gives -11 30 8 1/2 Vesc = (2 × 6.7 × 10 m3 kg-1 s-2 × 1.989 × 10 kg/(6.95 × 10 m)) = 6.2 × 105 m/s = approx. 620 km/s Or with the approximate values, -11 30 8 1/2 5 Vesc = (2 × 7 × 10 × 2 × 10 /(7 × 10 )) = 6.3 × 10 m/s = approx. 630 km/s 11. To compare the escape velocity of Mars and Saturn, the method is similar to the previous problem. The ratio simplifies to the expression, VSaturn = M Saturn RMars VMars M Mars RSaturn We then evaluate each expression using the appropriate mass and radius. From the
appendix, MMars = 0.1 MEarth, MSaturn = 95 MEarth, RMars = 0.5 REarth and RSaturn = 9.5 REarth. 1/2
Vesc(Saturn)/Vesc(Mars) = ( (95/0.1) x (0.5/9.5))
= ( (95/9.5) x (0.5/0.1) )1/2 = (10 x 5)1/2
= 7.1. This is a good example of a question where the numbers can be easily estimated by writing out the values and shifting terms around.
12. To calculate the ratio of the escape velocities from the Moon and Earth, start with the formula for escape velocity: 1/2 1/2 Vesc-Moon = [2GMMoon/RMoon] and V esc-Earth = [2GMEarth/REarth] . Next, divide the Moon’s escape velocity by Earth’s, 2GM Moon 2 R V Moon = Moon = 1 VEarth 2GM 2 Earth REarth 1
M Moon M Earth = RMoon REarth
M Moon REarth M Earth RMoon
We then evaluate each expression using the appropriate mass and radius; from the chapter, 81 MMoon = MEarth and REarth/RMoon = 3.8, so VMoon 1 = (3.8) = 0.22 VEarth 81 Vesc-Moon/Vesc-Earth = 0.22. In other words, the escape velocity of the Moon is about 1/5 of the escape velocity of Earth. 13. To find out if the pitcher can throw a ball fast enough to escape from Sinope, we need to find the escape velocity of Sinope. To do that, we use the escape velocity formula, V = (2GM/R)1/2. Putting in the values for the mass and radius of Sinope, we find that V = [2 × 6 × 1016 kg × 6.7 × 10-11 m3kg-1s-2/(1.8 × 104 m)]1/2 = [2 × 6 × 6.7 × 1016-11-4/1.8 (m/s)2]1/2 = [44.67 × 10]1/2 m/s = 21 m/s which is smaller than the speed of the pitch. Thus the ball can escape.
Answers to Test Yourself 1. (a),(b),(c),(d) All show a mass tending to remain at rest or in uniform motion. 2. (d) A body moving along a curved path is not in uniform motion. The speed is constant but the direction changes so the velocity is not constant—it must be being accelerated. The acceleration must be produced by a force. 3. (c) Newton’s work explained why Kepler’s laws worked with a physical reason: gravity. 4. (e) mass remains the same in both places. 5. (c) Newton’s 3rd law, the propellant’s action down has the net results of an action up for the rocket. 6. (a) True. Earth exerts the same gravitational force on you that you exert on Earth. 7. (d) If the distance between the two bodies is increased by a factor of 4, since gravity is an inverse square law the force is decreased by a factor of 4 × 4=16. 8. (a) Orbital velocity goes as one over the square root of the radius. 161/2 = 4, so the orbit would be 4 times slower. 1/2
9. (b) 6 times larger. Vesc = (2GM/R) . The radii are the same but one planet is 36 times the mass of the other. Thus, the escape velocity from the more massive planet is 6 times greater than that from the less massive body.
Chapter 4
Light and Atoms
CHAPTER 4 LIGHT AND ATOMS Notes: This material is crucial for understanding much of the latter parts of the text. However, some material can also be covered later, introduced with particular topics. For example, the sections on spectra and the Doppler shift can be given only a brief mention at this point and then treated more fully when discussing stars. The material on radiation in the atmosphere can be discussed with Earth. The subject matter is tough for most students but lends itself to many demos that can enhance understanding. Use a small electromagnet made from nail, wire, and a battery to illustrate the link between electricity and magnetism. Use coil attached to milliammeter and a bar magnet to illustrate the link of magnetism to electricity. Use a comb charged by rubbing on a sweater and bits of paper to illustrate electrostatic attraction that holds electrons to the nucleus. A rheostat on a lamp shows Wien’s law as the filament changes color. Answers to Thought Questions 1. We associate red with hot and blue with cold most likely because humans learned that red and yellow flames from fires were hot, and hot objects like coals get hot enough (a few thousand Kelvin) to glow red or orange. Objects hot enough to glow blue would probably be vaporized and are encountered much less often. On the other hand, lakes and oceans are blue and more likely to be cold, and frozen objects and ice often have blue hues. Perhaps also there is the association because people’s faces turn blue when they are cold, but flush red when they are hot. 2. Many night-vision cameras use IR detectors because all objects, particularly living things, radiate energy at IR wavelengths (they glow) regardless of whether there is visible light around. Warm-blooded animals are warmer than their surroundings and do actually glow in the infrared. 3. Just seeing a red object through a telescope would not be enough information to apply Wien’s law and deduce its temperature. The red object might be only be reflecting light (like the planet Mars), or it might be an ionized hydrogen cloud producing red line emission. (Also, even if the object is glowing—say a red giant or red dwarf star—to accurately deduce the temperature with Wien’s law requires obtaining at least a few different wavelength measurements to find the peak wavelength of the emission.) 4. Atoms do not emit a continuous spectrum because they can only emit or absorb energy (light) in discrete amounts corresponding to the differences between their energy levels. 5. Microwaves are absorbed by water molecules, causing them to vibrate and heat up. The filling of a Pop-Tart contains more water molecules than the crust. The microwaves are absorbed by the water in the filling, heating it up more than the crust.