Chapters 2, 3 and 4: Problem Keys
CHAPTER 2 - GRAVITY AND MOTIONKey TermsAcceleration
Escape Velocity
Inertia
Law of Gravity
Mass
Newton's First Law of Motion
Newton's Second Law of Motion
Newton's Third Law of Motion
Surface Gravity
Answers to Thought Questions1. The cinder block has more inertia than the balloon because of its greater mass. Even when they are both traveling at 1 m/s, it is harder to stop the cinder block because you have to exert a greater force on it.
2. 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 kicking the block would hurt your foot as badly as if you'd kicked it on the Earth's surface.
3. STUDENTS NEED TO DRAW A SKETCH
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.
Answers to Problems1. Since
F(Moon-Earth) = G MMoon MEarth R2Moon-Earth,
the gravitational force exerted by the Moon on the Earth is the same as that exerted by the Earth on the Moon.
2. The escape velocity for the Earth is given as follows:
Vesc = (2GM/R)1/2, where G=7x10-11 m3/kg/s2, M(Earth)=6x1024 kg and R(Earth)=6x106m.
Inserting these values in the equation, we find that
Vesc = (2GM/R)1/2 = (2 x 7x10-11 x 6x1024 / 6x106)1/2 =1.18x104 m/s
or
Vesc = approx. 12 km/s
3. The escape velocity for the Sun may be found using the formula for the escape velocity
Vesc = (2GM/R)1/2 with the Sun's mass and radius inserted. This gives
Vesc = (2 x 7x10-11 x 2x1030/ 7x108)1/2 = 6.3x105 m/s = approx. 630 km/s
4. To compare the escape velocity of Mars and Saturn, we write out the expression for the escape velocity for each planet. Thus,
Vesc(Mars) = [2GM(Mars)/R(Mars)]1/2,
Vesc(Saturn) = [2GM(Saturn)/R(Saturn)]1/2.
Next, we divide the expression for Saturn's escape velocity by the expression for Mars' escape velocity to get,
Vesc(Saturn)/Vesc(Mars) = [M(Saturn)/R(Saturn)]1/2/[M(Mars)/R(Mars)]1/2,
We then evaluate each expression using the appropriate mass and radius. Thus,
M(Mars) = 0.1 M(Earth), M(Saturn) = 95 M(Earth), R(Mars) = 0.5 R(Earth) and
R(Saturn) = 9.4 R(Earth).
Vesc(Saturn)/Vesc(Mars) = (95 x 0.5/9.4 x 0.1)1/2 = 7.1.
5. To compare the escape velocity of the Earth and Moon, we use the same approach as in the previous problem.
Since M(Earth)/M(Moon) = 81 and R(Earth)/R(Moon) = 3.8,
Vesc(Earth)/Vesc(Moon) = [M(Earth)/M(Moon) x R(Moon)/R(Earth)]1/2
= (81/3.8)1/2 = 4.6
6. STUDENTS NEED TO DO THIS.
The weight of a body is just the gravitational force exerted on it. You can therefore find your weight on the Earth or on the Moon from their surface gravities. The Earth's surface gravity is 9.8 m/s2 while the Moon's is 1.7 m/s2. The Moon's is thus 9.8/1.7 = about 5.8 times smaller than the Earth's. Thus your weight on the Moon is your weight on the Earth divided by 5.8.
7. Jupiter's distance from the Sun is about 5 AU.
The orbital velocity, v, of a small mass around a much larger one can be found from the formula in the Chapter, namely, v = (GM/R)1/2 , where M is the Sun's mass and R is Jupiter's distance from the Sun. Substituting in numbers then gives
v(Jupiter) = (6.7x10-11 x 2x1030/5 x 1.5x1011)1/2 = 1.3x104 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
P = 2R/v = 2 x 5 x 1.5x1011 / 1.3x104 = 3.6x108 seconds. Since there are approximately 3.2x107 seconds in 1 year, P = about 11 years.
8. The Milky Way Galaxy has a mass of about 1011 M(Sun), and the Sun is 2.6x1020 m from the center of the galaxy.
v = (GM/R)1/2 = (6.7Ex10-11 x 2x1030 x 1011/2.6x1020)
v = 2.3x105 km/s
P = 2R/v = 2 x 2.6x1020/2.3x105 = 7.1x1015 s.
Given that 1 year is about 3.2x107 sec, the period in years is 7.1x1015 s/3.2x107,
or the Sun makes one orbit of the galaxy in 2.2 x108 years.
9. To find out if the pitcher can throw a ball fast enough to escape from Cochise, we need to find the escape velocity of Cochise. To do that, we use the escape velocity formula, V =(2GM/R)1/2. Putting in the values for the mass and radius of Cochise, we find that
V = (2x9.6x1016 kg x 6.7x10-11 m3-kg-1-seconds-2/2.0x104 m)1/2
= [2x9.6x6.7x1016-11-4/2.0(m/seconds)2]1/2
= [64.3x10]1/2 m/seconds
= 25 m/second
which is smaller than the speed of the pitch. Thus the ball can escape.
10. The modified form of Kepler's third law states that M = 4r3/GP2, where a is the orbital radius (assuming it is circular) and P is the orbital period. In this problem a = 3x1011 meters and that P is 3 years. Before we can solve the problem, however, we need to convert P in year to P in seconds. Given that 1 year is about 3.2x107 seconds, P = 3x3.2x107 seconds = 9.6x107 seconds.
Inserting the values for a and P, gives
M = 4 (3x1011 m)3/[6.67x10-11 m3-kg-1-s-2 (9.6x107 s )2]
= 4 x27x1033-(-11)/[6.67x(9.62x107x2 )] kg
= 1.74x1044-14 kg
= 1.74x1030 kg
Since the Sun's mass is 2x1030 kg, the mass of 57 Fungaloid is
1.74x1030kg/2x1030 kg/solar mass = 0.87 solar masses.
Answers to Self-Test1. (e) All the examples 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. The resultant acceleration must be produced by a force.
3. (d) If the distance between the two bodies is increased by a factor of 4, the force of gravity is an inverse square law so the force is decreased by a factor of 4x4=16.
4. (a) True. The Sun and the Earth exert the same gravitational force on each other.
5. (c) Vesc = (2GM/R)1/2, the radii are the same but one planet is 25 times the mass of the other. Thus, the escape velocity from the more massive planet is 5 times greater than that from the less massive body.
Planetarium Exercises1. The Earth turns from West to East making the Sun and stars appear to move in the opposite direction.
2. Hopefully the students will see a close correspondence between the planetarium program and the real sky. One common problem people have is the large change in scale between the computer monitor and the real sky. The result is when looking for constellations in the real sky they underestimate their apparent size.
3. The stars near Polaris do not set. They complete circular paths about the North celestial pole each day. They appear to move counterclockwise. This implies the Earth turns from West to East. In the Southern Hemisphere the stars appear to turn clockwise about the South celestial pole.
4. By comparing the sunrise and sunset times the students will be able to find the number of daylight hours at summer solstice. For people living at latitude of 40 degrees North this is about 14.6 hours. This means that at the time of Winter solstice the number of daylight hours is about 9.4 (i.e., 24 ? 14.6). This is true because the Sun?s distance from the celestial equator at Summer solstice is the same as that at Winter solstice.
5. The results will depend on the latitudes the students pick for the sites. For latitude of 50 degrees North the times are 15.9 and 8.1 hours for June 21 and December 21 respectively. For a latitude of 30 degrees North the numbers are about 13.7 and 10.3 hours.
Group Exercises1. In this exercise the students will verbalize and perhaps find new ways to explain this 3-dimensional concept.
2. The goal would be to steer the ship in such a way as to keep Polaris at a constant angle above the Northern horizon.
3. On the day of an equinox the height of the sun in degrees is 90 ? your latitude. It is 23.5 degrees higher than this at summer solstice and 23.5 degrees lower than this at the winter solstice. For someone living at 40 degrees North latitude the sun has a height of 50 degrees at the time of equinox, 73.5 degrees at the summer solstice and 26.5 degrees at the winter solstice.
4. The students should learn just how difficult it is to prove the heliocentric theory and why our ancestors struggled with this idea.
5. Observations such as these were used to mark the seasons and became the basis of the calendar. This is also something that modern people pay little attention to.
6. This exercise will make the students much more understanding of the difficulties these ideas presented to our ancestors.
7. In June the Sun is near the summer solstice. This places it about 23 degrees above the celestial equator. For the Sun not to cast a shadow requires the observer to be at latitude of 23 degrees North.
CHAPTER 3 - LIGHT AND ATOMSKey Terms
Absorption
Atmospheric Window
Blackbody
Conservation of Energy Continuous Spectrum
Dark-line or Absorption-Line Spectrum
Doppler Shift
Electromagnetic radiation
Electromagnetic spectrum
Electromagnetic wave
Elements
Emission
Emission-Line Spectrum
Energy Levels
Excited
Frequency
Infrared
Light
Nanometer
Protons
Quantized
Spectroscopy
Ultraviolet
Visible spectrum
Wavelength
Wave-particle duality
White light
Wien's Law
Answers to Thought Questions1. 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.
2. Each element or ion produces a distinct pattern of spectral lines corresponding to its energy levels/electronic structure.
3. Taking a spectrum of Venus would show absorption lines and bands corresponding to the atoms and molecules in its atmosphere.
4. Adding water or carbon dioxide to our atmosphere would affect the heat loss from the planet because these molecules have absorption bands in the infrared through to cm wavelengths, which is where most of the incident sunlight is re-radiated from the ground. The Earth would become warmer in what is usually called the greenhouse effect.
5. The filling of a Pop-Tart contains more water molecules than the crust, so that the microwaves will be absorbed by the water in the filling, heating it up more than the crust.
6. Body temperature is about 300K. Using Wien's Law as given in the book,
m= 3x106 / T = 3x106/300 = 104 nm.
Note: Because 1 nm = 10-3 micrometers, m= 10-3x104 = 10 micrometers.
This wavelength is usually said to be in the mid-infrared. Rattlesnakes have organs which detect heat/IR radiation and so they can detect the presence of prey at night.
7. Many night-vision cameras use IR detectors because all objects, particularly living things, radiate energy at IR wavelengths regardless of whether there is visible light around.
8. The blue gas flame of a Bunsen burner is an incandescent gas and would therefore produce an emission line spectrum.
9. The ozone molecules absorbs UV radiation from the Sun. That extra energy heats the gas, raising the temperature of the ozone layer in comparison to the layers above and below it.
Answers to Problems1. To find how long it takes light to travel a given distance, divide the distance by the speed of light. Since 1 AU = 150x106 km = 1.5x1011 m, and the speed of light, c = 3x108 m/s,
time = distance/velocity = 1.5x1011 m / 3x108 m/s = 500 seconds or 8.3 minutes.
2. Signals travel to the spacecraft on Mars at the speed of light. Thus, if we send a command, the lander will not receive the signal until the signal has gotten there, a time equal to the light-travel time. However, the round trip will be twice that figure before you know what is going on!
The average orbital radius of Mars is 1.52 AU, and thus the closest approach to the Earth is when Mars is 0.52 AU away (technically, when Mars is at opposition). The light travel time between Earth and Mars is t = D/c. Substituting in numbers,
D=.52 x the number of meters in an AU = .52x1.5x1011 = 7.8x1010 m
Thus, t = 7.8x1010 m/3x108 m/s = 260 s or 4.2 min. on average.
The total time from signal being sent to receiving notification of its arrival will be twice that.
3. STUDENT MUST MAKE DRAWING
4. Using Wien's Law as given in the book, we can relate temperature to a wavelength of maximum emission.
If m= 3x106 /T= 3000 nm, then cross multiplying by T and dividing by m gives
T = 3x106/3000 = 1000 K. That is, the temperature of the light bulb filament is 1000 K.
5. Using our version of Wien's law, we have that m= 3x106 /T. Given that T = 2000 K, then m= 3x106 /2000=
6. Using our version of Wien's Law, we have m= 3x106 /T. If T =300, then m= 3x106 /300 = 104 nanometers = 10 micrometers, an infrared wavelength. Because we can't see infrared wavelengths, we can't see the earth's glow (as is obvious given that the ground doesn't glow in the dark).
Answers to Self-Test1. (e) Radio waves with wavelengths between 100 and 500 m.
2. (c) x-ray photons are the most energetic.
3. (d) Bright emission lines come from hot, rarefied gases.
4. (e) T = 3x106/450 = 6667 K or about 7000 K.
5. (a) If the lines are shifted to longer wavelengths (redshifted), the object is moving away from us.
Group Exercises
1. The best time to look for Earth satellites is a few hours after sunset or a few hours before sunrise. At these times the Earth’s shadow projects toward the Eastern horizon (after sunset) or the Western horizon (before sunrise) so a satellite passing overhead is seen in a dark sky background but it is high enough to be illuminated by the Sun. Satellites in low Earth orbits must have higher orbital speeds. This, and the fact they are closer, makes them appear to move across the sky much faster than satellites in high orbits. In general, large satellites appear brighter because they reflect more sunlight due to their greater surface area. The reflectivity of the satellite also contributes to its apparent brightness. Communications satellites need to pass over the most populated regions of the Earth. This is best accomplished with orbits that lie near the plane of the Earth’s equator. Satellites with polar orbits are good for mapping (and spying) as they eventually pass directly over all portions of the Earth. Geosynchronous satellites are located at a height where their orbital periods are equal to the Earth’s rotational period. From the Earth they appear stationary and dish antennas do not need to move to receive their signals.
2. This is an application of Newton’s Third Law of Motion. By throwing objects (e.g., shoes, coat, and wallet) in one direction you will slide in the opposite direction.
3. The other factors are the mass of the planet and its radius. For a fixed size, the more massive planet has the greater escape velocity. For a given mass, the smaller planet has a greater escape velocity.
4. In order to conserve angular momentum the rotation speed must increase as mass (arms in this example) move inward. The nebula from which the solar system formed had angular momentum but its rotation was initially very slow because the nebula was large. As it was drawn together by gravity and became smaller it spun faster to conserve angular momentum.
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CHAPTER 4 - TelescopesKey Terms
Adaptive Optics
CCD
Diffraction
Dispersion
Interferometer
Light-Gathering Power
Moon Illusion
Reflectors
Refraction Refractors
Resolving Power
Scintillation
Seeing
Answers to Thought Questions1. Binoculars have larger lenses than your eyes and hence they collect more light and allow you to see dimmer objects.
2. At night, we would choose the binoculars with 50-mm lenses over the 35-mm lenses because the larger lenses collect more light.
3. A 2-m telescope on Earth never reaches the limit of its resolving power of
= 0.02 (nm)/D(cm) = 0.02 x 500 / 200 = 0.05 arcsec.
Because we are observing through the turbulent atmosphere the best "seeing" conditions only allow resolutions of a few tenths of an arcsecond.
4. Looking down a beach on a hot day shows the image shimmering as the hot air currents rise and distort the light trying to reach our eyes. In the same way, turbulent atmospheric blobs drifting across our line of sight cause the star images to be blurred as the light is directed in and out of the aperture of the telescope. The "seeing" is a measure of how distorted the image appears.
5. When the pencil is perpendicular to the water's surface, it looks straight. As you tilt the pencil from the perpendicular, the pencil looks more and more bent. Similarly, light rays entering out atmosphere are more bent (refracted) when they enter it at angles far from the perpendicular.
Answers to Problems1. Pupil of the eye has a diameter D1 = 5 mm or 0.5 cm.
Small reflector has an aperture D2 = 10 cm = 100 mm.
Area A1 = D12, A2 = D22.
A1/A2 = (D1/D2)2 = 1/400
Therefore, the collecting power of a 4" reflector is 400 times greater than that of your eye.
2. EXPERIMENT FOR STUDENTS
3. Theoretically, = 0.02 /D = 0.02 x 500 / 0.5 = 20 arcsec.
So, if the diameter of the crater is 1 arcminute, you should be able to resolve it.
Answers to Self-Test1. (d) DA=3, DB=1, LGP=(DA/DB)2 = 3x3/1 = 9.
2. (c) Finer detail.
3. (d) No x-rays can penetrate to the bottom of the atmosphere.
4. (d) Interferometers greatly increase the ability to see fine detail.
5. (a) Since =0.02/D, you would have to make the mirror larger.
Planetarium Exercises1. It takes light about 8.3 minutes to travel from the Sun to the Earth.
2. The average light travel time from the Moon to Earth is 1.28 seconds
Additional Readings Fishman, Gerald J. and Dieter H. Hartmann. “Gamma-Ray Bursts.” Scientific
American 277 (July 1997): 46.
Hajian, Arsen R. and J. Thomas Armstrong. A Sharper View of the Stars.
Scientific America 284 (March 2001): 56.
Malin, David F. "A Universe in Color." Scientific American 269 (August 1993): 72.
Posted by astrohintz
at 6:31 AM CST
Updated: Friday, 17 February 2006 8:48 AM CST
Chapter P: Problem Key
Key Terms
Astronomical unit
Atom
Dark matter
Electric charge
Electric force
Electron
Galaxy cluster
Gravity
Light-year
Local group
Local supercluster
Milky Way Galaxy Neutron
Nuclear force
Nucleus
Planet
Powers-of-ten notation
Proton
Satellite
Scientific method
Scientific notation
Solar system
Star
Strong force
Universe
Weak force
Answers to Problems
1. RSun = 7x105 km. REarth = 6.4x103 km. We can determine how much bigger the Sun is than the earth by dividing the Earth's radius into the Sun's. Therefore RSun/ REarth= 7x105 km/ 6.4x103 km = 7/6.4 x(105/103) = 1.09x(105-3) = 1.09x(102) = 109.
2. The average distance between the Earth and the Sun is called the astronomical unit. 1 AU = 1.5x108 km. The distance traveled by a light ray in 1 year is called a light year (ly). 1 ly = 1x1013 km. We can find the number of AU in a light year by dividing the number of km in a light year (1013) by number of km in an AU (1.5x108). Thus,
Number of AU in a light year = 1013/(1.5x108) = 1013-8/1.5 = 105/1.5 = 6.7x104 = about 68,000.
3. If the Earth had a radius of 0.2 cm (2 mm), the Sun would have a radius of 100 x 0.2 = 20 cm. The Solar System is 8600 x R(Sun) and so,
R(solar system) = 8600 x 20 cm = 1.7x105 cm = 1.7x103 m, since there are 100 cm in 1 m.
R(Milky Way)= 1x108x R(solar system) = 1.7x1011 m = 1.7x108 km, since there are 1000 m in 1 km.
4. Using the scale in Q.3, the radius of the Solar System (to Pluto) is 40 AU = 1720 m.
In our model, 1 AU therefore = 1720 / 40 = 43 m
Planet a (AU) a (m) R (mm)
------------------------------------------------------------------
Mercury 0.39 16.8 0.76
Venus 0.72 31.0 1.90
Earth 1.00 43.0 2.00
Mars 1.52 65.4 1.06
Jupiter 5.2 223.6 22.4
Saturn 9.55 410.6 18.9
Uranus 19.18 824.8 8.02
Neptune 30.06 1293 15.5
Pluto 39.44 1696 0.36
5. The Milky Way has a radius of 50,000 ly = 5x104 ly. We scale this to 2 cm, and let the Local Group have a scaled radius of x. The Local Group's true radius is about 1.5x106 ly. Thus we can form the proportion that x/1.5x106 ly as 2 cm/5x104 ly. Thus,
x = 1.5x106x2 cm/5x104 = 300/5 cm or 60 cm, or about the size of a standard bed pillow.
If we let the scaled size of the local supercluster be y, then y/40x106 ly = 2 cm/5x104 ly. Thus, y = 2 cmx40x106 ly/5x104 ly = 16x102 cm = 16 m.
The visible Universe would be 2 cmx15x109 ly/5x104 ly = 6x105 cm = 6 km, or about 4 miles.
6. (4x108)3/(5x10-6)2 = 43x(108)3/(52x(10-6)2)=64x1024/(25x10-12)=2.56x1036.
7. t=D/v. For light, v=c, so t=D/c. Pluto is about 40 AU from the Sun and 1 AU =1.5x108 km.
Thus, Pluto is 40x1.5x108 = 6x109 km from the Sun. The speed of light is 3x105 km/s. Thus, t = 6x109/3x105 = 2x104 seconds. Divide by 60 to get the number of minutes = about 330 minutes.
8. To find the number of times one thing is smaller than another, divide the larger by the smaller. Thus the hydrogen atom is 10-6 m/10-10 m = 10-6-(-10) = 104 times smaller than the bacterium.
9. (3x104)2/(4x10-6)1/2 = 32x104x2/41/2x10-6x1/2 = 9x108/2x10-3 = 4.5x108-(-3) = 4.5x1011.
10. Given an object's speed and distance from another, we can find how long it took them to move apart by using the relation D=vt, where D is the distance, v is the speed, and t is the time. Solving for t gives, t=D/v.
Now inserting values for v and D, we find t = 3x108 ly/6,000 km-seconds-1. Notice that units are mixed here. We have ly on top and km below. Thus, we need to express ly in km. From the appendix we can find that 1 ly = 9.5x1012 km. Thus,
t = 3x108 lyx9.5x1012 km/ly/6,000 km-seconds-1 = (3x9.5/6)x108+12-3 seconds = 4.8x1017 seconds.
To express this in years, we need to divide by the number of seconds in a year = about 3.2x107. Thus, t = 4.8x1017 seconds/3.2x107 seconds/yr = 1.5x1010 yrs. This is one way that astronomers deduce the age of the Universe.
Answers to Self-Test
1 (b) The light year is a unit of distance.
2 (c) Your cosmic address is Earth, Solar System, Milky Way, Local Group.
3 (e) All of them are.
4 (c) Jupiter's diameter is about 10 times larger than the Earth's diameter.
5 (e) All except (d). Choice (d) is a matter of opinion.
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Chapter 1
Goals
Basic ideas of the heavens as seen by the naked eye: Features of the celestial sphere. Definition of celestial equator, poles. Definition of angular diameter. Ecliptic, Zodiac. Cause of seasons. Cause and appearance of eclipses and phases of Moon. A sense of the date and contributions of Aristarchus, Eratosthenes, Aristotle, Ptolemy; the methods they used, and approximately when they lived. Contribution of Kepler, Copernicus, Galileo and Newton and about when they lived. Kepler's laws and their use. The Kelvin Temperature scale.
This chapter introduces some important ideas that will be needed later, such as parallax (also described in Chapter 12), angular size (also discussed in Chapter 6), and Kepler's third law. Nevertheless, it may be omitted if these ideas are dealt with as needed.
If pressed for time, omit details of methods used by ancient astronomers, angular diameter, or naked eye astronomy. The material on the celestial sphere, zodiac, and so forth, appear nowhere else apart from Essay 1.
Key Terms
Angular size
Celestial equator
Celestial poles
Celestial sphere
Constellations
Ecliptic
Ellipse
Epicycle
Equinoxes
Focus
Geocentric theories
Heliocentric theories
Kepler's Laws
Lunar eclipse
Model
Parallax
Phases
Retrograde motion
Rotation axis
Semimajor axis
Solar eclipse
Solstices
Zodiac
Answers to Thought Questions
1. If you were standing on the Earth's equator, you would see the north celestial pole on the horizon (and the south celestial pole too). You cannot see the north celestial pole from Australia, 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 travel about 30 degrees per month across the sky.
4. The basic proofs are as follows:
- The Earth casts a circular shadow on the Moon during a lunar eclipse. Only a sphere will cast a circular shadow from any illumination angle.
- As a traveler moves south, stars previously hidden below the southern horizon come into view.
- Ships disappear over the horizon gradually, the mast being the last part to disappear from view.
The newer proofs come from modern technology (I think the last flat Earth society existed until about 1930). Some modern proofs are:
- Supersonic aircraft fly high enough to see the curvature of the Earth.
- Spacecraft send back pictures of the spherical Earth.
5. The position of sunrise along the eastern horizon changes during the year because the Earth's axis is tilted at 23.5 degrees to the plane of its orbit (the ecliptic) and the Earth maintains this same tilt throughout the year. At the equinoxes (March 21 and September 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.
6. At the winter solstice (Dec. 21), the Sun lies 23.5 degrees south of the celestial equator on the sky. It will therefore rise south of the east point on the horizon and set south of the west point. At the summer solstice (June 21), the Sun lies 23.5 degrees north of the celestial equator. It will therefore rise north of the east point and set north of the west point.
7. If the stars were very much closer than they really are, Aristarchus would have been able to demonstrate the stellar parallax caused by the Earth's orbital motion around the Sun.
8. The phases of Venus are caused by Venus orbiting the Sun, so keeping all distances and periods constant, it wouldn't matter whether Earth was orbiting the Sun or the Sun (and Venus) were orbiting the Earth.
Answers to Problems
1. The Sun is 36 degrees from straight overhead, so by the parallel line theorem, the angle between you, the center of the planet and the missile silo is also 36 degrees.
36 / 360 = 1/10 of a complete circle. You are 1/10th of the circumference of Myrmidon from the missile silo. If the distance to the silo is 1,000 miles, then the circumference of the planet is 10,000 miles.
The circumference = 2 x radius, therefore
Radius = 10,000 miles / (2) = 1,592 miles.
2. Historical question. Student must come up with own answers.
3. If P = 64 years, you can use Kepler's 3rd Law to estimate its distance from the Sun.
P2 = a3, where P=period in years and a=average orbital radius in AU. We can find a by taking the cube root of both sides to get P2/3 = a, or
a = P2/3 = (64)2/3 = 16 AU
4. Aliens have a 1Mo star and are 4 AU from it. According to Kepler's 3rd law
P2 = a3
P=square root of (43) = (43)1/2 = 8 years.
5. The Andromeda galaxy has an angular diameter of 5 degrees at a distance of 2.2x106 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,
L = 2 x 2.2x106 ly x 5 degrees / 360 degrees = 1.92x105 ly
6. A shell of gas has an angular diameter A = 0.1 degrees and a linear diameter L = 1 ly. We can find its distance D by reversing the procedure in the previous problem. We begin by writing out the angular diameter formula L = 2DA/360. We next solve it for D by dividing both sides by 2A and multiplying both sides by 360 to get
D = 360L/(2A). Next inserting values for L and D we find
D = 360 x 1 ly/( 2 x 0.1) = 573 ly.
7. This problem is a modern version of the method Eratosthenes used to measure the size of the 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 10x24 = 240 km. The radius, R, of the asteroid is related to its circumference, C, by C=2R. Thus R = C/2240/2= 38 km
8. 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. Using your pocket calculator, you will find that a = 25 AU. If the planet's orbit is circular, then that is also the planet's orbital radius.
9. 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 take the square root of 16 (=4) and then cube it to get 64.
Answers to Self-Test
1. (d) The north celestial pole is directly overhead.
2. (b) Retrograde motion causes planets to stop their regular eastward motion with respect to the stars and move westwards for a time.
3. (a) The Earth is spherical.
4. (e) Venus orbits the Sun.
5. (a) Kepler's 3rd Law relates a planet's orbital period to the size of its orbit around the Sun.
Planetarium Exercises
1. The purpose of this exercise is to get the students to install the software on their computers and to become familiar with its operation.
Group Exercises
1. For some students this may be the first time they have really paid attention to the night sky in a thoughtful way. The effect of light pollution is one of the obvious ways the sky has changed.
2. This is a continuation of the first exercise that now includes a light pollution experiment.
3. This may be the first time your students have actually seen the solar system drawn to scale. They are usually shocked by how empty it is.
Additional Readings:
Aveni, Anthony F. "Emissaries to the Stars: The Astronomers of Ancient Maya."
Mercury 24 (January/February 1995): 15.
Gingerich, Owen. The Eye of Heaven: Ptolemy, Copernicus, Kepler. New York:
American Institute of Physics, 1993.
Gurshtein, Alexander. "When the Zodiac Climbed the Sky." Sky and Telescope 90 (October 1995): 28.
Hetherington, Barry. A Chronicle of Pre-Telescopic Astronomy. New York: John Wiley and Sons, 1996
Posted by astrohintz
at 1:34 PM CST
Updated: Monday, 6 February 2006 4:05 PM CST
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