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PREVIEW: THE COSMIC LANDSCAPE
The Earth, Our Home
The Moon
The Planets
The Sun
The Solar System
A Sense of Scale
The Astronomical Unit
The Milky Way Galaxy
The Light Year
Galaxy Clusters and the Universe
Gravity
Atoms and Other Forces
The Scientific Method
Goals
This Chapter summarizes our knowledge of the overall structure of the Universe.
Students should come away from it with a sense of the hierarchical structure of the
Universe and how the Earth fits into it. Basic inventory of Universe (planets, stars,
galaxies, etc... general appearance and features and relative sizes). The chapter also
introduces powers-of-ten notation, vocabulary and astronomical distance units (AU and
light year) that are used throughout the text.
The video "Powers of Ten" is an excellent visual summary of the material. Some
narration by the instructor may help students better understand the numbers that flash on
the screen during the video.
Key Terms
Planet
Satellite
Star
Solar System
Astronomical Unit
Milky Way
Galaxy
Gravity
Universe
Atoms
Nucleus
Electron
Proton
Electric Charge
Light year
Local Group
Galaxy Cluster
The Local Supercluster
Electric Force
Strong or Nuclear Force
Weak Force
Scientific Method
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 0.1 cm (1 mm), the Sun would have a radius of 100 x 0.1 =
10 cm. The Solar System is 8600 x R(Sun) and so,
R(solar system) = 8600 x 10 cm = 8.6x104 cm = 8.6x102 m, since there are 100 cm in 1
m.
R(Milky Way)= 1x108x R(solar system) = 8.6x1010 m = 8.6x107 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 = 860 m.
In our model, 1 AU therefore = 860 / 40 = 21.5 m
Planet
a (AU)
a (m)
R (mm)
-----------------------------------------------------------------Mercury 0.39
8.4
0.38
Venus
0.72
15.5
0.95
Earth
1.00
21.5
1.00
Mars
1.52
32.7
0.53
Jupiter
5.2
111.8
11.21
Saturn
9.55
205.3
9.45
Uranus
19.18
412.4
4.01
Neptune 30.06
646.3
3.88
Pluto
39.44
848.0
0.18
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
are 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)
3 (e)
4 (c)
5 (e)
Your cosmic address is Earth, Solar System, Milky Way, Local Group.
All of them are.
Jupiter's diameter is about 10 times larger than the Earth's diameter.
All except (d). Choice (d) is a matter of opinion.
CHAPTER 1 - HISTORY OF ASTRONOMY
1-1 Prehistoric Astronomy
The Celestial Sphere
1-2
1-3
1-4
1-5
Constellations
Motions of the Sun and Stars
Daily or Diurnal Motion
Annual Motion
The Ecliptic
The Seasons
The Ecliptic's Tilt
Solstices and Equinoxes
The Planets and the Zodiac
The Moon
Eclipses
Early Ideas of the Heavens: Classical Astronomy
The Shape of the Earth
The Size of the Earth
Distance and Size of Sun and Moon
Extending Our Reach: Measuring the Diameter of Astronomical Objects
The Motion of the Planets
Ptolemy
Islamic Contributions
Asian Contributions
Astronomy in the Renaissance
Copernicus
Tycho and Kepler
Galileo
Isaac Newton and the Birth of Astrophysics
The Growth of Astrophysics
New Discoveries
New Technologies
The Nature of Matter and Heat
The Kelvin Temperature Scale
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.
Lecture Suggestions
The simple cardboard overhead projector planetarium will be helpful in
explaining rising and setting motions as well as retrograde motion. An orrery
(mechanical model of the Sun and Earth) will help illustrate that the constellations that
are visible change with the seasons. I set such a gadget on a table in the front of the room
and then mark off constellations on the wall with chalk. Moving the model Earth around
the model Sun then allows students to see how the stars visible change and how the Sun
"moves" through the Zodiac. You can also make decorative constellations on paper and
tape them to the wall. 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.
A tensor lamp or other bright light source and a tennis and ping-pong ball serve to
demonstrate features of eclipses. A volley ball or basket ball and a bright lamp can
illustrate lunar phases.
Key Terms
Celestial sphere
Model
Constellation
Celestial pole
Solstice
Zodiac
Retrograde motion
Phases
Parallax
Geocentric
Epicycle
Heliocentric
Celestial equator
Ecliptic
Equinox
Solar eclipse
Lunar eclipse
Angular size
Focus
Semi-major axis
Kepler's Laws
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 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.
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.
6. 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.
7. 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 = 2DA/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 = 2DA/360. We next solve it for D by
dividing both sides by 2A and multiplying both sides by 360 to get
D = 360L/(2A). 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 = 1/24 th 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=2R. Thus R = C/2240/2= 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.
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
ESSAY 1 - Backyard Astronomy
Learning the Constellations
Star Lore
Amateur Astronomy
Small Telescopes
Celestial Coordinates
Star Charts
Planetary Configurations
Your Eyes at Night
Goal
Introduce students to star lore and some terms used in naked eye and backyard
astronomy.
Key Terms
Asterism
Declination
Right Ascension
Altitude
Azimuth
Conjunction
Inferior Conjunction
Superior Conjunction
Transit
Opposition
Morning/evening star
Greatest Elongation
Synodic period
Dark Adaption
Averted Vision
Answers to Thought Questions
1. You would look for Mercury in the west, towards the sunset.
2. A "Morning Star", either Mercury or Venus, is seen close to the eastern horizon just
before sunrise.
3. If a planet is at opposition and you see it overhead it must be midnight because the
planet is 180 degrees from the Sun and rises at sunset.
4. You cannot see Mercury in the western sky at dawn because the Sun rises in the east
and Mercury can never get more than 28 degrees from the Sun.
5. Dark adaptation refers to the state of the human eye after 20 minutes or so in very low
light levels, when the pupil has widened and the retina has become more than a million
times more sensitive to low light than it is in daylight. The eye also becomes more
sensitive to blue light (which is why many astronomers use dim flashlights with red
cellophane over them).
6. The pupil of your eye gets bigger in dim light to let more light into the eye, in the
same way that a photographer adjusts the aperture of a camera to match the light levels.
7. Averted vision refers to the practice of looking at faint objects out of the corner of
your eye where there are more receptors for low light levels than at the center of the
retina.
8. Right ascension (R.A.) is an astronomical coordinate that locates objects east-west on
the sky, much like longitude locates objects east-west on the Earth. Technically, right
ascension is measured eastwards along the celestial equator from the vernal equinox
(point where the Sun crosses the celestial equator going north) to the hour circle (a great
circle in the equatorial system that passes through the celestial poles perpendicular to the
celestial equator) of the object.
9. Declination (Dec.) is an astronomical coordinate that locates objects north-south on
the sky, much as latitude locates objects north-south on the Earth. Technically,
declination measures the angular distance of an object from the celestial poles. The
celestial equator is 0 degrees, the north celestial pole is +90 degrees and the south
celestial pole is -90 degrees.
10. Azimuth can be defined in a few ways, but most of the time it refers to the angular
distance of an object from due north, measured eastwards up to 360 degrees.
Answers to Self-Test
1. (c) As a star travels across the sky, its azimuth changes but the R.A. and Dec. remain
the same.
2. (b) If a planet is at inferior conjunction, it is seen against the disk of the Sun, so it
must rise and set with the Sun.
3. (a) If Mercury is at greatest western elongation it is close to 28 degrees west of the
Sun, so that it rises just before the sun and is a morning star.
4. (d) Venus is an inferior planet (orbiting the Sun closer than the Earth) and so can be at
inferior conjunction, i.e., it can pass between the Earth and the Sun.
5. (e) When your eye is dark adapted, its pupil is dilated and it is much more sensitive to
light than during the day.