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Explorations: Introduction to Astronomy 9th Edition by Thomas Arny & Stephen Schneider | SOLUTIONS MANUAL (Complete 18 Chapters).

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Explorations: Introduction To Astronomy 9th Ed

Explorations: Introduction to Astronomy 9th Edition by Thomas Arny & Stephen Schneider. ISBN 9896. SOLUTIONS MANUAL. TABLE OF CONTENTS: Preview: The Cosmic Landscape Chapter 1: The Cycles o f the Sky Chapter 2: The Rise of Astronomy Essay 1: Backyard Astronomy Chapter 3: Gravity and Motion Chapter ...

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,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 ( = 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

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