Exam (elaborations)

Solutions for Explorations: Introduction to Astronomy, 10th Edition Arny (All Chapters included)

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Course
Astronomy

Institution
Astronomy

Complete Solutions Manual for Explorations: Introduction to Astronomy, 10th Edition by Thomas Arny, Stephen Schneider ; ISBN13: 9781266673375. (Full Chapters included Chapter 1 to 18). Problems included. Chapter 1: The Cycles of the Sky. Chapter 2: The Rise of Astronomy. Essay 1: Backyard Astron...

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10th edition
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Explorations:
Introduction to Astronomy
10th Edition
by Thomas Arny

Complete Chapter Solutions Manual
are included (Ch 1 to 18)

** Immediate Download
** Swift Response
** All Chapters included

,Preview The Cosmic Landscape

PREVIEW: THE COSMIC LANDSCAPE

Note:
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.

The problems in this chapter are a good place to have students practice working
numerical problems by hand to estimate solutions, rather than always reaching for the calculator.

Answers to Thought Questions
1. Earth, Solar System, (optional: galactic spiral arm), Milky Way Galaxy, Local Group, Virgo
Supercluster, (optional: cluster of superclusters), the Universe.

2. Student responses could be examined based on (1) is the idea testable/answerable by science,
(2) did they provide a clear hypothesis and if appropriate, discuss any obvious or existing
data that provide a starting point, (3) did they provide good questions to try to answer and
good tests to try; (4) did they discuss the implications of failed tests for the hypothesis. It
may be interesting to let students present these ideas and discuss whether the testing of the
hypothesis is in line with the ideas of the scientific method.

3. This is a somewhat open-ended question, but any theory of a new force would need to pass
all existing experimental tests with equal or better agreement than the old theory and provide
explanations and pass experimental tests (including predictive tests) for phenomena at size
scales, energy ranges, or in some other region that are not explained by existing theory. A
new theory could replace a previous theory, if, for example, it extended the physical range of
parameters to which the old theory applied. This has happened many times in physics. A
good example of this would be how the general theory of relativity explains gravity over a
broader range of circumstances than Newton’s universal law of gravity, and can, for
example, explain the orbit of Mercury where Newton’s laws fail. Similarly, a combined
theory of electromagnetism explains the phenomena of electricity and magnetism.

Therefore, any scientific explanation for ghosts or psychic powers must be consistent with, or
provide a better explanation for, existing theories of electromagnetism, gravity, special relativity,
and so on. One could only postulate a hypothesis that explained ghosts as manifestations of a
new part of the electromagnetic spectrum if the postulate, in addition to passing the usual
scientific tests for falsification, prediction, etc., also did not contradict existing results. Similarly,
one could not postulate a mechanism for the exchange of ideas telepathically via some kind of
radio transmission between organisms and simultaneously claim faster-than-light communication
was possible, at least not without providing some credible and testable reason to believe that the
special theory of relativity is incorrect! To consider a different issue, frequently, claims of ghost
sightings refer to changes in temperature related to where the ghosts appear. To formulate a
scientifically grounded theory, one would need to produce hypotheses that are consistent with
other known rules of thermodynamics.

In short, there is plenty of room to propose new hypotheses, but they cannot contradict known
science (unless they are able to convincingly supersede existing theories).

1

, Preview The Cosmic Landscape

Answers to Problems
1. R Sun = 7 ×105 km . R Earth
= 6.4 ×103 km . To show that the Sun is about 100 times the size of
Earth, divide the Sun’s radius by Earths:
( )
7 ×105 km / 6.4 ×103 km .
R Sun / R Earth =
We can use scientific notation to easily show that this is almost 100, because .4 is
approximately 1. Therefore,
R Sun / R Earth ≈ 105 km /103 km =
105 /103 =105−3 = 102 =100 .
(The exact answer is 109 times.)

2. Using the given values, divide 1 ly = 1×1013 km by 1 AU = 1.5 ×108 km to find the number of
AU in one light-year.
( )
.5 ×108 =1013−8 /1.5 =105 /1.5 = (10 /1.5) ×104 = 6.7 ×104 = about 67, 000 .
(The exact answer using unrounded values is about 63,240 AU.)

3. There are several ways to solve this problem. One easy way is to compare Earth’s radius or
diameter to the Moon’s radius or diameter. The Moon’s diameter is about ¼ Earth’s. Since C
= π × D, then the circumference is also ¼ Earth’s. So if a volleyball is about 68 cm in
diameter, the Moon model would be 0.25 × 68 cm = 17 cm. The Moon’s diameter is
= C / π 17= cm / π 5.4 cm .

A slightly more accurate calculation using the values of Earth and Moon’s size from the
Appendix (the Moon’s radius is 1738 km; Earth is 6378) shows the Moon’s radius to be 0.27
times Earth’s; therefore, the diameter and circumference are also 0.27 times Earth’s; this gives a
circumference for the model of the Moon of 18.4 cm and a diameter of 5.8 cm.

Either way, the model of the Moon would be a little smaller than a tennis ball or about the same
size as a small orange.

4. Use the formula time = distance/velocity . For light, V = c= 3 ×105 km / s . The distance to
= 1.5 ×108 km , so…
Eris in the Appendix is 67.7 AU, and 1 AU
= d 67.7 AU ×1.5 ×108 km / AU ≈ 100 ×108= km 1010 km .
The speed of light, c, is 3 ×105 km / s . So,
(
1010 km / 3 ×105 km / s =
t= )
3.33 ×104 s .
Divide by 60 s / min to get 564 minutes.

5. The Milky Way has a diameter of 100,000 ly, which we will represent as 2cm. According to
Table P.1, the Local Group is about 50 times bigger than the Milky Way, with a radius of
2.5 ×106 ly and diameter of 5 ×106 ly(= 50 ×100, 000 ly) . Therefore, the Local Group would
be about 100 cm in diameter.
For an exact calculation, Size (Local Group) / Size (Milky Way) = constant , so
5 × 106 ly x 5 × 106 ly
= so x = 5 × 106−5 × 2 cm =
× 2 cm = 5 × 10 × 2 cm =
100 cm
105 ly 2cm 105 ly

2

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