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SCI 103 Jaquin’s Sections
10/20/09
Second Astronomy Exam:
The Copernican Revolution and Gravity.
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Match the list of KEY TERMS below with their definitions by writing the correct term in
front of the definition.
Conjunction
Greatest elongation
Inferior Conjunction
Key Terms
Opposition
Retrograde Motion
Quadrature
Geocentric Model
Superior Conjunction Occam’s Razor
Definitions
1) ___ Quadrature __ The points in the orbit of an outer planet when it appears to be at
a 90 angle with respect to the Sun.
2) __ Conjunction ___ The appearance of two celestial bodies in approximately the
same direction on the sky.
3) __ Inferior Conjunction __ The appearance of an inferior planet when it appears
approximately in front of the Sun.
4) __ Superior Conjunction __ The appearance of an inferior planet when it appears
approximately behind the Sun.
5) _ Retrograde Motion _ The westward shift of a planet against the background stars.
6) _ Opposition __ The configuration of a planet when it rises as the Sun is setting.
7) _ Greatest elongation __ The position of an inner planet when it lies farthest from
the Sun on the sky.
8) Circle the seven planets of the ancient world from the alphabetic list presented below.
Earth
Jupiter
Mars
Mercury
Moon
Neptune
Pluto
Saturn
Sun
Uranus
Venus
9) In one sentence describe how these “planets” appeared by the naked eye to be
different from all the other stars.
Planets mover relative to the stars on long
time scales and change their brightness.
10) In a few sentences, describe the difference in apparent motion between the Superior
and Inferior planets
Inferior Planets
Take 1 year on average to cycle the zodiac
Superior Planets
Take more than 1 year on average to cycle
the zodiac
Retrograde at opposition
No Maximum elongation
Retrograde at (inferior)conjunction
Exhibit a maximum elongation
11) Which of the times listed below represents the time Saturn requires to cycle once
around the zodiac?
A) 365.25 days
D) 29 years
B) 687 days
E) 250 years
C) 12 years
12) Which of the angles listed below represents the maximum elongation of Mercury?
A) 23½ 
C) 45
D) 180
B) 28
13) Which of the statements listed below best represents the apparent relationship between the
Sun and the Superior Planets?
A) The Superior Planets are never seen at opposition to the Sun.
B) The Superior Planets have a maximum elongation and appeared “tied” to the Sun.
C) The Superior Planets only go retrograde when in opposition to the Sun.
D) The Superior Planets only go retrograde when in conjunction to the Sun.
14) Which of the statements listed below could ONLY be the planetary configuration known as
conjunction? (This question is tricky – sketch out the options if you have too.)
A) The Earth and the Sun are on the same side of the planet.
E
S
P (Conjunction)
or
S
E
P (opposition)
E
S (opposition)
B) The Earth and the planet are on the same side of the Sun.
E
P
S (Conjunction)
or
P
C) The Sun and the planet are on opposite sides of the Earth.
S
E
P (Opposition)
or
P
E
S (opposition)
S
E (Conjunction)
D) The Earth and the planet are on opposite sides of the Sun.
E
S
P (Conjunction)
or
P
The figure below shows four identical stars and four planets of various masses in circular orbits
of various sizes. In each case the mass of the planet is given in Earth masses and the orbital
distance is given in Astronomical Units (AU). Note that the sizes of the stars and the orbital
distances have not been drawn to scale.
Two Earth Masses
One Earth Mass
Three Earth Masses
1 AU
A
A
One Earth Mass
2 AU
1 AU
2 AU
C
C
BB
D
D
15) Which of the following is the best possible ranking for the period of the orbit of these planets
from shortest to longest? (Be careful. This is a tricky problem.)
A)
B)
C)
D)
E)
D<C<B<A
A = C < B =D (orbital velocity is independent of mass of orbiting satellite)
A=B<C<D
A < C < B <D
A=B=C=D
16) In a sentence or two explain how the modern Copernican model of the Universe
explains why inferior planets have a maximum elongation.
Inferior planets have a maximum
elongation because the orbit the Sun
inside the Earth’s orbit. Since they
are inside the orbit of the Earth, they
will always appear to be relatively
close to the Sun.
17) In a sentence or two explain how the modern Copernican model of the Universe
explains the occurrence of retrograde motion coincident with opposition and
brightening for the superior planets.
Superior planets orbit outside the Earth’s orbit and move more
slowly than the Earth. When the Earth passes between the
superior planet and the Sun (at opposition) the superior planet
will be closest to the Earth and will appear the brightest.
18) Kepler’s first two Laws of Planetary Motion contradicted the Aristotelian/Ptolemaic
Model of the Universe in two fundamental ways. What are Kepler’s first two Laws
of Planetary Motion and how were they anti-Aristotelian?
Kepler’s 1ST law states that planets orbit, not on circles, but on ellipses with the
Sun, not at the center, but at one focus of the ellipse. Aristotle stated that the
Planets move on circles with the Earth at the center.
Kepler’s 2ND law states that planets do not move at a constant speed , as
Aristotle had postulated, but move faster when they are closer to the Sun and
more slowly when farther away from the Sun.
19) Please choose one of Galileo’s telescopic observations of either the Moon, the Sun,
the moons of Jupiter, or Venus and briefly describe what he saw and how it
contradicted the Aristotelian Model of the Universe.
Observation of the Sun
Galileo observed Sun spots that he interpreted as
blemishes on the Sun and the sunspots changed.
This violated Aristotle’s precept of perfect and
unchanging celestial matter.
Observation of the Moon
Galileo observed terrestrial landforms on the
Moon. This violated Aristotle’s precept of the
distinction between terrestrial and celestial matter.
Observation of Jupiter
Galileo observed four satellites orbiting Jupiter.
This violated Aristotle’s precept of that all
motion was around the centered Earth.
Observation of Venus
Galileo observed Venus going through phases
that were correlated with its angular size
indicating that Venus orbited the Sun. This
violated Aristotle’s precept of that all motion
was around the centered Earth.
20) The Universal Gravitational constant G is an extremely small number. What does it
mean that G is so small? What would the universe, or daily life, be like if G were a
number closer to one? Answer in a few sentences below.
The meaning of a very small value for G is that the force of gravity is an
intrinsically weak force that is appreciable only whey at least one of the masses
is at least planetary in scale. If G were a larger number closer to one, then that
would indicate that gravity would be a much stronger force and the attraction
between even low mass objects like cars and people would be a noticeable
force. Navigating yourself down a crowded hallway or crossing a busy street
would be much more challenging if each person and auto were pulling you
towards them or it.
21) Newton’s law of gravity is an example of an inverse square law. In a few sentences,
describe what an inverse square law is and give a numerical example.
An inverse square law describes the way a force or other field varies with
distance from the source. Inverse square laws have field strengths that diminish
rapidly with distance being inversely proportional to the square of the distance
from the source. For example, the Earth at 1 AU from the Sun feels a certain
gravitational pull toward the Sun. If the Earth were moved out to 5 AU the
gravitational pull from the Sun would diminish not by a factor of 5 but by a
factor of 25. If the Earth were moved in to .2 AU the gravitational pull from
the Sun would grow not by a factor of 5 but by a factor of 25.
Problems: Please solve the following three problems.
1. Using the tables from the appendix of your text, calculate the orbital velocity of the
moon Phobos around the planet Mars.
GM
This is an orbital velocity problem so we will use the relation v 
, where G is
R
the gravitational constant of 6.657 x 10-19 in MKS units, M is the mass of the central
body (Mars in this problem) expressed in kilograms and R is the orbital radius of
the satellite (Phobos in this problem) in meters. This orbital radius is equivalently
to its distance from the central body (Mars).
From the tables provided
The mass of Mars: M = 6.42 x 1023 kg
The orbital radius of Phobos: R = 9.38 x 103 km = 9.38 x 106 m
Now we can calculate:
6.67 10 11 6.42 10 23   2,136 m
GM

R
s
9.38 10 6 
The orbital speed of Phobos is 2,136 m/s or 2.136 km/s.
v
2. The globular star cluster Omega Centauri is shown to the
right in a negative image (stars are dark and the background
is white). Each of the dots on the image represents a star in
some ways similar to our Sun.
a. Estimate its angular diameter given that the image is 1
degree on a side.
1 degree
The globular cluster appears to be about ½ as wide as
the picture. So the angular size must be close to 0.5
degrees.
b. Assuming that this star cluster is 100 lyrs in diameter, calculate its distance from the
Earth.
A
L

This is an angular size problem so we will use
, where A is the angular

2D
360
size in degrees, L is the true width or diameter of the globular cluster and D is the
distance to the globular cluster. We will solve for D.
A
L
L 360  100ly 360 

D



 11,460ly
2 A 
2 0.5 
360  2D
The Omega Centauri globular sluster is about 11, 459 light years away.
3. The mass of the planet Jupiter will play an important role in our future studies of ours
and other solar systems. Calculate the mass of Jupiter using the orbital parameters of its
satellite Europa. See the attached appendix of the text for relevant parameters.
4 2 R 3
 2 , where G is the
G
P
gravitational constant of 6.657 x 10-19 in MKS units, P is the orbital period of the
satellite expressed in seconds, and R is the orbital radius of the satellite in meters.
This is a central mass problem, so we will use M 
From the tables provided
The orbital period of Europa: 3.551 days = 3.068 x 105 seconds
The orbital radius of Europa: R = 670.9 x 103 km = 670.9 x 106 m
Now we can calculate:

 
670.9 10 6
4 2 R 3
4 2
M
 2 

G
P
6.67 10 11
3.068 10 5
The mass of Jupiter is 1.90 x 1027 kilograms.



3
2
 1.90 10 27 kg
Astronomical Formulas
A
L

360 2D
GM
vObital 
R
2GM
R
4 2 R 3
M
 2
G P
Circumfere nce  2R
vEscape 