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Second Astronomy Practice Exam Solutions:
1. The ancient Ptolemaic astronomers had deduced an order for the
planets as one proceeded away from the Earth towards the stellar
sphere. See figure to the right. On which of the following
apparent properties of the planets was this order based upon?
a. Apparent Magnitudes of the Planets (i.e. their brightness)
b. Maximum elongation angles
c. Periods of retrograde motion
d. Time to cycle the zodiac
e. Their mythological hierarchy
2. On the diagram to the right label the deferent, epicycle and equant.
Epicycle
Equant
Deferent
3. Where is a planet on its epicycle when it goes retrograde? Why does it go retrograde only when it is on
that part of its epicycle? Answer in a sentence or two.
In the geocentric Ptolemaic model, a planet only goes retrograde when it is on the inside of its
epicycle because that is when its faster epicycle motion is in the opposite direction of the slower
eastward deferent motion.
4. Why was an equant, or similar device, necessary in the Ptolemaic model? Illustrate with an example.
Answer in a few sentences.
The equant, or similar device, necessary in the Ptolemaic model because, as we know now, planets
did not move at constant speeds, so that they would spend more time on one half of their orbit
than on the other have. As an example, the Sun spends 1 week longer above the Celestial Equator
than it does below the Celestial Equator. The Equant shifted the center of the Suns motion away
from the Earth to mimic what we know now as the changing speed of the Earth around the Sun.
5. In 1543 Copernicus proposed that the Earth was a planet. What motions did Copernicus attribute to the
Earth? Be complete in your answer. Answer in a few sentences.
Copernicus attributes two motions to the Earth; (1) a eastward rotation on an axis that is tilted
23½° once every 23h 56m 4.09s and (2) an eastward revolution of the Earth around the Sun on a
circular orbit and constant speed.
6. 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 in the heliocentric Copernican model because the
inferior planets have smaller orbits than the Earth’s orbit around the Sun. Thus no matter where
the Earth and the inferior planet are in their respective orbits, an Earth-based observer always
has to look generally towards the Sun to find the inferior planet. An Earth-based observer could
never see the inferior planet at opposition because that would require looking away from the Sun.
So inferior planets have a maximum elongation because their orbits are smaller than the Earth’s
orbit.
7. 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.
The occurrence of retrograde motion coincident with opposition for the superior planets is just an
illusion caused by the faster orbiting Earth passing between the Sun and the slower orbiting
Superior planet. The definition of opposition is having the Earth between the Sun and planet. As
the eastward moving Earth is passing the superior planet at opposition, the superior planet
“appears” to move westward much the way a car on the freeway will appear to move backward as
you pass it in the fast lane. The retrograde motion is not real, only an illusion caused by observing
the moving planet from a moving platform (I.e. the Earth). The superior planet brightens because
the Earth is closest to the superior planet when we are between the Sun and the superior planet at
opposition.
The figure to the right illustrates Copernicus’ original heliocentric
model of the Solar System. As you can see it is quite complex
compared to the simpler model we use today. What two flaws in
the original Copernican model created this complexity? Answer in
a sentence.
The two flaws in the original Copernican model created this
complexity are the assumptions of (1) circular orbits and (2)
constant speeds of planets on their orbits.
8. Kepler’s first two Laws of Planetary Motion contradicted the
Aristotelian/Ptolemaic Model of the Universe in two
fundamental ways. State Kepler’s first two Laws of Planetary
Motion and how were they anti-Aristotelian? Use appropriate
vocabulary. Answer in a few sentences.
Kepler’s 1st Law is the law of non-circular orbits meaning that planetary orbits are not circles
centered on the Sun, but are ellipses with the Sun not at the center but at one focus. This is antiAristotelian because Aristotle postulates that all motions of celestial bodies is circular with the
Earth at the center.
Kepler’s 2nd Law is the law of non-constant speeds meaning that planets move not at a constant
speed around its elliptical. Rather, the planets move a bit faster when they are near perihelion
and a bit slower when they are near aphelion. This law is historically characterized by the
expression “Equal areas in equal times”.
The figure below is an ellipse. The axes are marked in units of AUs. The position of the Sun is marked.
Answer the questions that appear below the figure by filling in the blanks.
10
8
6
4
2

0
-10
-8
-6
-4
-2
0
2
4
6
8
10
-2
-4
-6
-8
-10
9. What is the semi-major axis of this ellipse in AU? 8 AU
10. What is the perihelion distance of this orbit in AU 3AU
11. What is the aphelion distance of this ellipse in AU? 11AU
12. What is the eccentricity of this ellipse? e 
c 3 AU

 0.375
a 8 AU
13. The asteroid 1620 Geographos discovered on September 14, 1951 at the Palomar Observatory by
Albert George Wilson and Rudolph Minkowski.. Its orbital semi-major axis is 1.24 AU. What is its
orbital period in days? Show your work to solve the problem below.
3
Use Kepler’s 3rd Law: Pyr2  aAU
3
Pyr2  aAU
 1.243  1.907
 Pyr  1.907  1.381 years  1.381 years 
365.241 days
 504.3 days
1 year
14. The same asteroid 1620 Geographos has an eccentricity of 0.34. What is its perihelion distance? Why
would this asteroid be of special concern for astronomers?
Use the two relations from the “abc’s of Ellipses: e 
c
and a  rP  c to express rp as a function of e and a.
a
c a  rp

a
a
 r p  a  ae  a  (1  e)  1.24 AU  1 - 0.34  0.818 AU
e
Asteroid 1620 Geographos has a perihelion distance of 0.818 AU and, thus, this asteroid crosses the
Earth’s orbit (at 1.0 AU). Earth crossing asteroids are interesting for astronomers for the potential of
a collision with the Earth.
15. Please choose one of Galileo’s telescopic observations of the Moon, the Sun, or Jupiter and briefly
describe what he saw and how it contradicted the Aristotelian Model of the Universe. Be complete in
your answer. Answer in a few sentences.
Observation Target
What Galileo Saw
The Moon, as seen through a telescope, has
definite landforms like mountains, valleys,
plains, and craters.
How he Interpreted it
Living in Italy, Galileo was
familiar with landforms like
mountains, valleys, plains and
volcanic craters (calderas).
Seeing these features on the
Moon made him believe that the
Moon was made of terrestriallike matter (rocks and stuff).
The Sun, as seen through a telescope and an
appropriate filter, will show dark spots on
its surface that we now call sunspots. These
spots were seen to change with time;
changing shape disappearing and
reappearing over a several day period.
Galileo interpreted these spots in
two ways. First they were
blemishes on the Sun’s surface –
imperfections. Second, they
were not a constant part of the
Sun but changed over time.
Jupiter looks like a star to the naked-eye.
However, through even a small telescope
Jupiter resolves itself into a disk. Around
this disk, after several weeks of observing,
Galileo had charted four tiny “stars”
orbiting Jupiter. These objects were called
the Medicean stars (after Cosimo Medici the
ruler of Galileo’s Florence) for centuries,
until the modern name of the Galilean
Satellites was adopted. We now recognize
these objects as the four largest moons of
Jupiter. Two of them are larger than our
own Moon and the other two are larger than
the planet Mercury.
Galileo observed Venus over several
months. He saw that Venus had phases
similar to the Moon’s phases and that these
phases were correlated with its distance
from the Earth: Full when far and Crescent
when close.
Galileo interpreted that these
stars or satellites were orbiting
Jupiter. That’s all he needed to
see.
Moon
Sun
Jupiter
Venus
As explained in the next
question, Galileo concluded that
Venus must orbit the Sun.
These observations could not be
explained using the geocentric
model of Ptolemy. See the next
question for more details.
How this contradicted Aristotle.
Aristotle postulated that all objects
above the atmosphere were not
made of terrestrial matter but out
another element called celestial
matter. If the Moon was apparently
made of terrestrial matter as
Galileo concluded, it could not be
celestial matter and it called into
question the very existence of
celestial matter.
Aristotle postulated that the
celestial matter was perfect –
unchanging and eternal. If these
sunspots were indeed blemishes
then the Sun was not perfect
celestial material and since the
spots changed over time, the Sun
was not an unchanging eternal
body. Again, the very existence of
celestial matter is called into
question by Galileo’s observations.
Aristotle and Ptolemy had all
motion around a fixed and central
Earth. There was no contingency
for “stars” to orbit a planet in that
geocentric model. Thus the
foundational premise of the
Aristotelian model that all motion
is around the Earth is nullified.
Aristotle and Ptolemy had all the
planets circling the Earth. To
definitively prove that one of those
planets circled the Sun called into
question the Aristotelian postulate
that any planets orbited the Sun.
Thus another mortal wound to the
natural philosophy of Aristotle.
The seventeenth century was not
kind to Aristotle.
16. The figure below is a reproduction of Galileo’s record of observations of Venus from Il Saggiatore
[The Assayer] Rome, 1623. What is it about Galileo’s Venus observations that was so damaging to the
Aristotelian/Ptolemaic Model of the Universe? Answer in a few sentences.
Galileo’s observations of Venus going through phases that were correlated
to its distance and planetary configurations could only be interpreted as a
result of Venus orbiting the Sun. The figure to the right contrasts the model
predictions of the Ptolemaic (geocentric) and Copernican (heliocentric)
systems. Next to the figure of model preditions is an actual photographic
representation of the phases of Venus. In the Ptolemaic model Venus would
never appear in the near full phases, but only as a crescent and it would not
vary much in apparent diameter (angular size) since it was always in front of
the Sun. However, the actual photographs of Venus pahses show it in
phases near full and in those phases Venus must be far away because it has
a smaller apparent diameter. The photographic evidence is much more
consistent with the Copernican mode that predicts that Venus will be in a
crescent phase and large when on the same side of the Sun as we are and
near full phase and small when on the opposite side of the Sun. The
observations of Galileo were a lethal wound to the Ptolemaic geocentric
model of planetary motion because they could only be interpreted with
Venus orbiting the Sun.
17. The Universal Gravitational constant G is an extremely small number equal to 6.6710-11 in mks units.
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 significance or meaning of G being so small is that gravity is an intrinsically weak force. Unless
Mm
at least one of the masses in the Universal Law of Gravity formula F  G 2 is astronomical in scale,
R
the force F will be too small to be perceived. If the numerical value of G were closer to 1, then
gravity would be a substantial and readily perceptible force between everyday objects…like two
people passing each other in the hallway. If G ≈ 1, then as you walked straight down the hallway and
pass close by another person headed in the opposite direction you would both be pulled toward each
other and diverted from your straight line path. Life would very much more interesting and complex
if G were about equal to 1.
18. If Neptune was 1.0 A.U. from the Sun (instead of 20 A.U.), would the gravity force between Neptune
and the Sun be less or more than it is now? By how many times?
The gravity force between Neptune and the Sun will be more than it is now by how 400 times
because it will be 20 times closer and gravity follows an inverse square relationship with distance.
19. A star designated as BD +48 738 is known to have a planet orbiting it. The mass of the planet is about
289 Earth masses and orbits exactly 1.00 AU from the star. The mass of the star is 0.74 times the mass
of the Sun. Which of the statements below regarding the time this planet takes to orbit the star true?
Circle the correct answer.
a.
b.
c.
d.
The planet takes one year to circle the star because it is 1 AU from it.
The planet takes longer than one year to circle the star because it is so massive.
The planet takes less than one year to circle the star because it is so massive.
The planet takes longer than one year to circle the star because the star is less massive than
the Sun.
e. The planet takes less than one year to circle the star because the star is less massive than the Sun.
(Note: the orbital velocity of a satellite around a central object does not depend on the mass of the
satellite.)
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.
20. Which of the following is the best possible ranking for the period of the orbit of these planets orbiting
from shortest to longest?
Two Earth Masses
One Earth Mass
Three Earth Masses
1 AU
A
A
B
One Earth Mass
C
C
B
a.
b.
c.
d.
e.
2 AU
1 AU
2 AU
D<C<B<A
A = C < B =D
A=B<C<D
A < C < B <D
A=B=C=D
21. The image to the right is of an asteroid named Ida and its tine satellite
named Dactyl (That’s right – an asteroid with a moon!) What would and
astronomer need to know about this Ida-Dactyl system to calculate the
mass of the asteroid Ida? Answer in a few sentences.
D
D
22. An asteroid is observed by a student astronomer with
their telescope. The asteroid is at opposition on January
1 and then again at opposition exactly 17 months later.
See the figure. Calculate the orbital period of the
asteroid in years.
First use your knowledge of the Earth’s orbit to
determine the value of the angle labeled α:
Knowing that the Earth requires 12 months to
complete one orbit:
360  
360

17 months 12 months
1st Opposition


α

2nd Opposition
Solve this for the angle α and you find that α = 150°.
Next, assuming that the asteroid is in a circular orbit at constant speed, use the proportion
150
360

17 months x months
Solving this for x yields 40.8 months which is 3.4 years.
1. The Moon appears to cycle around the zodiac once every________.
A. 365.25 days
D. 24 hours
B. 29.5 days
E. 23 hours 56 minutes
C. 27.3 days
2. Which of the statements below about the apparent path of the Moon through the stars is the true
statement?
A. The Moon drifts eastward through the stars on the ecliptic following the path of the Sun.
B. The Moon’s path through the stars is westward along the ecliptic completing once cycle every
365.25 days.
C. The Moon’s path through the stars is eastward closely following the ecliptic, but the Moon
does not return to exactly the same place in the sky after one sidereal period as the Sun
does.
D. The Moon’s path through the stars is generally eastward, but is interrupted by occasional
retrograde motion.
3. Put the phases of the Moon in chronological order, from earliest to latest, starting with waxing crescent.
A
B
Earliest: __D___,
C
__E___,
D
__A___,
___C__,
E
___B__ : Latest
4. Lunar phase today is Waning Crescent. The moon is in the zodiac sign Cancer.
A. Approximately how days from today to the next Full Moon? 2 weeks plus a few days ≈17 days
B. Approximately how many days from today will the Moon again be in the zodiac sign of
Cancer? About 27.3 days later.
5. The sketch to the right shows the Moon in a certain phase.
A. Name the phase of the Moon shown. Waning Gibbous
B. Estimate the number of days till the next
Full Moon
About 26 Days
C. Circle the position of the Moon on the diagram below that corresponds to the phase shown
above.
Sun Light
Earth
6. The position of the Sun and Moon are show on the 360 Mercator view of the sky below.
The Moon is about 9 hrs of RA
east of the Sun. 9 hrs of RA
corresponds to 135° east of the
Sun. So the moon is past 1st
Quarter (90° east of the Sun)
but not past Full Moon (180°
from the Sun) and must be in
the …
Sun’s position in
8 months

Moon’s position
in 3 weeks



12 hrs
8 hrs
4 hrs
0 hrs
20 hrs
16 hrs
12 hrs
A. In what phase would the Moon be seen in given its position in the map above?
Waning Gibbous Phase
B. Label the approximate position of the Sun in 8 months from the position shown.
8 months
 24 hr RA  16 hr RA east of its current position.
12 months
So the Sun' s RA in 8 months will be about 24 hr RA
C. Label the approximate position of the Moon in 3 weeks from the position shown.
3 weeks
 24 hr RA  18 hr RA east of its current position.
4 weeks
So the Moon' s RA in 8 months will be about 17 hr RA
7. The Moon is shown near setting in the image below. The line to the left is the celestial equator and the
upper line is the ecliptic. The gray bar represents the horizon.
A. In what constellation did the Moon rise earlier? Cancer
B. Where along the horizon did the Moon rise earlier? (Circle the correct response below)
N of E,
S of E,
due E,
N of W,
S of W or due W
C. In what constellation will the Moon set on the next day? Leo
8. In a few sentences below define the lunar sidereal period and the
lunar synodic period and explain why they are different time
periods. You may refer to the figure to the right in your
response.
The lunar sidereal period is the time for one rotation of the
Moon around the Earth, about 27.3 days (from point A to
point B on the diagram).
The lunar synodic period is the time between consecutive
identical moon phases (e.g. new moon to next new moon),
about 29½ days (from point A to point C in the diagram).
These two time scales are different because the Earth is revolving around the Sun as the Moon
revolves around the Earth. So after one lunar sidereal period the Earth-Moon system has moved
slightly around the Sun necessitating another almost 2 days for the Moon to come back into
alignment with the Sun.
9. Circle the seven planets of the ancient world from the alphabetic list presented below.
Earth
Jupiter
Mars
Neptune
Pluto
Saturn
Mercur
y
Moon
Sun
Uranus
Venus
10. In one sentence describe how these “planets” appeared to be different from all the other stars. (Two
differences are required for full credit)
Planets appear star-like to the naked eye, However, there are two important differences; (1) these
planets appear to move relative to the stars, and (2) they can change in their brightness.
11. Which of the times listed below represents the average apparent sidereal period of Mars?
A. 365.25 days
D. 29 years
E. 250 years
B. 687 days
C. 12 years
12. Which of the angles listed below represents the maximum elongation of Venus?
A. 23½ 
C. 45
B. 28
D. 180
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. In the figure below label the positions of a planet at opposition, conjunction, quadrature and maximum
elongation. The position of the Earth (  ) and Sun (  ) are shown on the figure.
Maximum
Elongation
Quadrature
Superior
Conjunction
Opposition
Conjunction

Inferior
Conjunction

Quadrature
Maximum
Elongation
15. Use the graph provided below, on which an imaginary planet’s motion has been plotted over several
months, to answer the next question.
May 15th
Path of the
imaginary
planet
March 21st
April 22nd
March 1st
March 31st
April 12th
April 5th
80
100 120
140
160 180 200
60
55
50
45
40
35
30
25
20
15
10
5
220 240 260
16. For how many days would this planet have appeared to move with retrograde motion?
A. 10 days
A) 17 days
B) 32 days
B. 12 days
C. 15 days
Use the two images below, which were obtained from the Solar Heliospheric Observer (SOHO) spacecraft, to
answer the following question. The Sun is located behind the circle drawn on the mask and the bright object to
the right is a planet. The images were taken approximately four days apart with the earlier picture on top.
Eastward
Westward
Image obtained on May 12
Image obtained on May 16
17. The planet is approaching which of the planetary configurations listed below?
A. Maximum Elongation
C. Opposition
D. Quadrature
B. Conjunction
18. 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.


P
or
Superior Conjunction



P
Opposition
NO
B. The Earth and the planet are on the same side of the Sun.

P

or
Inferior Conjunction


P
Opposition
NO
C. The Sun and the planet are on opposite sides of the Earth.


P
or
Opposition


P
Opposition
NO
D. The Earth and the planet are on opposite sides of the Sun.
P


Superior Conjunction
or


P
Superior Conjunction
YES

P