Download Study Guide for 1ST Astronomy Exam

2ND Astronomy Exam Spring 2015
Name Solutions
Class Meeting Time_________________
1. 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.
The position of the Sun and Moon are show on the 360 Mercator view of the sky below.
Position of
Moon in two
weeks.



12 hrs
8 hrs
4 hrs
0 hrs
20 hrs
16 hrs
12 hrs
2. In what phase would the Moon be in given its position in the map above?
First Quarter
3. On the All Sky Map above label the approximate position of the Moon in 2 weeks from the position shown.
The five phases of the Moon shown below are for Question #4.
A
B
C
D
E
4. If the moon is in the full phase today, how many of the moon phases shown above (A-E) would the moon go
through during the next 13 days?
a. none
b. only one
c. two
d. three
e. more than three
The sketch to the right shows the Moon in a certain phase.
5. Name the phase of the Moon shown. Waxing Gibbous
Illuminated
Portion
6. Estimate the number of days till the next Full Moon: 3 Days
7. Circle the position of the Moon on the diagram below that corresponds to the
phase shown above.
Sun Light
Earth
8. During which month of the year will the full moon be at its greatest altitude at transit? December
(Hint: Draw the All Sky Map, put the Moon at the point where it would have the highest altitude and then determine the date given that the moon must be full.)
The full moon is in
opposition to the Sun, so the
Sun must be near the winter
solstice for the full moon to
be high in the sky. So the
date is December.

Moon must be near the
summer solstice to have a
large maximum altitude.
12 hrs
8 hrs

4 hrs
0 hrs
20 hrs
16 hrs
12 hrs
9. 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 of 27.3 days is the time required for the Moon to orbit
the Earth once or equivalently to complete one cycle around the zodiac). The
lunar synodic period of 29½ days is the time required for the once cycle of lunar
phases (e.g. new moon to next new moon). These two time periods are different
because the Earth is revolving around the Sun as the Moon orbits the Earth. In
the figure, the lunar sidereal period is the time from A to B. During that 27.3
days, the Earth has moved to a different position relative to the Sun and the Moon
must rotate from B to C to again be aligned (i.e. new moon) with the Sun.
10. If the Moon rose today at about 10:00 p.m., at what time will it rise tomorrow (approximately)
a. 8:00 p.m.
b. 9:00 p.m.
c. 10:00 p.m.
d. 11:00 p.m.
e. 12:00 p.m.
11. Describe the apparent motion of the planets on a diurnal and long-term timescale in a few sentences.
On a diurnal time scale the planets appear to move westward with the stars as if they were attached to the celestial sphere.
However, all the planets, in fact, fall slightly behind the stars on average. Therefore, on a longer timescale of months to years
the planets appear to drift eastward through the stars closely staying within the zodiac. All planets (except the Moon and Sun)
periodically and briefly undergo retrograde motion where the stop their normal long term eastward drift through the stars
and appear to move westward through the stars.
12. What distinguishes an inferior planet from a superior planet in regards to their apparent motions?
Mercury and Venus are known as inferior planets. They take about 1 year on average to cycle the zodiac. They have a
maximum elongation and go retrograde only at inferior conjunction.
Mars, Jupiter and Saturn are known as superior planets. They take longer than a year to cycle the zodiac. They have no
maximum elongation restriction and go retrograde only at opposition. They also brighten at opposition.
13. An asteroid is discovered and after many months of observations it is observed that the asteroid has a maximum
elongation of 35○. What can be said about this asteroids distance from the Sun?
(Note: no calculations are necessary to answer this question correctly.)
Planet
Elongation Restrictions
Mercury
28 Max. Elong.
Venus
45 Max. Elong.
The asteroid must orbit the Sun between the orbits of Mercury and Venus because the asteroids maximum elongation is
between that of Mercury and Venus.
14. What were the motions attributed to the Earth in the Copernican Heliocentric model and how did they explain
retrograde motion?
Two motions were attributed to the Earth.
1. The Earth revolved eastward around the Sun once every 365.241 days.
2. The Earth rotated eastward once every 23h 56m 4.09s on an axis tilted 23½° relative to its orbital plane.
15. In the Aristotelian/Ptolemaic Geocentric model the planet Venus was “leashed” to the Sun so that if could never
appear more than 46○ ahead of or behind the Sun. How did Copernicus explain this restricted motion of Venus
in his Heliocentric model?
Since Venus has a maximum elongation (Mercury does as well) Venus can never be seen at opposition.
To be seen at opposition a planet must be farther away from the Sun than the Earth. So, Copernicus
concluded that Venus is never farther from the Sun than the Earth and, therefore must orbit closer to the
Sun than the Earth does. So Copernicus explained this restricted motion of Venus in his heliocentric
model by having Venus (and Mercury) orbit closer to the Sun than the Earth does.
16. A new dwarf planet is observed by a student astronomer with their telescope. The dwarf planet is at opposition
on January 1 and then again at opposition exactly 25 months later.
See the figure. Calculate the orbital period of the dwarf planet in
years.
1st Opposition
In 25 months, the Earth goes around the Sun twice plus a bit
more. So the angle α can be determined from
720  
360
. Solving for α yields α = 30°. Therefore,

25 months 12 months
the asteroid moves just 30° in its orbit in those 25 months. So to
30
360
find the orbital period of the asteroid use
.

25 months x months
Solving for the period x yields x = 300 months or 25 years.



2nd Opposition
17. Using Kepler’s 3rd Law and your calculated period from the previous question, calculate the semi-major axis of
this dwarf planet’s orbit around the Sun.
3
PYr2  aAU
2
3
Yr
2
3
 aAU  P  25  8.55 AU
The semi-major axis of this dwarf planet’s orbit around the Sun is 8.55 AU.
18. When Galileo observed the Moon in the early 1600’s he saw something no one else had ever seen. What did
Galileo see when he turned his telescope to the Moon and how did he interpret what he saw?
Galileo saw landforms similar to those on the Earth when he looked at the Moon through his telescope. As written in the Rice
University Galileo Project web page (linked from Hot Tips)
“The telescope delivered the coup de grace to attempts to explain away the Moon's spots and to the perfection of the heavens in general. With his
telescope, Galileo saw not only the "ancient" spots, but many smaller ones never seen before. In these smaller spots, he saw that the width of the
dark lines defining them varied with the angle of solar illumination. He watched the dark lines change and he saw light spots in the
unilluminated part of the Moon that gradually merged with the illuminated part as this part grew. The conclusion he drew was that the changing
dark lines were shadows and that the lunar surface has mountains and valleys. The Moon was thus not spherical and hardly perfect.”
Galileo interpreted his lunar observations as indicating that the Moon was not the perfect celestial body that Aristotle
demanded it be and that it was similar to the terrestrial material that Earth was made of.
The planet orbit shown in the drawing at right obeys
Kepler’s 2nd Law. Each lettered position represents
the location for the planet during a particular day in
a year. At which lettered position would the planet
move the least during one day?
B
A
C
A
D
E
19. 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 Moon has phases,
but the phases of the Moon are not correlated with its distance from
Earth (which is nearly constant) and, more importantly, the Full Moon occurs when the Moon is in opposition
while the Full Venus occurs when Venus is in superior conjunction. These observations were a lethal wound to
the Ptolemaic geocentric model of planetary motion because they could only be interpreted as Venus orbiting
the Sun, not the Earth.
20. Describe Kepler’s first two laws of planetary motion and state how they were contrary to the previously held
ideas of Aristotle and Ptolemy.
Kepler’s 1st Law states that planets orbit the Sun on ellipses with the Sun, not at the center of the ellipse, but at
one focus of the ellipse. Aristotle and Ptolemy insisted that planets move on circles.
Kepler’s 2nd Law states that as a planet's approaches its farthest point from the Sun and as it approaches its
closest point to the Sun its orbital speed changes. When a planet is farthest from the Sun is speed is slightly
slower than its average speed in its orbit. When a planet is closest to the Sun is speed is slightly faster than its
average speed in its orbit. Aristotle and Ptolemy insisted that planets move at constant speeds.
The figure below is an ellipse. The axes are marked in units of AUs. The position of the Sun is marked with the
. Note that the origin is at one focus of the ellipse and not at the center of the ellipse. Answer the questions
that appear below the figure by filling in the blanks.
6
5
4
3
2
1
0
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-1
-2
-3
-4
21. What is the semi-major axis of this ellipse in AU? 4 AU
-5
-6
22. What is the perihelion distance of this orbit in AU 2 AU
23. What is the aphelion distance of this ellipse in AU? 6 AU
24. What is the eccentricity of this ellipse? e 
c 2 AU

 .5
a 4 AU
25. What component of the Universal Law of Gravity explains the fact that when you walk past another person in
the hallway you do not feel the gravitational force between the two of you?
The component of the Universal Law of Gravity that explains the fact that when you walk past another
person in the hallway you do not feel the gravitational force between the two of you is the gravitational
constant G. The gravitational constant G is an exceedingly small number equal to 6.67 × 10-11 N∙m2/kg2.
Because G is so small the gravitational force between two objects is only perceptible if at least one of the
masses is astronomical in scale.
26. The image to the right is of an asteroid named Ida and its tiny satellite
named Dactyl (That’s right – an asteroid with a moon!) What would an
astronomer need to know about this Ida-Dactyl system to calculate the mass
of the asteroid Ida? Answer in a few sentences.
An astronomer would need to know both the orbital radius and orbital
period of the moon Dactyl around Ida or, equivalently, the orbital
radius and the orbital speed of Dactyl around Ida to calculate the mass
of Ida.
The four planets below orbit around a star identical to the Sun. The mass of each planet and its distance from
the star are indicated.
4 Earth Mass
1 Earth Mass
4 Earth Mass
1 Earth Mass
1 AU
A
2 AU
B
1 AU
C
2 AU
D
27. Which of the four planetary systems above has the strongest gravity between the Sun and Planet? C
28. Which of the four planetary systems above has the longest orbital period? B & D
29. How many times stronger is the force of gravity on the planet in system A compared system B? 4
(Since gravity is proportional to the product of the star’s and planet’s masses.)
30. How do the forces of gravity in system A compare to that in system D? Answer in a sentence.
In system D the planet is twice as far from the star so gravity will be ¼ as strong as in system A.
However, the planet is system D is 4 times more massive than the planet in system A which would make
gravity 4 times stronger than in system A. So, the weakening of gravity due to the greater distance in
system D is cancelled exactly by the greater mass of the planet in system D. Thus the force of gravity on
the two planets is the same.
Astronomy Formula and Constants Sheet for Exams
Conversions
Formulas
A
L

2D
360 
Ellipse Formulae
2a  rP r A
e
c
a
Main Sequence Lifetime t 
M
1010 yr
L