Exam 13
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University of Nevada, Las Vegas *
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104
Subject
Astronomy
Date
Apr 3, 2024
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14
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1
Name:
Intro Astro: Stars & Cosmology: Exam 1
Instructions:
There are 72 multiple-choice problems each worth 1 mark for a total
of 72 marks altogether. Choose the
BEST
answer, completion, etc.
Read all responses carefully.
NOTE
long detailed responses won’t depend on hidden
keywords: keywords in such responses are bold-faced capitalized.
This
is
a
CLOSED-BOOK / NOTES / WEB / OUTSIDE-WORLD-
ACCESS
exam.
NO
cheat sheets allowed.
NO
outside sources of any kind allowed.
You can use only what is inside your mind.
Calculators and cell phones are permitted
ONLY
for calculations. Remember, it is shameful to cheat—you are trying to steal from
your fellow students.
The exam is out of 72 marks altogether and is a 75-minute exam.
1. “Let’s play
Jeopardy
! For $100, the answer is: Stonehenge and many other prehistoric
monuments suggest that the makers were doing this.”
What is/are
, Alex?
a) special relativity calculations
b) orbital physics
c) simple alignment
astronomy
d) casting horoscopes
e) receiving alien visitors from outer
space
2. Stonehenge demonstrates that some prehistoric people:
a) could predict eclipses.
b) knew the northernmost rising location of the Sun.
c) knew nothing of astronomy.
d) knew more than the ancient Greeks about the universe.
e) suffered from back pain.
3. The ancient Babylonians were using a sexagesimal (number) system as early as circa
1800 BCE. We do not know why, but it may well have been to save labor in division.
The many whole number factors of 60 (1 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60 for a
total of 12 factors) simplifies many division problems. The sexagesimal system seems
to have been used consistently only for mathematical and astronomical purposes. For
everyday use, the Babylonians often or maybe mainly used other systems including
the ubiquitous decimal system: counting on fingers is as old as the hills so to say. In
the last centuries BCE the sexagesimal system was taken over into astronomy. Using
a large base number with a lot of factors has advantages.
But one needs a lot of
symbols for all the numerals unless one uses some subsidiary base which is what the
Babylonians did: 10. In any case 10 as a base has nothing very special to recommend
it, except for the old (very old) finger exercise.
As a non-finger exercise, subtract 61
◦
43
′
14
′′
from 120
◦
41
′
03
′′
. Recall that
′
stands
for arcminutes and
′′
for arcseconds.
HINT:
If sexagesimal subtraction seems too
tricky, you can try sexagesimal addition to recover 120
◦
41
′
03
′′
.
a) 182
◦
24
′
17
′′
.
b) 58
◦
57
′
49
′′
.
c) 58
◦
31
′
14
′′
.
d) 59
◦
51
′
49
′′
.
e) 58
◦
51
′
14
′′
.
2
4. The ancient Greek Presocratic philosophers:
a) may have hypothesized a spherical Earth in order to explain the daily rotation
of the celestial sphere. But it is equally likely that they thought that a spherical
Earth was proven by the axioms of geometry.
b) may have hypothesized a spherical Earth in order to explain the daily rotation
of the celestial sphere. Thales of Miletus then used the spherical Earth theory to
predict a solar eclipse.
c) may have hypothesized a spherical Earth because they thought the Earth needed
to be spherical in order to be in balance at the center of the cosmos. Aristotle
(384–322) later summarized empirical arguments for the spherical Earth. These
included the varying celestial locations of the stars and planets relative to the
horizon as one moved north-south and the fact that the Earth’s shadow on the
Moon in a lunar eclipse was always round. However,
ARISTOTLE
went on to
affirm that Greece must be on top of the spherical Earth because the ground in
Greece was nearly level.
d) may have hypothesized a spherical Earth because they thought the Earth needed
to be spherical in order to be in balance at the center of the cosmos. Aristotle
(384–322) later summarized empirical arguments for the spherical Earth. These
included the varying celestial locations of the stars and planets relative to the
horizon as one moved north-south and the fact that the Earth’s shadow on the
Moon in a lunar eclipse was always round.
However,
PTOLEMY
went on to
affirm that Greece must be on top of the spherical Earth because the ground in
Greece was nearly level.
e) may have hypothesized a spherical Earth because they thought the Earth needed
to be spherical in order to be in balance at the center of the cosmos. Aristotle
(384–322) later summarized empirical arguments for the spherical Earth. These
included the varying celestial locations of the stars and planets relative to the
horizon as one moved north-south and the fact that the Earth’s shadow on the
Moon in a lunar eclipse was always round.
5. A major obstacle that ancient Greek astronomers had in trying to determine the
nature of the Solar System was:
a) the eastward motion of the planets.
b) the inability to measure any distances beyond Pluto.
c) the inability to measure any distances beyond the Moon.
d) the lack of all theoretical biases.
e) the lack of geometrical skills.
6. Aristotelian cosmology:
a) consisted of perfect eternal cubes rotating about the Earth.
b) put the Earth at the center of the cosmos. The planets and fixed stars were located
on sets of solid spheres that rotated about the Earth. The celestial phenomena
were eternal exactly repeating motions. Beyond the sphere of the fixed stars was
a
CHAOS
of primordial material in which were embedded other finite cosmoses.
c) put the Earth at the center of the cosmos. The planets and fixed stars were located
on sets of solid spheres that rotated about the Earth. The celestial phenomena
3
were eternal exactly repeating motions. Beyond the sphere of the fixed stars was
NOTHING
, not even empty space. The universe was finite.
d) was
DISCARDED
by everyone in the medieval Islamic period. It put the Earth
at the center of the cosmos. The planets and fixed stars were located on sets of
solid spheres that rotated about the Earth. The celestial phenomena were eternal
exactly repeating motions. Beyond the sphere of the fixed stars was
NOTHING
,
not even empty space. The universe was finite.
e) was never seriously considered again after Ptolemy’s time.
7. In modern times (which here we mean to be after circa 1450), who first proposed the
heliocentric theory of the solar system?
a) Nicolaus Copernicus (1473–1543).
b) Thomas Digges (c. 1546–1595).
c) Tycho Brahe (1546–1601).
d) Galileo Galilei (1564–1642).
e) Isaac Newton (1643–1727).
8. A key reason (perhaps the most important reason) that led Copernicus to propose the
heliocentric Solar System was to:
a) get rid of uniform circular motion.
b) appease the Sun god.
c) answer Galileo’s insult.
d) get a prediction of the relative positions of the planets.
e) prove that the universe was infinite.
9. Apparent retrograde motion is:
a) the
westward
motion of a
star
on the sky.
b) the
westward
motion of a
planet
on the sky.
c) the
eastward
motion of a
planet
on the sky.
d) the
eastward
motion of a
star
on the sky.
e) the result of an inter-planetary collision.
10. If a planet has a mean distance from the Sun of 9 astronomical units, what is its
orbital period in years?
a) 28 years.
b) 3 years.
c) 9 years.
d) 81 years.
e) 27 years.
11. Galileo’s discovery of the moons of Jupiter:
a) had no bearing on the debate over the Copernican theory.
b) meant that the Earth was the center of Jupiter’s orbit.
c) explained the full phases of Venus.
d) meant that the Earth was not the physical center of all motion in the Solar
System, and that Earth could have a moon and still be on an
EPICYCLE
.
e) meant that the Earth was not the physical center of all motion in the Solar
System, and that Earth could have a moon and still be a
PLANET
.
12. Could heliocentrism be proven to be physically correct in circa 1610? By physically
correct we mean showing that the planets geometrically orbited the Sun
AND
that
this structure (i.e., the structure of the Solar System) was derivable from physical law:
i.e., that the planets ”physically” orbited the Sun.
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a) Yes. Galileo’s telescopic discoveries were proof.
b) Yes. Kepler’s 3 laws of planetary motion were proof.
c) Yes. Galileo and Kelper’s work combined constituted proof.
d) No. Only Newton’s physics published 1687 constituted proof.
e) No. Only Einstein’s general relativity published 1915 constituted proof.
13. “Let’s play
Jeopardy
! For $100, the answer is: This person’s work made astronomy in
a sense and to a degree an experimental science in that he/she showed that the same
physics applies on Earth and in the heavens.”
Who is
, Alex?
a) Marguerite de Navarre (1492–1549)
b) Giordano Bruno (1548–1600)
c) Galileo Galilei (1564–1642)
d) Isaac Newton (1643–1727)
e) Johann Sebastian Bach (1685–1750)
14. It is somewhat traditional or at least not unusual to begin a book or course on
with a philosophical/historical/poetical statement.
a) agronomy
b) astronomy
c) metallurgy
d) proctology
e) tautology
15. “Let’s play
Jeopardy
! For $100, the answer is: An American astronomer of the 2nd
half of the 20th century, famous for planetology and science popularization—arguably
the most well known scientist of his time.”
Who is
, Alex?
a) Albert Einstein (1879–1955)
b) George Gamow (1904–1968)
c) Richard Feynman (1918–1988)
d) Carl Sagan (1934–1996)
e) Brian Greene (1963–)
16. The scientific method can be schematically described as a/an:
a) square of theorizing and experiment/observation.
b) integrative process.
c) reductive process.
d) a cycle of theorizing and experiment/observation.
e) a pointless pursuit.
17. In scientific theorizing, a rule that has come to be accepted is
“Pluralitas non est
ponenda sine necessitate,”
i.e., “Plurality is not to be posited without necessity”
which was stated by Medieval scholastic philosopher John Duns Scotus (c.1266–
1308). The rule is called Occam’s razor after another Medieval scholastic philosopher
William of Occam (c.1287–1347), who said something similar.
In fact the general
idea was expressed by none other than Aristotle (384–322 BCE): “We may assume
the superiority
ceteris paribus
(other things being equal) of the demonstration which
derives from fewer postulates or hypotheses.” The essential idea in modern jargon is
don’t introduce into theories hypotheses that:
a) are needed.
b) both needed and not needed.
c) are not needed.
d) that neither needed nor not needed.
e) not purely philosophical.
18. “Let’s play
Jeopardy
! For $100, the answer is: This
17TH CENTURY FRENCH
scientist gave an early (and probably the first) statement of the falsifiabilty doctrine—
which has been much debated, but at least with qualifications has been accepted by
5
many. His statement is as follows:
To prove a hypothesis, it is not sufficient to show that all known phenomena can
be derived from it. On the other hand, if the hypothesis leads to a single wrong
prediction, it is false.
—yours truly’s own free translation.
As it stands, this statement is open to some qualifications and objections—but that’s
another story.”
Who is
, Alex?
a) Blaise Pascal (1623–1662)
b) Isaac Newton (1643–1727)
c) Charles Darwin (1809–1882)
d) Louis Pasteur (1822–1895)
e) Carl Sagan (1934–1996)
19. Physics can be briefly defined as the science of:
a) human relations.
b) sports and leisure.
c) matter and motion.
d) matter and rest.
e) light.
20. “Let’s play
Jeopardy
! For $100, the answer is: ‘Just so’ in physics.”
What is
, Alex?
a) a story by Rudyard Kipling
b) essential
c) eternal
d) fundamental
e) infernal
21. Astronomy includes both
and fundamental physics.
a) psychology.
b) applied physics.
c) other than physics.
d) fundamental physics.
e) indifferent physics.
22. “Let’s play
Jeopardy
! For $100, the answer is: In the opinion of the intstuctor, it is any
important theory that applies to reality in some form. Such theories are in some sense
and to some degree independent of other theories including the true fundamental
theory of physics.
They emerge from reality and are like Platonic ideals.
Another
view is that it is a theory that applies to a complex system but not to that system’s
components. It emerges from the complexity. The two views arn’t all that far apart
if you define complexity broadly enough.”
What is a/an
theory, Alex?
a) convergent
b) emergent
c) divergent
d) specious
e) faux
23. Evolution by survival of the fittest is used in computer calculations to find optimum
solutions to problems where the solutions are treated as breeding entities. The best
known of such techniques is called the:
a) genetic algorithm method.
b) scientific method.
c) method.
d) no-name method.
e) son of the method.
24. In the multiverse paradigm, it is posited that the 2nd law of thermodynamics must
absolutely everythere in the multiverse—in all the pocket universes and
6
all the regions between them—even though the multiverse outside our pocket universe
may have different physics in most respects from our pocket universe.
a) hold
b) not hold
c) be impossible
d) be infernal
e) be notorious
25. In the broadest sense,
is the study of all extraterrestrial phenomena
and also terrestrial phenomena that fall into the same categories as extraterrestrial
phenomena.
a) agronomy
b) antimony
c) astronomy
d) antiquity
e) antigone
26. Although one can quibble, most would agree that astronomy is the best candidate for
being:
a) the oldest empirical science.
b) the newest empirical science.
c) of little interest.
d) the same as astrology.
e) the queen of the sciences.
27. “Let’s play
Jeopardy
! For $100, the answer is: It is a notation in which one expresses
a number by a prefix number (usually in the range 1 to 10, but not including 10)
multiplied explicitly by 10 to the appropriate power.”
What is
, Alex?
a) British notation
b) scientific notation
c) metric notation
d) tensy notation
e) Irish notation
28. The quantity 2
.
9979
×
10
10
cm
/
s is the same as:
a) 29979000 cm
/
s.
b) 29979000000 cm
/
s.
c) 2
.
9979
×
10
10
km
/
s.
d) the speed of sound.
e) 2
.
9979 cm
/
s.
29. The metric or SI unit prefix symbols M- and m- stand for:
a) mega (factor of 10
6
) and milli (factor of 10
−
6
).
b) mega (factor of 10
6
)
and milli (factor of 10
−
3
).
c) kilo (factor of 10
6
) and milli (factor of 10
−
3
).
d) kilo (factor of 10
6
) and milli (factor of 10
−
6
).
e) merger (factor of 10
9
)
and melba (factor of 10
−
6
).
30. Ex-planet Pluto’s mean distance from the Sun is about:
a) 0.387 AU.
b) 1.0 AU.
c) 39.54 AU.
d) 67.781 AU.
e) 700 AU.
31. The lookback time to an object 300 light-years away is:
a) 3 years.
b) 10 years.
c) 30 years.
d) 100 years.
e) 300 years.
32. “Let’s play
Jeopardy
! For $100, the answer is: The approximate length of a standard
university lecture as pointed out by Nobel-prize-winning physicist Enrico Fermi.”
What
, Alex?
a) is a century
b) seems like a century
c) is a centi-century
d) is a micro-century
e) is a milli-century
33. “Let’s play
Jeopardy
! For $100, the answer is: It is a matter of point of view that this
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temperature state can actually be reached. A small enough sample can be put in the
state, but some say if it is small enough to be put in the state, then it is too small
to be considered to have a temperature since temperature is an average property of
matter. In any case, where the state is is exactly known from limiting processes.
What is
, Alex?
a) absolute zero
b) the freezing point of water
c) the boiling point of water
d) the triple point of water
e) absolute hot
34. Two simple math formulae that everyone
should know are for the amount
A
accumulated at constant rate
R
in time
t
and the inverse formula for the time
t
to accumulate amount
A
at constant rate
R
. The formulae are, respectively:
a)
t
=
A/R
and
A
=
Rt
.
b)
t
=
AR
and
A
=
R/t
.
c)
A
=
R/t
and
t
=
AR
.
d)
A
=
Rt
2
and
t
=
A/R
2
.
e)
A
=
Rt
and
t
=
A/R
.
35. Pluto is about 40 astronomical units from the Sun. One astronomical unit is about
1
.
5
×
10
13
cm. The speed of light is 3
.
00
×
10
10
cm
/
s. The light travel time from the
Sun to Pluto is:
a) 2
×
10
3
s or about half an hour.
b) 2
×
10
3
s or about 5.5 hours.
c) 3
.
6
×
10
3
s or one hour.
d) 2
×
10
4
s or about 5.5 hours.
e) 2
×
10
4
s or about 8 hours.
36. The Moon’s orbital period (i.e, the sidereal month) is 27.321661547 days (J2000).
What is the Moon’s orbital angular velocity relative to the observable universe (i.e.,
the fixed stars in the traditional expression)?
a) 12
.
19
◦
/
day.
b) 12
.
50
◦
/
day.
c) 13
.
18
◦
/
day.
d) 15
.
19
◦
/
day.
e) 27
.
32
◦
/
day.
37. A curve on a linear plot that decreases as 1 over the square of the horizontal axis
coordinate represents a/an
function.
a) linear
b) inverse-square
c) quadratic
d) logarithmic
e) perpendicular
38. An inertial frame is a reference frame with respect to which all laws of physics are
referenced (at least in any ordinary sense), except general relativity which tells us
what an exact inertial frame is.
What the simplest exact inertial frame is is a/an
that is
NOT
rotating with respect to the observable universe (i.e., the
bulk mass-energy of the observable universe).
a) accelerating frame
b) rotating frame
c) free-fall frame
d) non-rotating
e) oscillating frame
39. The orbits of the planets are best described as:
a) highly elongated ellipses.
b) perfect circles.
c) ellipses, most of them very nearly circular.
d) triangles.
e) 36-sided polygons.
8
40. “Let’s play
Jeopardy
! For $100, the answer is: This condition of astro-bodies means
that they show no parallax to unaided-eye observations for any movements about the
Earth’s surface.”
What is their
, Alex?
a) closeness relative to the size of the Earth
b) remoteness relative to the size
of the Earth
c) spherical nature
d) reflectivity
e) sensitivity
41. The celestial sphere is:
a) an imaginary remote sphere (centered on the
EARTH
) on which all the celestial
bodies are located.
b) a solid sphere (centered on the Earth) on which all the celestial bodies are located.
c) an imaginary remote sphere (centered on the
SUN
) on which all the celestial
bodies are located.
d) the path of the Sun on the sky.
e) cause of eclipses.
42. “Let’s play
Jeopardy
! For $100, the answer is: They are the extensions of the Earth’s
axis out to the celestial sphere.”
What are
, Alex?
a) zenith and nadir
b) horizon and nadir
c) the north and south celestial
poles (NCP and SCP)
d) the celestial equator and the eliptic
e) the
ecliptic pole and the celestial axis
43. What is zenith? What is nadir?
a) The point directly to the east; the point directly below.
b) The point directly above; the point directly below.
c) A kind of television; a kind of refrigerator.
d) The point directly above; the point directly west.
e) The name of the spring equinox point; the name of the fall equinox point.
44. Polaris is on the horizon. You are:
a) near the equator.
b) in New York City.
c) in Las Vegas.
d) near
the north pole.
e) over the rainbow.
45. Three astonomical location methods are by using modern constellations, horizontal
coordinates, and:
a) Cartesian coordinates.
b) polar coordinates.
c) equatorial coordinates.
d) vertical coordinates.
e) miscellaneous coordinates.
46. What does “to transit the meridian” mean? It means that an object:
a) passes through the zenith.
b) crosses the meridian of
GREENWICH
due to the rotation of the Earth.
c) crosses the meridian (i.e., the
LOCAL MERIDIAN
) due to the rotation of the
Earth.
d) is in conjunction with the Sun.
9
e) is in opposition (to the Sun).
47. What is right ascension (RA)?
a) The point directly below.
b) The point directly above.
c) The angular position of an object measured north or south from the celestial
equator.
d) The angular position of an object measured east or west from the celestial equator.
e) The azimuthal angular position of an object measured east from the vernal (or
spring) equinox.
48. In the summer of the northern hemisphere:
a) the northern hemisphere day side is tilted toward the Sun.
b) the northern hemisphere day side is tilted away from the Sun.
c) the southern hemisphere day side is tilted toward the Sun
d) the Earth is nearest the Sun.
e) the Earth is at 0.7 astronomical units from the Sun.
49. Retrograde motion (or in modern astro-jargon apparent retrograde motion) is when
a planet moves
on the celestial sphere.
a) westward
b) eastward
c) northward
d) southward
e) outward with the expansion of the universe
50. Sometime circa 1650 BCE, the ancient Babylonian astronomers discovered the Venus
cycle. Their record of this is called (by moderns) the
Venus Tablet of Ammisaduqa
:
King Ammisaduqa was Hammurabi’s grandson.
The Venus cycle is just a result of
the synodic period of Venus being about 583.92 days and the Earth’s orbital period
being about 365.25 days (a Julian year). The synodic period is the time for the planet
to return to the same position in the sky relative to the Sun. Five synodic periods
is 2919.60 days and 8 Julian years is 2922.00 days. These times are nearly equal. So
8 years from any day, Venus will be back where it was on that day relative to the
Sun (because 8 years is 5 synodic periods) and relative to the fixed stars (because
the Sun comes back to the same position relative to the fixed stars every year). It is
now easy to predict Venus’ position approximately to the past or future from 8 years
of observations. A lot of ancient astronomical prediction skill comes down to using
approximate cycles like the Venus cycle. It’s not rocket science.
If Venus is nearly in same place relative to the Sun every 8 years, where is it
relative to the fixed stars at the same intervals?
a) Far off the ecliptic.
b) 30
◦
farther east on the ecliptic every Venus cycle period.
c) 30
◦
farther west on the ecliptic every Venus cycle period.
d) It’s unpredictable.
e) Nearly the same place.
51. The stars in a constellation are:
a) in orbit about the Earth.
b) all about the same age.
c) at about the
same distance from the Earth.
d) usually unrelated, except that they are
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10
close in angular position as seen from the Earth.
e) members of the Solar
System.
52. All
historical
cultures
eventually
arrived
independently
at
the
same
set
of
constellations.
a) Yes.
b) For short periods of time.
c) Every other Thursday.
d) No. They all started with the same set of constellations, but as time passed
they varied them to arrive at very different sets.
e) No.
53. “Let’s play
Jeopardy
! For $100, the answer is: This person was the first to understand
the planetary motions using a physical theory that very adequately accounted for
terrestrial motions.”
a) Ptolemy (circa 100–175 CE).
b) Nicolaus Copernicus (1473–1543).
c) Isaac Newton (1642/3–1727).
d) Richard Feynman (1918–1988).
e) Stephen Hawking (1942–2018).
54. What is the difference between speed and velocity?
a) Velocity is the rate of change of speed.
b) There is no difference.
c) The difference is merely theoretical, not practical.
d) Both measure the rate of change of position with time: velocity specifies direction
as well as magnitude of the rate of change; speed specifies only magnitude.
e) Both measure the rate of change of position with time:
velocity specifies
acceleration as well as magnitude of the rate of change of position; speed specifies
only magnitude of rate of change of position.
55. Newton’s force law for gravitation for the magnitude of the force is
F
=
GM
1
M
2
r
2
.
a) The force is
ALWAYS ATTRACTIVE
and is felt only by the mass designated
M
2
. The distance between the centers of the two masses is 2
r
. This force law
strictly holds for
CUBICAL BODIES
.
b) The force is
USUALLY ATTRACTIVE
and is felt by both masses
M
1
and
M
2
. The distance between the centers of the two masses is
r
. Because
r
2
appears
in the denominator, the force law is an
INVERSE-CUBE LAW
. This force
law strictly holds only for
POINT MASSES
: a
TOTALLY DIFFERENT
FORCE LAW
applies to
SPHERICALLY SYMMETRIC BODIES
.
c) The force is
ALWAYS ATTRACTIVE
and is felt by both masses
M
1
and
M
2
.
The distance between the centers of the two masses is
r
. Because
r
2
appears in
the denominator, the force law is an
INVERSE-SQUARE LAW
. This force
law strictly holds only for
POINT MASSES
: a
TOTALLY DIFFERENT
FORCE LAW
applies to
SPHERICALLY SYMMETRIC BODIES
.
d) The force is
ALWAYS ATTRACTIVE
and is felt by both masses
M
1
and
M
2
.
The distance between the centers of the two masses is
r
.
Because
r
2
appears in the denominator, the force law is an
INVERSE-SQUARE LAW
.
This force law applies to all
POINT MASSES
and also to
SPHERICALLY
11
SYMMETRIC BODIES
. For nonspherically symmetric bodies, the force of
gravitation
VANISHES
.
e) The force is
ALWAYS ATTRACTIVE
and is felt by both masses
M
1
and
M
2
.
The
distance
between
the
centers
of
the
two
masses
is
r
.
Because
r
2
appears in the denominator,
the force law is an
INVERSE-
SQUARE LAW
. This force law applies to all
POINT MASSES
and also to
SPHERICALLY SYMMETRIC BODIES
outside of those bodies. For two
NONSPHERICALLY SYMMETRIC BODIES
, the force of gravitation
can be calculated by finding the force between each pair of small parts (one of
the pair from each of the two bodies) using the point-mass force law in its vector
formulation. The forces between all the pairs can be added up vectorially to get
the net force between the bodies.
56. The force of gravity between two bodies is proportional to the inverse square of
the distance between the centers of the two bodies either exactly or approximately
depending on nature of the bodies.
At 10 Earth radii, the Earth’s gravity force is
times its gravity force on its surface.
a) 1
/
10
b) 1
/
20
c) 20
d) 1
/
100
e) zero
57. Astronauts in orbit about the Earth are weightless because:
a) gravity vanishes in space.
b) gravity becomes repellent in space.
c) they are in free fall. They are perpetually falling away from the Earth.
d) they are in free fall.
They are perpetually falling toward the Earth, but keep
missing it.
e) they are in free fall. But they reach terminal speed due to air resistance and this
hides any effects of acceleration.
58. The formula for kinetic energy is
KE
=
1
2
mv
2
,
where
m
is mass and
v
is velocity. If velocity is doubled, kinetic energy changes by a
multiplicative factor of:
a) 2.
b) 4.
c) 1
/
2.
d) 1
/
4.
e) 1 (i.e., it is unchanged).
59. An everyday example of the 2nd law of thermodynamics is that:
a) heat flows from
HOT TO COLD BODIES
(at least at the macroscopic level)
provided there is no refrigeration process or absolute thermal isolation in effect.
b) heat cannot flow at all.
c) heat flows from
COLD TO HOT BODIES
(at least at the macroscopic level)
provided there is no refrigeration process or absolute thermal isolation in effect.
d) heat and coolness are both fluids.
e) heat is a fluid and coolness is relative absence of that fluid.
60. The mean lunar month is 29.53059 days. How many days are there in a year of 12
mean lunar months and approximately how many years on a lunisolar calendar before
you need to insert a 13th lunar month in a year (an intercalary month) in order to keep
12
the lunisolar calendar roughly consistent with the solar year (i.e., keep the months in
the seasons they are supposed to be in)?
a) 29.53059 days and every twelfth of a year.
b) 365.25 days and every 3 solar years.
c) 354.367 days and every
3 SOLAR YEARS
. Note you won’t get perfect
consistency with an every-3-solar-year insertion since your luni-solar calendar
will be short about
33 DAYS
after 3 solar years and a mean lunar month is only
29.53059 days.
d) 365.25 days and every 300 years.
e) 354.367 days and every
4 SOLAR YEARS
. Note you won’t get perfect
consistency with an every-4-solar-year insertion since your luni-solar calendar
will be short about
33 DAYS
after 4 solar years and a mean lunar month is only
29.53059 days
61. The Earth is at:
a) the geometrical center of the Moon’s
ELLIPTICAL
orbit.
b) the geometrical center of the Moon’s
ECLIPTICAL
orbit.
c) both foci (i.e., focuses) of the Moon’s elliptical orbit.
d) the perigee of the Moon’s orbit.
e) one of the foci (i.e., focuses) of the Moon’s elliptical orbit.
62. “Let’s play
Jeopardy
! For $100, the answer is: A lunar time period that is 27.321661
days long.”
What is the
, Alex?
a) lunar month
b) lunar sidereal month
c) lunar anomalistic month
d) lunar draconitic month
e) lunar pathetic month
63. The mean lunar month is 29.53059 days.
The
ANGULAR VELOCITY
of the
Moon relative to the Sun (
NOT
relative to the fixed stars) is
a) 12
.
19
◦
per day.
b) 13
.
18
◦
per day.
c) 29
.
531
◦
per day.
d) 360
◦
per day.
e) 12
.
19
◦
.
64. The standard lunar phases in time sequence are:
new moon,
waxing crescent,
, waxing gibbous, full, waning gibbous, 3rd quarter,
.
a) quarter lit; waning crescent
b) quarter full; morning crescent
c) 1st quarter; waning crescent
d) half lit; morning crescent
e) quartic; Mornington Crescent
65. At sunrise, you see the Moon in the eastern sky. It is:
a) partially eclipsed.
b) a waxing crescent.
c) a full moon.
d) a gibbous moon.
e) a waning crescent.
66. The Sun is setting; the Moon is 180
◦
away from the Sun on the sky. The Moon is:
a) setting too.
b) half-full.
c) a crescent.
d) being eclipsed.
e) rising and it is full.
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13
67. The Moon is a crescent—the horned Moon. Which way, in a rough sense, do the horns
point relative to the Sun?
a) Toward the Sun.
b) Away from the Sun.
c) They can have any
orientation depending on the time of year.
d) They can have any orientation
depending on the time of day.
e) Perpendicular to the line from the Moon
to the Sun.
68. Currently the tidal locking effect of the Moon on the Earth is increasing the length
of the Earth’s solar day by 1
.
70
×
10
−
3
seconds per century.
This rate is probably
not constant due to many perturbations. However, assuming it is constant, how long
until the solar day is 1 second longer than at present?
centuries.
a) 160
b) 454
c) 588
d) 674
e) 1012
69. For lunar eclipses (any of partial, total, annular, or penumbral) to occur, the Moon’s
orbital nodes do
NOT
have to be exactly on the Earth-Sun line: i.e., the line drawn
through the centers of Earth and Sun. This is because the light-emitting body, the
eclipsing body, and the eclipsed body all have finite sizes. The eclipse season is the
period during which nodes are sufficiently close to an exact nodal alignment that an
eclipse is possible.
The eclipse season for the Moon (for partial and total eclipses,
and
NOT
including penumbral eclipses) is about 24 days: 12 days before exact nodal
alignment and 12 days after. Why is there
NOT
a partial or total lunar eclipse during
every lunar eclipse season?
a) Lunar eclipses can only happen very near exact
FULL MOON
. If the Moon is
just past an eclipsable
FULLISH MOON
when a lunar eclipse season begins, it
will only get back to an eclipsable
FULLISH MOON
only somewhat less than
29.5 DAYS
later and so miss the eclipse season.
Consequently, there doesn’t
have to be either of a total or partial lunar eclipse in every lunar eclipse season
albeit usually there
IS
.
b) Lunar eclipses can only happen very near exact
NEW MOON
. If the Moon is
just past an eclipsable
NEWISH MOON
when a lunar eclipse season begins, it
will only get back to an eclipsable
NEWISH MOON
only somewhat less than
29.5 DAYS
later and so miss the eclipse season.
Consequently, there doesn’t
have to be either of a total or partial lunar eclipse in every lunar eclipse season,
and, in fact, there usually
IS NOT
.
c) Lunar eclipses can only happen very near exact
NEW MOON
. If the Moon is
just past an eclipsable
NEWISH MOON
when a lunar eclipse season begins, it
will only get back to an eclipsable
NEWISH MOON
only somewhat less than
22 DAYS
later and so miss the eclipse season. Consequently, there doesn’t have
to be either of a total or partial lunar eclipse in every lunar eclipse season albeit
usually there
IS
.
d) Lunar eclipses can only happen very exact
FULL MOON
. If the Moon is just
past an eclipsable
FULLISH MOON
when a lunar eclipse season begins, it will
only get back to an eclipsable
FULLISH MOON
only somewhat less than
29.5
DAYS
later and so miss the eclipse season. Consequently, there doesn’t have to
be a either of a total or partial lunar eclipse in every lunar eclipse season, and,
in fact, there usually
IS NOT
.
e) The Bos Domesticus effect in which the Sun sort of dodges the Earth happens
14
frequently near nodal alignment. This often prevents lunar eclipses.
70. In a partial eclipse of the Sun (as defined by
EARTHLINGS
), an Earthling in a
partially eclipsed part of the Earth sees the photosphere of the Sun with a bite out of
it. In a partial eclipse of the Moon (as defined by
EARTHLINGS
), how does the
Sun appear to a Moon-dweller (i.e., a Lunarian, Selenite, or even Subvolvan) on the
Earth-facing side of the Moon?
a) A disk with a bite out of it in
ALL
cases.
b) A disk with a bite out of it
OR
it could be totally eclipsed. It depends on the
location of the Selenite on the Moon.
c) The Sun is totally eclipsed
ALWAYS
.
d) It will look exactly like the Earth.
e) It would be nighttime on the Moon and the Selenite wouldn’t see the Sun in any
case.
71. Why is the corona visible to the unaided eye only during a total solar eclipse?
a) It is behind the photosphere of the Sun ordinarily, and thus cannot be seen
ordinarily.
b) The Moon’s shadow usually hides it.
c) Only during total eclipses is it compacted by magnetic fields.
d) It is too faint to be seen when any significant part of the photosphere of the Sun
is visible.
e) Only a total solar eclipse is long enough to let it stand out.
72. The Saros cycle is an approximate cycle of:
a)
ALL
eclipse phenomena with a period of 6585 days which equals
EXACTLY
18 Julian years plus 9 days.
b)
ALL
eclipse phenomena with a period of 6585.3213 days which equals about 18
Julian years plus 11 days.
c)
SOLAR
eclipse phenomena with a period of 6585.3213 days which equals about
18 Julian years plus 11 days.
d)
SOLAR
eclipse phenomena with a period of 6585 days which equals
EXACTLY
18 Julian years plus 9 days.
e)
SOLAR
eclipse phenomena with a period of 6585 days which equals
EXACTLY
18 Julian years plus 9 days that was discovered by Thales (c. 624–c. 546 BCE).