Download Solutions to the 1 st Astronomy Exam

Introductory Astronomy SCI-103 Spring 2016
Exam #1
Name____Solutions____
Class Meeting Time________________
Grade Summary
HW#1 _________________
HW #2 ________________
Exam #1 _______________
Grade so far….__________
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1. On the image of the celestial sphere below, assuming that the observer is located in Bangkok, Thailand (14 N), draw
or label the following: I won’t draw all the answers to this question but I’ll describe the answer.
a. The position of the zenith: the point in the celestial sphere directly above the observer.
b. The meridian: the line on the celestial sphere that starts at the north point on the horixon, progresses up
the celestial sphere to the zenith and then down the celestial sphere to the south point on the horizon
effectively dividing the local observer’s sky into western and eastern halves.
c. The horizon: the line where the horizon meets the celestial sphere forming a circle around the observer
where the celestial sphere appears to sir on the horizon.
d. Polaris (the NCP really)
e. The SCP
f. The Celestial Equator: intersects the horizon due east and due west with a max altitude = 76◦
NCP
g. The path of a northern star
W
h. The path of a southern star
14◦
S
N
SCP
E
The portion of the celestial sphere appears to the right.
Answer the three questions that follow.
2. What do the dashed lines represent?
.Constellation boundaries in lines of RA and Dec.
3. The solid lines that connect the stars (e.g. Alkaid to
Mizar to Alioth etc.) identify the Big Dipper. Is it true
that the Big Dipper is a constellation? Explain your
answer.
The Big Dipper is an asterism, not a constellation. A constellation is a well-defined internationally-agreed-upon
area of the sky with boundaries of RA and Dec. There are 88 official constellations that cover the sky
completely. Every piece of sky is part of a constellation. An asterism is an easily recognized pattern of stars,
There are hundreds of asterisms many are culturally specific
4. What do the other solid lines on the map represent?
The solid lines represent lines off Right Ascension and Declination.
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Draw a letter during class
5. You are observing from the location on the world map below designated as A? . Draw the apparent motion of
the stars looking North, East and South in the boxes provided below. Include. all relevant time and quantitative
information.
You’ll want to review HW #1 and #2 . I’ll list the points I was looking in each direction below, but I’m not drawing in
the pictures. You have similar sketches in the HW #1 and #2 solutions and from class notes.
Points looking North
 Stars appear to circle counterclockwise around Polaris (NCP
60
really) once every 23h 56m 4.09s
 Polaris (NCP really) has an
altitude equal to the observer’s
50
latitude above (or below if the
latitude is in the southern
hemisphere).
40
 For observers at northern
Altitude
latitudes the circumpolar region
where stars never set is bounded
30
○
by declination (90 -Obs. Lat.)
and all stars looking north are
20
above the horizon for more than
12 hours.
 For observers at southern
10
latitudes the northernmost visible
star has a declination of
(90○+Obs. Lat.) and all stars
looking north are above the
horizon for less than 12 hours.
North
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Points looking East
 The Celestial Equator intersects the
horizon due East and rises slanted
toward the south (right) at an angle
relative to the vertical equal to the
observer’s latitude.
 Stars appear to follow paths parallel to
the celestial equator rising along the
eastern horizon and slanting to the
south.
 All stars looking east are above the
horizon for about 12 hours.
Points looking South
 Stars appear to circle clockwise around
the SCP once every 23h 56m 4.09s
 The SCP has an altitude equal to the
observer’s latitude below (or above if the
latitude is in the southern hemisphere).
 For observers at southern latitudes the
circumpolar region where stars never set
is bounded by declination (-90○-Obs.
Lat.) and all stars looking south are above
the horizon for more than 12 hours.
 For observers at northern latitudes the
southernmost visible star has a
declination of (Obs. Lat.- 90○) and all
stars looking south are above the horizon
for less than 12 hours.
East
You are observing from Melbourne, Australia (38S). Answer
the following questions. You may use the figure below to
help answer the questions. It is really easier to answer this
question after you complete the drawing as shown and label
all the angles.
6. What would be the maximum altitude of the Celestial
Equator? What would its azimuth be at max. altitude?
South
Zenith
Circumpolar
Region
The northernmost visible star has declination of 56.
So a star at 70 would never be visible.
0◦ Dec
-50◦ Dec
◦
-90 Dec
SCP
The maximum altitude of the Celestial Equator is 56
degrees above the northern horizon.
7. Would a star at declination +70 be visible? If yes
what would be its maximum altitude?
CE
56◦
S
-56◦ Dec
34◦
34◦
56◦
56◦ Dec
-34◦ Altitude
70◦ Dec
NCP
90◦ Dec
8. Would a star at declination -50 be visible? If yes
what would be its maximum altitude?
This star at -50 dec would indeed be visible – not circumpolar but visible. It would have a maximum altitude
of 74up from the southern horizon.
9. How long would a star at 0 declination be above the horizon?
Trick question: A star at 0 dec is on the celestial equator and all stars on the celestial equator are abover the
horizon for 12 hours for all observers.
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10. In about 300 B.C.E. Aristotle wrote “Of these fixed poles the one is always visible, being above our heads in the
northern region of the sky, and is called the Arctic Pole; the other is always hidden beneath the earth to the
south and is called the Antarctic Pole.” Why did Aristotle not mention that Polaris is a fairly bright star located
near the Arctic Pole? Answer in a couple of sentences.
Aristotle did not mention that Polaris is a fairly bright star located near the Arctic Pole because at that time there
was no bright star near the north celestial pole and Polaris was far from the NCP. The effect of precession – the
slow reorientation of the Earth’s rotation axis over a 26,000 year period – has moved all the stars on the
celestial sphere over the past 2,000 years. Now, Polaris is nearest the NCP, but 2,300 years ago there was no
bright star near the NCP.
11. The date is April 15 and the Sun is rising 27 away from due east. What is the Sun’s rising azimuth?
The tricky part of this question is knowing whether the Sun rose 27 away from due east towards the north or
towards the south? To answer which side of due East the Sun rose you compare the date to the date of the
equinoxes. If the date is after Mar 22 and before Sep 22 then the Sun is north of the celestial equator and
therefore will rise and set north of east and north of west. On the other hand and if the date is after sep 22 and
before mar 22, the Sun is south of the celestial equator and therefore will rise south of east and south of west.
Since the date in this problem is Apr 15, after Ma2 22 and before Sep 22, the Sun in in the northern celestial
hemisphere and will rise north of east. So the azimuth of Sun rise must be 90○-27○ = 63○.
12. Which of the following locations is farthest from the north celestial pole? To be farthest from the NCP you
must have a declination farthest from +90○. The RA is irrelevant.
d. RA = 18 h
dec = +30
a. RA = 0 h
dec = -60
b. RA = 6 h
dec = -30
e. RA = 24 h
dec = +60
c. RA = 12 h
dec = +0
13. Which of the following coordinates in your local horizon system (in Syracuse, NY at 43 N) is closest to the
horizon? To be closest to the horizon you must have an altitude that is closest to0○. The azimuth is irrelevant.
a. Azi. = 0,
Alt. -30
d. Azi. = 270, Alt. =90
b. Azi. = 90, Alt. =30
e. Azi. = 360, Alt. = 15
c. Azi. = 180, Alt. = 60
14. Define the solar day and the sidereal day and explain why they
are different. Answer in three sentences
The excerpt below is from AstronomyNotes: Imagine that at noon
there is a huge arrow that is pointing at the Sun and a star directly in line
behind the Sun. The observer on the Earth sees the Sun at its highest
point above the horizon: on the arc going through the north-zenith-south
points, which is called the meridian. The observer is also experiencing
local noon. If the Sun were not there, the observer would also see the
star on the meridian.
Now as time goes on, the Earth moves in its orbit and it rotates from
west to east (both motions are counterclockwise if viewed from above
the north pole). One sidereal period later (23 hours 56 minutes) or one
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true rotation period later, the arrow is again pointing toward the star. The observer on the Earth sees the star on the
meridian. But the arrow is not pointing at the Sun! In fact the Earth needs to rotate a little more to get the arrow lined
up with the Sun. The observer on the Earth sees the Sun a little bit east of the meridian. Four minutes later or one
degree of further rotation aligns the arrow and Sun and you have one solar day (24 hours) since the last time the Sun
was on the meridian. The geometry of the situation also shows that the Earth moves about 1 degree in its orbit during
one sidereal day. That night the Earth observer will see certain stars visible like those in Taurus, for example. (Notice
that the Earth's rotation axis is still pointed toward Polaris.) A half of a year later Taurus will not be visible but those
stars in Scorpius will be visible. (Again, notice that the Earth's rotation axis is still pointed toward Polaris.) The extra
angle any planet must rotate on its axis to get the Sun back to the meridian equals the angle amount the planet moved
in its orbit in one sidereal day.
The Sun is located in the constellation of Sagittarius on as indicated on the map below.
Ophiuchus

East
South
15. In what constellation will the Sun be found 6 hours after the time pictured above?
a. Aquarius
d. Ophiuchus
b. Capricornus
e. Scorpius
f. Libra
c. Sagittarius
West
g. Virgo
The apparent diurnal motion of the Sun is westward with the stars. The Sun, therefore, appears in front of the
same stars all day long.
16. At what two azimuths does the celestial equator intercept the horizon? Answer in a sentence.
The celestial equator intersects the horizon due east and due west at azimuths 90 and 270 degrees for all
observers.
17. How does the Sun move with respect to the stars during the day? ...during the year? Answer in a few sentences.
The apparent diurnal motion of the Sun is westward with the stars. The Sun, therefore, appears in front of the
same stars all day long. However, the Sun actually falls slightly behind the stars each day by about 4 minutes (3
min and 55.91sec) Thus over an annual timescale, the Sun appears to move eastward through the stars
following a path called the ecliptic completing one cycle every 365.241 days. The ecliptic is a great circle
inclined to the celestial equator by 23½ degrees. The ecliptic intersects the celestial equator at 0 hr RA and 12
hr RA and is farthest north of the celestial equator at 6 hr RA and farthest south of the celestial equator at 18 hr
RA. The four mentioned points are the spring equinox (≈Mar 22), fall equinox(≈Sep 22), the summer solstice
(≈Jun 22) and the Winter solstice (≈Dec 22).
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18. What angle does the stars nightly path make with respect to the eastern and western horizons?
sentence.
Answer in a
The angle the stars nightly path makes with respect to a vertical line along the eastern and western horizons is
equal to the observer’s latitude. So the angle the stars nightly path makes with respect to the eastern and
western horizons is the compliment of the observer’s latitude or (90○ – Obs. Lat.).
19. How does maximum altitude of the celestial equator, depend on the latitude of the observer? Answer in a few
sentences.
The celestial equator appears hinged to the horizon at exactly due east and due west for all observers. The
altitude of the celestial equator depends on the observers latitude. For observers near the equator the celestial
equator passes nearly through the zenith – a very high altitude. For observers near either geographic pole the
celestial equator has a very low maximum altitude in the south for northern observers and in the north for
southern observers. The rule is that the maximum altitude of the celestial equator when it crosses the meridian
is given by (90○-Obs. Lat).
20. Why do astronomers prefer using right ascension and declination as opposed to using altitude and azimuth?
Answer on a sentence or two.
Astronomers prefer using right ascension and declination as opposed to using altitude and azimuth because RA
and Dec are independent of the observer’s location. The position of an object in the sky as seen by two widely
separated observers will have different altitude and azimuth (they are local coordinates), but the RA and Dec
will be the same for each observer (celestial coordinates).
21. Estimate the approximate altitude and azimuth of the star shown in the figure to the right.
Azimuth = _between 180○ (due south) and 270○ (due west), I would guess the azimuth of the star to be about
225○.
Altitude = The altitude is closer to the zenith than the horizon so the altitude must be between 45○ and 90○. I
would guess the azimuth is about 60○.
W
S
N
E
22. What reference point is a celestial object on when it is at its highest position above the horizon? Answer in a
sentence.
The reference point of a celestial object when it is at its highest position above the horizon is the meridian
because all astronomical objects reach their highest altitude when they transit the meridian.
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23. The star Mintaka in the constellation of Orion has a declination of nearly 0○
dec.
1. At what azimuth or direction on the horizon would this star reach its
maximum altitude as seen by an observer in the northern hemisphere?
Since Mintaka has a dec of nearly 0○ it must sit nearly on the celestial
equator. As seen by a northern observer, the celestial equator
reaches its maximum altitude when it transits the meridian above the
southern horizon. So Mintaka would reach its maximum altitude
looking due south at 180○ azimuth.
2. At what azimuth or direction on the horizon would this star set as seen
by an observer in the northern hemisphere?
Since the celestial equator intersects the horizon due east (90○ azimuth) and due west (270○ azimuth) for
all observers, Mintaka would set nearly due west at nearly 270○ azimuth.
24. Assume this image was taken in the Earth’s
southern hemisphere. What direction is the
camera looking at the time the image was
taken? Be specific.
An observer in the southern hemisphere would
be able to see the SCP above the southern
horizon and the NCP would appear below the
northern horizon. In the photograph it must be
the NCP that is below the horizon on the right
portion of the image. So the center of the
photo is in the NW direction and stars would
be setting.
25. Assume this image was taken by a photographer
in Arizona. What direction is the camera looking
at the time the image was taken? Be specific.
This image is facing either east or west as
evidenced by the nearly straight star trails. Since
we know that the image was taken in the northern
hemisphere, star trails rising in the east should
slant to the right towards the south and star trails
setting in the west should slant to the left coming
down from the south. Since these star trails slant
from the left we must be looking west. This is
supported by the slight curve of star trails as seen
in the upper right corner of the image. The NCP
must be further off the image to the upper right.
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Astronomy Formula and Constants Sheet for Exams
Conversions
Formulas
A
L

2D
360 
Main Sequence Lifetime t 
M
1010 yr
L
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