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Astronomy Assignment #09: The Celestial Sphere and the Apparent Motion of
the Stars and Sun
Your Name____Solutions______
1. What motion of the Earth produces the apparent motion of the stars around us?
The rotation of the Earth on its axis, once every 23h 56m 4.09s, creates the apparent diurnal motion of
the stars, Sun, Moon and planets. On a longer time scale, the apparent motion of the stars, Sun, Moon
and planets is caused by the combined revolution of the Earth around the Sun, once every 365.241 days,
and the actual motion of the celestial object. We observer celestial objects that have their own particular
motion from a moving platform – the Earth – which has two motions – a rotation on its axis and a
revolution around the Sun
2. How many degrees is 30 arc minutes? How many degrees is 10 arc seconds?
There are 60 arc minutes in one degree. So there 30 arc minutes is equivalent to 0.5 degrees.
There are 3600 arc seconds in one degree (6060), so to convert 10 arc seconds into degrees do the
following:
1 degree


10 arc seconds  10 arc seconds  
 0.00278 degrees
 3600 arc seconds 
3. How do the positions of the celestial equator, celestial poles, zenith, and meridian depend on the latitude
of the observer?
The position of the zenith is directly overhead (90 Altitude) for all observers – it does not depend on
the observer’s latitude.
The position of the meridian also does not depend on the observer’s latitude. The meridian is a
conceptual line that runs from the north point (0 Altitude, 0 Azimuth) through the zenith and to the
south point (0 Altitude, 180 Azimuth).
The position of the Celestial Poles does depend on the observer’s latitude. The Rule #1 is that the
altitude of the NCP above the north point equals the observer’s latitude. So an observer at 40 N
latitude would see the NCP 40 degrees above the north point. An observer at 40 S latitude would “see”
the NCP 40 degrees below the north point (-40 altitude). Obviously, the NCP cannot be seen if it is has
a negative altitude.
The SCP is opposite the NCP, so the altitude of the SCP is the opposite of the observer’s latitude. So an
observer at 40 N latitude would “see” the SCP 40 degrees below the south point (-40 altitude).
Obviously, the SCP cannot be seen if it is has a negative altitude. An observer at 40 S latitude would
see the SCP 40 degrees above the south point.
4. During a night, how do the stars move? What angle does their nightly path make with respect to the
horizon? How does it depend on latitude?
During the course of a night the stars appear to move westward, rising somewhere along the eastern
horizon (except for the circumpolar stars that never rise because they are always above the horizon) and
later setting at a corresponding azimuth on the western horizon.
Stars that rise in the vicinity of due East (i.e. near the celestial equator) and later set in the vicinity of
due West, rise and set at a slant angle (slanted toward the south) relative to the vertical equal to the
observer’s latitude. An observer at 40 N latitude would see stars rise along the eastern horizon slanted
40 toward the south and set along the western horizon slanted (from the south) at a 40 slant angle. For
an observer at the equator (0 latitude) the rising stars along the eastern horizon and the setting stars
along the western horizon would have 0 degree slant and, thus, rise and set vertically. An observer at
40 S latitude would see stars rise along the eastern horizon slanted 40 toward the north and set along
the western horizon slanted (from the north) at a 40 slant angle.
5. What reference point is a celestial object on when it is at its highest position above the horizon?
A celestial object is at its highest position above the horizon when it is transiting the meridian; that is
when it crosses the imaginary line (the meridian) that divides the sky into eastern and western halves.
6. Why do observers in the northern hemisphere see celestial
objects above the celestial equator for more than 12 hours?
When an observer in the northern hemisphere is looking north,
they will see that celestial objects are above the horizon for more
than 12 hours. This is because the center of their circular motion
is the NCP which is above the horizon. So the path of a star on
its ccw circular motion around the NCP is mostly above the
horizon with only a short path below the horizon. Thus, for
northern observers, the northern star spends most of its time
above the horizon because most of its path is above the horizon.
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7. For northern hemisphere observers, which celestial object would be above the horizon for the greatest
amount of time: one that is on the celestial equator, one that is 30° above the celestial equator, one that is
70° above the celestial equator, or one that is 40° below the celestial equator? Which one would be
above the horizon the greatest amount of time for southern hemisphere observers? Explain your answer.
Let’s assume that the northern observer is at latitude +45. This will allow us to make more quantitative
conclusions. Construct the simplified 2-D celestial Sphere diagram as shown below. Remember the
altitude of the NCP equals the observer’s latitude and the celestial equator is perpendicular to the line to
the NCP.
40 dec Zenith
70 dec
Celestial Equator
0 dec
NCP, +90 dec
45
45
-40 dec
45 Altitude
45 Altitude
Horizon
As the diagram above indicates, the stars will follow the solid black lines on their diurnal cycles. So we
can see that the 70 dec star is circumpolar and is above the horizon for 24 hours a day. The 40 dec star
is not circumpolar but is above the horizon more than 12 hours a day since most of its path is above the
horizon. The 0 dec star is above the horizon for 12 hours a day since half of its path is above the
horizon. Finally, the -40 dec star is above the horizon less than 12 hours a day since little of its path is
above the horizon.
If you were to construct a similar diagram for an observer at -45 latitude, you would find that the object
with the southernmost declination (e.g. -40 dec) had the longest path above the horizon and would thus
be above the horizon for the longest time.
8. How does the Sun move with respect to the stars during the day? ...during the year?
On a diurnal (daily) timescale the Sun does not appear to move with respect to the Stars. It appears to
move westward with the stars on a daily basis. However, it is actually moving eastward through the
stars but so slowly that it is imperceptible on a daily basis. Over a longer timescale, like a year, the Sun
appears to drift eastward through the stars on a path called the ecliptic once every 365.241 days.
9. Why does everyone have 12 hours of daylight on the equinoxes?
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On the equinoxes the Sun is on the celestial equator (0 dec) and acts like a stars on the celestial equator
for that day(s). The celestial equator intersects the horizon for all observers exactly due east and due
west and is half above and half below the observer’s horizon. Thus, while the Sun is on the celestial
equator its path will be half above and half below the celestial horizon and it will spend 12 hours aobe
the horizon for all observers.
10. Why is the length of daylight in the northern hemisphere so short on
December 21?
On Dec 21 the Sun is at the Winter Solstice on the Ecliptic at -23½ dec. It is
thus acting like a southern star. Southern stars rise S of E, reach a low max
altitude, rest S of W and spend less than 12 hours above the horizon because
most of their diurnal path is below the horizon (See Figure to right). Since
the Sun is acting like a southern star on that day it will share the
characteristics of the southern stars – having most of its diurnal path below
the horizon. So the winter Sun is only above the horizon for less than 12
hours.
11. When will the Sun be at its highest altitude in the year in Los Angeles or Seattle? How about Singapore
(on the Equator)? Why?
The Sun will be at its maximum altitude for all observes in the northern hemisphere when it has the most
northern declination on the Summer Solstice (+23½ dec) on June 21 or 22. Thus in Los Angles (34 N)
and Seattle (48 N) the Sun will be at its maximum altitude on noon of June 21 or 22.
In Singapore (1 N), which is really close to the equator, the Sun will reach its maximum altitude
when it is on the celestial equator on Mar 21 or 22 and Sep 21 or 22. This is because the celestial
equator is at the zenith for an observer on the Earth’s equator.
For southern observers, the Sun will highest in the sky when it is at the farthest southern declination on
the winter solstice (-23½ dec) on Dec 21 or 22.
12. On what date is the Sun above the horizon the shortest amount of time for the
Southern Hemisphere? Why?
For southern observers, the Sun will highest in the sky when it is at the farthest
southern declination on the winter solstice (-23½ dec) on Dec 21 or 22 since the
southern stars are at the zenith for observer’s below the Earth’s equator.
So for southern observers, the Sun will lowest in the sky when it is at the farthest
northern declination on the summer solstice (-23½ dec) on Dec 21 or 22 since
the southern stars are at the zenith for observer’s below the Earth’s equator.
13. At what two azimuths does the celestial equator intercept the horizon?
The celestial equator intersects the horizon at due east and due west for all observers.
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14. If a star's position at 10 pm is 110° azimuth and 40° altitude, will its azimuth be greater or less at 11 pm?
If the star is still east of the meridian at 11 pm, will its altitude be greater or less than it was at 10 pm?
First assume you are in the northern hemisphere. Explain your answer. Then
assume you are in the southern hemisphere and explain your answer.
As you can see, using a figure on the right from the UNL Rotating Sky
Explorer Module, a star at 110 azimuth and 40 altitude is still rising above
the eastern horizon. In another hour, at 11 p.m., that star will have a higher
altitude and a slightly larger azimuth.
If you are at a southern latitude, the image changes to the figure at the right
from the UNL Rotating Sky Explorer. Here you can see that an hour later the
star will have increased its altitude, but its azimuth will have decreased.
15. Why do astronomers prefer using right ascension and declination as opposed to using altitude ad
azimuth?
Altitude and azimuth are local coordinates, while RA and dec are celestial coordinates.
An object a the zenith (90 alt) in one location will not be at the zenith in another location. However,
the celestial coordinates are the same for an object at all locations. So, astronomers prefer the celestial
coordinates of RA and dec because, if they know how to transform those celestial coordinates to local
coordinates of alt and azi, they can find an object using its celestial coordinates at any location on the
Earth
16. What is the azimuth of any object when it crosses the meridian at any time of year in the southern sky?
The azimuth of any object when it crosses the meridian at any time of year in the southern sky is either
180 azimuth.
17. If a star has a RA of 5 hours and crosses the meridian at 10:45 pm, what is the RA of a star that crosses
the meridian at 1:00 am? Explain your answer.
The time difference (moving forward in time) between 10:45 p.m. and 1:00 a.m. is 2h 15m. So the 1:00
a.m. star must be 2h and 15m behind the 10:45 p.m. star. Thus the 1:00 a.m. star must have a RA of 7h
15 m.
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18. What is the Sun's altitude when it crosses the meridian in Syracuse, NY (+43 N) and its declination is
+23.5°?
Zenith
Sun at
+23½  dec
Celestial Equator
0 dec
NCP, +90 dec
23½
47
47Altitude
43Altitude
You should be able to construct the simplified 2-D celestial Sphere you see above. Remember the
altitude of the NCP equals the observer’s latitude and the celestial equator is perpendicular to the line to
the NCP. You can see from the diagram that the Sun’s altitude will be 43+23½=76½  when the Sun
is on the Summer Solstice.
19. What is the altitude of the NCP at Fairbanks, Alaska (lat. = 65° N)?
The Rule #1 is that the altitude of the NCP above the north point equals the observer’s latitude. So an
observer at 65 N latitude would see the NCP 65 degrees above the north point.
20. How do the positions of the equinoxes and solstices with respect to the horizon depend on the latitude?
Since the Spring and Fall Equinoxes are on the celestial equator, their position on an observer’s horizon
is always either due east of due west.
For northern observers, the Summer Solstice will intersect the horizon north of east and north of west
while the Winter Solstice intersects the horizon south of east and south of west.
21. What is the maximum altitude of the Sun on the vernal equinox for people on the equator? What is the
Sun's azimuth and right ascension at that time?
For observers on the Earth’s equator, the celestial equator runs through the zenith (Can you prove that
using a simplified 2-D celestial sphere diagram?). Thus the Sun will be at its highest altitude on the
celestial equator at 90 altitude on the Vernal Equinox. The RA of the Sun on the Vernal Equinox is 0 h
RA.
What is the azimuth of the Sun for an observer on the Earth’s equator on the Vernal Equinox is an
interesting question. Since the Sun is at the zenith for such an observer on that date, the azimuth of the
Sun is not well defined. If an object is exactly at the zenith, then it has no well defined azimuth – or it
can be all azimuths. Hmm…
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22. What will the Sun's declination be on the following dates: June 21, March 21, September 22, and
December 21?
Identifier
Spring Equinox
Summer Solstice
Fall Equinox
Winter Solstice
Date
Mar 21 or 22
Jun 21 or 22
Sep 21 or 22
Dec 21 or 22
Dec of Sun
0 dec
+23½  dec
0 dec
-23½  dec
23. If the Sun sets 10° away from due West on
October 20, what is the sunset azimuth?
To answer this question you first must know if the declination of the Sun on Oct 20 is north or south of the
celestial equator. The Sun is below the celestial equator on Oct 20 since the date is after the Fall Equinox
but before the Winter Solstice. So the Sun is acting like a southern star (i.e. it has a southern declination).
We know (or better know) that southern stars appear to rise south of east and set south of west. So if the
Sun on Oct 20 sets 10 away from due west it must be 10 to the south of west and therefore at an azimuth
of 260.
24. If the Sun rises 12° away from due East on April 19, what is the sunrise azimuth?
To answer this question you first must know if the declination of the Sun on Apr 19 is north or south of the
celestial equator. The Sun is above the celestial equator on Apr 19 since the date is after the Spring Equinox
but before the Summer Solstice. So the Sun is acting like a northern star (i.e. it has a northern declination).
We know (or better know) that northern stars appear to rise north of east and set north of west. So if the Sun
on Apr 19 rises 12 away from due east it must be 12 to the north of east and therefore at an azimuth of
78.
25. What causes precession?
Precession is the slow wobble of the Earth’s rotation axis over a 25,800 year period during which the tilt
of the axis is constant, but the orientation of the tilt migrates around the line
perpendicular to the Earth’s orbital plane. Precession is caused by the Sun’s
uneven gravitational pull on the slightly non-spherical Earth.
26. How does precession affect the positions of the stars?
Since the “addresses” of the stars in RA and dec are tied to the Earth (i.e. the
celestial poles are the extension of the Earth’s rotation pole and the celestial
equator lies directly above the Earth’s geographic equator), as the Earth
reorients its rotation axis due to precession all the stars address change
slightly from year to year. The effect is small enough for astronomers to
ignore it for 50 year intervals after which they recalculate all the stellar coordinates and issue new star
maps. We are current using stellar coordinates for the 2050 epoch indication that these stellar
coordinates are sufficiently accurate to be used until the year 2050 C.E.
27. Which star is the current pole star? Which star was the pole star 2,000 years ago? Which star will be the
pole star 8,000 years from now?
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Polaris is the current pole star. 2,000 years ago the pole star was Kochab in Ursa Minor (see page 42 in
AstronomyNotes). 8,000 years from now the pole stars will be Deneb in the constellation Cyngus.
28. Which is used for our clocks---sidereal day or solar day? Why?
Our clocks use the solar day (24 hours) because our natural cycle is driven by the day/night cycle. If our
clocks used the sidereal day then noon would no longer correspond to the time the Sun transits the
meridian (except on one day a year) and you would find that “noon” on the clock would once every 4
years occur at midnight!
29. Why is there a difference between the sidereal day and solar day?
The reason for the difference between the sidereal and solar day is that the
Earth both rotates on its axis and revolves about the Sun simultaneously.
The sidereal day is the time required for the Earth to rotate once on its axis
– 23h 56m 4.09s. The solar day is the time between successive transits of
the Sun across the merdian (non-to-noon) 24 h (on average). These two
“days” are not the same because as the Earth rotates through on sidereal
day it also moves along its orbit a small amount. Thus to realign with the
Sun the Earth must rotate just a little (about 4 min) more.
30. Mars rotates once every 24.623 hours (its sidereal day) and it orbits the Sun once every 686.98 solar
earth days. Show how to find out how long a solar day is on Mars.
To answer this question, first find out how many degrees Mars moves around its orbit in one sidereal
day. You can use a proportion to find this:
24.623 hr
x

686.98 days 360
1.026 days
x

686.98 days 360
 1.026 days 
  360  0.538
 x  
 686.98 days 
So during the course of one Martian sidereal day, Mars moves 0.538 around its orbit. Thus, after
completing one rotation on its axis (i.e. one Martian sidereal day), Mars must yet rotate an additional
0.538 to realign with the Sun. How long a time period is required for Mars to rotate 0.538? Use
another proportion to determine that time period.
x
0.538

24.623 hr
360
 0.538 
  24.623 hr  0.0368 hr  2.21 minutes
 x  
 
 360 
So Mars must rotate another 2.21 minutes to realign with the Sun after one sidereal day. Thus the
Martian Solar Day must be equal to 24.623 hours + 2.21 min = 24.660 hours or 24 h 39m 35s.
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Instructor Assigned Topic:
Sketch the apparent motion of the stars, following the example in class, looking in the four cardinal directions
(N, E, S & W) for the following four locations : The North Pole (90 N), Syracuse, NY (43 N), The Equator
(0 N), Punta Arenas, Chile (53 S)
The apparent motion of the stars from the North Pole (90 N)
●Stars complete one cycle in 23h 56m 4.09s
(The sidereal day.)
●The entire sky is circumpolar.
●Polaris (NCP really) is at the zenith.
●Stars are above the horizon for 24 hours a day.
●Stars complete one cycle in 23h 56m 4.09s
(The sidereal day).
●The Celestial Equator is coincident with the
horizon
●Stars are above the horizon for 24 hours a day.
●Stars complete one cycle in 23h 56m 4.09s
(The sidereal day).
●The Celestial Equator is coincident with the
horizon.
●The SCP is at the nadir.
●Stars are above the horizon for 24 hours a day.
●Stars are above the horizon for 24 hours a day
●Stars complete one cycle in 23h 56m 4.09s
(The sidereal day).
●The Celestial Equator is coincident with the
horizon.
●Stars are above the horizon for 24 hours a day.
Celestial Equator
Celestial Equator
Celestial Equator
Celestial Equator
North
East
South
West
The apparent motion of the stars from Syracuse, NY (43 N)
●Stars complete one cycle in 23h 56m
4.09s (The sidereal day.)
●Polaris (NCP really) is at 43altitude
●Stars are above the horizon for more
than 12 hours.
●Stars reach a high max. altitude.
.
●Stars complete one cycle in 23h 56m 4.09s (The
sidereal day).
●The Celestial Equator intersects the horizon due east
and rises at a slant angle of 0.
●Stars are above the horizon for 12hrs.
●Stars reach a high max. altitude.
●Stars complete one cycle in 23h 56m 4.09s
(The sidereal day).
●The SCP is -43 altitude (Below the horizon)
●Stars are above the horizon for less than
12hrs.
●Stars reach a low max. altitude.
●Stars complete one cycle in 23h 56m 4.09s
(The sidereal day).
●The Celestial Equator intersects the horizon
due east and rises at a slant angle of 43.
●Stars are above the horizon for 12hrs.
●Stars reach an intermediate max. altitude.
Circumpolar Boundary
Celestial
Equator
Polaris
(NCP really)
43
North
43
43
Celestial
Equator
43
43
East
South
43
43
West
The apparent motion of the stars from the Equator (0 N)
●Stars complete one cycle in 23h 56m
4.09s (The sidereal day.)
●Polaris (NCP really) is at 0altitude
(on the horizon).
●Stars are above the horizon for  12
hours.
●.
●Stars complete one cycle in 23h 56m 4.09s (The
sidereal day).
●The Celestial Equator intersects the horizon due east
and rises at a slant angle of 43.
●Stars are above the horizon for 12hrs.
●Stars reach an intermediate max. altitude.
●Stars complete one cycle in 23h 56m
4.09s (The sidereal day).
●The SCP is 0 altitude (On the
horizon)
●Stars are above the horizon for
12hrs.
There are no circumpolar stars
43
altitude
Celestial
Equator
Polaris (NCP really)
0 altitude
North
●Stars complete one cycle in 23h 56m 4.09s
(The sidereal day).
●The Celestial Equator intersects the horizon
due east and rises at a slant angle of 43.
●Stars are above the horizon for 12hrs.
●Stars reach an intermediate max. altitude.
Celestial
Equator
SCP
East
South
West
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The apparent motion of the stars from Punta Arenas, Chile (53 S)
●Stars complete one cycle in 23h 56m 4.09s (The
sidereal day.)
●Polaris (NCP really) is at -53altitude (Below the
horizon).
●Stars are above the horizon for less than 12 hours.
●.Stars reach only a low maximum altitude.
●Stars complete one cycle in 23h 56m 4.09s (The
sidereal day).
●The Celestial Equator intersects the horizon due east
and rises at a slant angle of 43.
●Stars are above the horizon for 12hrs.
●Stars reach an intermediate max. altitude.
●Stars complete one cycle in 23h 56m
4.09s (The sidereal day).
●The SCP is 0 altitude (On the
horizon)
●Stars are above the horizon for
12hrs.
●Stars complete one cycle in 23h 56m 4.09s
(The sidereal day).
●The Celestial Equator intersects the horizon
due east and rises at a slant angle of 43.
●Stars are above the horizon for 12hrs.
●Stars reach an intermediate max. altitude.
Circumpolar Boundary
SCP
Altitude = 53
Celestial
Equator
Celestial
Equator
53
North
53
East
53
53
53
South
53
53
53
West
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