ASTR 3130, Majewski [FALL 2012]. Lecture Notes
ASTR 3130 (Majewski) Lecture Notes
THE CELESTIAL SPHERE: BASIC PRINCIPLES
REFERENCE: Birney Chapters 1, 2 and 4. Kaler Chapter 1, 2, 6.
Woodcut image displayed in Flammarion's `L'Atmosphere:
Meteorologie Populaire (Paris, 1888)'. The actual
origin of the woodcut is unknown. Courtesy University of
Oklahoma History of Science Collections.
In this lecture I give a very basic review of the workings of the
heavens: the celestial sphere as a concept, the consequences
of the Earth's rotation on its axis and the Earth's revolution around
the Sun, and how these affects vary for different observers on the Earth.
More advanced topics will be discussed later.
This material is basic to all the lab work, but also to the overall general
understanding of the workings of the night sky you must have when you
leave this class. Many students will have seen much of this material before,
but if any of these concepts are unfamiliar or confusing, please make sure to do
the readings.
Everything appears to move around us as though we were at the center of the
Universe.
The Celestial Sphere is an imaginary concept that is a useful tool
for understanding some workings of the sky:
 Sphere of infinite radius with Earth at center.
 Stars on "surface of sphere".
Of course, this is NOT the way things really are (though many
ancient cultures actually thought so).
Some concepts of spheres that are useful to have in thinking about the
celestial sphere:
 A great circle on a sphere is any circle that
shares the same center point as the sphere itself.
 A great circle exactly divides the surface area of a sphere in half.
 Every great circle intersects every other, different great circle at
antipodal (i.e. exactly opposite) points and this divides each of the great circles into
pairs of arcs of identical length.
 A small circle is any circle drawn on the sphere
that does not share the same center as the sphere.
 Any two points on the surface of a sphere, if not exactly
opposite one another, define a unique great circle going through them. A
line drawn along the great circle is actually the shortest distance between
the two points.
It often seems surprising to people used to looking at flat
maps that the shortest distance between two points takes them over
unexpected places on air flights. The loxodrome is a
straight line constant heading on the Mercator Projection and
the great circle, although appearing longer than the
loxodrome, is actually the shortest route between New York and
London.
 Since there are an infinite number of great circles on a sphere,
there are an infinite number of great circles possible.
 Any coordinate system we would define on a sphere has a basis on
selecting some primary great circle  obviously we have an infinite
variety of coordinate systems we can select based on the infinite
number of possible great circles  but in reality only a few of them
are sensible to adopt.
 For any great circle we can define a pole as a line
perpendicular to the great circle's plane and through the center of the
sphere. Every great circle has two poles.
 In fact, even after we select a primary great circle to define
a coordinate system, we have not yet fully defined a unique coordinate system
since there is no uniqueness of rotation about the pole until we select
a zeropoint along the primary great circle.
That is, a given great circle defines a family of coordinate systems
with the same pole and equator, and these are distinguished by selection of
a secondary great circle.
 A secondary great circle is a great circle that goes through the
poles of a primary great circle.
In astronomy, we have four principal coordinate systems defined for the celestial sphere,
each defined by a particular pair of primary and secondary great circles:
coordinate system

principal great circle

"prime meridian" 
secondary great circle

coordinates

horizon or observer's

observer's horizon

northsouth meridian

altitude, azimuth

equatorial or celestial

projection of Earth's equator

through head of Aries  vernal equinox

right ascension,
declination, δ

ecliptic

plane of Earth's revolution

through head of Aries  vernal equinox

ecliptic longitude, λ
ecliptic latitude, β

Galactic

plane of the Milky Way

through Galactic center

Galactic longitude l
Galactic latitude b

Note that for both the equatorial and ecliptic coordinate systems, the
definition of the "prime meridian" or prime secondary great circle is
given by one of the points where the equatorial and ecliptic great circles
cross in the sky  known as the "vernal equinox".
This is also the position of the Sun in the sky on the first day of spring.
Note that we can transform from one coordinate system to any other coordinate system by use of
Euler angle rotations about axes in a Cartesian reference frame.
For example, if we convert each spherical coordinate system defined above to
its corresponding Cartesian coordinates (e.g., with the z axis defined by
the pole of the spherical system, etc.), then the coordinates
x_{1}, y_{1}, z_{1} in system 1 can be converted
to the coordinates x_{2}, y_{2}, z_{2}
in the second system using rotations:
 : rotation about z axis
 β: rotation about y axis
 γ: rotation about x axis
This transformation can be done with matrix math with Eulerian matrix operators
that look like this:
A. REVIEW OF THE OBSERVER'S REFERENCE FRAME
Recall from previous discussions:
We define your:
 apparent horizon as the horizon obstructed by trees, mountains, etc.
 true horizon as the horizon you would have in the absence of obstructions
Note that in these lectures, when the word "horizon" is used
we will always mean the horizon that pertains in the case that no
landscape exists (i.e., no trees, buildings, mountains, etc.).
This ideal or true horizon is
typically rarely seen (where can you see the true horizon?). Your
apparent horizon is defined by the ground obstructions
around you.
The true horizon is the primary great circle of the observer's coordinate system,
and divides sky in half (half of the sky is above the true horizon).
 zenith as that point directly above you, defined as the
pole of the horizon.
 nadir as that point directly below you  opposite zenith.
 North point as that point on the horizon in direction of geographic north.
 vertical circle a great circle that contains the zenith and is perpendicular
to the horizon.
 meridian as that vertical circle that contains the zenith, nadir, the
north celestial pole, and the due north and south points on the
horizon. Your meridian, which is a secondary great circle, also
divides your sky in half.
It is sometimes useful to refer to objects in the sky with reference to
your horizon, zenith and compass points. Indeed, two angles commonly used
for uniquely specifying the placement of an object in your sky at some
moment of time are the angles of:
 altitude, i.e. how far above the horizon to look
for an object, from zero degrees at the horizon to 90 degrees at
the zenith.
 azimuth, i.e., the direction towards the horizon
one must face to look up from the horizon to the object.
In this system we start from 0 degrees for the north meridian,
then 90 degrees for due east, etc.
One often finds this system of coordinates used on instruments with
mainly terrestrial applications, such as military spotting scopes and (at
least in terms of azimuth) airport runway designations.
Because astronomers prefer to observe objects as close to the zenith as possible,
they tend to work with zenith distance = (90^{o}  altitude), rather
than altitude.
While the observer's system is obvious and easy to use for these types of
terrestrial applications, there are drawbacks to its use in astronomical
contexts:
 Objects in sky constantly moving  altitude and azimuth
changing.
 Celestial motions generally not simple in, consistent with,
this reference system.
 Each observer has own reference system (note the frequent use
of word "your" above!).
THUS: This type of reference system can never be universal,
i.e., specifying the position of an object complicated  altitude and
azimuth not constant for an object over all times for all observers.
B. THE EQUATORIAL REFERENCE SYSTEM
Even the ancients realized that a less directly personal, more
universal system of the heavens existed.
Because much of the apparent motion of objects in the sky can be
attributed to the daily rotation of the Earth, it is more useful to
contemplate a sky reference system that relates to the entirety of the
Earth (not just your specific view of the sky).
Because the relative positions of stars and constellations appear to be
fixed, many ancient cultures pictured them to be firmly fixed or painted
onto a crystal sphere that surrounded and rotated about the Earth.
Of course, stars vary greatly in their distance. Yet
for purposes of understanding the sky, the concept of a Celestial
Sphere  a fictitious crystal sphere of immense radius onto which
stars are fixed  has pedagogical usefulness.
BUT REMEMBER:
 The Celestial Sphere is only a learning device.
 It is not the Sphere that rotates about the Earth, it is the
Earth that is spinning "within the Celestial Sphere".
 The center of the Celestial Sphere is the center of the
Earth.
 Even though we now know that this ancient model of a
stationary Earth surrounded by a rotating sphere of stars is
incorrect, we still use this model because it is a convenient way
to predict the motions of the stars and planets relative to a
location on the Earth.
There are a number of reference points shared by the Celestial Sphere and
the sphere that is the Earth:
 The center of the Celestial Sphere is the center of the
Earth.
 The rotational axis of the Earth defines the
apparent rotational axis of the Celestial Sphere.
Consequently, the poles of the Earth define the Celestial
Poles of the Celestial Sphere.
 The equatorial plane of the Earth is perpendicular to the
rotational axis of the Earth. Therefore the plane of the
Celestial Equator, which is perpendicular to the Celestial
Poles, must be coincident with the Earth's equatorial plane.
THOUGHT PROBLEM TIME OUT:
 If you are standing at the North Pole of the Earth, what part
of
the celestial sphere do you see at your zenith ?
 If you are standing on the Earth's equator, what part of
the celestial sphere do you see at your zenith ?
 Do you see a trend here?
We have now seen how the North Celestial Pole and the South Celestial Pole
are "extensions" of the Earth's North Pole and South Pole, and also how
the
Earth's equator can be extended out into the sky to define a Celestial
Equator.
We can continue this extrapolation of earth reference points into the sky,
and
define a Celestial Coordinate System that is an extension of
the Earth's
coordinate system of longitude and latitude. The following reviews that
system:
Recall:
 Earth's North and South Poles at latitudes 90^{o}
and +90^{o}.
 Earth's equator is latitude 0^{o}.
 Longitude reckoned from 0 to 360^{o} starting at
Greenwich, England (by convention).
 Note alternative longitude angle unit is the time zone
, for which the Earth has 24
zones, each nominally 15^{o} wide (ignoring
geopolitical variations).
Astronomers continue this longitude/latitude analogy in the definition of
the Celestial or Equatorial Coordinate System.
Imagine the lines of longitude and latitude on the "globe" lifted off and
onto the Celestial Sphere; we then have analogous lines on the Celestial
Sphere:
 The equivalent of the "latitude" small circles on Earth are the
"declination" small circles on the celestial sphere.
 The declination coordinate is symbolized by the Greek letter
delta, δ
Analogous to latitude, the celestial poles
have declination +/ 90^{o}, and the celestial equator has declination of
0^{o}.
 The equivalent of Earth's "lines of longitude" are the "lines of right
ascension" on the celestial sphere.
 BUT unlike lines of longitude, whose positions are fixed to the spinning Earth,
the lines of right ascension are fixed onto the heavens.
 The right ascension coordinate is symbolized by the Greek letter
alpha.
 The tradition in astronomy (though not always followed) is that
Right Ascension angles are reckoned in units of 0 to 24 hours ,
mimicking the "time zone" idea for longitude.
However, one does sometimes see the right ascension in units of degrees.
EARTH (GLOBE) SYSTEM 
CELESTIAL SPHERE SYSTEM 
North, South Poles 
North, South Celestial Poles 
Equator 
Celestial Equator 
Latitude (90^{o}
to 0^{o} to +90^{o}) 
Declination (90^{o}
to 0^{o} to +90^{o}) 
Longitude (0 to 360^{o},
or 24 time zones) 
Right Ascension (0 to 24 hours) 
Advantages of Celestial Coordinate system:
 Celestial Coordinate System is fixed with respect to the
stars.
 Coordinates of most celestial objects remain approximately
constant over time and for all observers.
 Easy to specify unique and universally understandable
positions for objects.
 System automatically accounts for the rotation of Earth  it
ignores it...
 ... Unfortunately, the observer cannot ignore the fact that she
sits on a rotating platform. We will return to this problem shortly.
Telescope Mountings
 But this does raise the point that we can define telescope mountings
that correspond to each type of coordinate system defined above.
 A telescope mounting based on the horizon system is the socalled
altitudeazimuth mounting.
 Because of the "drawbacks" of the horizon coordinate system
mentioned above, the altaz mounting is more complicated because we
must always drive the telescope continuously in both axes to follow
objects in the sky as the Earth turns, and the rate of
motion is constantly changing  requires a computer.
Image from http://astronomy.trilobytes.com.au/scope.htm.
 We can design a much simpler telescope mounting which requires tracking in
only the hour angle direction (around the polar axis) and at only one
rate (one rotation per about 24 hours).
Socalled "German Equatorial Mounting" of a
telescope has one axis pointed toward the celestial pole and another
towards the celestial equator. One only need turn the
telescope along polar axis to counteract the Earth's rotation.
Image from http://wwwistp.gsfc.nasa.gov/stargaze/Ssky.htm.
C. MORE ON RIGHT ASCENSION AND DECLINATION
We assign objects on the celestial sphere coordinates of right ascension.
In both cases it is traditional to use sexagesimal units:
 For declination:
 circle is 360 degrees
 1 degrees = 60 minutes of arc = 60 arcminutes = 60'
 1 arcmin = 60 seconds of arc = 60 arcseconds = 60"
 Note that since π radians = 180^{o}, then
we find that there are 206,265 arsceconds/radian. THIS IS A USEFUL NUMBER
TO MEMORIZE  IT COMES UP ALL THE TIME IN ASTRONOMY.
 The declination of an object, e.g., the star Vega, can be written in
ever increasingly precise coordinates as:
 38^{o}
 38^{o} 46' (more precisely)
 38^{o} 46' 24" (even more precisely)
 Note that the latter may also be written as 38^{o} 46.4',
since 0.4' = 24".
 Another way to write this is 38:46:24, but note that
you need to know from context that this means declination,
since we can write right ascension in same way.
 For right ascension:
 circle is 24 hours
 1 hour = 60 minutes of time = 60^{m}
 1 minute of time = 60 seconds of time = 60^{s}
 The right ascension of an object, e.g., the star Vega, can be written in
ever increasingly precise coordinates as:
 18^{h}
 18^{h} 36^{m} (more precisely)
 18^{h} 36^{m} 36^{s}
(even more precisely)
 18^{h} 36^{m} 36.1^{s}
(even more precisely)
 Note that the latter may also be written as
18^{h} 36.6^{m},
since 0.6^{m} = 36^{s}.
 Another way to write this is 18:36:36.1, but note that
you need to know from context that this means right ascension,
since can write declination in same way.
 BE CLEAR: DON'T CONFUSE ANGLE UNITS WITH UNITS OF TIME!!
The actual angular size of a minute of
time is almost never the same as the angular size of
an arcminute , and the actual angular size of a
second of time is not the same as the angular size
of an arcsecond .
 The size of RA angles as measured in units of time
are actually variable in size depending on the
declination.
 On the Celestial Equator, 1 hour of RA corresponds to
360^{o}/24=15^{o}.
A figure showing the convergence of lines of
RA at the poles, and the narrowing of the definition
of an hour of right ascension with declination.
From http://www.astro.princeton.edu/PBOOK/strategy/mapgrid.gif.
(NOTE THAT, AS SHOWN, THIS FIGURE IS NOT "SKY RIGHT"  if viewing the
celestial sphere from the outside, then the RA lines should increase to the right
(i.e., to the East)).
 To account for the fact that lines of RA converge at
the poles (see above figure),
and therefore narrow with declination (DEC),
the actual size of RA time units varies as
follows:
 1 hour = 15^{o} cos(δ)
 1 minute of time = 15 cos(δ) arcmin
 1 second of time = 15 cos(δ) arcsec
 Note, since in general a minute of time is larger than an arcminute
and a second of time is larger than an arcsecond, it is traditional
to write the coordinates of an object with one more significant digit in
right ascension than in declination. This keeps the precision in
each coordinate about the same.
For example, Vega:
 If you write 38:46 for Dec, write RA=18:36.6
 If you write 38:46:24 for Dec, write RA=18:36:36.1
Judging angular sizes in the sky:
Angles subtended by a quarter at distance D
 1 degree @ D ~ 56 in
 1 arcmin @ D ~ 270 feet
 1 arcsec @ D ~ 3 miles
Bowl of "Big Dipper" ~ 5 x 10 degrees in angular size
"Handy" measuring scale:
1 degree = width index finger @ arm's length
10 degrees = closed fist @ arm's length
20 degrees = distance thumb to little finger on outstretched hand @ arm's length
Example: If diameter of Sun or Moon is 0.5 degree, should be able to cover wit h
index finger @ arm's length. Try it!
Human eye can resolve angles as small as 12 arcmin in daytime, (recall!) progressively
degrades with decreased lighting.
Second "star" from end of "Big Dipper" handle is two stars (Arabic
"Alcor and Mizar"  i.e. "horse and rider"), separated by 11'. Known to Native
American and other cultures as a test of visual acuity.
The Hubble Space Telescope can resolve
angles as small as 0.1" ; i.e. can resolve a quarter @ distance of 30
miles
D. HOW DO THE OBSERVER'S REFERENCE SYSTEM AND THE CELESTIAL SPHERE RELATE?
For most observers, not trivially.
But first let's discuss some things that are always true for all observers.
We have now defined three interesting great circles in the sky, two that
are part of the observer's location and a third defined by the earth's rotation:
 horizon
 meridian
 celestial equator
These three great circles have an interesting symmetry in that the poles
of each great circle land on
one of the other great circles.
Now recall that any great circle divides the sphere into equal halves and that
every great circle intersects the other at antipodal points. Thus we
have these rules that apply for any observer:
 At any given time, exactly one half of the sky is
visible from anywhere (ignoring horizon or other obstructions).
 The horizon intersects the meridian so exactly one half of the
meridian is above the horizon.
 The meridian intersects the horizon at the north and south points,
and divides the visible sky exactly in half (so there are
east and west, visible and invisible quadrants of the sky).
 The intersection points of the celestial sphere and the horizon obviously
must be on the horizon and must be 180^{o} opposite
one another. Thus the celestial equator always crosses the horizon
exactly at the due East and due West points on horizon.
Other than the points made in the above statements,
the interaction between the equatorial
and observer's systems can be complicated.
Some simple examples to demonstrate what does change for observers at
different places on the Earth.
 First realize that an observer standing on the surface of the
Earth has about 1/2 their view of the Celestial Sphere blocked by
the Earth.
 The observer's horizon plane bisects the sky.
The horizon plane is a plane tangent to the Earth's surface at the
observer's feet. At any time 1/2 of the Celestial Sphere is
above your horizon, and 1/2 is below it. (Keep in mind that
in terms of the size of the universe  Celestial Sphere  the
radius of the Earth is negligible and so the horizon
plane really is bisecting the sphere).
 It is helpful to imagine a specific observer's local
"sky bubble" as a function of their position on Earth. Let us
return to questions posed earlier:
For a polar observer (in this case, North Pole):
 The North Celestial Pole is at the zenith.
 The Celestial Equator is exactly on and around the horizon.
For an equatorial observer:
 The North (and South) Celestial Pole lies on the horizon.
 The Celestial Equator goes through the zenith.
The more difficult (but more typical) situation is an observer at some
other latitude.
For an observer at latitude "LAT":
 The North Celestial Pole (when LAT > 0^{o}) is at an
angle LAT above the north horizon on the meridian.
(For an observer with LAT < 0^{o}, the South
Celestial Pole is at an angle LAT above the south horizon on the
meridian.)
 The Celestial Equator crosses the meridian
(90^{o}LAT) above the south horizon and LAT degrees
below zenith (to the south).
(For an observer with LAT < 0 degrees, the Celestial Equator
crosses the meridian (90^{o}LAT) above the north
horizon and LAT degrees below zenith (to the north).
It is important to understand the socalled "meridianal slice" geometry:
When a star is on the meridian, there is a simple relation between its
declination, its altitude and your latitude.
 Based on the diagram, the altitude of a star above
the southern horizon (SALT in the diagram above) is given by the
following expression, where LAT is your latitude:
SALT = DEC + (90^{o}  LAT)
 LAT = 38^{o} at Charlottesville, so here SALT = DEC +
52^{o}.
 Objects with DEC = LAT have altitudes of
90^{o}, i.e. they cross through your zenith.
 The celestial equator (DEC = 0^{o}) has SALT =
(90^{o}  LAT).
SALT is a useful concept for determining how high in the sky a
planet or the Moon, for instance, can be on a given night.
 As an example, the maximum declination the Moon can ever
achieve is +28.5^{o}. The maximum altitude we can ever
see the Moon in Charlottesville above the southern horizon is
then:
SALT = 28.5^{o} +
(90^{o}  38^{o}) =
80.5^{o}.
All of the above examples were simplified by only considering the plane of
the meridian. If we take into account the appearance of the Celestial
Sphere for the full sky, we have something resembling this:
In the figure:
 Various altitude gradations are marked along the meridian
(22.5^{o}, 45^{o}, 67.5^{o}).
 The arcs with arrows show lines of constant declination.
Note that since stars do not change their declination (both the
stars and the declination lines are fixed on the Celestial
Sphere) the declination arc lines also show the paths of rising
and setting stars across the sky.
 NOTE: The Celestial Equator intersects the horizon
exactly at the due East and due West compass points for
the observer.
 The above fact holds for any observer on the
Earth.
 You can prove this to yourself by considering: (1)
Any great circle on a sphere intersects with any other
great circle on a sphere always at two points
exactly 180^{o} apart. Thus, every great circle
exactly divides every other great circle exactly in half.
(2) The Celestial Equator and your horizon both
represent great circles on the Celestial Sphere. (3)
Thus the Celestial Equator must divide your horizon
exactly in half, and it does so at the due East
and due West points on the horizon.
 NOTE: The meridian is also a great circle that divides your
horizon exactly at the due North and due South compass points for
any observer (but this is how we defined the meridian!).
 Approximately what latitude does the figure represent?
E. CONSEQUENCES OF THE EARTH'S ROTATION ON ITS AXIS
First, important to clarify two patterns of motion in the sky:
 ROTATION is used to describe a spinning motion
about an axis.
 E.g., the Earth spinning once a day on its polar
axis.
 REVOLUTION is the term we use to describe orbital
motion.
 E.g., the Moon orbits (revolves around) the Earth
about once a month.
 Earth orbits (revolves around) the Sun once a year.
Because the Earth rotates once every 24 hours on its polar axis, the
Celestial Sphere appears to sweep across the sky (e.g., as defined by
your meridian) approximately once every 24 hours.
From our perspective on a spinning platform (the Earth), we see the
Celestial Sphere appear to rotate about the Celestial Poles.
Star trails around South Celestial Pole in a 10 hour
exposure.
To understand the repercussions of the diurnal Earth rotation on the
appearance of the sky, let's first return again to the two simple cases:
 For a polar observer (in this case, North Pole):
 Recall that the North Celestial Pole is at the
zenith for this observer.
Thus, the sky appears to spin around the zenith
for a polar observer.
 Recall that the Celestial Equator is exactly on and
around the horizon.
Thus, objects on the Celestial Equator appear to
spin around the horizon.
 Note one other important fact: The polar observer
can only ever see 1/2 the entire Celestial Sphere
.
The following movie shows what an
observer standing on the pole of the Earth would see as they look towards
the horizon, and then towards their zenith.
 For an equatorial observer:
 Recall that the North (and South) Celestial Pole
lies on the horizon.
Thus, the sky appears to spin about the
northsouth line on the Earth.
The following
movie shows what an observer standing just North
of the
equator of the Earth would see as they
look towards the horizon facing North, and then
towards their zenith.
Time exposure taken by Mr. Majewski on his trip to
Tanzania (which lies just south of the equator) and looking
towards the Southern horizon. Note how the stars are circling
about the South Celestial Pole, which lies just above
the Southern horizon. The stars of the Southern Cross
are to the lower left.
 Recall that for someone on the Earth's equator
the Celestial Equator goes through the
zenith and it goes through the due East and due
West points on the horizon (the latter always the case
for all observers on Earth).
Thus, objects on the Celestial Equator must rise
straight out of the due East point, head straight
up the sky towards the zenith, then head straight
back down to meet the horizon at the due West
horizon point.
The following
movie shows what an observer
standing on the equator of the Earth would see as
they look towards the horizon facing East.
 Note one other important fact: The equatorial
observer can see both the North and South Celestial Poles,
and, after one daily rotation, is able to "see" the
entire Celestial Sphere.
 As usual, the more difficult (but more typical) situation is
an observer at some other latitude.
For an observer at latitude "LAT":
 The North Celestial Pole (when LAT > 0^{o})
is at an angle LAT above the north horizon on the
meridian.
Now consider what happens to a star that is
located at an angle less than LAT from the North
Celestial Pole...these stars never set
below the horizon.
 But note also that the South Celestial Pole (when
LAT > 0^{o}) is at an angle LAT below the south
horizon on the meridian.
Now consider what happens to a star that is
located at an angle less than LAT from the South
Celestial Pole...these stars never rise
above the horizon.
 The Celestial Equator crosses the meridian
(90^{o}LAT) above the south horizon and LAT
degrees below zenith (to the south).
All stars between (90^{o}LAT) and
(90^{o}LAT) spend some time above the
horizon.
We can define several groups of stars based on the above rise/set
characteristics for stars of different declination.
 We define the region with (90^{o}LAT) <
declination < 90^{o} the North Circumpolar
Region. These objects are above the horizon 24
hours a day (whether we can see them or not  e.g., due to the haze of
the bright Sun).
 E.g., for Charlottesville at LAT = 38^{o},
the North Circumpolar Region is all declinations above
52^{o}, and this includes the constellations Ursa
Minor, Ursa Major, Draco and Cassiopeia.
 The following movie
shows what an observer standing at the latitude of
Charlottesville would see over the course of a
day when they look towards the North horizon, and then
towards their zenith.
 Note the circumpolar constellations in the star
trails image shown earlier in this web page.
 We define the region with (90^{o}LAT) >
declination > 90^{o} the South Circumpolar
Region. These objects are below the horizon 24
hours a day.
 E.g., for Charlottesville at LAT = 38^{o},
the South Circumpolar Region is all declinations below
52^{o}, and so there are constellations (like
Chamaeleon, Volans, Hydrus, etc.) that we can never see
from our vantage point on Earth.
 As we mentioned above, the Celestial Equator great circle
divides the horizon great circle in half, but the reverse is true
too!
 This means that objects on the Celestial Equator:
 rise due East
 are above the horizon 12 hours a day
 set due West
 are below the horizon 12 hours a day
 This also implies that objects with
0^{o} < declination < (90^{o}LAT):
 rise North of due East
 are above the horizon more than 12 hours a
day
 set North of due West
 are below the horizon less than 12 hours a
day
 and that objects with 0^{o} > declination >
(90^{o}LAT):
 rise South of due East
 are above the horizon less than 12 hours a
day
 set South of due West
 are below the horizon more than 12 hours a
day
 The following movie shows what an
observer
standing at the latitude of Charlottesville would see over the course of
many hours when they look towards the East horizon. Note:
 that stars are rising at an angle to the horizon (can you
figure out what that angle is??).
 that stars above the Celestial Equator rise North of due
East.
 that stars below the Celestial Equator rise South of due
East.
 The following movie shows what an
observer
standing at the latitude of Charlottesville would see over the course of
many hours when they look towards the South horizon. Note:
 that stars with very low declinations (near
[90^{o}LAT]) are only above the horizon briefly, and
always near due South.
THOUGHT PROBLEM TIME OUT:
 You know from experience that in the summer the Sun is
above the horizon more than 12 hours in a day (i.e., the daytime
is longer than the nighttime). What does this imply about the
declination of the Sun during summer? Think before playing the
movie:
The following movie
shows what the Sun looks like over the course of a summer
day to an observer at the latitude of Charlottesville
looking towards the South horizon.
 You know from experience that in the winter the Sun is above
the horizon less than 12 hours in a day (i.e., the nighttime is
longer than the daytime). What does this imply about the
declination of the Sun during winter? Think before playing the
movie:
The following movie
shows what the Sun looks like over the course of a winter
day to an observer at the latitude of Charlottesville
looking towards the South horizon.
 The word "equinox" comes from the Latin for "equal night" 
in the sense that the day is equal to the night in duration. What
does this imply for the declination of the Sun on the day of the
Vernal Equinox (Mar. 21)? The Autumnal Equinox (Sep. 23)? Think
before playing the movie:
The following movie
shows what the Sun looks like over the course of March 21
to an observer at the latitude of Charlottesville looking
towards the South horizon.
F. MORE CONSEQUENCES OF EARTH'S ROTATION: PASSAGE OF TIME AND THE
HOUR ANGLE
Let's review the Celestial Sphere coordinate system:
The declination system is intuitive because it mimics the latitude system
of the Earth.
But what is the rationale for using units of time for the right ascension?
 The Earth rotates once every 24 hours on its polar axis, so the
Celestial Sphere appears to sweep across your meridian
approximately once every 24 hours.
 When a celestial object moving from East to West due to the diurnal
motion of the Earth crosses your meridian, we say that object
transits .
 Imagine your meridian to be like an hour hand on a clock  then
lines of right ascension sweeping past your meridian tell you what time it
is...
The following movie shows the
passage of time via the changing right ascension lines passing the
meridian of a Charlottesville observer looking toward the South
horizon.

...except that the true rate at which the Earth spins on its axis is 23
hours, 56 minutes  just short of a 24 hour human clock!
 We call the 24 hour period that we are used to a solar
day .
A solar day is the period it takes for the Sun to go
from one transit to the next transit.
 We call the 23 hour, 56 minute period of the Earth's
spin period a sidereal day .
A sidereal day is the period for a star to go from
one transit to the next transit.
 The reason for the two different days (transit periods)
will be made clear later...
Thus, the use of units of time for the right ascension coordinate
system is based on the notion of the sky as a sidereal clock :
 The sidereal time is the imaginary line of right
ascension transiting your meridian at any given instant.
 Of course, there are an infinite number of such lines in the
sky, distinguished by arbitrarily fine gradation of units of time
(hours, minutes of time, seconds of time).
 Because the spin of the Earth is counterclockwise (eastward)
as seen from above N pole ==> apparent rotation of sky is
westward.
 Therefore, because we want units of time to increase:
 the RA must increase as we look to the East of the
meridian (because these RA lines will transit in the
future)
 and RA must decrease to the West of the meridian
(because these RA lines have already transited earlier in
the day).
 Like the choice of the Prime Meridian that sets
longitude=0^{o} on the Earth, the choice of
RA=0^{o} on the sky was set by matter
of convention.
 The zero point of the RA system is at the point where
the ecliptic plane crosses the equator in the
constellation Aries. We discuss the ecliptic
more later.
Aside: In the "old days" many observatories has a "transit telescope".
This is a telescope that can move in declination but is constrained to
move only through the meridian. The purpose of these telescopes were to
time the passage of celestial objects and were used to set local time
(e.g., local noon as the time the Sun transits).
Transit telescope from 1775 at the Osservatorio Astronomico di
Brera in Milan, Italy.
At the Leander McCormick Observatory you can see a transit telescope like
that shown.
We define the hour angle, or HA, as the amount of time before or
after transit for any given object.
 Thus HA is a unit of time (time from transit), in units of
hours, minutes, seconds.
 Of course, HA is also an angle, being the angle in
the sky between your meridian and the line of RA corresponding to
the object of interest.
 Unlike the RA for an object, which is always fixed, the HA of
an object is constantly changing (increasing).
 Because the HA of an object specifies the time from transit,
mathematically, the HA at any given instant is given by:
HA = LST  RA
where ST=sidereal time.
Thus,
 objects to the west of the meridian have HA > 0
 objects to the east of the meridian have HA < 0.
Examples:
 A star with HA = 2 crossed the meridian (i.e. transited) two
hours ago.
 A star with HA = 1 will be on the meridian in one hour.
 An HA of 23 hours is equivalent to HA = 1.
 Note that to avoid confusion over sign, astronomers will often
include an "E" or a "W" in the time to indicate east or west of the meridian,
respectively.
For example,
 an hour angle of 1:23 could be written as "1E23".
 an hour angle of +3:07 could be written as "3W07".
 Through HA you can determine the time of best
visibility for any object, which corresponds to transit or
HA=0 (recall the airmass curves from before).
 THOUGHT PROBLEM TIME OUT: For an object on the Celestial
Equator, what is the HA for an object rising? Setting?
 The HA for rising and setting objects is given by 1/2 the
time an object is above your horizon.
 THOUGHT PROBLEM TIME OUT: For what objects can you actually
observe them at HA=+/12 hours (ignoring daylight
considerations)?
G. CONSEQUENCES OF EARTH'S REVOLUTION ABOUT THE SUN
 Earth is a planet moving in orbit around the Sun
 Orbit is nearly circular (distance to Sun varies only 3%)
 Orbit lies in a plane called the "ecliptic" plane. Seen
faceon, the orbit is technically an ellipse but deviates only
slightly from a circle. Seen edgeon, the orbit is a line.
 Earth orbits Sun in 365.25 days (one year) moving at ~66,000 mph.
Motion is counterclockwise as seen from above N. pole
 The night side of Earth is that opposite the Sun. So the
stars visible at night are those "opposite" the Sun. See figures
above.
 So, in May, the constellations Scorpio, Leo and Virgo will be
prominent in the night sky, while in November, they lie in the
direction of the Sun and therefore are not visible because of the
glare of the Sun illuminating the Earth's atmosphere.
 In November, Aries, Taurus and Gemini will be prominent in
the night sky, while in May they lie in the direction of the Sun
and therefore are not visible because of the atmospheric
glare.
 Warning! this drawing, and most others you
will see in this course and in astronomy texts, is grossly distorted
and not to scale! In reality, the Earth's orbit is 100 times the
diameter of the Sun; the Earth is 100 times smaller than the Sun; the
stars are vastly distant from the Earth's orbit; and the stars in a
given constellation are not necessarily near one another in space.
We are viewing Earth's orbit here from an angle such that it looks
highly elliptical, whereas it is nearly circular seen faceon.
Obviously, no one could produce or sensibly view a figure like this
drawn to actual scale.
 The apparent path that the Sun takes through the sky over the course
of a year is known as the ecliptic .
 The constellations that lie along the ecliptic are called the
zodiac . There are twelve "officially recognized" zodiacal
constellations even though there are 13 actual constellations the
ecliptic crosses (Ophiuchus is "left out").
 IMPORTANT!: The ecliptic is not coincident with the Celestial
Equator!
 The ecliptic is inclined by 23.5^{o} from the
celestial equator.
 The ecliptic and celestial equator define two separate great
circles in the sky. One of the places they cross is called the
"Vernal equinox"  this is the position the Sun has on Mar. 21
every year, in the head of Aries.
 Recall the movie
showing the motion of the Sun as we face South from
Charlottesville over the course of a winter day.
The changing position of the ecliptic over the course
of one daytime (i.e. when the Sun is up) can be seen.
 Recall the
movie showing the motion of the Sun as we face South
from Charlottesville over the course of a summer day.
The changing position of the ecliptic over the course
of one daytime (i.e. when the Sun is up) can be seen.
 Recall the movie
showing the motion of the Sun as we face South from
Charlottesville over the course of Mar 22.
The changing position of the ecliptic over the
course of this one day (during daytime) can be seen.
 Thus, half of the ecliptic is North of the celestial equator
and half is South of the celestial equator.
 Thus, the Sun spends half of the year above the celestial
equator and half of the year below it.
 This movie shows the
position of the noontime Sun as we face South from Charlottesville
over the course of 3 years. The changing position of the ecliptic
at noontime over the course of the three years
is also shown.
 THOUGHT PROBLEM TIME OUT:
Recalling the length of time objects below and above the celestial
equator spend above the horizon, is it now clear why summer
daylight is longer than 12 hours and winter daylight is shorter
than 12 hours in duration?
THE SIDEREAL AND SOLAR DAY
We defined two types of day:
 A solar day is the time between two successive
transits of the Sun.
It is what human clocks are based upon = 24 hours
long.
 A sidereal day is the time between two successive
transits of the same star.
The sidereal day is 23 hours 56 minutes long.
Why is the solar day longer than the sidereal day?
 In the figure showing the Earth orbit from above,
the person stands on one point of the Earth shown on two
consecutive sidereal days.
(Note: all angles
greatly exaggerated!)
This person's meridian points to the same distant
star on the two days.
 But note, during the passage of the sidereal day, the Earth
has advanced in orbit around Sun (the amount is greatly
exaggerated in the figure.)
The angle the Earth advances in its orbit in one day
is approximately equal to (360^{o}/year)/(365.25
days/year) = ~ 1^{o}.
 But in the passage of that one sidereal day, the Sun has not
yet transited.
In order to do so, Earth must rotate an
extra amount, which is, by Euclidean geometry (angle A =
angle B in the figure), equal to the ~ 1^{o}
from above.
 Thus, the Earth must turn about 361^{o} every
solar day !
The extra 1^{o} rotation takes
(360^{o} daily rotation)/(23 hours 56 minutes) =
4 minutes of time. PROVE THIS TO YOURSELF!
 Thus, the solar day, corresponding to 361^{o} Earth
rotation is 4 minutes longer that a sidereal day, which
corresponds to a 360^{o} Earth rotation.
Now turn things around, to a human time frame.
 In a solar day, since the Earth has turned
361^{o}, the background stars will have actually
advanced in position to the West
by ~1^{o} with respect to the Sun.
 The Sun will now have moved in RA to the East with respect to
the background stars.
 The sidereal clock will appear to run 4 minutes fast per day,
or 2 hours fast per month.
 Thus, every 12 months, the sidereal clock will
repeat in its cycle, and so every year on the same day of
the year, the same stars transit at the same time.
 The one day of the year when the solar and sidereal
clocks are synchronized is the day of the Autumnal Equinox
(Sep. 22), when the Sun has an RA of 12 hours, and so
when the Sun transits the sidereal and the solar time is
12 hours.
THOUGHT PROBLEMS TIME OUT:
 Do you see now why certain constellations are seasonal
(i.e., "winter" or "summer") constellations?
 The moon and planets more or less lie along the ecliptic
plane. For a planet that is up at midnight (say transiting about
12 hours after the Sun), what kind of declination will this
planet have in winter? In summer? Use the movies above to help
if you are stuck.
Great circle/small circle figure from
www.ncgia.ucsb.edu/education/curricula/giscc/units/u014/figures/figure05.gif.
Great circle route figure from
www.ncgia.ucsb.edu/education/curricula/giscc/units/u014/figures/figure06.gif.
All other material copyright © 2002,2008,2012 Steven R. Majewski. All
rights reserved. These notes are intended for the private,
noncommercial use of students enrolled in Astronomy 313 and Astronomy 313o at the
University of Virginia.
