ASTR 3130, Majewski [FALL 2012]. 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 zero-point 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 north-south 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 x1, y1, z1 in system 1 can be converted to the coordinates x2, y2, z2 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 = (90o - 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 -90o and +90o.

• Earth's equator is latitude 0o.

• Longitude reckoned from 0 to 360o starting at Greenwich, England (by convention).

• Note alternative longitude angle unit is the time zone , for which the Earth has 24 zones, each nominally 15o 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 +/- 90o, and the celestial equator has declination of 0o.

• 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 (-90o to 0o to +90o) Declination (-90o to 0o to +90o)
Longitude (0 to 360o, or 24 time zones) Right Ascension (0 to 24 hours)

• 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 so-called altitude-azimuth mounting.

• Because of the "drawbacks" of the horizon coordinate system mentioned above, the alt-az 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).

### 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 = 180o, 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:

• 38o

• 38o 46' (more precisely)

• 38o 46' 24" (even more precisely)

• Note that the latter may also be written as 38o 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 = 60m

• 1 minute of time = 60 seconds of time = 60s

• The right ascension of an object, e.g., the star Vega, can be written in ever increasingly precise coordinates as:

• 18h

• 18h 36m (more precisely)

• 18h 36m 36s (even more precisely)

• 18h 36m 36.1s (even more precisely)

• Note that the latter may also be written as 18h 36.6m, since 0.6m = 36s.

• 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 360o/24=15o.

##### 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 = 15o 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 1-2 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 equator and the horizon obviously must be on the horizon and must be 180o opposite one another. Moreover, the meridian must equally divide the visible parts of the celestial equator. Thus, with these constraints, 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 > 0o) is at an angle LAT above the north horizon on the meridian.

##### (For an observer with LAT < 0o, the South Celestial Pole is at an angle -LAT above the south horizon on the meridian.)
• The Celestial Equator crosses the meridian (90o-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 (90o-|LAT|) above the north horizon and |LAT| degrees below zenith (to the north).

It is important to understand the so-called "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 + (90o - LAT)

• LAT = 38o at Charlottesville, so here SALT = DEC + 52o.

• Objects with DEC = LAT have altitudes of 90o, i.e. they cross through your zenith.

• The celestial equator (DEC = 0o) has SALT = (90o - 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 +28.5o. The maximum altitude we can ever see the Moon in Charlottesville above the southern horizon is then:

SALT = 28.5o + (90o - 38o) = 80.5o.

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 (22.5o, 45o, 67.5o).

• 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 180o 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 north-south 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 > 0o) 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 > 0o) 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 (90o-LAT) above the south horizon and LAT degrees below zenith (to the south).

All stars between (90o-LAT) and -(90o-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 (90o-LAT) < declination < 90o 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 = 38o, the North Circumpolar Region is all declinations above 52o, 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 -(90o-LAT) > declination > -90o the South Circumpolar Region. These objects are below the horizon 24 hours a day.

• E.g., for Charlottesville at LAT = 38o, the South Circumpolar Region is all declinations below -52o, 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 0o < declination < (90o-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 0o > declination > -(90o-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 -[90o-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=0o on the Earth, the choice of RA=0o 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 (currently in the basement).

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 face-on, the orbit is technically an ellipse but deviates only slightly from a circle. Seen edge-on, 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 face-on. 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 23.5o 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 (360o/year)/(365.25 days/year) = ~ 1o.

• 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 ~ 1o from above.

• Thus, the Earth must turn about 361o every solar day !

The extra 1o rotation takes (360o daily rotation)/(23 hours 56 minutes) = 4 minutes of time. PROVE THIS TO YOURSELF!

• Thus, the solar day, corresponding to 361o Earth rotation is 4 minutes longer that a sidereal day, which corresponds to a 360o Earth rotation.

Now turn things around, to a human time frame.

• In a solar day, since the Earth has turned 361o, the background stars will have actually advanced in position to the West by ~1o 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.