Hours, minutes, and seconds are defined identically in both solar and sidereal measure
A solar day is the amount of time that passes for the sun to cross the same celestial meridian
For any location on earth, the observer's meridian is the curved line connecting the point on the horizon due south of the observer's location with the zenith (the point directly overhead)
This line is often simply referred to as the meridian
In the following illustration, the observer is at p, facing south
So, if you were to pick an outdoors location and observed when the sun crossed the meridian, one solar day (exactly 24 hours) would pass before it did so again.
One solar day corresponds to the rotation of the earth by 360.986 degrees—perhaps not what you were expecting
As the earth revolves around the sun, it has to rotate more than 360° for the observer's meridian to realign with the sun
4 more minutes are required for the earth to rotate that 'extra' .986 degrees
And as it's rotating for 4 more minutes, it's also travelling in its orbit around the sun for those 4 'extra' minutes
Sidereal time
A sidereal clock measures ‘star time’
Compared to the definition for a solar day, a sidereal day is much easier to understand geometrically
A sidereal day is the time that passes between one meridian crossing and the next for a star (other than the sun)
Unlike the sun, any other star in the sky is so far away that its light always appears to come from the same direction in the sky
One sidereal day is equivalent to 23 hours and 56 minutes
That's how long it takes the earth to rotate exactly 360.0°
A sidereal day is four solar minutes shorter than a solar day because it doesn't need that extra 4 minute of rotation to follow the relatively nearby sun
Remember that hours, minutes, and seconds are defined exactly the same in both solar and sidereal time--it's only the length of a day that's different
Seeing the difference between solar and sidereal time
I've created the following app, Solar and Sidereal Time, to demo how each system records time as the earth revolves about the sun
In the app,
Our viewpoint lies above the north poles of the earth and the sun
The earth moves counter-clockwise about the sun, and rotates counter clock-wise (west to east) about its polar axis
The blue line represents the direction of light from a star in the direction of the First Point of Aries (the point in the sky where the sun crosses the celestial equator on or near March 21)
The yellow line represents the direction of light from the sun,
VE, JS, AE, and DS represent the vernal (March) equinox, June solstice, autumnal (September) equinox, and December solstice respectively
As the earth (the blue disk with the red meridian) rotates, any meridian crosses the solar 0-hour line (on the yellow ring with orange numbers) exactly once per solar day
From the point of view of an observer on the ground, the center of the sun crosses her meridian (the line extending from due south of her position on the horizon to the point directly overhead)
On a desktop browser, you can click, hold, and dragthe blue disk (the earth) around the circle to see the difference between solar and sidereal elapsed time at different points in the earth's orbit around the sun
You can also drag with your finger on touch screen devices
The slider at the top controls the animation speed (moving all the way to the left pauses the animation)
A solar day is 4 minutes longer than a sidereal day (defined below)
That extra 4 minutes of rotation (.986°), multiplied by 365 days, equals approximately 24 hours of 'extra' rotation
That's why a solar year contains approximately one more sidereal day than it does solar days
A solar year takes 365.2422 solar days or 366.2284 sidereal days to complete
And that's why GPS satellites appear to set 4 minutes earlier each solar day
Their orbits are fixed with respect to the background stars—not the sun
Fun fact: a solar clock and a sidereal clock read the same time on the autumnal equinox (around Sept. 22)
Now that we know how sidereal time works, we can understand how celestial coordinates work
Celestial coordinates provide a fixed (or inertial) coordinate system independent of the earth's rotation
By fixed we actually mean very slow moving relative to the earth's rotational motion and orbital motion about the sun
Nothing is really stationary in space!
Defining celestial reference planes
The celestial equator
For an observer, the celestial equator is an arc across the southern sky
The celestial equator intersects the observer’s horizon at his/her due east point
It is highest at the point it crosses the observer’s meridian
It extends to and intersects the horizon at the observer’s due west point
At the meridian, the celestial equator has a height above the horizon equal to 90 degrees minus the observer’s latitude
An object’s declination is its angle above or below the celestial equator
Our GIS lab has a latitude of 42 deg, 41 min, 14 sec N (42.6872 degrees)
For our latitude, the celestial equator has a height of 90 – 42.6872 or 47.3128 degrees
An hour circle is a celestial meridian
The zero-hour circle of right ascension corresponds to the direction of the vernal equinox (first point of Aries), the meridian passing through the point in the sky at the moment the center of the sun crosses the celestial equator from north to south (as seen from the surface of the earth)
This sun simulator is a good model of the sun's apparent motions across the sky for any day and time over a year
I believe credit for this simulator goes to Kevin M. Lee of the Nebraska Astronomy Applet Project but attribution information is hard to find on this page
Right ascension is measured from the hour circle that intersects the celestial equator at the ‘vernal equinox’
When the center of the sun passes through this point, it marks the beginning of spring in the northern hemisphere
For us, the absolute location of the 0h circle is no more important than that of the prime meridian—both are arbitrary starting points
The important thing is at sidereal time X hours Y minutes Z seconds, the right ascension line marked with the same time will be on your meridian (will go from your zenith down to your southern horizon)
An object with a right ascension equal to the current local sidereal time will be due south of your position
An object with a right ascension 1 hour later (or 15 degrees larger) than the local sidereal time will appear south of your position in 1 hour
Every ‘fixed’ celestial object (as opposed to the planets or other objects with fast apparent motion relative to the earth) has a fixed declination and right ascension
Think of the sky as being a rigid bowl with the stars painted inside it in fixed locations
Now imagine that a coordinate grid, like latitude and longitude, is drawn inside the bowl
The stars do not move with respect to the grid—they will always have the same coordinates
The sun, moon, and planets do change in both declination and right ascension over relatively short periods of time
Their positions are updated in almanacs for dates and times of day
Star positions do of course change over very long periods of time but that’s not important now
Sidereal hour angles are given in angular measurement
Like RA, there are 360° or 24 hours of rotation but, unlike right ascension, their coordinates increase from west to east
Because a sidereal day is about 4 minutes shorter than a solar day, stars will appear to rise and set approximately 4 minutes earlier each day
Just like our SVs
Since SV orbital paths appear 'fixed' (at least over periods of several days) relative to the background stars, we can specify orbital planes in celestial coordinates
Celestial coordinates resemble latitude and longitude but do not rotate with the earth
Points on the orbital planes can be specified in declination and right ascension (RA)
Using declination and right ascension to find an object in the sky
Tables of astronomical objects give their celestial coordinates
The beginning of spring is defined as the time at which the sun appears to cross above the celestial equator (declination 0) at right ascension 0 (the First Point of Aries)
Using the ephemeris, enter the approximate date for the first day of spring this year (the vernal equinox) and enter a time of day (0 hours, 0 minutes, and 0 seconds is midnight on that day)
The time of day is UTC using a 24 hour clock
UTC can be found by adding 5 hours to Eastern Standard Time or 4 hours to Eastern Daylight Time
For example, enter March 21, 2013, 0 hours, 0 minutes, 0 seconds and click the Get Ephemeris button
Now check the position of the sun at that time in declination and right ascension
It should be relatively close to 0 for each coordinate
Keep adjusting the date and time as necessary to get the declination and RA of the sun to exactly 0.
