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The equatorial coordinate system isn't too complicated. However, for people new to this, it seems rather intimidating at first.

Is there an easy way to explain it?

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3 Answers 3

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A system projected into the sky

Pretend all the stars are painted on the inside of a large ball, and you are at the center of the ball. The imagined ball is called the Celestial Sphere. If the Earth wasn't blocking the lower half of your view, and the Sun wasn't making the blue sky so bright, you would be able to look at the stars in any direction.

First, the rotation of the Earth causes the imagined ball of stars to appear to move around the sky. If you lay on your back, (in the norther hemisphere,) with your head to the north, the stars slowly slide by from left to right. There are two points however, where the stars don't appear to move: These two points are directly above the Earth's north and south poles. These two points, projected out into the sky are labeled the North, and South, Celestial Poles.

Next, the Earth's equator is projected out into the sky and labeled the Celestial Equator.

Finally, sorting out the east-west positioning, (ie left-right if you lay on your back, head to the north), requires an arbitrary selection: There are an infinite number of circles one can imagine drawing on the sky. (For example, small circles the size of the moon, and also many much bigger circles.) The largest circle you can draw is called a "great circle", from geometry. Point at the North Celestial Pole, (way above and behind your head if you face south in the northern hemisphere,) and draw a line straight south to the horizon, then down below your feet through the South celestial pole and back up to the North Celestial Pole. You have just drawn one Great Circle, through the Celestial Poles, all the way around the Celestial Sphere.

But that Great Circle is just one of an infinite number of similar ones. One could start at the North Celestial Pole and draw a Great Circle a little to the left or the right. So astronomers have chosen one specific Great Circle and labeled one side of that circle as "zero degrees" -- or, "we will start measuring angles from here."

So how do we use this system?

"right ascension" is measured rightward, (from your left to your right if you face south in the northern hemisphere,) from the zero-degrees half of that chosen great circle. So given a right ascension, one can find the corresponding half of a great circle. For a given right ascension, you have half of circle selected from the North Celestial Pole, to the South Celestial Pole. (This is called a line of equal longitude.)

"declination" is measured from the Celestial Equator. The Celestial Equator cuts through the middle of your line of longitude. From the Celestial Equator it is 90° northward(+) to the North Celestial Pole, and 90° southward(-) to the South Celestial Pole. So we can measure from +90° (northward) to -90° (southward) along that line of longitude, from the Celestial Equator. (Don't be confused by "positive declinations"; 45° of declination is northward from the Celestial Equator.)

Notice that the system does not depend on where you are standing on Earth, nor on the time of day. If you have a Right Ascension and a Declination, you have an unambiguous spot specified on the Celestial Sphere. The system is actually very simple, but the rotation (time of day) of the Earth, and where you are geographically located needs to also be figured in.

What's over my head right now?

Face south, (if you're in the northern hemisphere.) What is your latitude -- geographically, how far north are you from the Earth's equator? If you are 40° north, then Celestial declination of 40° north is directly over your head. Declination of 50° is 10° further northward from directly over your head, etc. But for that declination, you have a great circle -- left-to-right across the whole sky that has that same declination.

Right ascension you simply have to look up. That arbitrarily chosen, zero-degree line, appears to spin around the sky every day at the Earth rotates. Also, over the course of a year, the Earth's orbit about the sun makes an extra apparent turn of the sky. (What is overhead mid-winter, is directly under your feet mid-summer!) So there's an offset -- what right ascension is overhead on this date at midnight, and how far (in hours) are we from midnight right now? Combining those two you can figure out what Right Ascension is over head at any time, on any date.

(potential editors: I've intentionally not used acronyms for NCP, RA, etc as that makes it more complicated for people learning this material.)

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The equatorial coordinate system is very similar to the system used on a globe or on maps. To specify a point on a sphere or a globe you need only two numbers. These are the longitudes and latitudes.

On a globe of the earth you have longitudes from -180º to +180º and latitudes from -90º to +90º. The south pole lies at -90º latitude, the north pole at +90º latitude.

The sky uses the same system, projected onto the imaginary celestial sphere. The sky's equator is at the same position as the earth's equator. Only projected onto the (infinitely large) sky sphere. You can find lots of very nice pictures of this on the web.

However, since the earth rotates, the sky's coordinate system would rotate also, if it only were a synchronized projection. So every star would have a coordinate that depended on the time of day!

To solve this, the equatorial coordinate system is fixed on the equinox points. Since the earth is tilted, the eclipitc (i.e. the earth's orbit around the sun) has two intersections with the sky's equator. These are the equinox points. The sun will be located there during the days of spring and autumn when day and night are of equal length. These points on the sky sphere are relatively fixed and well suited for pinning down the coordinate system.

The sky's coordinate system is not using the degree notation for both longitude and latitude. Instead we use the terms Right Acension (i.e. longitude), which goes from 0h to 24h, similar to a day on earth. And the other is Declination (i.e. latitude), which goes from -90º to +90º. This is similar to the earth's traditional coordinate system.

So every fixed star can be assigned a coordinate in the sky which is independent of the time of day or year. For example, the star Sirius (α CMa) has the following coordinates:

RA 06h 45m 08.9173s Dec −16° 42′ 58.017″

These can be programmed into computerized telescope mounts, which can then find and track the star as it moves with the earth's rotation.

This is a rather simplified explanation. One should read up on equatorial mounts for telescopes, which utilize this coordinate system in a very elegant manner.

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The easiest way to think of equatorial coordinates is by extending lines of latitude/longitude (the geographic coordinate system we usually learn in school) out into space. The earth's equator becomes the celestial equator, and the north and south poles become the celestial north and south poles, respectively. By doing this you have a fairly simple way to describe the location of an object in space.

Each location is described by a pair of coordinates:

1) Declination (dec), which is measured in degrees (-90 for the south celestial pole, 0 for the celestial equator, and +90 for the north celestial pole). Partial degrees are usually described in terms of minutes (') and seconds (''). Think of declination as latitude.

2) Right Ascension (RA), which is usually measured in hours, minutes, and seconds. Think of this as terrestrial longitude extended out into space, then measuring it in hours makes a bit more sense (time zones). You could just as easily measure it in degrees, too, where 1 hour is 15 degrees. RA is measured from the point where the sun crosses the celestial equator at the March equinox (located in Pisces).

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