<p>The coordinate system you are using is not actually the equatorial coordinate system that is expressed in right ascension and declination but the coordinate system that uses Hour-angle and declination. This system is not fixed with respect to the stars (or the vernal equinox).
The hour-angle is the angle between the intersection of the celestial equator and your local meridian (the line from south to the zenith). It can be calculated by:
$$H = \Theta - \alpha$$
where $\Theta$ is the <a href="http://en.wikipedia.org/wiki/Sidereal_time" rel="nofollow">local sidereal time</a> and $\alpha$ is the right ascension. The declination is the same in both systems.</p>

CC BY-SA 3.0
2014-10-13T15:55:19.487Z
2014-10-13T15:55:19.487Z
Dieudonné
<p>I've just started studiyng Astronomy.</p>

<p>I understood how I would measure Right Ascention and Declination: for the first one I'd trace an Hour cirlce to the Celestial Equator, and then, compute the angle from this point and the Vernal Equinox; while the second one is the angle subtended by the object and the intersection of the Hour Circle with the Celestial Equator.</p>

<p>So, during the night, stars will change their Right Ascension value? Does this means this coordinate system is not fixed with respect to the stars?</p>

<p>If so, why people store the RA/Dec to find objects?
For instance, I know  the RA/Dec of the NGC 5585 Galaxy ( <a href="http://en.wikipedia.org/wiki/NGC_5585" rel="nofollow">http://en.wikipedia.org/wiki/NGC_5585</a> ), but how can I use this information, if the RA changes during the night? </p>

<coordinate>
Why objects are uniquely defined by their right ascention and declination?
Nikolai Gogol
# Why objects are uniquely defined by their right ascention and declination?

(Asked by: username_0 on 2014-10-11)

I've just started studiyng Astronomy.

I understood how I would measure Right Ascention and Declination: for the first one I'd trace an Hour cirlce to the Celestial Equator, and then, compute the angle from this point and the Vernal Equinox; while the second one is the angle subtended by the object and the intersection of the Hour Circle with the Celestial Equator.

So, during the night, stars will change their Right Ascension value? Does this means this coordinate system is not fixed with respect to the stars?

If so, why people store the RA/Dec to find objects?
For instance, I know the RA/Dec of the NGC 5585 Galaxy ( <http://en.wikipedia.org/wiki/NGC_5585> ), but how can I use this information, if the RA changes during the night?

**Tags:** coordinate

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## 1 answers

### Answer by username_1 on 2014-10-13. Score: 0

The coordinate system you are using is not actually the equatorial coordinate system that is expressed in right ascension and declination but the coordinate system that uses Hour-angle and declination. This system is not fixed with respect to the stars (or the vernal equinox).
The hour-angle is the angle between the intersection of the celestial equator and your local meridian (the line from south to the zenith). It can be calculated by:
$$H = \Theta - \alpha$$
where $\Theta$ is the [local sidereal time](http://en.wikipedia.org/wiki/Sidereal_time) and $\alpha$ is the right ascension. The declination is the same in both systems.


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