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?


2 Answers 2


Stars do not change their right ascension or declination values as the Earth turns.

The position of the stars in the night sky (called azimuth and elevation) will change, but that's because the lines of right ascension also move.

However, right ascension and declination do not uniquely define an object. Over time, stars move with respect to the sun (called "proper motion"), and the direction of the Earth's north pole (currently pointing near Polaris, the North Star) also changes (called "precession").

  • $\begingroup$ Oh..So, does the Vernal Equinox moves during the day? I mean, Since it is the origin of the coordinate system, it must rotate as stars do? $\endgroup$ Oct 12, 2014 at 16:07
  • $\begingroup$ Yes, it does. You may want to use a program like stellarium to visualize how the stars (and the vernal equinox) trace a full circle in the sky every 23h56m (the sidereal day). $\endgroup$
    – user21
    Oct 12, 2014 at 17:33

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 and $\alpha$ is the right ascension. The declination is the same in both systems.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .