From my newly and developing understanding of the J2000 coordinate system it is defined as the point at which the sun appears to cross the celestial equator from the Southern to the Northern hemisphere at noon on January 1, 2000. This is the vernal equinox in the image below of the celestial sphere.

enter image description here (Wikipedia Image Reference)

Say for example the celestial sphere in the image above was taken at the exact moment of the J2000 epoch. If it even makes sense to ask, at what point on the Earth does the +X axis of the J2000 coordinate frame intersect the Earth's surface?

  • 2
    $\begingroup$ This is a sensible question. space.stackexchange.com/questions/51083/… should get you most of the way there (ECEF coords are Earth-Centered, Earth-Fixed) $\endgroup$
    – Erin Anne
    Commented Sep 8, 2022 at 1:32
  • $\begingroup$ This is a frame challenge comment, both in the sense of a StackExchange "frame challenge" and challenging a physics-based frame. J2000, like ICRF, has its origin at the solar system barycenter. In that sense, the answer to your question is hardly ever. Perhaps you are instead asking about an Earth-centered inertial frame whose axes are co-aligned with those of the J2000 frame (or almost equivalently, an ICRF frame). $\endgroup$ Commented Sep 9, 2022 at 7:46
  • $\begingroup$ @DavidHammen do you have a reference on that? My recollection (which is admittedly REAL bad these days) was J2000 being ECI; Wikipedia lists both it and M50 as examples of ECI systems on en.wikipedia.org/wiki/Earth-centered_inertial. Also the lack of common origin isn't reflected in the StackExchange question I linked earlier. $\endgroup$
    – Erin Anne
    Commented Sep 10, 2022 at 2:05
  • $\begingroup$ @ErinAnne Page 7 of naif.jpl.nasa.gov/pub/naif/toolkit_docs/Tutorials/pdf/… , for example. $\endgroup$ Commented Sep 10, 2022 at 4:03

1 Answer 1


I started this as a comment, but ran out of room. It doesn't actually answer the question, but explains why the question can't be answered in a meaningful way.

I'm sure you're aware that Jan 1 isn't the time the equinox happens, but the J2000 reference frame is based on the Earth's orientation at that time. By the time the actual equinox comes around the Earth will have precessed by then, and where the Sun actually crosses the equator, won't be the J2000 equinox.

So, honestly, it's better to just accept that the x-axis is where it is "by definition". It's a computed value rather than an actually observed one. The computation only includes precession, not nutation, polar motion, nor delta T. So, even if Jan 1 was the equinox, the Sun still wouldn't be at exactly 0,0 RA/Dec when it crossed the equator.

In fact, the ICRS has replaced the J2000 system. It has been aligned to the J2000 system to within a very tight margin. But it is not defined by the equinox, instead there are very distant radio sources which are assigned RA/Dec values, and those coordinates are correct by definition. Whatever happens to the Earth isn't relevant.

If you still want to know where the 0,0 RA/Dec point was on Jan 1, 2000, this can be done by computing the Geographic Position of 0,0 at Jan 1, 2000 12:00:00 UT. The geographic position of any object, at any time can be computed by:

$$ \begin{align*} \theta &= \delta \\ \phi &= \alpha - GST \end{align*} $$

$ \theta $: Latitude
$ \phi $: Longitude
GST: Greenwich Sidereal Time (Mean or Apparent depending on accuracy needs)
$ \alpha $: Right Ascension
$ \delta $: Declination

Since we're looking for the 0,0 point, $\theta=0$ and $\phi=0$. GMST at that time was about 18.697374, which equates to a lat/lon of:

+00° 00' 00.00" N

+079° 32' 21.76" E

Here is a page which computes the geographic position of the Sun and many stars for any time. It uses Mean sidereal time rather than apparent, so the actual position will be off by a bit.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.