5
$\begingroup$

I am new to this topic so correct me if I am wrong. But as I understand it, the vernal equinox point is a point on the celestial sphere where the ecliptic and celestial equator meets. When the sun's sub-solar point is directly falling on this point that moment in time is the vernal equinox. So, if we take a projection of it on earth, this point has to resolve to a particular latitude and longitude.

I know the latitude is 0°, but what is the longitude for this point? I am assuming that it's definitely not the 0° lat, 0° long.

Note: I know that this point can be slowly moving away to right or left due to Earth's precession. I would like to know where this point is atleast for this year.

$\endgroup$
3
  • 3
    $\begingroup$ It's not moving slowly ,either. The geocentric longitude of the subpoint of the Vernal Equinox changes by 15° per hour, and since there aren't an integral number of solar days per year, the longitude of the solar subpoint at the instant vernal equinox can vary widely on consecutive years. $\endgroup$
    – notovny
    Sep 11, 2023 at 10:18
  • 2
    $\begingroup$ This page gives the geographic position of the J2000 origin. Note, that the "of date" origin is constantly changing due to precession and nutation. celestialprogramming.com/snippets/geographicPosition.html $\endgroup$ Sep 11, 2023 at 21:52
  • 1
    $\begingroup$ @GregMiller Hi! Thanks again. Big fan of your work. I took parts of your JS code and converted it to C++ in my project (RA/Dec to Alt/Az finder) last year. Here it is -> github.com/omkarium/starf $\endgroup$ Sep 12, 2023 at 10:56

1 Answer 1

10
$\begingroup$

Let's take the easy way and let's look up the time for the equinox this year: my astronomical yearbook tells me that the autumn equinox happens at exactly 8:50h MESZ on 23 September 2023. MESZ is UTC+2, thus ignoring time equation the meridian passage of the Sun is at 12h for 30° East.

We can now calculate at which longitude the Sun is in the meridian at this time 8:50h, or rather vice versa: 8:50h is 3:10h before 12h where the Sun is in the meridian. So we have to walk East in longitude which corresponds 3:10h of rotation. One hour corresponds to 15° and thus 10 minutes corresponds to 1/6 of an hour or $15°/6 =2.5°$. So adding up, the Sun is in the merdian the autumn equinox 2023 at longitude $\lambda = 30°E + 3*15°E + 2.5°E = 77,5°E$.

Now, this was all calculated with the mean sun. One needs the true Sun position for the sub-solar point. This is expressed by the equation of time which varies over the course of the year by several minutes and is visible by the analemma if you take a combined photo of the Sun at always exactly the same time-of-day (as your clock shows) throughout the course of a year. The reason for this correction is that Earth moves differently fast at in its orbit around the Sun during perihel compared to aphel, but rotates around its axis at constant speed. A positive number on the equation of time means that the Sun is ahead of the mean solar time. On September 23rd this is +7.4 minutes which gives another $7.4/60 * 15° = 1.85° $ degrees - but West as the Sun is ahead of the mean solar time our clock shows.

Thus adding this to the longitude we got for mean solar time, we get a longitude for the autumn equinox 2023 of 75.65° East.

$\endgroup$
1
  • 4
    $\begingroup$ @MikeG thanks a lot Mike. I made an edit to take this into account. Not insignificant :) $\endgroup$ Sep 11, 2023 at 13:26

You must log in to answer this question.

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