# Measurement precision of celestial events

The local weatherman declared that fall starts at exactly 10:21am (local time). Ignoring the fact that seasonal boundaries have multiple definitions (meteorological, astronomical, etc.), what he perhaps means to have said is that the fall equinox occurs at that moment.

Presumably, that is the moment at which Earth's axis of rotation is exactly perpendicular to the Sun-Earth line.

My question is: to what degree of precision can this be measured? The weatherman's statement was to the minute. But is it actually measured to a finer precision? If so, plus or minus how many seconds?

What about the the solstices? Can they be measured to the same degree of accuracy? Are there factors which uniquely affect the accuracy of either measurement?

• I'm pretty sure the Earth's axis is never perpendicular to the Earth-Sun line (the angle is always right around 23.5 degrees), but the solstices and equinoxes occur when the Sun's ecliptic longitude is 0, 90, 180, and 270.
– user21
Sep 20 '16 at 14:20
• Earth's axis is never perpendicular to its orbital plane, but at the equinoxes, the Sun crosses Earth's equatorial plane, which is perpendicular to its axis. Sep 20 '16 at 16:08
• @barrycarter Yes, the Earth's axis is at a fixed orientation in space (if you ignore precession, nutation, etc.), which is inclined relative to the orbital plane. But twice a year, it does become perpendicular to a line drawn between the the centers of Sun and Earth (remember, it's the Sun-Earth line that is moving, not the Earth's axis) Sep 27 '16 at 23:32
• I still don't think that's true. A line drawn between the Sun and Earth will remain in Earth's orbital plane (roughly) and thus at a ~23.5 degree angle with respect to the Earth's axis.
– user21
Sep 28 '16 at 0:32
• Actually, I see your point. If you're standing at the center of the Earth on the equinox, the Sun is above the equator which is perpendicular to the line from the center of the Earth to the North Pole. @MikeG Can you help me w/ this one? :)
– user21
Sep 28 '16 at 1:49

This doesn't answer your question, but has two screenshots, so that's something :)

If you run http://ssd.jpl.nasa.gov/?horizons with these settings: part of the output looks like this: This shows the Sun crosses the equinox point between the 8th and 9th second of 14:21 (which is 10:21 your local time). HORIZONS will even let you compute at half second intervals but not beyond that.

If you use the http://naif.jpl.nasa.gov/naif/tutorials.html library, you can compute the time down to the nanosecond or better, but this doesn't really help because we don't know the Sun's position, the Earth's position, or even the direction the North Pole is pointing to that level of precision.