Distance between Earth’s ecliptic path and the celestial equator throughout the year

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MetricRocketeer

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Hi TRF colleagues,

At each equinox (vernal and autumnal), Earth’s ecliptic crosses the celestial equator. Thus, for us in the Northern hemisphere, at the March (or vernal) equinox Earth transits the celestial equator on its ecliptic path from south to north. Then at the September (or autumnal equinox) Earth crosses the celestial equator as it proceeds from north to south.

Therefore, at each equinox, the distance between Earth and the celestial equator is zero. So far so good.

Here is my question, please.

What is the distance between Earth and the celestial equator at other times during the year? What is that distance right now, for example? What is that distance at the December and the June solstices? I should be able to figure that last one out, but I am blanking on how to do it.

Thank you for your consideration.

Stanley
 
I'm liking this because I want to see the answer. Although, I'm not entirely sure that I fully understand the question.
 
I think you might have your terminology confused. The celestial equator is just Earth's equator, projected out into the sky and used as a useful device for talking about where things are in the night sky, same as latitude and longitude lines on a map of Earth. Asking what the "distance between Earth and the celestial equator" is a meaningless question.
 
Also, what happens at equinoxes is that, relative to Earth, the sun crosses the point at which the celestial equator and ecliptic meet. The celestial equator and ecliptic are defined relative to Earth, so talking about Earth crossing them is meaningless.
 
Antares is trying to tell you that your first post mis-states the situation. Both the celestial equator and the ecliptic are earth-centered coordinate systems. The Earth never leaves the plane of the ecliptic, nor the celestial equator. You can figure the angular difference at a certain date using the sun as reference - but it's angles. The distance you'd project those angle out to in order to find a 'height' is always zero.

This is different from the case of a satellite around the earth which is in an inclined orbit - in that case you can project the position of the satellite in its ellipse down onto the plane of the -earth's- equator, and figure how high above the earth's equatorial plane the satellite is.

But in the earth-sun case we 1) don't use solar-centered coordinates. 2) The sun's equatorial plane is the same as ecliptic - so there's no difference between the two at any radial distance. First approximation, at least. 3) You -can- define an alternate 'ecliptic' plane based on the weighted mass*inclinations of the planets, which shifts from a sun-earth plane to a (mostly) sun-Jupiter plane. The Earth can be 'above' and 'below' that plane - but that's not the coordinate system you were asking about.
 
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Hi TRF colleagues,

The three posts above are screenshots from Stellarium, which is superb and highly educational planetarium software. This software is free, but so far I have made a $50 donation to it.

Now, the first post above (#8) shows the astronomical situation at the coming March equinox (20 Mar 2025) 09:02 UTC. Notice that the Sun is at the intersection of the celestial equator (in blue) and the ecliptic (in orange).

The second post (#9) shows the astronomical situation at the coming June solstice (21 Jun 2025) 02:42 UTC. Notice that the Sun on the ecliptic is above the celestial equator.

And the third post above (#10) shows the astronomical situation at the coming September equinox (22 Sep 2025) at 18:20 UTC. Notice that the Sun is again at the intersection of the celestial equator and the ecliptic.

I did not show the coming December solstice, which will occur on 21 Dec 2025 at 15:03 UTC. At that time, the Sun on the ecliptic will be below the celestial equator.

So, I am asking genuine questions, because I am unclear on the answer. But when the Sun is above or below the celestial equator, what does that mean? And when the Sun on the ecliptic intersects the celestial equator, what does that mean?

Thank you.

Stanley
 
It means that at (local) noon, on either equinox, if you're standing on the equator, the sun will be directly overhead. And that at noon on the summer solstice, the sun is directly overhead if you're standing on the tropic of cancer. And the Tropic of Capricorn for the other.

Some globes used to have that line traced out.

An analemma for a location on the earth's equator would also trace out the changing angle. I think I've seen globes with those, too.
 
Hey - the sun is more tilted than I thought. 7.25° (from the earth defined ecliptic)

Earth's inclination to the invariable plane is (currently) 1.58°

Jupiter's
Designations
Inclination1.303° to ecliptic 6.09° to Sun's equator 0.32° to invariable plane
Longitude of ascending node100.464°
Time of perihelionJanuary 21, 2023
Argument of perihelion273.867°
 
An analemma for a location on the earth's equator would also trace out the changing angle.
The analemma has a figure 8 shape, the top (north) loop is smaller than the bottom (south) loop.

So, to Stanley's question, the two curves are not a factor of a sine apart, but it's close.

My go-to for this is the nautical almanac. Look at the changes through the year of
local apparent noon, time and declination.

Read about the Equation of Time: https://en.m.wikipedia.org/wiki/Equation_of_time

And here's the Nautical Almanac published by US Naval Observatory

https://thenauticalalmanac.com/TNARegular/2025 Nautical Almanac.pdf

Also, somewhere on the USNO site, you can download the software to print your own nautical almanac.

If anyone gets extremely nerdy into this stuff, pick up a copy of the book Positional Astronomy from the NO or maybe NHO.
 
It means that at (local) noon, on either equinox, if you're standing on the equator, the sun will be directly overhead. And that at noon on the summer solstice, the sun is directly overhead if you're standing on the tropic of cancer. And the Tropic of Capricorn for the other.
Thank you, @Charles_McG for that answer.

But explain this to me, please. I don't disagree with what you said. But can you determine those facts by looking at the Stellarium images that I posted? How would one do so?
 
Thank you, @Charles_McG for that answer.

But explain this to me, please. I don't disagree with what you said. But can you determine those facts by looking at the Stellarium images that I posted? How would one do so?
It's the distance between the curves - but you're trying to read the angles from a flat projection of a sky map. It's likely much harder than putting a ruler across the points and moving it to a scale. The mm to degrees scale will vary across parts of the flat map.
 
Thank you, @Charles_McG for that answer.

But explain this to me, please. I don't disagree with what you said. But can you determine those facts by looking at the Stellarium images that I posted? How would one do so?
Look at my post above. Calculate it. Don't try to get it from a flat screen. Sheesh.
 
Thank you, @Charles_McG for that answer.

But explain this to me, please. I don't disagree with what you said. But can you determine those facts by looking at the Stellarium images that I posted? How would one do so?
Remember that the celestial equator is just Earth's equator projected out into space. The ecliptic is the line that the sun and planets move on as they traverse the sky so the sun is on the ecliptic at all times. At the equinox, the sun would be directly over the Earth's equator and therefore on the celestial equator.

As for the Tropics of Cancer and Capricorn, the sun being directly over them at the solstices is how they are defined - the northernmost and southernmost latitudes at which the sun can be directly overhead.
 
At the equinoxes, the angle is zero.
At the solstices, the angle is 23.439281º, the tilt of the earth’s axis.

You have to realize, like @Antares JS was trying to state, that it’s the tilt of the earth’s spin axis relative to the plane established by the earth’s orbit around the sun that defines the Ecliptic and the Celestial Equator. Since in inertial space, that is defined by the positions of the stars, the Earth’s axis always points to the same place (ignoring the precession of that spin axis), it creates two planes that open and close against each other as the year progresses from the perspective of a viewer on the earth’s surface.
 
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I'm not sure if I'll help or add to the confusion. Isn't the ecliptic plane the plane of the earth's orbit about the sun? And the celestial equator is, as stated by others, is simply the projection of the earth' equator, or equatorial plane, into space. These are ALWAYS at a 23° angle (a tidge more) to one another due to the tilt of the earth's axis of rotation, and that inclination does not rotate relative to the sun, it is fixed in space. The equinoxes and solstices come about every 90° of the earth's orbit around the sun, so that one or the other occurs every 3 months.

What those sky charts above mean, I really can't tell.
 
Look at my post above. Calculate it. Don't try to get it from a flat screen. Sheesh.
Hi @cls,

I would very much like to calculate the distance between the two curves. At the moment, however, I don't understand how to use the equation of time, which you say would yield the answer.

The calculations for my Post #8 and Post #10 — the astronomical situation at the two equinoxes — are of course trivial. There, the two curves intersect, so the distance between them equals zero, and apparently a distance of zero translates to an angle of 90°. (And why does the equation of time mean that a distance of zero results in a 90° angle?)

But what about my Post #9 — the picture showing the astronomical situation at the June solstice? Would you please go through the calculations for the equinox, which result in an angle of approximately 23.4º.

Thank you very much.

Incidentally, I tremendously appreciate all my TRF colleagues who have weighed in on this challenging but enlightening issue.

Stanley
 
At the equinoxes, the angle is zero.
At the solstices, the angle is 23.439281º, the tilt of the earth’s axis.
Angle between what and what?

The two planes (the celestial equator and the ecliptic) always intersect at that angle. The intersection line runs through the center of the earth.

From any particular spot on earth, the celestial equator doesn't move; its position in the sky depends on your latitude. If you are at a pole, it's on the horizon all around you. If you're on the equator, it goes directly overhead. If you're at 39° north latitude, it will rise to 51° above the southern horizon.

From any particular spot on earth, the plane of the ecliptic will shift throughout the day. For example, imagine you're on the equator and it's an equinox. At noon, the ecliptic passes directly overhead. Six hours later, the earth has spun 90° and it's shifted 23° to south, so it's 67° above your southern horizon. Six more hours and it's back to directly overhead at midnight, then 67° above the northern horizon at 6 AM.

This pattern continues throughout the year, but shifts slightly each day, such that on the solstices, it's directly overhead at 6:00 AM and PM, and 67° above the horizon at noon and midnight.

If you move off the equator, the angles shift by your latitude. If you're at 5° north, it will range from 62° above the south horizon to 72° above the northern horizon. If you're at 39° north, it will range from 28° above the southern horizon to 74° above the southern horizon (never crossing directly overhead).

A consequence of this daily cycle is that (outside the tropics) in the winter when the sun's arc through the sky is low during the day, the ecliptic is high at night. In the summer, the sun goes high and the ecliptic is low at night. In winter, the full moon takes a high arc through the sky, and in summer it stays lower

Incidentally, if you have a clear sky, you can easily visualize the ecliptic in the sky now. Tonight, the moon, Mars, and Jupiter are making an obvious line, and if you follow it, it also runs through Venus. Tomorrow the moon will rise later and be farther from Mars and Jupiter, but you should still be able to see the line made by Mars, Jupiter, and Venus in the early evening, and the moon, Mars, and Jupiter later.
 
Sorry, wasn’t complete in my description…you know, we’re all describing the correct things, just from different perspectives. My angles are the apparent angles between the sun (the ecliptic) and the celestial equator at noon, on the equator.

Let me try this…
OP: So, I am asking genuine questions, because I am unclear on the answer. But when the Sun is above or below the celestial equator, what does that mean? And when the Sun on the ecliptic intersects the celestial equator, what does that mean?
The Stellarium images you’ve posted represent projections of Earth-centric things onto the sky, where the stars are the reference. It is showing you the apparent angles between things, there is no “above” or “below”. Stellarium is telescope software and telescopes are pointed only using angles, so everything it shows is a projection showing angular differences or time.

Perhaps you could use a different visualizer, “Eyes on the Solar System” at https://eyes.nasa.gov/apps/solar-system/#/home.
 
I would very much like to calculate the distance between the two curves. At the moment, however, I don't understand how to use the equation of time, which you say would yield the answer.
please study the wiki page for equation of time. the section :Major Components" explains the 2 major factors affecting the calculation: eccentricity, and obliquity. then, take a look at the section below it, "Mathematical description". yes, it's complicated!! but it needs to be correct. how correct? you don't have -precision instruments, so ... within a degree? that's a lot, the full moon is a half a degree of arc.


depending on how one counts, there are 2 or 3, or 5 or 7 different astronomical coordinate systems. as several posters noted, we're not really communicating because we are not all using the same system.

personally I use nautical coordinates because intuitively a Ptolemaic (pre-Copernican) universe is what I see. but I suspect other folks prefer geocentric equatorial. here's a good chart: https://en.wikipedia.org/wiki/Astronomical_coordinate_systems

perhaps what you really want is to understand declination, how high to the north the sun goes each day. depends on your time, or right ascension, or local hour angle. but if you are happy knowing the maximum each day at "noon", just look it up in the nautical almanac.

I hope some of this helps.
 
In navigation we make the assumption that the earth is the center of the universe and doesn't move. All things move around us. The celestial equator then doesn't move. It's extended out from earth's equator. The suns path in our celestial sky (relative to the stars) is the "path of the ecliptic". That reaches it's maximum angles north or south at the solstices, which is approximately 23.5°. (There is some slight variation). So your original question is to be measured in ANGULAR DISTANCE, or 23.5°, not linear distance. When the sun crosses the celestial equator, that's the equinoxes.
 
From any particular spot on earth, the celestial equator doesn't move; its position in the sky depends on your latitude. If you are at a pole, it's on the horizon all around you. If you're on the equator, it goes directly overhead. If you're at 39° north latitude, it will rise to 51° above the southern horizon.

From any particular spot on earth, the plane of the ecliptic will shift throughout the day. For example, imagine you're on the equator and it's an equinox. At noon, the ecliptic passes directly overhead. Six hours later, the earth has spun 90° and it's shifted 23° to south, so it's 67° above your southern horizon. Six more hours and it's back to directly overhead at midnight, then 67° above the northern horizon at 6 AM.
Hi TRF colleagues,

I have learned a lot from all the posts on this topic.

I found this post by @Azamiryou particularly instructive. The entire message was useful, especially the portion that I quoted.

Stanley
 
please study the wiki page for equation of time. the section :Major Components" explains the 2 major factors affecting the calculation: eccentricity, and obliquity. then, take a look at the section below it, "Mathematical description". yes, it's complicated!! but it needs to be correct. how correct? you don't have -precision instruments, so ... within a degree? that's a lot, the full moon is a half a degree of arc.


depending on how one counts, there are 2 or 3, or 5 or 7 different astronomical coordinate systems. as several posters noted, we're not really communicating because we are not all using the same system.

personally I use nautical coordinates because intuitively a Ptolemaic (pre-Copernican) universe is what I see. but I suspect other folks prefer geocentric equatorial. here's a good chart: https://en.wikipedia.org/wiki/Astronomical_coordinate_systems

perhaps what you really want is to understand declination, how high to the north the sun goes each day. depends on your time, or right ascension, or local hour angle. but if you are happy knowing the maximum each day at "noon", just look it up in the nautical almanac.

I hope some of this helps.
That's some fun stuff. hope I have time to peruse it better. Thanks!
 
Hi TRF colleagues,

At each equinox (vernal and autumnal), Earth’s ecliptic crosses the celestial equator. Thus, for us in the Northern hemisphere, at the March (or vernal) equinox Earth transits the celestial equator on its ecliptic path from south to north. Then at the September (or autumnal equinox) Earth crosses the celestial equator as it proceeds from north to south.

Therefore, at each equinox, the distance between Earth and the celestial equator is zero. So far so good.

Here is my question, please.

What is the distance between Earth and the celestial equator at other times during the year? What is that distance right now, for example? What is that distance at the December and the June solstices? I should be able to figure that last one out, but I am blanking on how to do it.

Thank you for your consideration.

Stanley
Well i believe that earth is biggest around the equator as thats why the force of gravity is less strong at the equator
 

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