Longitude of ascending node of the Earth

TheBro95

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Hello everyone!

I am a Sixth Form student and I am writting a programme in Mathematica to calculate the position of bodies of the solar system and other stuff.
The problem is that I am not able to understand why the longitude of the ascending node of the Earth's orbit is 348.74 degrees. Why isn't it 0?
The ascending node for the Earth is the vernal point, isn't it? And the longitude of the ascending node is the angle between the vernal point and the ascending node.
I am very confused and I would really appreciate if anybody could help me understand it despite my poor knowledge of astronomy.

Thank you very much for your help!
 

sorindafabico

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The longitude of the ascending node is the angle between the vernal point and the ascending node.

500px-Orbit1.svg.png
 

TheBro95

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Thank you for your posts but I still have the doubt. I do know what the longitude of the ascending node is, but the problem is that kind of paradox, I mean:

If the longitude of the ascending node is the angle between the vernal point and the ascending node, and the ascending node of the orbit of the earth is the vernal point, then they are the same and the longitude of the ascending node should be 0, why the hell is it -11.26064 instead of 0??

I cannot understand it at all :facepalm:
 

eldalie

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If the longitude of the ascending node is the angle between the vernal point and the ascending node, and the ascending node of the orbit of the earth is the vernal point, then they are the same and the longitude of the ascending node should be 0, why the hell is it -11.26064 instead of 0??

Hello everybody, I'm having the same problem of TheBro95...

maybe is something due to precession of equinoxes?

so the 0 point of the vernal point could be the "true" vernal point, and the -11.26064 could be the "mean" vernal point (corrected for earth axis nutation)?

I'm not sure...
:(
 

TheBro95

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Thank you very much for your post eldalie.
I also thought that could be the answer, because I have managed to do good calculations supposing that the longitude of the ascending node of the Earth is the angle between the point of Aries which points to the Aries constellation and the current (J2000) point of Aries due to precession of equinoxes.
But still I would like to really know what the answer is, I cannot find any answer on the internet :S
 

Kinetics

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This is a good question - I was wondering about this too.

Does the -11 means -11 degrees

It cannot have moved that far in only 12 years. It takes 26,000 years to precess 360 degrees.

The problem is most things show the Earth not having any ascending node as they use the the Earth's orbit to determine the orbit plane hence they measure inclination as 0.000005 etc so not much information on this subject

http://upload.wikimedia.org/wikipedia/commons/f/f0/Seasons1.svg
 
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ADSWNJ

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djw

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Here is my answer to what was a very astute question, posed almost ten years ago. Essentially, the question was: why is the longitude of the ascending node of the Earth's orbit 348.74 degrees, rather than zero? The questioner clearly understood that Longitude angle (a.k.a. Right Ascension) is measured in the reference plane (taken to be Earth's orbital plane), between "the" reference direction (i.e. "the" vernal equinox, seen from Earth) and "the" ascending node. The questioner also clearly understood that, in the case of Earth's orbit, the ascending node is the vernal point direction and, by definition, the reference direction, so the longitude angle should be zero. And that, in principle, is true, hence the befuddlement. [NB I write "the" in quotes, because all "the" parameters change gradually with time, which is basically what complicated matters.]

Consider these data, reproduced from

https://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html

Earth Mean Orbital Elements (J2000)

Semimajor axis (AU) 1.00000011
Orbital eccentricity 0.01671022
Orbital inclination (deg) 0.00005
Longitude of ascending node (deg) -11.26064
Longitude of perihelion (deg) 102.94719
Mean Longitude (deg) 100.46435

Key to answering the original question is to note that the inclination is given as 0.00005 deg, not zero. This means that we are dealing with two planes, inclined only very, very slightly to one another. Such a small inclination means that the nodes - both ascending and descending - could be almost anywhere; their longitude, almost any angle between 0 and 360 deg. Their position becomes increasingly sensitive and (here is the main point) increasingly irrelevant, as the inclination decreases.

Precisely why it is that the above J2000 (simplified) orbital elements have this particular small inclination between the reference plane and "the" orbit plane, and this particular longitude, I am not entirely certain. I suspect it will be to maximize the usefulness ("average accuracy") in the current epoch of the static, Keplerian orbits that the above, constant orbital elements will produce. What I can say with confidence is that for school / college / amateur computations, the above elements can be used to satisfactorily propagate orbits. An informative exercize would be to investigate the sensitivity of the propagated orbits to changes in this longitude. For more serious calculations, known rates-of-change w.r.t. time of these elements are available. Even dealing with the above data properly requires some care, and I am happy to explain in more detail any of this. [As an aside, note that 360 - 11.26 = 348.74, which reconciles the two different longitude numbers mentioned above].

Incidentally, I came across this question and associated (growing) thread yesterday, almost ten years after it was posed, and so impressed was I by such a question from a then high-school 6th former that I got up early today to "sign up" on this forum, just so I could give what I think is a reasonable and long-overdue answer.

D.J. Walker, Wirral, UK.
29th Oct. 2022
 

Ajaja

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Key to answering the original question is to note that the inclination is given as 0.00005 deg, not zero. This means that we are dealing with two planes, inclined only very, very slightly to one another. Such a small inclination means that the nodes - both ascending and descending - could be almost anywhere; their longitude, almost any angle between 0 and 360 deg. Their position becomes increasingly sensitive and (here is the main point) increasingly irrelevant, as the inclination decreases.
Yes, for example if you try to get elliptical osculating elements for Earth for an arbitrary date from Horizons System ( https://ssd.jpl.nasa.gov/horizons/app.html#/ ) using J2000 reference frame you get almost random LAN between 0 and 360 deg. :)
 
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