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The present northern direction to the nearest 0.5 degrees with degree zero being the start of January and measured counterclockwise (same as Earth's orbit relative to celestial north pole).

If the assumptions below do not allow a precision of 0.5 degrees then I imagine this question is unanswerable without specifying a specific date and time, in which case the specific date and time would be 1 January 2017, 00:00:00 hours.

Assumptions: (1) That the Earth's orbit around the sun is a perfect circle (I know it's very slightly elliptical but intuitively imagine that won't significantly affect the precision I need). (2) That the Sun's center is at the center of Earth's circular orbit (I know the Sun's center isn't exactly but don't imagine that will affect the accuracy I need either).

Note 1: I've found three general answers from three different sites, but, two of them appear to me to conflict and the other is too imprecise.

Note 2: This is my first post to this site and I did not plow through the hundreds of questions related to these tags. So if this question has already been answered I apologize and please be merciful and point me in right direction.

Thank you

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  • $\begingroup$ I'm not sure I follow what you're asking. It's more common to compare the plane of the Earth's axis to the plane of the Sun's equator, cause those are only a few degrees off and the Sun's equator is a simple 90 degrees to the Sun's axis. You can do it either way, but I would err towards the traditional comparison of orbital plane to equatorial plane. Also, it might be easier to see what you mean if you'd post your sites to point out more clearly where you see a conflict and imprecision. $\endgroup$
    – userLTK
    Dec 7, 2016 at 12:47
  • $\begingroup$ Thank you userLTK for your response. Apologies for any ambiguity; accordingly I've edited the title of my question. I hope the edit is useful. I understand your point re the more common comparison; for my purposes, the Earth's actual orbital plane around the Sun is most useful. Apologies also for not providing links to the sites I mentioned. I had these recorded on a HD that wasn't backed up and failed months ago sometime after first pursuing this question. I will attempt to find them and provide them thru edit of my question's body. $\endgroup$
    – emansnas
    Dec 7, 2016 at 16:38

2 Answers 2

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Well, Wikipedia says that it's 7.25°to the ecliptic and gives a reference.

Looking there clarifies the value in 3 dimensions:

North Pole of Rotation

Right Ascension: 286.13
Declination : 63.87

Since those coordinates are relative to the Earth’s axis, being a practical measurement system for Earth ground observers to use in pointing their telescopes, you need to combine that with Earth's orbit.

As a useful visualization, mark that point on a model of the celesial sphere. Compare that with the ecliptic pole: the difference between thise points is what you want to know.

I expect you’ll find they are separated by 7.25°, but by marking a model sphere you’ll understand the actual direction.

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  • $\begingroup$ Thank you JDługosz for your useful response. I remember reading the Wikipedia article you mention some time ago but did not notice/pursue the reference you cite. A related problem: I am unclear re the relationship of the ecliptic plane to the celestial plane - the sources I've read seem to conflict. Those two sources would be Wikipedia's article: Ecliptic, and [link](hyperphysics.phy-astr.gsu.edu/hbase/eclip.html‎). This problem pertains to the plane of orientation of the Sun's axial tilt. $\endgroup$
    – emansnas
    Dec 7, 2016 at 17:07
  • $\begingroup$ Post another question. $\endgroup$
    – JDługosz
    Dec 7, 2016 at 17:11
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This isn't really a great answer, but it has an image:

enter image description here

Planetarium programs like Stellarium let you view the sky from different planets, including the Sun. As above, the Sun's "north pole" is in Draco, very close to NGC 6552 (note it's declination puts it about 2.5 minutes away from the Sun's north pole).

Although Stellarium is fairly accurate, this answer shouldn't be regarded as canonical.

REMINDER: Do not observe directly from the surface of the Sun. You will be burnt to a crisp ;)

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  • $\begingroup$ Thank you @barrycarter for your interesting answer. I was unaware of the Stellarium program, I find that knowledge to be of potential value. Re your REMINDER: Indeed, I imagine one would in fact be vaporized into one's ~$4.50 worth of elemental components :) As time allows, I've been doing additional research and hope to calculate the answer to my specific question before too long. If so, I will post my answer here on SE. During the process I'm discovering what I think may turn out to be some interesting aspects of the Sun's axial orientation. If so, I will also post those here also. $\endgroup$
    – emansnas
    Dec 10, 2016 at 8:09

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