<p>I wanted to ask this question even though I realized the correct geometry in the middle of typing it. So I will answer this myself.</p>

<p>First of all, <strong>the sub-lunar point</strong> never exceeds 28.545 N/S. If you're standing at the sub-lunar point, the Moon will be directly overhead. The Moon never wanders outside this latitude bracket.</p>

<p>But that does not mean the Moon's shadow can't wander outside it. Here's one way it could happen.</p>

<p><a href="https://i.sstatic.net/rjrKl.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/rjrKl.png" alt="enter image description here"></a></p>

<p><strong>Edit:</strong> this is my attempt to draw something approximately to scale. I think I got the Earth/Moon sizes about right, but the Moon-Earth distance should around 4x greater. And the solar rays are of course not parallel but I don't think it's possible to draw or perceive tiny angles for this scale.</p>

<p>The green line is the ecliptic. The Sun and Moon are supposed to be aligned along a horizontal line, but for some reason I drew the solar rays half and half beside the green line which is deceptive. I'll try to edit the pic when I can.</p>

<p>In technical parlance, the Moon could be slightly above or slightly below its <a href="https://en.wikipedia.org/wiki/Lunar_node" rel="nofollow noreferrer">Node</a>. If the Moon was <strong>exactly</strong> at its node, then the umbra would absolutely be centered somewhere between 28.5 N/S (actually 23.4 N/S because the Sun never goes beyond that). But as you can see, the farther it gets from its node, the steeper the shadow impacts Earth and the farther towards a pole it gets. Too far from the node, and the shadow will not intercept Earth at all and we won't have an eclipse.</p>

<p>If you draw a line from the center of the Moon to the center of the Earth, you can see that the sub-lunar point is definitely within 28.5 degrees N/S. But the shadow <strong>does not</strong> necessarily follow that imaginary line. The shadow always follows the <strong>line from the Sun to the Moon.</strong></p>

<p>You could even imagine a very extreme case where the shadow falls on "the other side" of the pole. I think that may be the case <a href="https://en.wikipedia.org/wiki/Solar_eclipse_of_August_12,_1942" rel="nofollow noreferrer">here</a> since that eclipse looks really short.</p>

CC BY-SA 3.0
2017-08-22T23:20:20.543Z
@barrycarter Yep, but declination is a measure of absolute coordinate sky system. In a relative coordinate system (altitude? i think), the sun and moon could be any elevation above the horizon, right down to the horizon itself. But yeah I see your point that I should **not** have added 23.4 and 5.1 degrees in my original thought process. I think the 2nd paragraph of my answer is still correct in that the sublunar point is bound by 28.5 N/S. About the drawing, they are ofc not parallel, it was just my attempt to draw something approximately to scale. I should edit with some info on that.
DrZ214
CC BY-SA 3.0
2017-08-21T20:15:26.540Z
2017-08-23T19:22:31.857Z
2017-08-23T19:22:31.857Z
DrZ214
<p>The ecliptic (Earth's orbital plane) is inclined 23.4 degrees. This is the same as Earth's axial tilt. The Moon's orbit is inclined 5.145 degrees to the ecliptic.</p>

<p>Therefore, the maximum latitude where a solar eclipse produces an umbra <strong>should</strong> be 28.545 N and 28.545 S.</p>

<p>But this is clearly not the case. Wikipedia has a huge list of solar eclipses with charts. <a href="https://en.wikipedia.org/wiki/Solar_eclipse_of_October_12,_1939" rel="nofollow noreferrer">Here's one</a> that goes very near the south pole. How is this possible?</p>

<p><strong>Edit:</strong> Actually I missed something. Since the Moon and Sun must be in the same place, and the Sun can never exceed 23.4 degrees declination, solar eclipses must be bound by 23.4 N/S. I should not have added the two angles. Thanks to user:berrycenter for pointing it out.</p>

<orbit><the-moon><solar-eclipse><ecliptic>
Why do some solar eclipses' umbra cross arctic/antarctic regions?
DrZ214
# Why do some solar eclipses' umbra cross arctic/antarctic regions?

(Asked by: username_0 on 2017-08-21)

The ecliptic (Earth's orbital plane) is inclined 23.4 degrees. This is the same as Earth's axial tilt. The Moon's orbit is inclined 5.145 degrees to the ecliptic.

Therefore, the maximum latitude where a solar eclipse produces an umbra **should** be 28.545 N and 28.545 S.

But this is clearly not the case. Wikipedia has a huge list of solar eclipses with charts. [Here's one](https://en.wikipedia.org/wiki/Solar_eclipse_of_October_12,_1939) that goes very near the south pole. How is this possible?

**Edit:** Actually I missed something. Since the Moon and Sun must be in the same place, and the Sun can never exceed 23.4 degrees declination, solar eclipses must be bound by 23.4 N/S. I should not have added the two angles. Thanks to user:berrycenter for pointing it out.

**Tags:** orbit, the-moon, solar-eclipse, ecliptic


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## 1 answers

### Answer by username_0 on 2017-08-21. Score: 7 | Accepted answer

I wanted to ask this question even though I realized the correct geometry in the middle of typing it. So I will answer this myself.

First of all, **the sub-lunar point** never exceeds 28.545 N/S. If you're standing at the sub-lunar point, the Moon will be directly overhead. The Moon never wanders outside this latitude bracket.

But that does not mean the Moon's shadow can't wander outside it. Here's one way it could happen.

[![enter image description here](https://i.sstatic.net/rjrKl.png)](https://i.sstatic.net/rjrKl.png)

**Edit:** this is my attempt to draw something approximately to scale. I think I got the Earth/Moon sizes about right, but the Moon-Earth distance should around 4x greater. And the solar rays are of course not parallel but I don't think it's possible to draw or perceive tiny angles for this scale.

The green line is the ecliptic. The Sun and Moon are supposed to be aligned along a horizontal line, but for some reason I drew the solar rays half and half beside the green line which is deceptive. I'll try to edit the pic when I can.

In technical parlance, the Moon could be slightly above or slightly below its [Node](https://en.wikipedia.org/wiki/Lunar_node). If the Moon was **exactly** at its node, then the umbra would absolutely be centered somewhere between 28.5 N/S (actually 23.4 N/S because the Sun never goes beyond that). But as you can see, the farther it gets from its node, the steeper the shadow impacts Earth and the farther towards a pole it gets. Too far from the node, and the shadow will not intercept Earth at all and we won't have an eclipse.

If you draw a line from the center of the Moon to the center of the Earth, you can see that the sub-lunar point is definitely within 28.5 degrees N/S. But the shadow **does not** necessarily follow that imaginary line. The shadow always follows the **line from the Sun to the Moon.**

You could even imagine a very extreme case where the shadow falls on "the other side" of the pole. I think that may be the case [here](https://en.wikipedia.org/wiki/Solar_eclipse_of_August_12,_1942) since that eclipse looks really short.

> **Comments:**
> **username_0**: @barrycarter Yep, but declination is a measure of absolute coordinate sky system. In a relative coordinate system (altitude? i think), the sun and moon could be any elevation above the horizon, right down to the horizon itself. But yeah I see your point that I should **not** have added 23.4 and 5.1 degrees in my original thought process. I think the 2nd paragraph of my answer is still correct in that the sublunar point is bound by 28.5 N/S. About the drawing, they are ofc not parallel, it was just my attempt to draw something approximately to scale. I should edit with some info on that.  

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