<p>The question is:</p>

<blockquote>
  <p>If you stand on the equator of the Moon, at midday your position occurs when the Moon is located at the line of the node. What is the altitude of the sun at that moment?</p>
</blockquote>

<p>Information: This is when the Moon is located at the line of node (line of node is a line that occurs from crossing between the ecliptic plane and the Moon's orbital plane, which makes an angle of approximate 5 degrees). The equatorial plane of the Sun makes an angle of 1 degrees, 32 arc minutes and 40 arc seconds.</p>

<p>The solution is 88 degrees, 27 arc minutes and 20 arc seconds through 90 degrees.</p>

<p>But I don't understand why the answer isn't 1 degree, 32 arc minutes and 20 arc seconds, because the ecliptic plane where the sun is located is above the horizon or celestial equator. equal to its altitude.</p>

CC BY-SA 3.0
2015-01-03T21:29:08.543Z
Think about what your proposed answer means for a bit. Does it make any sense? (The answer is no.) Your answer means that at local noon on the Moon's equator you'll see the Sun just 1.5 degrees above the horizon. On Uranus, maybe. On the Moon? No. The Sun is 1.5 degrees shy of vertical, not 1.5 degrees above the horizon.
David Hammen
CC BY-SA 3.0
2015-01-08T03:24:22.783Z
Your comment was very helpful. I understand why the altitude should be 88.5 now,but why the solution is 88.5 through 90 degrees? Isn't it should be one altitude, not in a range?
user9686
CC BY-SA 3.0
2015-01-11T05:47:10.887Z
Does the line of note relate to the result in range?
user9686
<the-moon><the-sun>
How do you find the altitude of the Sun if you are on the Moon?
user9686
# How do you find the altitude of the Sun if you are on the Moon?

(Asked by: username_0 on 2015-01-03)

The question is:

> If you stand on the equator of the Moon, at midday your position occurs when the Moon is located at the line of the node. What is the altitude of the sun at that moment?

Information: This is when the Moon is located at the line of node (line of node is a line that occurs from crossing between the ecliptic plane and the Moon's orbital plane, which makes an angle of approximate 5 degrees). The equatorial plane of the Sun makes an angle of 1 degrees, 32 arc minutes and 40 arc seconds.

The solution is 88 degrees, 27 arc minutes and 20 arc seconds through 90 degrees.

But I don't understand why the answer isn't 1 degree, 32 arc minutes and 20 arc seconds, because the ecliptic plane where the sun is located is above the horizon or celestial equator. equal to its altitude.

**Tags:** the-moon, the-sun

> **Comments:**
> **username_1**: Think about what your proposed answer means for a bit. Does it make any sense? (The answer is no.) Your answer means that at local noon on the Moon's equator you'll see the Sun just 1.5 degrees above the horizon. On Uranus, maybe. On the Moon? No. The Sun is 1.5 degrees shy of vertical, not 1.5 degrees above the horizon.  
> **username_0**: Your comment was very helpful. I understand why the altitude should be 88.5 now,but why the solution is 88.5 through 90 degrees? Isn't it should be one altitude, not in a range?  
> **username_0**: Does the line of note relate to the result in range?  

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## 0 answers

