<p>If the Moon's orbit were fixed, the area covered by the Moon would just be $360^{\circ}*(0.5^{\circ}+2*1^{\circ})$. The 5.3 degree inclination relative to the ecliptic (or some other inclination relative to the equator) would not be a factor.</p>

<p>What is important to this question about the coverage is that the orbit changes over time. In particular, the position of the ascending node (where the Moon's orbit crosses the ecliptic) moves <em>along the ecliptic</em> with a period of 18.6 years (Wikipedia article on <a href="https://en.wikipedia.org/wiki/Lunar_precession" rel="noreferrer">Lunar precession</a>). </p>

<p>In a plot of the sky using ecliptic coordinates (so that the ecliptic is a straight line), the 5.3 degree inclination of the Moon's orbit to the ecliptic is shown as "year X" in my figure. The precession changes the orbit to positions such as year Y and year Z in my figure, and during the entire 18.6 year cycle, the entire band within 5.3 degrees of the ecliptic is covered by the Moon. In other words, the ascending node moves through all 360 degrees of ecliptic longitude. This results in the term 2*5.3 in your equation. (Note: I do not know if the precession is to the "right" or "left", so do not interpret my figure to mean that year Y is later than year X. Year Y could be earlier than X. I'm sure someone will correct me!)</p>

<p><a href="https://i.sstatic.net/XTqeu.png" rel="noreferrer"><img src="https://i.sstatic.net/XTqeu.png" alt="ecliptic projection"></a></p>

<p>If the same plot were made in equatorial coordinates (so that the equator is a straight line), the band covered by the Moon is of course the same on the sky. However, the precession along the ecliptic limits the ascending node of the Moon's orbit on the equator to a small range or right ascension; the ascending node does not travel around the entire equator. Representing that band using equatorial coordinates and determining the area inside the bands is much more difficult.</p>

<p><a href="https://i.sstatic.net/Md9RL.png" rel="noreferrer"><img src="https://i.sstatic.net/Md9RL.png" alt="equatorial projection"></a></p>

CC BY-SA 4.0
2018-09-14T16:54:48.157Z
2018-09-14T16:54:48.157Z
JohnHoltz
<p>I was solving a task that said (paraphrasing):</p>

<blockquote>
  <p>What percentage of the night sky can the Moon cover during the entire
  year, when observed from all points on Earth (what percentage of the
  night sky can be studied with the occultation method)?</p>
</blockquote>

<p><br>
This is a task I read from a book, and the solution states that the Moon covers a "ribbon" of width $\alpha$ where $\alpha = 0.5 ^{\circ} + 2*1^{\circ} + 2*5.3 ^ {\circ} = $ angular diameter of the Moon + 2*the Moon's daily parallax + 2*the inclination of the Moon's orbit to the ecliptic; and length $360 ^{\circ}$.<br>
Why is this like this, I understand the diameter and parallax part, but I couldn't visualize the problem very well. Why is the third number the inclination to the ecliptic and not to the equator?</p>

CC BY-SA 4.0
2018-09-13T14:18:23.943Z
I suspect they're using the ecliptic as a reference plane for all calculations, that's all.
Carl Witthoft
CC BY-SA 4.0
2018-09-13T15:34:10.507Z
@CarlWitthoft I doubt it. If one were to look at the equator the way this solution looks at the ecliptic, and use the equator as a reference plane they would certainly obtain different results (since only the inclination would be different, and, in layman's terms, the sky is the same sphere, only "tilted"). It seems the ecliptic must be used and I'm asking why.
Tosic
CC BY-SA 4.0
2018-09-15T15:34:10.763Z
@barrycarter Yes, I meant the lunar parallax. "Moon's daily parallax" is the literal translation from my native language.
Tosic
<orbit><the-moon><apparent-motion>
What percentage of the celestial sphere can the Moon cover?
Tosic
# What percentage of the celestial sphere can the Moon cover?

(Asked by: username_0 on 2018-09-12)

I was solving a task that said (paraphrasing):

> What percentage of the night sky can the Moon cover during the entire
> year, when observed from all points on Earth (what percentage of the
> night sky can be studied with the occultation method)?

This is a task I read from a book, and the solution states that the Moon covers a "ribbon" of width $\alpha$ where $\alpha = 0.5 ^{\circ} + 2\*1^{\circ} + 2\*5.3 ^ {\circ} = $ angular diameter of the Moon + 2\*the Moon's daily parallax + 2\*the inclination of the Moon's orbit to the ecliptic; and length $360 ^{\circ}$.  
Why is this like this, I understand the diameter and parallax part, but I couldn't visualize the problem very well. Why is the third number the inclination to the ecliptic and not to the equator?

**Tags:** orbit, the-moon, apparent-motion

> **Comments:**
> **username_1**: I suspect they're using the ecliptic as a reference plane for all calculations, that's all.  
> **username_0**: @username_1 I doubt it. If one were to look at the equator the way this solution looks at the ecliptic, and use the equator as a reference plane they would certainly obtain different results (since only the inclination would be different, and, in layman's terms, the sky is the same sphere, only "tilted"). It seems the ecliptic must be used and I'm asking why.  
> **username_0**: @barrycarter Yes, I meant the lunar parallax. "Moon's daily parallax" is the literal translation from my native language.  

---

## 1 answers

### Answer by username_2 on 2018-09-14. Score: 4 | Accepted answer

If the Moon's orbit were fixed, the area covered by the Moon would just be $360^{\circ}\*(0.5^{\circ}+2\*1^{\circ})$. The 5.3 degree inclination relative to the ecliptic (or some other inclination relative to the equator) would not be a factor.

What is important to this question about the coverage is that the orbit changes over time. In particular, the position of the ascending node (where the Moon's orbit crosses the ecliptic) moves *along the ecliptic* with a period of 18.6 years (Wikipedia article on [Lunar precession](https://en.wikipedia.org/wiki/Lunar_precession)).

In a plot of the sky using ecliptic coordinates (so that the ecliptic is a straight line), the 5.3 degree inclination of the Moon's orbit to the ecliptic is shown as "year X" in my figure. The precession changes the orbit to positions such as year Y and year Z in my figure, and during the entire 18.6 year cycle, the entire band within 5.3 degrees of the ecliptic is covered by the Moon. In other words, the ascending node moves through all 360 degrees of ecliptic longitude. This results in the term 2\*5.3 in your equation. (Note: I do not know if the precession is to the "right" or "left", so do not interpret my figure to mean that year Y is later than year X. Year Y could be earlier than X. I'm sure someone will correct me!)

[![ecliptic projection](https://i.sstatic.net/XTqeu.png)](https://i.sstatic.net/XTqeu.png)

If the same plot were made in equatorial coordinates (so that the equator is a straight line), the band covered by the Moon is of course the same on the sky. However, the precession along the ecliptic limits the ascending node of the Moon's orbit on the equator to a small range or right ascension; the ascending node does not travel around the entire equator. Representing that band using equatorial coordinates and determining the area inside the bands is much more difficult.

[![equatorial projection](https://i.sstatic.net/Md9RL.png)](https://i.sstatic.net/Md9RL.png)


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