# What percentage of the celestial sphere can the Moon cover?

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?

• I suspect they're using the ecliptic as a reference plane for all calculations, that's all. – Carl Witthoft Sep 13 '18 at 14:18
• @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 Sep 13 '18 at 15:34
• When you say "moon's daily parallax", do you mean different observers see the moon in different positions since the moon is so close to the Earth? EG, an observer may see the moon 1/2 degree north of its geocentric position? – barrycarter Sep 15 '18 at 14:47
• @barrycarter Yes, I meant the lunar parallax. "Moon's daily parallax" is the literal translation from my native language. – Tosic Sep 15 '18 at 15:34

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.