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Old January 12th 20, 03:35 AM posted to sci.astro
Barry Schwarz[_3_]
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Default What is the cycle for when the maximum amount of the Moon's apparent area is sunlit?

On Sat, 11 Jan 2020 20:43:08 -0000 (UTC), Harold Davis
wrote:

The Moon is defined as full when its ecliptic longitude is 180° distant
from the Sun's.

But what is the cycle for when the maximum proportion of the Moon's
apparent area is sunlit, and for observers at what latitudes of the Earth's
surface is this maximum realised?

Note that the plane of the Moon's orbit around the Earth is inclined at an
angle of 5° to the plane of the Earth's around the Sun. To get a handle on
why this affects the Moon's apparent fullness, imagine if it were 90°.
Given the inclination I am sceptical as to whether the proportion ever
reaches 50%.


Since the Sun's diameter exceeds the Moon's, the Sun always
illuminates slightly more than 50% of the Moon's surface. (To
visualize, draw the external common tangents from a larger circle to a
smaller one.) Using 98M miles as the distance, 432K miles as the
Sun's radius, and 1K miles for the Moon's, the Sun illuminates almost
7 miles into the opposite hemisphere.

Any observer a finite distance from the Moon can only ever see less
than 50% of the Moon's surface. The further away the observer is, the
closer the observable percentage is to 50. (To visualize, draw the
two tangents from an external point to a circle.) However, when the
Moon is in the plane of the ecliptic, an observer at the north pole
can see approximately 12 miles over the Moon's north pole into the
opposite hemisphere. So he can see some 5 miles beyond the
terminator. Similar geometry causes his view of the Moon's south pose
to fall short of the terminator.

However, for a person standing in line with the ecliptic, his view in
any direction falls 4 miles short of the terminator and that is the
maximum you seek.

When the Moon is within 4K miles of the ecliptic, an observer on Earth
at he same offset from the ecliptic will have the view described
above.

When the Moon is more than 4K miles away from the ecliptic, all
Earth-bound observers see some non-illuminated portion and therefore
do not see the maximum you want.

Not only is the Moon's orbit tilted with respect to the ecliptic, but
the direction of the tilt wobbles like a child's top. I have no idea
how to factor that into an equation that would compute the time
between one visible maximum and another..

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