The Moon is defined as full when its ecliptic longitude is 180$^{\circ}$ 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?


1) After first guessing (on this expert Q&A site) that the answer is "one lunar month" and asking me whether there was "a reason" why the said cycle might be different from it, user UhOh has requested clarification of the term "moon's apparent area", pointing out that the moon has a rough surface. The difference of the apparent perimeter of the Moon's disc from a circle does not have a bearing on the answer to this question that is of anything like the magnitude of the contribution made by the 5$^{\circ}$ tilt of the plane of the Moon's orbit around the Earth relative to the plane of the Earth's orbit around the Sun. Please ignore mountains and craters. Please assume either that the Moon is a sphere, or if you wish then take it to be a spheroid with an eccentricity of about 1/900. The question is about the relative movements of the three celestial bodies concerned. (If further clarification is required of any other term used, please let me know.)

2) Wouldn't it be great if there were a website where you could input your latitude and longitude and desired start date and end date and get a graph of the proportion of the moon's apparent disc that is sunlit from your location? But I don't think this exists (yet!)

  • $\begingroup$ Naively I would guess it's also one lunar month, at least on average. Is there some reason that it might be different? $\endgroup$ – uhoh Jan 11 at 6:51
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    $\begingroup$ If there wasn't a reason I wouldn't have asked. The plane of the Moon's orbit around the Earth is inclined at an angle of 5$^{\circ}$ to the plane of the Earth's around the Sun. To get a handle on why that affects the Moon's apparent fullness, imagine if it were 90$^{\circ}$. An article on the possible origins of the lunar inclination appeared in Nature in 2015. Given the inclination I don't think the proportion ever reaches 50%. The 5$^{\circ}$ angle is not completely constant. This is not a simple problem. $\endgroup$ – ruffle Jan 11 at 12:57
  • $\begingroup$ Local maxima occur once per synodic month, yes, but the cycle is more complicated. $\endgroup$ – ruffle Jan 11 at 13:42
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    $\begingroup$ okay, then how exactly should "the maximum proportion of the moon's apparent area is sunlit" be defined? The Moon isn't a sphere, it has a rough, contoured surface. And what if when all the orbital-mechanical details and approximations for the geography are taken into account one near-maximum has a 0.001% less illuminated proportion than the next? Would it no longer be considered a maximum? Or if 0.1% less? If you want a detailed answer you'll have to define your terms better and constrain the scope of the answer more clearly. $\endgroup$ – uhoh Jan 11 at 13:54
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    $\begingroup$ Thank you for the edit! I've reversed my down vote to up again and removed my vote to close. Asking for clarification is routine in Stack Exchange and the goal of clarification is to make it easier for people to figure out what kind of answer that the question's author is hoping for. Sometimes people are afraid to put a lot of work into writing in-depth answers to two-sentence questions to then receive a comment "that's not what I meant". Now that you've added much more detail, your question is much clearer and readers have a better idea what to take into account in an answer. $\endgroup$ – uhoh Jan 11 at 23:51

I asked the same question on the Usenet group sci.astro and received the following helpful reply from Barry Schwarz that goes some way towards an answer, reposted here with permission. But the question remains hard. I have not seen this cycle addressed in any astronomical textbook, although it may be in one that I haven't looked at.

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 pole 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 the 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.


The illuminated fraction k is related to the Sun-Moon-Earth phase angle i by

$$ k = \frac{1 + \cos i}{2} $$

At full moon, $i \approx |\beta|$ where β is the Moon's ecliptic latitude, so k is between 0.998 and 1.000. The Moon crosses the ecliptic twice each draconic month. Synodic (29.53 day) and draconic (27.21 day) months together have a beat cycle length of 346.6 days. Maxima of k occur twice as often, 173.3 days apart on average. The same interval separates eclipse seasons.

Using the DE430 ephemeris, Skyfield's almanac functions, and Matplotlib, here are illuminated fractions vs. dates of several full moons, compared with a sine wave of amplitude 0.001 and period 173.3 days.

plot: illuminated fraction vs. full moon date

This assumes a point of view at the center of the Earth. Far northern or southern observers would see the Moon displaced up to 1° south or north, biasing the mean topocentric β away from 0. In the plot of k vs. time, alternate maxima would shift left and right, and alternate minima would shift up and down.


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