How can the author of this article conclude that earth-shine on the moon is 100 times brighter than moon-shine on earth? [duplicate]

I don't understand the reasoning of the author of this article, on page 26. How can he conclude that full earth-shine on the moon is 100 times brighter than full moon-shine on earth?

• – ProfRob Feb 19 at 13:56
• That said, the article you quoted is not peer reviewed. Another factor is that the author appears to be giving a rough order of magnitude (ROM) calculation. In such a calculation, 32 rounds up to 100 while 30 rounds down to 10. – David Hammen Feb 19 at 14:13
• @Constantthin The Earth's albedo is about 0.3 to 0.4, depending on wavelength and source. Clouds can have an albedo of 0.8, ice, 0.9. So that's more than a factor of two swing. You are reading too much into this self-published article. – David Hammen Feb 19 at 14:30
• @Constantthin You are reading too much into a self-published, non-peer reviewed article. Also, ROM, done correctly, means to the nearest base ten order of magnitude. The cutoff points between rounding up versus rounding down are integral powers of ten times the square root of ten. – David Hammen Feb 19 at 14:38
• Exactly. I don't know how they got a factor of 100 because they don't say. They say it is a factor of 100 and not a factor of 30-40 because of "absorption and directional effects". I have no idea whether there is a big opposition surge or not. The fact is we never see the Moon directly illuminated by the Sun with a normal reflection because when the Sun Earth and Moon are in line, you get an eclipse. That is not the case for "full Earth", where normal reflection can almost be observed. So probably you get a bigger opposition surge. – ProfRob Feb 19 at 18:50

The author says that the difference between the factor of 100 and the geometric increase due to area and albedo is because of "atmospheric absorption and directional effects". The former is probably only about 20% or so (i.e. about 20% of moonlight is absorbed/scattered in the Earth's atmosphere). The latter effect could be quite big.

The reflectance of solar system objects shows an opposition surge at small phase angles.

It is difficult to see the full Moon at really small phase angles ($$<2$$ degrees) from Earth because you would get a lunar eclipse. However, it has been measured by spacecraft and according to Buratti et al. 1996, the brightness of the Moon increases by "more than 40%" between phase angles of just 4 degrees and zero.

When the Sun illuminates the Earth as seen from the Moon then it would be much easier to see at very small phase angles because the Moon's shadow is that much smaller.

I suspect this effect plus the ~20% absorption in the Earth's atmosphere is what the author of your reference is talking about.

• +1. For presenting an interesting, well thought out, plausible theory. – Constantthin Feb 20 at 0:47
• 40% + 20% = 60%. 160% of 41 = 65.6 times brighter. 100 - 65.6 = 34.4. The missing 34.4 times needs to be accounted for? – Constantthin Feb 20 at 12:56
• @Constantthin I don't know what the opposition surge looks like for the Earth, the 40% figure is for the Moon. As has been explained to you at length the albedo used by the author of your non peer-reviewed article is some sort of average and could easily be larger if the visible Earth is cloudy. The figure of 20% is also not a fixed quantity, assumes that the Moon is at zenith, and depends on your altitude and air quality. Your insistence on agreement to even two (let alone three) significant figures is not something I'm interested in pursuing. – ProfRob Feb 20 at 15:59
• “Since Gehrels' early studies, an opposition surge has been noted for most airless solar system bodies. No such surge has been reported for bodies with significant atmospheres.” (Wikipedia:Opposition Surge) – Constantthin Feb 20 at 23:40