<p>The author says that the difference between the factor of 100 and the geometric increase due to area and albedo is because of &quot;atmospheric absorption and directional effects&quot;. 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.</p>
<p>The reflectance of solar system objects shows an <a href="https://en.wikipedia.org/wiki/Opposition_surge" rel="nofollow noreferrer">opposition surge</a> at small phase angles.</p>
<p>It is difficult to see the full Moon at really small phase angles (<span class="math-container">$&lt;2$</span> degrees) from Earth because you would get a lunar eclipse. However, it has been measured by spacecraft and according to <a href="https://www.sciencedirect.com/science/article/abs/pii/S0019103596902250" rel="nofollow noreferrer">Buratti et al. 1996</a>, the brightness of the Moon increases by &quot;more than 40%&quot; between phase angles of just 4 degrees and zero.</p>
<p>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.</p>
<p>I suspect this effect plus the ~20% absorption in the Earth's atmosphere is what the author of your reference is talking about.</p>

CC BY-SA 4.0
2021-02-20T00:47:21.270Z
+1. For presenting an interesting,  well thought out, plausible theory.
Constantthin
CC BY-SA 4.0
2021-02-20T12:56:41.223Z
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
CC BY-SA 4.0
2021-02-20T15:59:18.943Z
@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
CC BY-SA 4.0
2021-02-20T23:40:56.130Z
“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
CC BY-SA 4.0
2021-02-19T19:01:17.593Z
2021-02-19T19:01:17.593Z
ProfRob
<p>I don't understand the reasoning of the author of <a href="http://webserver.dmt.upm.es/%7Eisidoro/tc3/Space%20environment.pdf" rel="nofollow noreferrer">this</a> 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?</p>

CC BY-SA 4.0
2021-02-19T13:43:21.843Z
@ProfRob. No, because the answer there proposed a figure of 41 times, which I think may be right. There is a big difference between 43 and 100.
Constantthin
CC BY-SA 4.0
2021-02-19T13:51:39.443Z
@ProfRob. No. My question  is not what the right figure is, but how the author is reasoning. Maybe his figure of 100 applies to a different measurement, that equals 41 in Apparent Brightness. If so, I don’t understand what it is? Or if he possibly could have made a mistake?
Constantthin
CC BY-SA 4.0
2021-02-19T13:51:51.537Z
OK, well the author says in the same paragraph that the difference is  "because of atmospheric absorption and directional effects." So I guess this is what you are asking about. Absorption might account for a ~20% increase.
ProfRob
CC BY-SA 4.0
2021-02-19T13:56:04.073Z
Related: https://astronomy.stackexchange.com/questions/41277/how-many-times-brighter-would-full-earth-shine-appear-to-people-standing-on-the?noredirect=1&lq=1
ProfRob
CC BY-SA 4.0
2021-02-19T14:04:39.890Z
@ProfRob. So you are saying that the 41 times answer didn’t take that in consideration?
Constantthin
CC BY-SA 4.0
2021-02-19T14:10:46.773Z
@ProfRob. You mentioned 20% increase. That couldn’t right, could it? The difference between 100 and 41 has to be more than 100%.
Constantthin
CC BY-SA 4.0
2021-02-19T14:11:35.213Z
My 41% figure used the average albedo of the Earth. Thick clouds are extremely reflective while oceans and forests are extremely absorptive. The brightness of the full Earth as seen from the Moon can vary by an order of magnitude, depending on conditions on the Earth and what part of the Earth one is seeing.
David Hammen
CC BY-SA 4.0
2021-02-19T14:13:26.780Z
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
CC BY-SA 4.0
2021-02-19T14:28:43.380Z
@David Hammen. According to ‘the Project Management Body of Knowledge’, a ROM has only a swing of max 50%.
Constantthin
CC BY-SA 4.0
2021-02-19T14:30:09.450Z
@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
CC BY-SA 4.0
2021-02-19T14:33:40.367Z
@David Hammen. If your figure is the average figure, then the swing can be max 41 either way. In these theoretical, non likely, scenarios the swing is 100% either way. However,  an Apparent Brightness of 100 is 143% greater than 41. There is a discrepancy here that is a bit mysterious.
Constantthin
CC BY-SA 4.0
2021-02-19T14:38:53.413Z
@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
CC BY-SA 4.0
2021-02-19T18:50:12.320Z
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
CC BY-SA 4.0
2021-02-19T23:29:02.363Z
One interesting piece of information I just dug up from the net is that a calm ocean has high reflectability, while a rough ocean has a low.
Constantthin
<the-moon><earth><brightness><visible-light>
How can the author of this article conclude that earth-shine on the moon is 100 times brighter than moon-shine on earth?
Constantthin
# How can the author of this article conclude that earth-shine on the moon is 100 times brighter than moon-shine on earth?

(Asked by: username_0 on 2021-02-19)

I don't understand the reasoning of the author of [this](http://webserver.dmt.upm.es/%7Eisidoro/tc3/Space%20environment.pdf) 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?

**Tags:** the-moon, earth, brightness, visible-light

> **Comments:**
> **username_0**: @username_1. No, because the answer there proposed a figure of 41 times, which I think may be right. There is a big difference between 43 and 100.  
> **username_0**: @username_1. No. My question  is not what the right figure is, but how the author is reasoning. Maybe his figure of 100 applies to a different measurement, that equals 41 in Apparent Brightness. If so, I don’t understand what it is? Or if he possibly could have made a mistake?  
> **username_1**: OK, well the author says in the same paragraph that the difference is  "because of atmospheric absorption and directional effects." So I guess this is what you are asking about. Absorption might account for a ~20% increase.  
> **username_1**: Related: https://astronomy.stackexchange.com/questions/41277/how-many-times-brighter-would-full-earth-shine-appear-to-people-standing-on-the?noredirect=1&lq=1  
> **username_0**: @username_1. So you are saying that the 41 times answer didn’t take that in consideration?  
> **username_0**: @username_1. You mentioned 20% increase. That couldn’t right, could it? The difference between 100 and 41 has to be more than 100%.  
> **username_2**: My 41% figure used the average albedo of the Earth. Thick clouds are extremely reflective while oceans and forests are extremely absorptive. The brightness of the full Earth as seen from the Moon can vary by an order of magnitude, depending on conditions on the Earth and what part of the Earth one is seeing.  
> **username_2**: 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.  
> **username_0**: @username_2. According to ‘the Project Management Body of Knowledge’, a ROM has only a swing of max 50%.  
> **username_2**: @username_0 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.  
> **username_0**: @username_2. If your figure is the average figure, then the swing can be max 41 either way. In these theoretical, non likely, scenarios the swing is 100% either way. However,  an Apparent Brightness of 100 is 143% greater than 41. There is a discrepancy here that is a bit mysterious.  
> **username_2**: @username_0 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.  
> **username_1**: 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.  
> **username_0**: One interesting piece of information I just dug up from the net is that a calm ocean has high reflectability, while a rough ocean has a low.  

---

## 1 answers

### Answer by username_1 on 2021-02-19. Score: 3

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](https://en.wikipedia.org/wiki/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](https://www.sciencedirect.com/science/article/abs/pii/S0019103596902250), 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.

> **Comments:**
> **username_0**: +1. For presenting an interesting,  well thought out, plausible theory.  
> **username_0**: 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?  
> **username_1**: @username_0 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.  
> **username_0**: “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)  

---

