<blockquote>
<p>How much brighter is full-earth-shine on the moon, than full-moon-shine on earth?</p>
</blockquote>
<p>On average, it's about <a href="https://www.wolframalpha.com/input/?i=%28%28earth+radius%29%2F%28moon+radius%29%29%5E2+*+%28%28earth+albedo%29%2F%28moon+albedo%29%29" rel="nofollow noreferrer">41 times brighter</a> -- in terms of luminosity, or a bit over four times brighter (<span class="math-container">$2.5 \log_{10} 41 \approx 4.03$</span>) in terms of the logarithmic response of the human eye.</p>
<p>There are two factors that come into play in the luminosity calculation: The much larger size of the Earth than the Moon (a factor of over 13, the square of the ratio of the Earth's radius to the Moon's radius), and the much larger albedo of the Earth than the Moon (a factor of a bit over 3).</p>
<p>This is an average value. If the full Earth view from the Moon happens to be centered on a cloudless Pacific Ocean, that factor of three due to albedo becomes a factor of less than one. On the other hand, if the full Earth view from the Moon happens to be centered on a very cloudy Earth, that factor of three due to albedo becomes a factor of five or more.</p>
<p>The Moon's albedo (0.12) is roughly the same as that of slightly aged asphalt. The Earth's albedo varies a lot. The Earth's oceans are rather dark when seen from space, as dark as freshly laid asphalt. Clouds can have an extremely high albedo, up to 0.9, or 18 times the albedo of water.</p>

CC BY-SA 4.0
2021-02-08T11:00:54.727Z
+1. Would the moon's lack of atmosphere be an influencing factor in the light reaching the moon surface at night, from a full earth? In other words, would it be easier, or more difficult, to read a newspaper under a full earth, if a moon atmosphere was in place?
Constantthin
CC BY-SA 4.0
2021-02-08T14:46:46.793Z
Given that we have full Earth, [sunglint](https://en.wikipedia.org/wiki/Sunglint) should increase albedo considerably.
Ruslan
CC BY-SA 4.0
2021-02-09T01:27:42.800Z
@Ruslan. Thx. I had to read the word “sunglint” a few times before I realised that it says “sunglint”, and not “sunlight” :)
Constantthin
CC BY-SA 4.0
2021-02-09T09:22:33.480Z
Wow. The moon looks sooo bright in the night sky (much brighter that "slightly aged asphalt"), so is that just an optical illusion due to the black background?
jcaron
CC BY-SA 4.0
2021-02-09T10:25:12.653Z
@jcaron - https://i.sstatic.net/OVZLJ.jpg
David Hammen
CC BY-SA 4.0
2021-02-09T11:23:30.937Z
Nice. Is the venue lit up by the earth, or the sun, in that picture?
Constantthin
CC BY-SA 4.0
2021-02-09T19:10:39.400Z
@Constantthin The photo is from [Apollo 12](https://en.wikipedia.org/wiki/Apollo_12#ALSEP). I couldn't 100% confirm the information (I probably didn't search for the right term), but they were lit by the sun at least at some point while on the moon (that's how the color video camera got [destroyed](https://en.wikipedia.org/wiki/Apollo_12#Lunar_surface_activities)), and given the length of a lunar day, I assume the whole mission (and that picture) was lit by the sun. It seems it was the case for every Apollo mission? I'm surprised I didn't find a question about that on the stack.
Luris
CC BY-SA 4.0
2021-02-09T20:34:11.323Z
@Constantthin All Apollo missions occurred under sunlit conditions. Landing occurred an Earth day or two after sunrise had occurred at the landing site (think of roughly an hour or two after sunrise on the Earth) to enhance the crew's ability to see shadows, and launch occurred an Earth day or two before lunar noon to avoid extreme lunar heating. The longest mission on the Moon was a bit over three Earth days long.
David Hammen
CC BY-SA 4.0
2021-02-09T20:36:19.197Z
@Luris: I added a linked question on this site in which an answer by uhoh contains this very image. I added the link under the question.
David Hammen
CC BY-SA 4.0
2021-02-10T02:33:24.427Z
@David Hammen. I suppose that temperature must have played a (small?) part in their choice also, considering temperature swings between +200 and -200?
Constantthin
CC BY-SA 4.0
2021-02-10T12:06:01.087Z
Now I'm wondering how bright the Full Moon would be if the Moon had a high albedo, eg if we covered it in chalk dust...
PM 2Ring
CC BY-SA 4.0
2021-02-10T13:16:21.570Z
@PM2Ring Chalk dust? Why not titanium dioxide? That would raise the Moon's albedo to nearly 1, or a factor of eight increase in luminosity. But that's only a bit over a factor of two in terms of perceived brightness; the human eye has a logarithmic response. I updated by answer to reflect the difference between luminosity and magnitude.
David Hammen
CC BY-SA 4.0
2021-02-10T15:30:16.270Z
Well, I did originally think of titanium dioxide, but I settled on chalk because it's less dense and a lot cheaper. :) Hopefully, I'd get a bulk discount on 94.8 billion (metric) tons of chalk: that'd give 1 mm depth over the whole Moon. But of course I can ignore most of the far side.
PM 2Ring
CC BY-SA 4.0
2021-02-13T12:08:28.383Z
So your answer is 4 times brighter, not 41 times?
Constantthin
CC BY-SA 4.0
2021-02-13T19:39:45.460Z
@Constantthin It depends on what one means by "brighter". If it's luminosity, then the answer is 41 (better, 42, because that's the answer to life, the universe, and everything). If it's magnitude, then the answer is 4 because magnitude is a $\log_{2.5}$ function.
David Hammen
CC BY-SA 4.0
2021-02-13T22:26:16.443Z
Is it not clear on my question which of the two I am asking for?
Constantthin
CC BY-SA 4.0
2021-02-13T23:01:01.860Z
@Constantthin No, it is not. Both luminosity and magnitude are valid measures of the concept of "brightness".
David Hammen
CC BY-SA 4.0
2022-05-05T06:25:03.360Z
The log measure should be stated as "four magnitudes brighter", not "four times brighter". Likewise, in a comment, you said "a factor of two in terms of perceived brightness", but that should be "an difference of two magnitudes in terms of perceived brightness". Once you have taken the log, the comparison is additive, not multiplicative. If A is three magnitudes brighter than B and B is three magnitudes brighter than C, then A is six magnitudes brighter than C. Your phrasing like "perceived as three 'times' brighter" would lead to thinking that A is perceived as *nine* 'times' brighter than C.
nanoman
CC BY-SA 4.0
2021-02-08T09:07:45.713Z
2021-02-10T13:14:12.003Z
2021-02-10T13:14:12.003Z
David Hammen
<p>The definition of brightness, according to Wikipedia:</p>
<blockquote>
<p>the term brightness in astronomy is generally used to refer to an object's apparent brightness: that is, how bright an object appears to an observer. Apparent brightness depends on both the luminosity of the object and the distance between the object and observer, and also on any absorption of light along the path from object to observer.</p>
</blockquote>
<p><strong>How many times brighter is full-earth-shine as seen from the moon's surface, than full-moon-shine as seen from the earth's surface?</strong></p>
<p>Different internet sources give different figures. I have seen figures as low as  3x (<a href="https://androidweatherapps.wordpress.com/tag/earth/" rel="nofollow noreferrer">here</a>), and as high as 100x (<a href="http://webserver.dmt.upm.es/%7Eisidoro/tc3/Space%20environment.pdf" rel="nofollow noreferrer">here</a>). What is the right figure?</p>
<p>1st reference to be found In the last paragraph of the article.</p>
<p>2nd reference to be found on page 26.</p>
<p>Ethan on <a href="https://www.forbes.com/sites/startswithabang/2017/03/18/ask-ethan-how-bright-is-the-earth-as-seen-from-the-moon/?sh=52d3dd4d769d" rel="nofollow noreferrer">this</a> website claims that the figure is 43x.</p>
<p>Besides an original answer, it would be good to get explanations on where the references made wrong calculations to get to their different conclusions.</p>

CC BY-SA 4.0
2021-02-08T06:02:57.180Z
what is ggr? The ration should be about 4x - just the geometric ratio. Add a bit albedo difference and it might depend on the actual cloudiness of the Earth at the time
planetmaker
CC BY-SA 4.0
2021-02-08T06:04:49.443Z
@planetmaker. You are right. will change it to x.
Constantthin
CC BY-SA 4.0
2021-02-08T06:33:27.960Z
@planetmaker the albedo difference will surprise you; don't dismiss it so quickly.
uhoh
CC BY-SA 4.0
2021-02-08T08:44:50.683Z
@planetmaker Your factor of about four is the ratio of the Earth's radius to the Moon's radius. You need to square this, resulting in a factor of 13.45 rather than 3.67. Then you need to multiply by a factor of 3.06  to factor in the much higher albedo of the Earth than that of the Moon.
David Hammen
CC BY-SA 4.0
2021-02-08T08:52:05.210Z
@DavidHammen yes... without looking it up, I grossly overestimated the lunar radius compared to Earth :) Thx for pointing it out
planetmaker
CC BY-SA 4.0
2021-02-09T20:35:28.990Z
Related: [Why does the Moon appear gray when passing between the Sun and the Earth?](https://astronomy.stackexchange.com/questions/16704)
David Hammen
CC BY-SA 4.0
2021-02-14T05:36:50.273Z
Nothing in this world is certain except death, taxes, and  OP necroing this post
user177107
CC BY-SA 4.0
2021-02-16T23:04:47.937Z
Consequently, there seems to be different ways of calculating this, which is confusing since they come to different conclusions. Because what I really want to know is how many times easier it is to read a newspaper on the moons surface on a night when a full earth is brightest, compared with doing it on the earth’s surface on a cloudless high pressure night under the full moon?
Constantthin
<the-moon><albedo>
How many times brighter would full-earth-shine appear to people standing on the moon, than full-moon-shine does to people on earth?
Constantthin
# How many times brighter would full-earth-shine appear to people standing on the moon, than full-moon-shine does to people on earth?

(Asked by: username_0 on 2021-02-08)

The definition of brightness, according to Wikipedia:

> the term brightness in astronomy is generally used to refer to an object's apparent brightness: that is, how bright an object appears to an observer. Apparent brightness depends on both the luminosity of the object and the distance between the object and observer, and also on any absorption of light along the path from object to observer.

**How many times brighter is full-earth-shine as seen from the moon's surface, than full-moon-shine as seen from the earth's surface?**

Different internet sources give different figures. I have seen figures as low as 3x ([here](https://androidweatherapps.wordpress.com/tag/earth/)), and as high as 100x ([here](http://webserver.dmt.upm.es/%7Eisidoro/tc3/Space%20environment.pdf)). What is the right figure?

1st reference to be found In the last paragraph of the article.

2nd reference to be found on page 26.

Ethan on [this](https://www.forbes.com/sites/startswithabang/2017/03/18/ask-ethan-how-bright-is-the-earth-as-seen-from-the-moon/?sh=52d3dd4d769d) website claims that the figure is 43x.

Besides an original answer, it would be good to get explanations on where the references made wrong calculations to get to their different conclusions.

**Tags:** the-moon, albedo

> **Comments:**
> **username_1**: what is ggr? The ration should be about 4x - just the geometric ratio. Add a bit albedo difference and it might depend on the actual cloudiness of the Earth at the time  
> **username_0**: @username_1. You are right. will change it to x.  
> **username_2**: @username_1 the albedo difference will surprise you; don't dismiss it so quickly.  
> **username_3**: @username_1 Your factor of about four is the ratio of the Earth's radius to the Moon's radius. You need to square this, resulting in a factor of 13.45 rather than 3.67. Then you need to multiply by a factor of 3.06  to factor in the much higher albedo of the Earth than that of the Moon.  
> **username_1**: @DavidHammen yes... without looking it up, I grossly overestimated the lunar radius compared to Earth :) Thx for pointing it out  
> **username_3**: Related: [Why does the Moon appear gray when passing between the Sun and the Earth?](https://astronomy.stackexchange.com/questions/16704)  
> **username_4**: Nothing in this world is certain except death, taxes, and  OP necroing this post  
> **username_0**: Consequently, there seems to be different ways of calculating this, which is confusing since they come to different conclusions. Because what I really want to know is how many times easier it is to read a newspaper on the moons surface on a night when a full earth is brightest, compared with doing it on the earth’s surface on a cloudless high pressure night under the full moon?  

---

## 1 answers

### Answer by username_3 on 2021-02-08. Score: 31 | Accepted answer

> How much brighter is full-earth-shine on the moon, than full-moon-shine on earth?

On average, it's about [41 times brighter](https://www.wolframalpha.com/input/?i=%28%28earth+radius%29%2F%28moon+radius%29%29%5E2+*+%28%28earth+albedo%29%2F%28moon+albedo%29%29) -- in terms of luminosity, or a bit over four times brighter ($2.5 \log\_{10} 41 \approx 4.03$) in terms of the logarithmic response of the human eye.

There are two factors that come into play in the luminosity calculation: The much larger size of the Earth than the Moon (a factor of over 13, the square of the ratio of the Earth's radius to the Moon's radius), and the much larger albedo of the Earth than the Moon (a factor of a bit over 3).

This is an average value. If the full Earth view from the Moon happens to be centered on a cloudless Pacific Ocean, that factor of three due to albedo becomes a factor of less than one. On the other hand, if the full Earth view from the Moon happens to be centered on a very cloudy Earth, that factor of three due to albedo becomes a factor of five or more.

The Moon's albedo (0.12) is roughly the same as that of slightly aged asphalt. The Earth's albedo varies a lot. The Earth's oceans are rather dark when seen from space, as dark as freshly laid asphalt. Clouds can have an extremely high albedo, up to 0.9, or 18 times the albedo of water.

> **Comments:**
> **username_0**: +1. Would the moon's lack of atmosphere be an influencing factor in the light reaching the moon surface at night, from a full earth? In other words, would it be easier, or more difficult, to read a newspaper under a full earth, if a moon atmosphere was in place?  
> **username_5**: Given that we have full Earth, [sunglint](https://en.wikipedia.org/wiki/Sunglint) should increase albedo considerably.  
> **username_0**: @username_5. Thx. I had to read the word “sunglint” a few times before I realised that it says “sunglint”, and not “sunlight” :)  
> **username_6**: Wow. The moon looks sooo bright in the night sky (much brighter that "slightly aged asphalt"), so is that just an optical illusion due to the black background?  
> **username_3**: @username_6 - https://i.sstatic.net/OVZLJ.jpg  
> **username_0**: Nice. Is the venue lit up by the earth, or the sun, in that picture?  
> **username_7**: @username_0 The photo is from [Apollo 12](https://en.wikipedia.org/wiki/Apollo_12#ALSEP). I couldn't 100% confirm the information (I probably didn't search for the right term), but they were lit by the sun at least at some point while on the moon (that's how the color video camera got [destroyed](https://en.wikipedia.org/wiki/Apollo_12#Lunar_surface_activities)), and given the length of a lunar day, I assume the whole mission (and that picture) was lit by the sun. It seems it was the case for every Apollo mission? I'm surprised I didn't find a question about that on the stack.  
> **username_3**: @username_0 All Apollo missions occurred under sunlit conditions. Landing occurred an Earth day or two after sunrise had occurred at the landing site (think of roughly an hour or two after sunrise on the Earth) to enhance the crew's ability to see shadows, and launch occurred an Earth day or two before lunar noon to avoid extreme lunar heating. The longest mission on the Moon was a bit over three Earth days long.  
> **username_3**: @username_7: I added a linked question on this site in which an answer by username_2 contains this very image. I added the link under the question.  
> **username_0**: @username_3. I suppose that temperature must have played a (small?) part in their choice also, considering temperature swings between +200 and -200?  
> **username_8**: Now I'm wondering how bright the Full Moon would be if the Moon had a high albedo, eg if we covered it in chalk dust...  
> **username_3**: @PM2Ring Chalk dust? Why not titanium dioxide? That would raise the Moon's albedo to nearly 1, or a factor of eight increase in luminosity. But that's only a bit over a factor of two in terms of perceived brightness; the human eye has a logarithmic response. I updated by answer to reflect the difference between luminosity and magnitude.  
> **username_8**: Well, I did originally think of titanium dioxide, but I settled on chalk because it's less dense and a lot cheaper. :) Hopefully, I'd get a bulk discount on 94.8 billion (metric) tons of chalk: that'd give 1 mm depth over the whole Moon. But of course I can ignore most of the far side.  
> **username_0**: So your answer is 4 times brighter, not 41 times?  
> **username_3**: @username_0 It depends on what one means by "brighter". If it's luminosity, then the answer is 41 (better, 42, because that's the answer to life, the universe, and everything). If it's magnitude, then the answer is 4 because magnitude is a $\log_{2.5}$ function.  
> **username_0**: Is it not clear on my question which of the two I am asking for?  
> **username_3**: @username_0 No, it is not. Both luminosity and magnitude are valid measures of the concept of "brightness".  
> **username_9**: The log measure should be stated as "four magnitudes brighter", not "four times brighter". Likewise, in a comment, you said "a factor of two in terms of perceived brightness", but that should be "an difference of two magnitudes in terms of perceived brightness". Once you have taken the log, the comparison is additive, not multiplicative. If A is three magnitudes brighter than B and B is three magnitudes brighter than C, then A is six magnitudes brighter than C. Your phrasing like "perceived as three 'times' brighter" would lead to thinking that A is perceived as *nine* 'times' brighter than C.  

---

