At lunar midnight (i.e. the new moon as seen from Earth), the Earth is in its full phase with its entire disk in sunlight, and it is the brightest object in the lunar sky. How bright is it, and how variable is its brightness?

More precisely, I'm interested in specific figures for the magnitude and in comparisons to similar objects. Would I be able to read in that light? To drive? Would I notice the variations in brightness? What are the main variations due to?

  • 1
    $\begingroup$ According to stellarium, the full Earth reaches a brightness of magnitude -16.21, but I have no idea how they came up with that calculation or how accurate it is. answers.yahoo.com/question/index?qid=20110309115214AAy4Rk8 may or may not be helpful. $\endgroup$
    – user21
    Commented Dec 4, 2014 at 23:36
  • $\begingroup$ Just putting this up for comparison, but Pluto-time is in the neighborhood of -19.2 (based on 1,000 times less bright than the sun from Earth), so, "full Earth" would be some 10-20 times less bright than Pluto time. You might be able to read by a full Earth, but just barely and your eye Dr. might recommend against it. nasa.gov/feature/… $\endgroup$
    – userLTK
    Commented May 12, 2017 at 16:22
  • $\begingroup$ You can read and drive by a full moon if your eyes are completely dark adapted. Due to the ubiquity of artificial light, most people have never experienced full dark adaptation,,, it takes about 1/2 hour. But most people will flick on a light within a few seconds. $\endgroup$
    – Woody
    Commented Nov 16, 2023 at 20:36

2 Answers 2


Let's take an average albedo for the Earth of 0.3 (it depends, which hemisphere is visible, how much cloud cover etc.). That means the Earth reflects 30% of the light incident upon it.

The flux $f$ falling on the Earth is given by $$ f_{\odot} = \frac{L_{\odot}}{4\pi d^2} = 1.369\times10^{3}\ Wm^{-2}$$ where $L_{\odot}=3.85\times10^{26}\ W$ from the Sun and $d= 1$AU.

The integrated luminosity from the illuminated hemisphere will be $$L_{earth} = 0.3\pi R^2 f = 5.2\times10^{16}\ W$$

So now we can compare this with the Sun. One hemisphere of the Sun radiates $1.93\times10^{26}\ W$, and produces a flux of $1.369\times10^{3}\ Wm^{-2}$ at 1 AU. Therefore the illuminated hemisphere of the Earth results in a flux of approximately $f_E=0.056\ Wm^{-2}$, assuming the average Earth-Moon distance of 384,400 km. This calculation assumes isotropic emission, but it is quite likely that the albedo is higher for light reflected through 180 degrees.

The Sun has an apparent magnitude of -26.74, so the magnitude of the "full-earth" at the moon is $$ m_{Earth} = 2.5\log_{10}\left( \frac{f_{\odot}}{f_{E}}\right) - 26.74 = \underline{-15.77}$$

The answer will of course vary with the albedo of the visible hemisphere, which in turn depends on the time of year and how much of the polar regions can be seen (e.g. http://www.climatedata.info/Forcing/Forcing/albedo.html ). Variations of a few hundredths seem possible, which will lead to apparent magnitude variations in $m_{Earth}$ of $\sim \pm 0.1-0.2$ mag. The albedo may also vary in detail with the exact angle at which the sunlight hits the Earth - an "opposition surge" in brightness, when the Sun-Earth and Moon are almost aligned is possible. The Earth-Moon distance varies from 363,000 to 405,000 km. This will lead to magnitude variations of $\pm 0.12$ mag.

A further way to check this is that the albedo of the Moon is 0.12 and it has a radius of 0.273 times that of the Earth. Therefore the Earth seen from the Moon ought to be $(0.3/0.12)\times(1/0.273)^2 = 33.5$ times brighter. This is 3.81 magnitudes brighter. The mean magnitude of the full Moon is -12.74 (maximum is -12.92), so the brightness of the "full Earth" should be -16.55 on average.

I am not sure why these figures don't agree; I suspect it is that the albedo for reflection when the Sun's light is normally incident on the Moon is quite a bit larger than 0.12. The so-called "opposition surge". If the Earth's albedo behaves in the same way, then the latter figure may be more accurate than my first calculation. My gut instinct is that the answer is somewhere between the two.

  • $\begingroup$ Thank you for the detailed calculation. What do these numbers compare to, though? $\endgroup$ Commented Dec 21, 2014 at 18:58
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    $\begingroup$ @EmilioPisanty 33.5 times brighter than the full moon. $\endgroup$
    – ProfRob
    Commented Dec 21, 2014 at 22:29

Ask Ethan: How bright is the Earth as seen from the Moon? has some detailed explanations, including discussions of lunar eclipses as seen from the moon. It doesn't include magnitude calculations, but it concludes that

a “full Earth” as seen from the Moon is about 43 times brighter than the full Moon is as seen from Earth. When the icecaps are larger and the cloud cover is greater — and also when the deserts are visible in the Sun — the Earth appears at its brightest, up to approximately 55 times brighter than the Moon.

log(43) to the base 2.512 is 4.1, which when subtracted from the full moon's mean magnitude of -12.7 yields magnitude -16.8. Using the "55 times brighter" figure would lead to a maximum brightness of -12.7 - 4.4 = -17.1.

It doesn't seem to cover the effects of the distance between the Earth and Moon, so it might get even brighter at perigee.


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