As is well known, and as you can see in the image below, the far side of the Moon is brighter than its near side. What is its overall albedo? And what would be the magnitude of a full moon if the (current) far side were facing us? sides of moon

  • $\begingroup$ Magnitude means nothing here unless you specify where from. From Earth, presumably but you need to make this clear. $\endgroup$ Commented Aug 9, 2023 at 16:46
  • $\begingroup$ Oh no! The far side cannot be seen from earth! I mean when facing the far side, so the moon is between you and earth. $\endgroup$
    – George Lee
    Commented Aug 9, 2023 at 20:40
  • $\begingroup$ I know the far side cannot be seen from Earth. This is why your question is poorly specified. Unless you define a distance then the magnitude question is impossible to answer - from a long way away is Moon will be very dim for instance. $\endgroup$ Commented Aug 10, 2023 at 10:34
  • $\begingroup$ ok, edited, I hope its clear enough now. $\endgroup$
    – George Lee
    Commented Aug 10, 2023 at 20:43
  • 1
    $\begingroup$ Related: astronomy.stackexchange.com/questions/41264/… $\endgroup$ Commented Aug 11, 2023 at 9:42

2 Answers 2


This is a bit rough, since I can't find the exact numbers anywhere. However the lunar highlands have an albedo that is double the lunar mare (about 0.2 compared to 0.1)(source). The far side is almost completely highland, so its albedo is about 0.2

But only about 1/3 of the nearside is lunar mare (source).

So the relevant calculation is $(1/3×0.1 +2/3×0.2)\div 0.2 =0.83$ That is the nearside is 83% of the far side in albedo. Or the far side is 20% brighter.

In terms of magnitude, a 20% increase in brightness corresponds to about 0.2 magnitudes brighter. The full moon has magnitude -12.7, so a far side full moon would have magnitude -12.9.

These are fairly rough figures, but the impression is that a "far-side" moon would be brighter, but not exceptionally brighter than the moon as we see it now.

  • $\begingroup$ This would make the albedo of the near side 0.167, while the highest number you can find for this is 0.137. Source: the-moon.us/wiki/Albedo. $\endgroup$
    – George Lee
    Commented Aug 10, 2023 at 22:01
  • $\begingroup$ Yes, this is all a bit rough, I think that's within the kind of error boundaries I imagined for this calculation. You could reduce the number "0.2" to 0.17, and see if that makes a difference $\endgroup$
    – James K
    Commented Aug 10, 2023 at 23:51
  • $\begingroup$ This study should be useful to you: articles.adsabs.harvard.edu//full/1968SvA....11.1034L/… $\endgroup$ Commented Aug 11, 2023 at 9:53
  • $\begingroup$ The LRO ( Lunar Reconnaissance Orbiter) has gathered a ton of lunar albedo data, eg the LAMP data, but I think it's mostly looking at UV albedo (because it's looking for water). lunar.gsfc.nasa.gov/dataproducts.html $\endgroup$
    – PM 2Ring
    Commented Aug 11, 2023 at 19:50
  • $\begingroup$ Yes, but it's local, not integrated over a whole hemisphere. One could do the integration, and perhaps someone has, but I can't find it. If someone does find it, I'd encourage them to write an answer. $\endgroup$
    – James K
    Commented Aug 11, 2023 at 21:07

We may be able to shed some light on this question by looking at the magnitude of the Moon from the Lunar Reconnaissance Orbiter (LRO).

Ideally, we'd like data from a satellite in a high altitude equatorial orbit, but the LRO is in a polar orbit (inclination ~85.2° to the lunar equator), and it has a rather low altitude. Its semi-major axis is ~1830 km, but the Moon's radius is ~1737.4 km. Wikipedia claims that the LRO is in an eccentric orbit, but that information is out of date, and its current eccentricity is ~0.00129, but admittedly its altitude deviates from that of a simple Kepler ellipse, due to the lunar mascons. The inclination of its orbit is currently quite stable, though.

The LRO has JPL Horizons ID -85. Its trajectory data was recently updated, on 2023-Aug-3. You can see its osculating orbit elements for 2023-Aug-1 18:30 TDB here. That time is close to the most recent full moon (2023-Aug-1 18:31:40 UTC).

The LRO has a sidereal period of almost 2 hours (1h 57m 2.9s). Here's a plot showing its distance from the centre of the Moon, covering 6 hours, with a 2 minute timestep.

LRO-Moon distance I made that plot using the script here. I have a magnitude plotting script here.

When it's full moon, the far side is dark. When it's new moon the far side is fully illuminated. So we can get a rough idea of the difference in albedo by looking at the magnitude of the Moon at those times.

Here's a Moon magnitude plot covering the same timespan (using the same timestep).

Moon magnitude, Full

And here is the corresponding plot for the previous new moon (2023-Jul-17 18:32:50 UTC).

Moon magnitude, New

The mean magnitude on the new moon is -21.466, which is slightly brighter than the full moon mean of -21.168. Interestingly, the peak magnitude at full moon is slightly higher than at new moon. The minima during the new moon are noticeably brighter that at full moon. I assume that's mostly due to earthshine.

However, Horizons doesn't say how it computes these magnitudes, and it gives this warning:

Moon's approximate apparent visual magnitude and surface brightness. When phase angle < 7 deg (within ~1 day of full Moon), computed magnitude tends to be about 0.12 too small.

  • $\begingroup$ You can plot LRO's orbit in 3D using my script at the end of this answer: astronomy.stackexchange.com/a/49823/16685 $\endgroup$
    – PM 2Ring
    Commented Aug 11, 2023 at 21:59
  • $\begingroup$ I understand that the magnitude plots include when the LRO was at the dark side of the Moon. So what causes the brightness of over -17? And does that make the data unreliable? $\endgroup$
    – George Lee
    Commented Aug 11, 2023 at 22:17
  • $\begingroup$ @George I'm not sure why Horizons gives such a high brightness over the whole orbit. It shouldn't be getting any sunlight when it's in the middle of the night side. It's altitude is ~100 km, so the Moon should completely block the Sun. $\endgroup$
    – PM 2Ring
    Commented Aug 12, 2023 at 0:27
  • $\begingroup$ So I'm afraid that we can't accept anything from that data as is. $\endgroup$
    – George Lee
    Commented Aug 14, 2023 at 16:11

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