Setting: we have a fantasy world which is a planet with much of the same conditions that can be seen on Earth, but with the difference that it is orbited by a moon with a mean diameter of 3 500 km. The moon has an albedo of 0.12 and its distance from our planet is 20 000 km at perigee.

The moon/planet system orbits a star that has an apparent magnitude of -27, when observed from the moon/planet system. Assume that we have the same atmospheric conditions as on Earth, and - if other parameters have to be taken into account besides the atmosphere - assume those are equal to Earth's as well.

I would like to learn two things here:

  1. What would the apparent magnitude of the moon in question be to an observer looking up at the night sky from the planet it orbits, when the moon is at perigee and in its full phase?
  2. What does the equation that gives you the solution to this problem look like?
  • $\begingroup$ 20,000 kilometers? That's ~1/19 times the distance from Earth to the Moon. There should be some interesting effects there. $\endgroup$
    – HDE 226868
    Jun 21, 2015 at 23:48
  • $\begingroup$ Indeed there will be, including much higher tidal range and more geologic activity. You can read a bit about it in my previous question link[/link]. $\endgroup$
    – fantasia
    Jun 21, 2015 at 23:52
  • $\begingroup$ By the way, please don't take offense at this, but this looks something like a homework problem. $\endgroup$
    – HDE 226868
    Jun 21, 2015 at 23:53
  • 2
    $\begingroup$ None taken. I should have paid a lot more attention to math in school, and especially my homework. When it comes to researching this particular problem, I have found out that it is quite easy to find examples of how to compare absolute magnitudes of stars with different apparent magnitude. For some reason, I haven't been able to find a single equation that solves my problem - although I admit I might very well have read it without understanding what I saw. $\endgroup$
    – fantasia
    Jun 21, 2015 at 23:58

1 Answer 1


There's a relatively easy way to calculate this number. Compare your moon's parameters to ours and multiple those differences

Start with out Moon (diameter 2,156 km) and compare to your moon (diameter 3,500). This means that the size difference in moons makes yours $ \left( \frac{3500}{2156} \right)^2 = 2.6 \times $ brighter due to the size difference.

Your moon is ~ $19 \times$ closer than ours, which would make it ~$ 19^2 = 361 \times$ brighter due to its distance.

Luna (our Moon)'s albedo runs around 7%, while yours comes in around 12%. Your moon comes in around $ \frac{12}{7} = 1.7 \times $ brighter due to its reflectivity.

Sun's brightness
Your world's star has an apparent magnitude of -27, while Sol's (our Sun's) apparent magnitude runs around -26.7. The difference in brightness from this is negligible.

When you combine these factors, you get the following:

$$ 2.6 * 361 * 1.7 = 1631 \times \text{brighter than our Moon, Luna}$$

$ m_{Luna} = -12.6$ (Moon's magnitude)
$ m_{Moon2} = \text{Unknown} $
$ \frac{F_{Moon2}}{F_{Luna}} = 1631 $ - Ratio of the two moon's brightness

The equation looks like this:

$$ m_{Moon2} - m_{Luna} = -2.5 log_{10} \left( \frac{F_{Moon2}}{F_{Luna}} \right) $$

Plug in the numbers:

$$ m_{Moon2} - -12.6 = -2.5 log_{10}\left(1631\right) $$


$$ m_{Moon2} = -2.5 \times 3.21 - 12.6 = -20.6 $$

So, the visual apparent magnitude of your Moon would have is about -20.6.

My above numbers were performed using our Moon as a comparison and factoring those differences. I did NOT derive the value directly from your Moon's values. You can do that as an alternative method of getting the same answer.

  • 1
    $\begingroup$ A neat way of solving the issue, and thanks for the equations as well. It is not a problem for me that the values are derived indirectly, since several of the other worlds existing in my saga will be relatively similar to Earth, meaning that I only have to alter the parameters you mention (including the primary star's apparent magnitude, when I am calculating for other planets than the one mentioned above), and then compute. $\endgroup$
    – fantasia
    Jun 22, 2015 at 9:54
  • $\begingroup$ Bravo. Excellent job. $\endgroup$
    – HDE 226868
    Jun 22, 2015 at 14:42

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