Obviously there's a limited part of earths surface where the moon is visible at any given time. I imagine the area of this surface would be a lot smaller than the area where the sun is visible at any given time, due to the fact that the moon is a lot closer to earth. But how big is that area exactly?

Or, to put it differently: Imagine you are standing outside your house and looking at the moon. At the exact same moment someone else does the same in another location. But how far apart can you (theoretically) be?


If we take 6371.0 km as the mean radius of Earth, an apogee of Moon of 405503 km, and a perigee of 363295 km, we get ratios of 6371.0 km / 405503 km = 0.01571 = sin 0.9002° resp. 6371.0 km / 363295 km = 0.01754 = sin 1.005°. So on both sides of Earth between $$40030\mbox{ km}\cdot 0.9002°/360°=100\mbox{ km}$$ and $$40030\mbox{ km}\cdot 1.005°/360° =112\mbox{ km},$$ with an average circumference of $2\pi\cdot 6371\mbox{ km}=40030\mbox{ km},~$ get lost of the 20015 km of the half circumference of Earth as zone of visibility.

The calculation is simplified to the simultaneous visibility of a point at the distance of the center of moon. More precisely one part of the moon can be visible from one observer, while an other part of the moon is visible from a second observer. This adds between 29.3 and 34.1 arc minutes or between 54.3 (apogee) and 63.2 km (perigee) to the diameter of the zone of visibility.

The visibility of the Moon is influenced by additional factors like air temperature, height of the observer above sea level or the geographic position by Earth's flattening. But this applies also for the visibility of the sun.

The corresponding calculations for the sun: With aphelion of Earth of 152098232 km and perihelion of 147098290 km, we get ratios of 6371.0 km / 152098232 km = 0.000041887 = sin 0.0024000° resp. 6371.0 km / 147098290 km = 0.000043311 = sin 0.0024815°. So on both sides of Earth between $$40030\mbox{ km}\cdot 0.0024000°/360°=0.26686\mbox{ km}=266.86\mbox{ m}$$ and $$40030\mbox{ km}\cdot 0.0024815/360°=0.27780\mbox{ km}=277.80\mbox{ m}$$ get lost of the 20015 km of the half circumference of Earth as zone of visibility.

(For a rough estimate you may take the 100 km loss of the moon visibility, and multiply it by the quotient of the distances Earth-Moon and Earth-Sun roughly equal to 400,000 km / 150,000,000 km = 0.002667, to get 266.7 m visibility loss of the sun.)

This is again simplified, and applies to the center of the sun. The appearent diameter of the sun varies between 31.6 and 32.7 arc minutes, adding between 58.6 (aphelion) and 60.6 km (perihelion) to the diameter of the zone of visibility for observers looking at different parts of the sun.

Depending on which scenarios to be compared, the lost area varies. As an example, a loss of a radius of the visible zone of the moon in comparison to the sun of 100 km corresponds to a roughly cylindrical area of $$100\mbox{ km}\cdot \pi\cdot 40030\mbox{ km}=12.6 \mbox{ million square kilometers}.$$

How far can two observers be apart at maximum, to see two differnt parts of the moon at the same time? $$20015 \mbox{ km} - 2\cdot 100 \mbox{ km} + 54.3 \mbox{ km} = 19869.3 \mbox{ km}$$ (apogee) for the average Earth radius. For the same part of the moon it's $$20015 \mbox{ km} - 2\cdot 100 \mbox{ km} = 19815\mbox{ km}.$$

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    $\begingroup$ It would be helpful to compare this to the visibility of the sun as alluded to in the question. Otherwise, great answer! $\endgroup$
    – called2voyage
    Feb 25 '14 at 17:32
  • $\begingroup$ Thanks! I'll try to add the corresponding calculations for the sun in a few hours; shouldn't be difficult. $\endgroup$
    – Gerald
    Feb 25 '14 at 18:00
  • $\begingroup$ Wow, thanks! So, if I've understood your answer correctly, two people watching the moon simultaneously can be a maximum of roughly 19800 km apart? Did not expect the zone of visibility to be that big! Please correct me if I've misunderstood your results. $\endgroup$
    – Magnus
    Feb 25 '14 at 21:55
  • $\begingroup$ You are correct! I've now added that part explicitely to the answer. $\endgroup$
    – Gerald
    Feb 25 '14 at 22:08

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