The moon seems gigantic and is extremely bright.

The planet BlinketyBlink has a mass of 1.6 Earth masses and a single large moon which orbits it at about 240 000 miles/386 242 km.

The moon Totengot is about the size of Ganymede, has an atmosphere similar to that of Venus and an albedo of 0.70, with a lurid green tinge to it.

On BlinketyBlink night is only ever dark at the time of the new-moon, otherwise it has a more cloudy-day appearance (not caused by clouds), becoming "cloudier" as the new-moon phase approaches, and then increasingly brighter as it moves towards the full-moon phase.

If a plain ol' human on the surface of BlinketyBlink looked up at the night-sky during the full-moon, would he/she see the stars as long as the skies are clear?


  • $\begingroup$ Location location location $\endgroup$
    – user6760
    Oct 3, 2017 at 4:39
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    $\begingroup$ How bright would this moon be? Are we assuming humans as we know them, or is e.g. making changes to the creatures' eyesight allowed? I'm putting this on hold for the moment, but with some more specifics, this should be answerable. (As shown by the two existing answers, as it stands this question can only be answered with what basically amounts to "it depends", which isn't all that useful.) $\endgroup$
    – user
    Oct 3, 2017 at 8:04
  • $\begingroup$ Saying "night is only ever dark at the time of the new-moon" is claiming two things, that the moon is always visible in the night sky and also that it moves through phases. You can't have both. $\endgroup$ Oct 6, 2017 at 9:26
  • $\begingroup$ @LioElbammalf sorry??? Don't get that one. If the moon goes through phases it' because of its position in relation both to the sun and earth. And because it moves around the earth it raises and set being visible at night and even during the day, the only time it is not visible is during the new moon. $\endgroup$ Oct 6, 2017 at 10:21
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    $\begingroup$ The only time the moon is visible for the whole night is the full moon, the rest of the time it is only visible for some of the night. So the night would be dark some of the time during other nights, not only the new-moon. So either you have partially dark nights and a new-moon or always bright nights and no phases (ie always a full moon, not sure how you would get that physics-wise though) $\endgroup$ Oct 6, 2017 at 17:12

4 Answers 4


This depends on how bright this extra bright moon is. Naturally it could block out the stars if it was bright enough, just as the Sun prevents us from seeing the stars in daytime on Earth. If the question is if a moon can get this bright, it depends on a number of factors such as how close it is to the sun, how bright its sun is (More sunlight means more reflected light.), how large the moon is, how close the moon is to its planet, and what the surface of the moon is made out of.

However I wouldn't doubt that it would be possible to block out many more stars, if not all, using the light from an extremely bright moon.


All else being equal, the brighness intesity is proportional to the albedo ($\alpha$) and the square of the radius ($r$) and inversersely proportional to the distance ($d$). The intensity ratio of Totengot to our moon is:

$$ \mathrm{intensity\ ratio} = \frac{(\alpha_{Totengot}/\alpha_{Luna}) \times (r_{Totengot}/r_{Luna})^2}{(d_{Totengot}/d_{Luna})^2} $$ substituting the values for the moon and Ganymede: $$ \mathrm{intensity\ ratio} = \frac{(0.7/0.136) \times (2634.1 /1737.1)^2}{(386242 /384399)^2} = 11.7 $$

So Totengot would be 11.7 times as bright as the moon (and appear to be about 1.5 times as big in the sky).

Whether this is bright enough to block the view of the stars depends on the amount of light scattereing in the atmosphere. I can't say what the scattering will be.


This very much depends upon the apparent brightness of the stars, the composition of the atmosphere as well as the albedo of the moon and the fraction of sky the moon fills.

A good solution is to think about sunrise and sunset with clear weather, and choose a specific time (in your location) and date as to precisely when the sky is equivalent to the full moon of your planet.

Then, by knowing the size and the albedo of the moon one can reasonably easily calculate the light equivalent.

Of course, if there are any stars as bright as Venus is to us, then you will see them just as we can see Venus in the twilight and the gloaming.

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    $\begingroup$ "depends upon the brightness of the stars" - and on the inverse of their distance. If the planet is close to the galactical core, there will be many more stars - like the Milkway occupying the whole sky - and they will be much brighter. $\endgroup$ Oct 3, 2017 at 13:47
  • $\begingroup$ @LuísHenrique, yes - I should have said apparent brightness, and you were correct to interpret this as ambiguously referring to absolute brightness. $\endgroup$
    – Konchog
    Oct 4, 2017 at 8:11

Other Factors Matter

Location and atmospheric moisture content matter. You said no clouds sure, but their can be varying levels of moisture in the atmosphere in different locations without there being clouds. This is important because this reduces the amount of light that can enter as well as reflect light trying to escape back in. This is a major reason observatories are built on mountains, so they have less atmosphere to peer through. This is especially concerning for a planet 1.6 times Earths mass, because arguably it could have a denser atmosphere.

Because this factor contributes HEAVILY to calculating apparent brightness of an object in the sky I cannot answer this question mathematically (not that I particularly want to).

My Answer:

Assuming this planet is the same distance from the sun as Earth and the observer is at sea level. I would reason that between the atmosphere and the moon's significantly greater albedo, there would be enough light pollution to block out the stars. However, I would expect bright planets (Venus like) would still be faintly visible.


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