How would you be able to calculate what magnitude of stars would be visible given a certain brightness of the moon. As I know that the visibility of stars are based on their compared magnitude to a source of light (the moon technically isn't a light source, I know), which is why no stars are visible during the day.

I know it is basically anything under positive 6-7 (ignoring light pollution), with our own moon. But I just am not sure how it would change if the moon had a different brightness.

So, how would I find this? Like, if I have a moon with a higher (technically lower, like, -10 compared to -9) apparent magnitude, how would I mathematically find the range of magnitudes for visible stars? Like, for the case of this question, as an example to show the maths, what about a moon with a magnitude of -18?

  • $\begingroup$ I think as the question is, it might be a better fit for the astronomy SE. One way to word similar questions in the future is to give more context in reference to the world you are building, and frame it as a question that relates to the world. This gives us more leeway to answer it as we will know what your purpose for the question is. $\endgroup$
    – Enthu5ed
    Commented Oct 15, 2023 at 12:54

2 Answers 2


Unfortunately, I don't believe what you're looking for exists.

Stars are invisible during the day to the average human eye due to Rayleigh scattering in the atmosphere. (We're going to ignore how this changes if, e.g., you use a telescope or electronics to sense something other than visible light.) Rayleigh scattering is what's responsible for that lovely blue glow we see at noon on a cloudless day. It's also what obscures the stars. All atmospheres scatter light. If you stood on a world without an atmosphere and looked up during the day, you'd see the stars quite easily.

It's worth noting that Rayleigh scattering is based on...

  • the chemical composition of the atmosphere,
  • the density of the atmosphere,
  • the thickness of the atmosphere,
  • and the "time of day" (angle of incidence of sunlight as it hits the atmosphere), which is a function of knowing where on a planet the observer is standing in relation to the rotation of the planet compared to the sun.

It's worth noting that light reflected from the Moon contributes little, if any, to the obscuring effect of Rayleigh scattering. I don't have the science to back this up. But I've seen cloudless days with the moon in the sky and cloudless days without it and the imperceptible star field remains imperceptible either way.

But, could you use the brightness of a moon to index the visibility of stars?

I'll give you the bad news later on. But we need to talk about moons. The reflected light of a moon is a function of its albedo. "Albedo" is a measurement of the reflectivity of an object. A moon's albedo is a function of...

  • the chemistry of the surface materials,
  • the size of the moon,
  • the distance of the moon from the planet.

Said simply, if you don't know the albedo of a moon, then you don't know how much light is being reflected, which means the perceived brightness of the moon can't be used to calculate star visibility.

BTW, as a reference to how complex this can get. Our Moon has an albedo of 0.07 while Venus has an albedo of 0.60. Nevertheless, at high noon we can see the closer and perceptually larger moon than the more distant and perceptually smaller Venus. Albedo is a measure of reflectivity, not brightness. An object can have a very high albedo and still have a low magnitude brightness.

Thanks to Rayleigh scattering, magnitude is a perception

Let's mention the numbers you're using. 6-7, -10, -9, and -18. I believe you're talking about stellar or astronomical magnitude. That magnitude, when perceived in open space, has a predictable absolute value. But when an atmosphere gets involved, the perceived magnitude is lower and fairly messy to calculate.

Think of it this way: thanks to Rayleigh scattering, you need a certain number of photons ("perceived magnitude") to get through the obscuring effect. Our Moon's albedo is low, but thanks to it being fairly close to us and taking up a sizeable chunk of our sky, it can be seen (has a lot of photons or a high "perceived magnitude") during the day. Venus, being so distant and taking up so little of our sky, simply isn't dumping enough photons (low "perceived magnitude") on us to be seen through the scattering.

The problem with trying to find a relationship between the perceived magnitude of an arbitrary star (e.g., +$\infty$ during the daylight hours on Earth) is that you need to know (simplistically)...

  • The color of the star.
  • The unaffected-by-atmosphere brightness (magnitude) of the star.
  • The effect of Rayleigh scattering on the star's brightness.
    • Meaning you need to know the chemical composition of the atmosphere
    • and its density
    • and it's thickness
    • and the "time of day" or angle of incidence of sunlight
  • The albedo of the moon.
  • The effect of Rayleigh scattering on the moon's brightness (see sub-list, above).
  • The unaffected-by-atmosphere brightness of a particular target star.
  • The effect of Rayleigh scattering on the target star's brightness (see sub-list, above).

And that ignores questions like which star? What's its color, size, and distance from the target planet? Etc.

What can I say, but "yuck."

Is there a way to simplify this?

Well... no.

The problem is that I suspect you're looking for a "realistic" way to "easily" decide whether or not stars are visible during the day given any arbitrary combination of star, planet, moon, and observer. It feels like we should be able to come up with a simple relationship. If we're in open space, celestial object magnitudes are reasonably absolute. Rayleigh scattering scatters all light. It's like a camera filter. As the scattering effect increases, the ability for any particular magnitude to be visible from the surface decreases. In other words...

If you can index the obscuring ability of Rayleigh scattering and if you know the unobscured magnitudes of everything in question, you can come up with a formula that does what you're looking for. A dial you can turn to identify the visibility of stars from the surface.

But that formula would only be useful if you ignore the specifics. What's creating a moon's albedo? What's causing a particular amount of Rayleigh scattering? The moment you want to know those values, the formula becomes way too complex to deal with easily. In fact, it gets pretty complex the moment you try to apply it to Earth and ask the question, "why can I usually see Venus before any stars?"

Well... Venus' albedo, time of year (angle of reflection), time of day (angle of sunlight incidence on Earth), yada yada yada....

So... to answer your question...

You don't want this. It's a level of "realism" that's unproductive for worldbuilding. I've read a couple of papers about Rayleigh scattering and I don't believe a simple formula can be derived given (e.g.) the ratios of atmospheric chemistry, density, and thickness and you'd need at least that just to get to the "simple" formula.

Therefore, I don't believe the formula you're looking for (in a form easily used) exists. I must recommend that you simply declare the visibility to be what you want it to be in each case you need it.

  • 1
    $\begingroup$ “If you stood on a world without an atmosphere and looked up during the day, you'd see the stars quite easily.” Well, maybe, but it’s much more likely that your eyes would be adjusted to the bright surroundings, rather than the dim stars $\endgroup$
    – Topcode
    Commented Oct 14, 2023 at 18:00
  • $\begingroup$ I already know things like the Magnitude of my moon and its albedo (both bond and geometric), and all such things. You have no need to explain all of that to me. If I did not know it, my question would be about how to find it. But I get the misunderstanding. But, understandable that it comes to Reyleigh Scattering (I also know what that is, you didn't need to explain it in such detail). Thank you for trying to help though, dissapointing that it can't be done without way too much information. $\endgroup$ Commented Oct 14, 2023 at 18:25
  • $\begingroup$ I also disagree that it is not productive in worldbuilding. I find it extremely productive as I have my stellar neightborhood planned out and need to see what stars are visible from which planets. I already know distance, coordinates, magnitude, etc. Which would help me make out what the night sky would look like on different planets. As I know a brighter moon would make the required magnitude of visible stars increase (again, technically decrease), which would make less stars visible, and I wanted to know how few would be. I personally find it pretty important. $\endgroup$ Commented Oct 14, 2023 at 18:28
  • $\begingroup$ JBH's fourth answer in a row advocating people to stop using science for worldbuilding. The first three-quarters of these answers are excellent, but then the conclusion is something I so intrinsically disagree with, I can't decide how to vote. $\endgroup$
    – KeizerHarm
    Commented Oct 14, 2023 at 18:31
  • 1
    $\begingroup$ Perfectly understanable, I am sorry if I came off as aggressive as well. You are right that I didn't clarify what I knew, I kinda just expected people to assume what I knew, which, looking back, was a bad idea. Your answer is still enlightening, even if the end result was ultimately dissapointing. Thanks for taking your time on me. $\endgroup$ Commented Oct 14, 2023 at 20:03

This is a bit of a digression.

It is sometimes possible to see Venus during the daylight. In fact seeing Venus during the daytime has been suggested as the cause of some UFO reports.

Venus, Jupiter, and Mars are the brightest and, therefore, the easiest planets to observe during the day. If you have clear skies and good eyesight, you can see them with the naked eye, but you need to know where to look.


At sunset, if your eyesight is sharp, you may even try to see Sirius with your own eyes, but the chances are really low.


As seen from Earth Venus has a maximum brightness of - 4.92 and a minimum brightness of - 2.98.

Objects brighter than magnitude - 4.00 can be seen in the day sky when the Sun is high.

As seen from Earth Jupiter has a maximum brightness of - 2.94 and a minimum brightness of - 1.66.

As seen from Earth Mars has a maximum brightness of - 2.94 and a minimum brightness of + 1.86.

Objects with brightness of magnitude -2.5 are supposed to be visible in the day when the Sun is less than 10d degrees above the horizon.

Sirius has an apparent magnitude of -1.47, Canopus -0.72, Alpha Centauri A & B - 0.27, Arcturus -0.04, and Vega + 0.03.


Stars can be observed during daylight with binoculars and telescopes if someone knows where to look. And I think I have read that a very few people claimed to have seen a star in the daytime with the naked eye.

In Habitable Planets for Man (1964) Stephen H. Dole calculated what is necessary for a world to be habitable for humans and for life forms with similar requirements.


On pages 13 to 19 Dole calculated the necessary atmospheric composition.

Earth's atmosphere has a pressure of 760 millimeters of mercury (mmHg). Dole Calculated that a breathable atmosphere requires between about 60 and 400 mmHg of oxygen and about 0.5 to 7.0 mmHg of carbon dioxide, and some nitrogen, and a little water vapor. The maximum pressure of nitrogen should be about 2,300 mmHg.

Thus a human habitable planet should have a atmospheric pressure between 0.092 and 3.565 of Earth's sea level atmosphere.

And that should cause a noticeable difference is how visible objects are in the skies of habitable planets.

I also note that few habitable planets are expected to have natural satellites or Moons many times brighter than Earth's Moon.

If a habitable world is itself a moon, orbiting a giant planet, then the giant planet will appear many times as large and bright in the planet's sky as the Moon appears from Earth.

For decades there has been a bit of controversy and discussion whether a large moon is necessary for planet to develop life and/or become habitable for humans. At the present time it is short of undecided.

So if it is possible for a planet to be habitable without a large moon, most habitable planets would have much smaller moons than Earth does.

Beside Earth, the only reasonably Earth like planet known to have moons is Mars, with 2 tiny moons. The Martian moons provide only a tiny fraction of the moonlight which Earth gets from the Moon, and so would have dim the stars in the Martian night sky much less than the Moon does on Earth.

And if large, bright moons aren't necessary for for a planet to be habitable, most human habitable planets should have much dimmer moons that have very little effect on the visibility of stars.

I can imagine a story in which a character from Earth, or an omniscient narrator, explains how the atmospheric composition, and the brightness level of the planet's moon(s), and the local density of stars in that part of the galaxy, affects the number of stars seen a night from the planet.

And I can also imagine a story where someone notices that the stars are visible that night, without thinking about how many stars, or how bright, or comparing them with the skies of other worlds.

  • $\begingroup$ Upvoted. As an aside, Venus is not only possible to see in the morning shortly after sunrise or in the evening shortly before sunset, if it is above the horizon; it is an obvious bright object in the sky, you cannot not-see it. In Romanian we call it the luceafăr (from Lucifer), the Light-Bearer. Seeing it in daytime when the sun is high up is rarely possible unaided, because it is always quite close to the sun. $\endgroup$
    – AlexP
    Commented Oct 16, 2023 at 16:52
  • $\begingroup$ Great answer, and I do have a question, what is the maximum amount of atmospheric water vapor I can use according to that book? As my planet is meant to have high water vapor in the air, but still be survivable. (It doesn't necessarily have to be pleasent, just survivable) $\endgroup$ Commented Oct 16, 2023 at 18:46
  • $\begingroup$ @DanceroftheStars: Assuming normal air, at 1 atmosphere pressure and 30 °C (86 °F), 100% humidity is 30.4 grams of water vapor per cubic meter of air, going to 39.6 g/m³ at 35 °C (95 °F). Higher ambient temperatures are not really survivable at 100% humidity. $\endgroup$
    – AlexP
    Commented Oct 17, 2023 at 13:20

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