# Time after sunset until star can be seen

Given a star's apparent location and apparent magnitude, how how many degrees below the horizon must the sun be for the star to be visible to the naked eye for an observer on Earth at a specified location and elevation? You can ignore all light pollution, though I understand there will be some error due to air quality/weather/atmospheric aberrations.

Is there such a formula? If not, why not and what methods of approximation are available?

(I figure the brightness of the afterglow of the sun at a certain angle above the western horizon can be calculated for a certain time and place, and if it is less than the apparent brightness of the star in question, the star will be able to be seen.)

• Not exacty an answer to your question, but a very nice article about daylight astronomy that shows you actually can see quite something during the day: sky.velp.info/daystars.php Aug 27, 2014 at 5:53
• MODTRAN models the light from the sky. Seems a bit more complicated than just a formula. See: modtran5.com and en.m.wikipedia.org/wiki/MODTRAN Aug 27, 2014 at 6:07
• I was given the following two references which allegedly deal with this question, which I don't know how to find. If someone is able to find and summarize their contents that would be helpful. 1) M. J. Koomen et al., "Measurements of the Brightness of the Twighlight Sky," J. Opt. Soc. of Amer. vol 42 (1952), pp. 353-356. 2) R. Tousy and M. J. Koomen, "Visibility of Stars and Planets During Twilight," ibid, vol. 43 (1953), pp. 177-183. Additionally, if someone knows how to follow the trail from those to newer follow up studies, please let me know, as those appear to be from the 1950s. Jun 1, 2021 at 1:48
• Seems like these two behind a paywall osapublishing.org/josa/abstract.cfm?uri=josa-42-5-353 osapublishing.org/josa/abstract.cfm?uri=josa-43-3-177 Jun 1, 2021 at 2:04
• I do not understand the Formula ... mlimit = 1.07 ⋅ θ − 10 from the previous answer? It was stated that, the Visual Limit was 6th magnitude. and rounding 1.07 to exactly "1.00" then ... -10 mag limit = 0° - 10 ( at sunset ) 0 mag limit = 10° - 10 ( 45 minutes after sunset ) 10 mag limit = 20° - 10 ( 90 minutes after sunset ) So how can it change from Visible Limit = 6th mag, to now Visible = 10th mag ?
– mvas
Jul 7, 2022 at 14:57

It depends on the star's brightness. Astronomical darkness (full night, essentially) begins when the sun is 18 degrees below the horizon. Then, you can see all of the ones you're going to be able to see.

You will see stars before that point; which ones and how many will depend on how bright they are relative to the brightness of the twilight sky.

• Well, yes, I knew that "it depends on the star's brightness"; I basically said so in the question. I asked for how to calculate for a given brightness and location. I would downvote if I had enough rep. Aug 27, 2014 at 13:13
• It's a decent answer. And he did answer the "degrees" part of the question. Aug 27, 2014 at 17:15
• Sorry, I answered from the practical point of view, when can you count on doing something useful with the observation, which is after astronomical darkness. You can't do much useful with an observation when the star is just barely visible. And you could quibble about visibility - naked eye or with a scope? Whose naked eye - different people can see different apparent magnitudes in full darkness, so it is likely different people will see stars appear differently. Perhaps it would help if you explained why you want to know the answer, what is the intended purpose... Aug 27, 2014 at 22:05
• and where. If there's a lot of artificial light around, you may never see any but the brightest stars. Sep 3, 2014 at 8:59
• @jwenting I said to ignore all light pollution. Once we have the formula I described you can add in light pollution as an error term. Feb 27, 2016 at 23:57

I think we can work this our pretty well with a few approximations.

To start, let's say that the observable limit of the naked eye under completely dark skies is 6th magnitude in $V$. Let's also say that the human eye has an angular resolution of 1 arcminute, which will subtend $\pi \cdot 1^2 \approx 3$ square arcminutes on the sky.

Actual dark skies on Earth still have some measurable sky brightness - which is one reason we can't see arbitrarily faint stars - of about $21.8 \ \mathrm{mag} \over \text{arcsec}^2$ in V-band. See Figure 1 here.

The integrated brightness, $m$, of the sky background, $S$, will be:

$$m = S - 2.5 \cdot \log(A)$$

where $A$ is the angular area. Note that the integrated sky brightness will just scale along with $S$, the surface brightness of the sky, since $A$ remains constant.

So, if for $S = {21 \ \mathrm{mag} \over \text{arcsec}^2}$ the observable limit is 6th mag in $V$, the limit for arbitrary $S$ will be:

$$m_{\text{limit}} = 6 - [21 - S(z)]$$

where $S(z)$ is the sky brightness when the Sun is at some zenith angle $z$.

From that reference I linked to, $S = 21.8$ at about $z = 105^{\circ}$ (Fig 5), and it goes up pretty much linearly to $S = 10$ in $V$ at $z = 94^{\circ}$. So:

$$S(z) = 1.07 \cdot z - 90.5$$

Thus the limiting visible $V$ magnitude for a star, as function of the Sun's zenith angle ought to be roughly

$$m_{\text{limit}}(z) = 1.07 \cdot z - 105.5$$

But you wanted the altitude of the Sun below the horizon, not away from the zenith, so subtract $90^{\circ}$ from $z$ in the above:

$$m_{\text{limit}} = 1.07 \cdot \theta - 10$$

where $\theta$ is how far the Sun is below the horizon in degrees.

Approximately, at any rate.

• Order of operations may require some parentheses for these formulas to work as intended. Jul 7, 2022 at 18:17