# 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 – agtoever Aug 27 '14 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 – agtoever Aug 27 '14 at 6:07

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. – Double AA Aug 27 '14 at 13:13
• It's a decent answer. And he did answer the "degrees" part of the question. – HDE 226868 Aug 27 '14 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... – Jeremy Aug 27 '14 at 22:05
• and where. If there's a lot of artificial light around, you may never see any but the brightest stars. – jwenting Sep 3 '14 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. – Double AA Feb 27 '16 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.