# Suppose an array of mirrors 50 km above the surface reflect light straight down. How long before the sun rise are they illuminated?

Suppose at a point in the earth’s future, a superstructure has been built at the northern 45 parallel. 50 kilometers above sea level, a large array of mirrors has been built. The mirrors are articulate, such that they always reflect light directly “downward” toward the base of the structure. How long before sunrise and after sunset would the mirrors be illuminated at the equinoxes and the solstices?

Context; I’m working on a society that lives in the shelter of biological technologies they have long since lost the keys to. I’ve run into a spot of math that is a bit beyond my level- the sunrise equation! Putting my head together with a friend, we reached crude numbers of 41 minutes at winter solstice and 68 at summer, but neither of us are astrophysicists, and I don’t totally trust our calculations. Any help would be appreciated!

• Don't know how large, but if large enough so that it circles the entire 45° parallel, the articulated mirrors superstructure could work together to collect light and reflect it at will anywhere on the dark side below it.
– user35577
Dec 14, 2022 at 9:23
• qq- I apologize if it wasn't clear from the question, but the mirrors I was imagining occupy only a limited area around a kind of tower, they do not wrap around the entire 45th parallel. For the purposes of the question, I am treating them as a singular postion, east-west wise. Dec 15, 2022 at 23:36

Here is an approximate geometrical solution.

Suppose your position at height $$h$$ above the Earth's surface is $$O$$, the center of the Earth is $$C$$, a point of the Earth's surface that you see at the horizon is $$B$$, and the radius of the Earth where you are is $$r$$. Then the sine of $$COB$$ is $$r / (r + h)$$, so if $$A$$ is a point on the "normal" horizon (perpendicular to the zenith) directly above $$B$$, this will also be the cosine of the complementary angle $$BOA$$. Therefore, the horizon has a dip angle of $$d_1=\arccos (r/(r + h))$$ below the "normal" horizon.

Projecting these points to the celestial sphere, if the center of the Sun sets below or rises above the dipped horizon at $$B$$, the north celestial pole is at $$P$$, and the direction towards north on the normal horizon is at $$N$$, then $$BAN$$ and $$ANP$$ are right angles, $$AB$$ equals the dip angle ($$d$$ say), $$NP$$ equals the latitude $$L > 0$$, and $$PB$$ equals 90° minus the Sun's declination ($$90^\circ - \delta$$, say.)

Solving using the spherical laws of sines and cosines gives

$$\cos BPN = \frac{\sin d + \sin L \sin \delta}{\cos L \cos \delta}.$$

To find the dip angle $$d$$ for a sunrise or sunset at the Earth's surface, you can use that the Sun's angular radius is approximately 16 arcminutes, depending on where the Earth is in its orbit, and also apply a refractive correction of about 34 arcminutes ([1], §1.1, p. 2.) This gives approximately $$d=d_0=50^\prime$$. To find the dip angle for a sunrise or sunset high above the surface, you could just take $$d=d_0+d_1$$ (but see below.) The difference in times then will be about the time it takes the Earth to rotate by an angle of $$\alpha= \arccos\frac{\sin d_0 + \sin L \sin \delta}{\cos L \cos \delta} - \arccos\frac{\sin (d_0 + d_1) + \sin L \sin \delta}{\cos L \cos \delta}.$$

Plugging in values of $$h=50 \ \rm km$$, $$r=6367.49 \ \rm km$$ at $$45^\circ N$$, $$L=45^\circ$$, $$d_0=50^\prime$$, and $$\delta=\pm 23^\circ 26^\prime$$ for the solstices or $$\delta=0$$ for the equinox gives estimates of $$\begin{array}{cl} \alpha=13^\circ 15^\prime & \hbox{for the summer solstice,}\\ \alpha=10^\circ 10^\prime & \hbox{for the equinox, and}\\ \alpha=11^\circ 36^\prime & \hbox{for the winter solstice.} \end{array}$$ Converting these to times gives approximately $$\begin{array}{cl} 53 \ \hbox{minutes} & \hbox{for the summer solstice,}\\ 41 \ \hbox{minutes} & \hbox{for the equinox, and}\\ 46 \ \hbox{minutes} & \hbox{for the winter solstice.} \end{array}$$

A major source of error in these estimates is the refractive correction. According to [1], §1.1, p. 2, it's conventional to use a correction of $$34^\prime$$, and this is what's often used in published tables of sunrises and sunsets, but it's not always accurate and can lead to errors of several minutes. The correction will surely also be different 50 km above the Earth's surface. [1] gives some more accurate refractive models, which might help to compute a better estimate. In its article on the sunrise equation, Wikipedia suggests that you take $$d_1=2.076 \sqrt{h}, \qquad \hbox{if d_1 is in arcminutes and h is in meters.}$$ Using this estimate would give

$$\begin{array}{cl} 58 \ \hbox{minutes} & \hbox{for the summer solstice,}\\ 44 \ \hbox{minutes} & \hbox{for the equinox, and}\\ 50 \ \hbox{minutes} & \hbox{for the winter solstice,} \end{array}$$ but I doubt that this formula is intended to be accurate at a height of 50 km.

Addendum: Here are some graphs showing how the time varies with the latitude and the Sun's declination. They are computed using the formulae above and $$d_1=\arccos (r/(r+h))$$.

In the first graph, the declination where the time is smallest, which, interestingly, is not zero (unless you are at the equator), is marked with an "x". References [2], [3], [4], [5] and [6] discuss the similar problem of finding when the shortest twilight occurs.

References:

[1]: "Evaluating the Effectiveness of Current Atmospheric Refraction Models in Predicting Sunrise and Sunset Times", Teresa Wilson, Ph. D. thesis, Michigan Technological University, 2018; DOI 10.37099/mtu.dc.etdr/697.

[2]: "The shortest twilight", B. G. Marsden and R. F. Griffith, The Observatory, Feb. 2000, vol. 120, pp. 62-66.

[3], [4], [5], [6]: "The Twilight", Orrin E. Harmon, Popular Astronomy, Sep. 1896, vol. 4, pp. 148-154; Oct. 1896, vol. 4, pp. 211-214; Nov. 1896, vol. 4, pp. 252-258; and Sep. 1897, vol. 5, pp. 257-263.

• Thanks David! Like I said in the OP, my math is fairly weak, so it’s taking me some time to follow your answer. The idea that, from equinox, the time of illumination for is longer for both summer AND winter solstice is boggling my mind. I’d like to chart out that relationship for other latitudes, and see how it varies. I’m assuming the solstices and equinoxes are the max/min points of variance. Dec 15, 2022 at 19:18
• Your question is similar to asking how long civil twilight is. If you go to www.timeanddate.com and look at their graphs of sunrise and sunset, for example this one for Belgrade, Serbia, you can see how this varies with season and latitude. Dec 15, 2022 at 19:55
• Civil twilight begins and ends when the center of the Sun is $6^\circ$ below the horizon, so it's like taking $d_1=5^\circ 10^\prime$ in the formulae above. Dec 15, 2022 at 20:20
• I think it's more complicated than this, because the time is not smallest at the equinox (unless you are at the equator.) The seasonal differences do seem to increase with latitude though. Dec 17, 2022 at 21:55
• I added some graphs of the seasonal and latitudinal variation to my answer. Dec 17, 2022 at 21:57

From a height of 50 km, the horizon is at about 800 km. At 45 degrees latitude, 800 km is about 10 degrees of longitude. The solar time 10 degrees of longitude away is about 40 minutes ahead or behind local time. (Because the sun moves about 15 degrees per hour.) Overall, the mirrors will become illuminated about 40 minutes before local sunrise, and will remain illuminated about 40 minutes after sunset.

(Because most of the time the sun does not rise and set true east and true west, the extra illumination time will be a little longer than the time difference corresponding to 10 degrees of longitude. I compensated for this assuming an extra 1 hour 30 minutes instead of 1 hour 20 minutes.) (Hat tip to @Litho.)

At equinoxes, the mirrors will be illuminated for about 13 hours and 20 minutes, with 10 hours 40 minutes of darkness; at the summer solstice the mirrors will be illuminated for about 16 hours 30 minutes, with only 7 hours 30 minutes of darkness; and at the winter solstice the mirrors will be illuminated for about 10 hours 30 minutes, with 13 hours 30 minutes of darkness.

• Thanks Alex! That’s pretty much the exact calculation we used without any correction for atmospheric distortion, and probably good enough for my purposes. What I can’t seem to wrap my head around is treating the movement of the sun across the sky as a constant. It seems to me that the vector of the earth’s rotation which is perpendicular to the position of the sun is very different at noon than at sunset or sunrise. In the case of the mirrors, I imagine this difference would be particularly exaggerated. Dec 14, 2022 at 4:30
• @IrvingWashington: The mirrors become illuminated when the sun rises 800 km east of the point of interest, and go into darkness when the sun sets 800 km west of the point of interest. The earth rotates 15 degrees per hour, sufficiently constantly so that it was the standard of time for millennia, so that sunrise 800 km east is 40 minutes earlier than at the point of interest. See how sundials take into account the equation of time. Note that the advance / delay depends on the day, not the time of day. Dec 14, 2022 at 9:27
• @AlexP The mirrors get illuminated when the nearest point of the day/night boundary on Earth's surface is 800 km away. If it's not an equinox, the boundary doesn't go in the north/south direction, so this nearest point is not in the eastern direction. It will be roughly in NE at summer solstice, and SE at winter solstice. The point at the same parallel where the sun rises at this moment will be >800 km away, so the time between the mirrors getting illuminated and the sunrise will be greater than 40 min. Dec 14, 2022 at 12:22
• @Litho: Actually, the point on the circle 800 km SE will be less than 10 degrees of longitude east ... Dec 14, 2022 at 14:02
• @AlexP That's true, but not relevant. The time to sunrise is determined by how far you need to go along your parallel until you are in the sunlit part of the globe — because this is how Earth's rotation moves you. And this distance will be more than 800 km. Dec 14, 2022 at 15:10