# Where is it possible to observe the Moon for 24 hours?

I wrote an astronomy olympiad yesterday and there was one task I couldn't figure out an answer to:

Estimate at what latitudes it is possible to observe the Moon for at least 24 hours. Under what conditions is it possible to observe the phenomenon at border latitudes?

Is it possible at all to observe the Moon for such a long time?

• This also helped: Why does the Moon never set in Svalbard, Norway? Nov 19, 2021 at 6:51
• Out of the box answer would be "Standing on the Moon, at any latitude". Likewise of course, orbiting the Moon, or in any number of orbits in space which are not blocked by the Earth, Sun, or any other objects. Nov 19, 2021 at 17:47

## Lunar standstill

Estimate at what latitudes it is possible to observe the Moon for at least 24 hours.

This will depend on the Moon's maximum declination, which varies with a period of 18.6 years (see lunar standstill). During a minor lunar standstill, the declination is 18.134° at most, during a major lunar standstill, it's 28.725° at most.

I wrote a small program to plot the position of the moon every half hour, during a year.

## Diagrams

Here's the polar diagram (azimuth + elevation) for 2015, as seen from the Equator. It was during the last minor standstill, so the moon was close to the celestial equator:

and the same diagram, for 2025. It will be the next major standstill, so the moon will have a much wider range:

This diagram is convenient, because you can directly see the required latitude if you want to see the moon for a whole day: it's the lowest elevation when the moon crosses the meridian.

In 2015, you'd need to be at least at 72° latitude (90° - minor lunar standstill):

And in 2025, you could be lower, at about 62° (90° - major lunar standstill):

In 2025, at 72°, you'd have many days during which the moon is visible for 24h:

But in 2015, at 62°, you'd have none:

The diagrams would have the exact same shape in the southern hemisphere, but flipped vertically.

## "Code"

For reference, the program was written in INSEL, which is similar to Simulink:

The diagram was created with Gnuplot:

unset border
set zeroaxis
unset xtics
set ytics ("90°" 0, "80°" 10, "70°" 20, "60°" 30, "50°" 40, "40°" 50, "30°" 60, "20°" 70, "10°" 80, "0°" 90)
set ytics axis nomirror
set rtics ("" 0, "" 10, "" 20, "" 30, "" 40, "" 50, "" 60, "" 70, "" 80, "" 90)
set rtics scale 0
unset raxis

set rrange [0:90]
set grid polar 10

set size square

set lmargin 2
set rmargin 5
set tmargin 5
set bmargin 2

set label "N"  at 0, 93
set label "S"  at 0, -94
set label "E"  at 91, 0
set label "W"  at -97, 0

set key at 120, 105

plot data_file using 1:2 title curve_name1,\
90 with lines notitle


## Checking with Stellarium

On the morning of 4th of April 2025, at 62°N, the moon will be visible due north, ready for a complete rotation above the horizon.

As on the 3rd of October 2015, at 72.5°N:

• I'd love to see the source of this program Nov 19, 2021 at 20:14
• @RossPresser: I've edited the answer. The "source" is a graphical programming model. I like how small it is. It was adapted from a diagram showing sun analemmas, similar to andrewmarsh.com/software/sunpath2d-web. Nov 19, 2021 at 20:29

Near the Poles is where unusual things occur. Like, in parts of Greenland, which is pretty close to the north pole, the sun never sets during summers, and the sun never rises during winter. The same way, during some parts of summers and winters over there, the full moon never sets and new moon never rises. But it's not only during winter solstice and summer solstice that you can see the moon all day long. In fact, every month, on some days, you can see it all day long. Also, moon's orbital inclination is around only $$5^\circ$$. Due to this low orbital inclination, it can be observed 24 hours long only in some places on the earth. But if its orbital inclination was greater, it could have been observed 24 hours long all around the globe.

Verifying with Timeanddate's night sky map

You can also verify if this is right or not by using Timeanddate's interactive night sky map. The link is here: https://www.timeanddate.com/astronomy/night/
When you go there, it will first show the map for your location. You can type the city you want to find the map of in the text box on the right. For example, I typed in Nuuk, which is Greenland's capital.

Image Credit: Night sky map / Timeandate

Then, I went to the night sky map of Nuuk and chose the date as 5th December 2021 (new moon day). As you can see in the image below, on the top left, it doesn't show the rise and set time. This is because it's almost winter solstice and it's a new moon day. Therefore, it doesn't rise at all.

Night sky map of 5th December 2021, Nuuk, Greenland. Image Credit: Night sky map / Timeandate

I then changed the date to 19th December 2021 (full moon day). As you can see in the image below, it doesn't show rise and set time because it's up all day! It shows the best time to see it too. This day, you could observe the moon all day long.

Night sky map of 19th December 2021, Nuuk, Greenland. Image Credit: Night sky map / Timeandate

Also, each month, there are days when the moon is up all day long, there are days when it rises and sets, and there are days when it's below the horizon the whole day.

To conclude, around the latitudes $$60^\circ$$ to $$90^\circ$$, few days every month, the moon can be observed 24 hours long.

• Correct, but as teachers say, "Show your work" :-) . It would be very nice to see the latitudinal extremes of the moon's orbit and the geometric sightlines (vs. latitude) which allow visibility. Using the sun as analogy is not good, because that depends on the Earth's rotational axis tilt. Nov 19, 2021 at 12:47
• Nitpicking: In the poles, during some parts of the summer, the full moon never rises and during some parts of the winter the full moon never sets. In the pole the moon rises and sets every month.
– Pere
Nov 19, 2021 at 13:32
• Regarding observing from the polar regions, it would be more accurate to say that there are days each month when the Moon is above the horizon all day long, days when it rises and sets, and days when it is below the horizon all day long. Depending on the phase of the Moon and whether the sky is bright, it may be difficult to see the Moon even if it is above the horizon. Nov 19, 2021 at 14:00
• The key bit of knowledge that's required but not mentioned here is the fact that the moon's orbital inclination is only about 5 degrees. If the moon were in a highly inclined orbit, it would be possible to see a 24-hour moon from anywhere on the planet. Nov 19, 2021 at 15:37
• I like (sarcastically) the "so, the final answer is" when no derivation of said answer takes place Nov 19, 2021 at 19:46

In addition to the previous answers, another point might be added. We are dealing with the Moon. The Moon - just like the Sun - has a movement eastward (in the right ascension) during the day which makes the distances between Moon Rises 24h 50m apart on average. So the simple answer taking into account only the Moon maximum declination will result in location (the location closet to the equator) where the Moon is seen for ~ 24h and 50m.

Now, in light of the other answers, lets take the extreme case that the declination of the moon is ~28.5 degrees. Moreover, In the settings where the Moon is on his max/min declination, we can safety assume the Moon movement eastward is even faster (for more info the reader might refer to the "equation of time" and "effect of Obliquity". In short it says that when the Sun/Moon at Solitices/standstill the days are longer, since the movement on the orbit is mainly in RA component).

Let's assume then, for the sake of the example, that in this setting we save 60m - so the time interval between subsequent Moon crosses of the Meridian is 25h. We are left to ask how much we can now come closer to the equator from latitude of 61.5? On what latitude the declination circle of $$\delta=28.5$$ cuts the horizon in 2 points where the distance between them (RA-wise) is exactly 1 hour (= 15 deg)? if we mark $$\phi$$ as the latitude and $$\alpha$$ RA for a LST of 0. we know that:

$$\sin(0) = \sin(\delta)*\sin(\phi) + \cos(\delta)*\cos(\phi)*\cos(\alpha)$$ or $$\cos(\alpha)= \tan(\delta)*\tan(\phi)$$

Let's plug $$\delta=28.5$$. The result is about ~$$61.3$$ in which the gap in the intersection points RA's with the horizon is about 15 RA degrees apart. So we can be about 0.2 deg closer to the equator and still see the Moon for 24h.

(*) This is only one facet however and of course far from being accurate.

In this answer (and other answers actually...) I assumed the declination component stays fixed during the day. According to my calculation however the movement in declination of the Moon cannot be neglected. Even in the best case scenario of minor Lunar standstill we will have about 0.3 deg movement in declination at minimum during period of 24h (we could have 0.15 each side of the limit. it looks to be confirmed here). This effect in general looks little stronger than the 24h 50m effect - So I believe we will need to go at the end little farther from the equator to see the Moon on for 24h. Though it hard to say with certainty since the movement of the Moon is full of inequalities.

• That's interesting. Can you find an exact location and time in order to check your assumptions? Stellarium is probably accurate enough. The fact that the moon crosses the meridian every 24h50min caused some weird bugs in a program I wrote. I was checking the moonrise and moonset times every day for a location in Europe. But those times are not always defined. Nov 21, 2021 at 15:31
• Oh, and you might be even closer to the equator if you use the exact max declination of 28.725° instead of ~28.5°. You'd be very close to 61° then. Nov 21, 2021 at 15:43
• @EricDuminil, it's little tricky, since during the day we have also movement in the declination component that I neglected to keep it simple (though at the standstill this component should be low, but still it seems about 0.5 degree movement). I'll try to come up with concrete example though to demonstrate my point later on.
– d_e
Nov 21, 2021 at 17:04
• Your last paragraph is totally correct. While playing with stellarium, I also noticed that's there s no guarantee at all that the moon rises while the declination is highest, and the declination will vary afterwards anyway. The lowest latitude I could find was 61.45° IIRC, probably due to this problem. Nov 27, 2021 at 9:22

The Artic and Antarctic circles are, by definition, the latitudes at which the sun is visible throughout the summer solstice. Keep in mind that "summer solstice" for the Antarctic circle is in December. Also, since the actual solstice isn't exactly centered on the "day" (that is, the middle of the solstice isn't generally at noon), the actual length of solar visibility is more complicated.

The latitude of these circles, 66° 32′, is 90° minus the inclination of the Earth, 23° 68′. At the summer solstice, the Earth's tilt points towards sun. If you imagine the y-axis as being perpendicular to the orbital plane with "up" being in the north-ish direction (the "ish" because of the tilt), and the x-axis pointing towards the sun, the Earth's axis is tilted 23 degrees clockwise from the vertical. As long as you travel less than 23 degrees counterclockwise from the North pole, you haven't traveled far enough to counteract the tilt, so you are still in sunlight (which direction counterclockwise represents changes over the course of the day; it's whatever longitude is at "midnight"). Since the North pole has a latitude of 90 degrees, this means the boundary is at a bit more than 66 degrees.

All of this is in terms of the sun, but similar logic applies to the Moon. However, the Moon's orbital plane isn't exactly the same as the Earth's, so the numbers aren't exactly the same. It differs by about five degrees, and it varies as to what direction that difference is with respect to the Earth's inclination. Thus, the latitude varies between 66-5 and 66+5 degrees, i.e. between 61 and 71 degrees.