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On Earth, are north and south polar days the same length? If not, at what latitudes north and south would they become equal? Or, what resources could I use to answer this question?

I am referring to the time from sunrise to sunset. Because of perihelion during the south polar day and aphelion during the north polar day I assume that the length of light exposure would vary. I seem to remember that in the middle latitudes the difference is compensated for by increased rotation speed at perihelion but that does not seem like it would apply to the polar day. (I am using day as the period with direct sunlight (including refraction) and not a calendar day).

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    $\begingroup$ Needs more information on what you're asking By 'day' do you mean the time between sunrise and sunset as viewed from a particular location on a particular calendar date? The time between successive solar noons on a particular calendar date, or possibly at different calendar dates? The period of time designated by 86400 SI seconds? Some other definition of day? $\endgroup$
    – notovny
    Feb 3, 2023 at 19:28
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Feb 3, 2023 at 21:28
  • $\begingroup$ The only time sunlight shines on both poles at the same time is near the equinoxes, where a piece of the Sun will be above the horizon for both places. It will slowly change from day light to night over the course of a calendar day or so. The dipping of the Sun below the horizon is caused by Earth's orbit, not its rotation, so it will be slower closer to aphelion and faster nearer parhelion. $\endgroup$ Feb 4, 2023 at 15:07
  • $\begingroup$ Apologies for the vagaries. I am referring to the time from sunrise to sunset. Because of perihelion during the south polar day and aphelion during the north polar day I assume that the length of light exposure would vary. I seem to remember that in the middle latitudes the difference is compensated for by increased rotation speed at perihelion but that does not seem like it would apply to the polar day. (I am using day as the period with direct sunlight (including refraction) and not a calendar day). $\endgroup$
    – Carex
    Feb 5, 2023 at 3:55
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    $\begingroup$ Earth's rotation experiences almost no seasonal variation (just long-term lunar momentum transfer), but at perihelion the orbital speed is increased. That's going the same way as the earth's rotation, so tropical days are apparently slower then. $\endgroup$ Feb 8, 2023 at 5:36

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The equinoxes are not exactly half a year apart due to the eccentricity of the earth's orbit: each year the sun spends 5 or 6 more days north of the equator than south of the equator.

Consequently, the period of continuous daylight at the north pole is correspondingly longer than the same at the south pole.

The equinoxes also represent a half-day shift between the tropical day and the sidereal day (ignoring Milankovitch drift).

The earth's tropical year is 31556926 seconds, during which it makes 366.2422 rotations - one more than the number of days - with each rotation taking 86164 seconds. The length of n consecutive tropical days (from noon meridian to noon meridian) is given by (n + α/2π) × 86164 where α is the angular progression along the earth's orbit around the sun (in radians).

In accordance with Kepler's law, α for a given day is inversely proportional to the earth-sun distance that day, so α[perihelion] is about 1.034 × α[aphelion], leading to about 8 seconds variation in the length of the tropical day.

Because the southern summer solstice is near perihelion, anywhere outside the circumpolar "continuous daylight" zone will, for a given latitude, have slightly longer summer days in the southern hemisphere.

Adding to this effect, the sun is about 3% closer and therefore 3% larger at the southern summer solstice, meaning that the first light of sunrise is a few seconds earlier, and the last light of sunset is a few seconds later.

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    $\begingroup$ Can you check or source your numbers? I get 186.4 days for northern and 178.8 for southern summer. $\endgroup$
    – James K
    Feb 5, 2023 at 7:52
  • $\begingroup$ My apologies, I mis-transcribed one of the equinox dates when calculating the days between them (I put 2 March instead of 21 March). So the difference is nearer to the 6 days you suggest. I will re-do the calculations and revise the answer later today. $\endgroup$ Feb 6, 2023 at 0:09
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(I have already written an answer but I felt it only complicates things; Hopefully, this new attempt will be better)

(In this answer's terminology it is the Sun that moves around the Earth.)


There are two factors that determined the time the Sun spends above the horizon:

  1. how much it moves Eastward in RA - the longer it goes the longer the day.
  2. How much it moves away or towards the celestial equator (i.e., toward declination 0): if it moves away it makes the day longer. if it moves towards the day is shorter. (Let's assume we deal only with Summer days where the declination is Negative in Southern Hemisphere and Positive if the North)
  • When the Sun moves faster (when near the perihelion) then it means both factors are amplified; but the amplified second-factor effect might have a positive or negative impact on the length of the day.

  • If that was not enough, One should not that those two factors are not independent: bigger movement in declination means smaller movement in RA and vice versa. this has nothing to do with the "speed amplifier" mentioned above.

  • There is yet another big catch: the influence of each factor varies as a function of the latitude we are in. in the Equator only factor (1) counts and at the poles only factor (2) counts. (this is because at the Equator the declination circles are perpendicular to the horizon, hence movement in declination won't change the altitude or the day length).

Let's try to look at a specific example our day will be 26 Jan in Southern latitude. when the Sun is quite fast (the perihelion 4 Jan). We should consider our two factors. First, we should note that the internal division (as we said there are dependent) makes the RA component stronger on this day than average. So far for Factor (1) the RA has very strong movement to make the day longer! (both the internal division and the speed - it could be seen in this graph, thanks to @PM2Ring - both dash lines are low)

But as for factor (2) it has a negative impact on the day length since after 21 Dec the declination goes closer to the equator. On one hand, this factor is mitigated, as we said, on account of the internal division, but on the other hand, it is amplified because the Sun is fast.

So we have conflicting influences of our two factors. On small Southern latitudes, clearly, the winner is the first factor resulting in a longer Summer day, but as we go closer to the South pole it becomes less obvious.

Let's take a look at Peter I Island (latitude 68 South) rise and set time in January. On 26 Jan the day length becomes shorter by 15 minutes or ~900 seconds. we are faster than the equivalent day of Norther Summer by about 2% that's about 18 seconds. The RA gain must be very small on this day (on the equator it makes the day no longer than 8 seconds; so in Peter I Island, this effect is probably around 1 second.). So overall on this day in Peter Island, the faster movement of the Sun actually makes the day a little shorter by several seconds.

The point being made in this long post is that this is not all about the speed of the Sun and there are other factors the determine the length of the day.

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    $\begingroup$ How does one pick an "equivalent day" between the northern summer and southern summer? Counting days off the solstice? Days off the equinox? Pro-rata days between equinoxes? Half a year apart? None of these results in picking dates that are separated by an integer number of days; even "half a year" is 182.6211 days. $\endgroup$ Feb 8, 2023 at 5:23
  • $\begingroup$ @MartinKealey, The day which the Sun rises at the same declination at absolute value. We have to move on longitude to make sure the declination at rise at the time matches our core selection at the South. $\endgroup$
    – d_e
    Feb 8, 2023 at 9:34
  • $\begingroup$ Thank you, The original discussion that prompted the question was in regards to solar forcing. So "MartinKealey"'s answer provided the clearest answer to the basic question but "d_e"'s answer is better for actually applying the concepts. Thus a much clearer understanding was provided by seeing both approaches to the question. $\endgroup$
    – Carex
    Feb 8, 2023 at 13:03

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