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I've been trying to find an answer to this question for some time now, and I seem to be missing something. I understand that the rotation and revolution of the Earth are not synchronized. I understand that solar days and years are mean averages of the actual fluctuations over a year. And I know that solar days and years are different than sidereal ones. However, if the sun appears to be in the exact point in the sky as that time the previous solar year, how does the earth reach this exact position while only being 5hrs, 48 min, 45 secs into its final rotation of the year? Wouldn't that difference of almost 6hrs or 18hrs mean it would be an entirely different time of day than the previous year? Can someone please tell me what I'm missing her?

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    $\begingroup$ Because, in astronomy, normal numbers are weird. $\endgroup$ Apr 11, 2023 at 12:14
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    $\begingroup$ I feel like all of the existing answers miss the point of the question, which might be better stated "why isn't the solar year defined to be the time it takes for the sun to return to the same position in the sky, relative to the Earth's surface, as that time the previous solar year?" (Which would necessarily be a whole number of solar days.) After all, the entire reason there's a difference between a sidereal day and a solar day is that we do use a "return to the same position in the sky relative to the surface" [sort of] definition for a solar day. $\endgroup$
    – zwol
    Apr 11, 2023 at 19:22
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    $\begingroup$ I'm the OP. This comment above all the others addresses the root of my question. That's a much better way to state what I'm asking. But what is the answer to this question? $\endgroup$
    – Jimmy Jam
    Apr 11, 2023 at 20:42
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    $\begingroup$ @RobbieGoodwin Days and years represent durations of very real cyclical physical phenomena that would continue whether humans existed or not - I'm perplexed how you can argue that the motion of the planets has "nothing to do with astronomy". The notion of weeks is an arbitrary human invention, but days and years are not. It is not a "human idea" that the earth takes some fixed amount of time to complete one rotation or orbit. $\endgroup$ Apr 12, 2023 at 16:35
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    $\begingroup$ @RobbieGoodwin The period of the earth's orbit is not evenly divisible by the period of the earth's rotation no matter what unit you choose to count the passage of time with. That ratio is not a whole number, whether humans exist or not. Solar days and solar years are defined by physical phenomena, not by arbitrary human choice. We didn't choose to have solar days and years be non-divisible, we observed it to be the case. $\endgroup$ Apr 13, 2023 at 13:01

12 Answers 12

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if the sun appears to be in the exact point in the sky as that time the previous solar year,

This statement means the sun is at (nearly) the same position relative to the stars that it was the previous year. Not that the sun is at the same point in the sky relative to the earth's surface.

So there are two motions: the sun across the field of stars, and the earth's direction to see things.

For the sidereal year (which is really close to the solar year), all we care about is where the sun has moved relative to the stars. The angle of earth's rotation so that your house is able to see the sun doesn't matter.

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Let‘s turn your question around: Why should they be in sync?

The only mechanism that redistributes angular momentum between two celestial bodies are tidal forces (not considering general relativity and crashing the bodies against each other. Angular momentum of the whole system must be conserved). In the long term this leads to locked rotation as in the cases of moon, Mercury and between Pluto and Charon. As you probably know, tides created by sun and moon are creating a drag on earth to go slower and earth’s rotation was quicker in the past. Note that this force in the end is going to force earth into a locked rotation around sun (cf. spring tide). So we will end up with a day length of one year (if the force is strong enough compared to other influences on angular momentum like meteors entering earth atmosphere).

Simply put: There‘s a mechanism to create a resonance between day length and year length, but the day length is going to become 1 year. That it doesn't match an integer number of days at the current point in time is coincidence.

Note that stronger resonances do exist between planet orbits (e.g. Jupiter‘s moons), which is like pushing gravitationally a child on a swing, but these forces do not couple to the angular momentum of the respective body and the tidal forces are much smaller in comparison.

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    $\begingroup$ There‘s no mechanism to create a resonance between day length and year length, except that your previous paragraph explained that there is, it's just too weak between Sun and Earth to lead to resonance within the lifetime of the solar system. $\endgroup$
    – gerrit
    Apr 11, 2023 at 11:19
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    $\begingroup$ Agree this answer seems to veer around in what the takeaway should be. It goes from "what mechanism could possibly be responsible" to "here is a mechanism that could do it and examples where it happens" and then concludes with "there is no mechanism, rotational and orbital period are unrelated", which rather contradicts everything else. $\endgroup$ Apr 11, 2023 at 15:08
  • $\begingroup$ @gerrit thanks for catching my lapse. Hope I fixed it now. $\endgroup$
    – Grimaldi
    Apr 12, 2023 at 12:48
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It isn't in the same place.

We can measure "one year" starting at any point in time, suppose we start at sunrise on January 1st at some convenient location (eg London). We know where the sun is; it is on the horizon. The time is about 8am.

One solar year later (there is a slight detail about the exact meaning of this which I shall gloss over) it is 1:48pm on January 1st, and the sun is not in the same place. Unless there has been a leap day, in which case it is about 1:48 on December 31st.

So you are totally right, after one solar year it is an entirely different time. We set our calendar to start each day at midnight and this means that if we didn't have leap days the calendar would quickly get out of sync with the seasons.

So why not define the solar year to be the time for the sun to reach the same place in the sky? Let's look at this in more detail. On Jan 1st 2022 at sunrise (in my location) it is 8:03:25 and the sun has an azimuth of 127.09 degrees (and is on the horizon). When is it in exactly the same position? Well not on Jan 1st 2023 (it has an azimuth of 127.13), nor on Jan 2nd (126.99) nor Dec 31st (127.28)

The sun doesn't come back to the same place after an integer number of days, it's pretty close but not exact In fact, what you would find is that after 4 years of 365 days each, the sun would be close to the 127.09 degrees on Jan 2nd. If you kept on going, after about 700 years, the sun would be rising at that azimuth in the middle of June!

It's just not possible to choose an integer number of days between which the sun returns to exactly the same position.

So the definition of the solar year is the time for the sun to get back to the same position relative to the equinox (the point in the sky where the path of the sun crosses the equator) This is the "tropical year" it doesn't have an integer number of days, but it does keep the seasons exactly aligned. If it is the start of spring, then 1000 "tropical years" later it will still be the start of spring (though a different time of day) And we then design a calendar to approximately align a variable number of days to the tropical year. This is convenient, it means (for example) farmers don't need to adjust for the shift in the calendar when planning the best time for planting.

This is not the only way of designing a calendar. Most notably the Islamic calendar strictly follows a cycle of twelve lunar months, and so a month such as Ramaḍān may occur in the northern hemisphere summer or winter.

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  • $\begingroup$ It seems to me that this is the only answer that a) understands the question and b) is correct. $\endgroup$
    – Mike Nakis
    Apr 11, 2023 at 8:47
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    $\begingroup$ Or possibly we have both misunderstood the question in the same way. $\endgroup$
    – James K
    Apr 11, 2023 at 9:17
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    $\begingroup$ @MikeNakis but JamesK does not answer the question "Why are there not a whole number of solar days in a solar year?". He just says "yes" to "Wouldn't that difference of almost 6hrs or 18hrs mean it would be an entirely different time of day than the previous year?" $\endgroup$
    – RonJohn
    Apr 11, 2023 at 16:22
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Take a bike and a race track. Mark one spot on the side of one of your tires.

Now place the bike on the starting line, making sure the mark is precisely at the lowest point of the wheel (where it touches the ground), precisely on the line.

Do one lap, then stop on the starting line.

The chances the mark on your tire will be aligned with the line at that point are quite small (it will happen only if the path you have taken around the track is an exact multiple of the circumference of your wheel).

(aside: on a race track the exact distance around the track will depend on where on the track you ride, so you can get very different results from one lap to the next)

The same happens for the Earth around the Sun. It spins, and a given meridian on the surface of the Earth will point towards the Sun every day (like the mark on the tire will be touching the ground each time the wheel makes one turn), and at the same time, it orbits around the Sun, like your bike doing a lap around the track. No reason for the same meridian to be towards the Sun exactly one year later.

You can see that either as "365 days later, when the reference meridian points towards the Sun, we haven't made exactly one lap around the Sun (not back in the same place on the orbit)" or "exactly a tropical year later, our reference meridian does not point towards the sun", which really are the same thing (the spot on your tire touches the ground somewhere other than the start line, or when you cross the start line, the spot on your tire is not on the ground).

This answers your specific question:

However, if the sun appears to be in the exact point in the sky as that time the previous solar year, how does the earth reach this exact position while only being 5hrs, 48 min, 45 secs into its final rotation of the year? Wouldn't that difference of almost 6hrs or 18hrs mean it would be an entirely different time of day than the previous year?

At a given location, the Sun will not appear at the same exact point in the sky as that time the previous solar year. Either it appears at the same exact point in the sky as that time the previous year (i.e. 365 days later in most cases), but the Earth hasn't gone through a full orbit around the sun (we haven't crossed the start line on the racetrack yet), or if you are exactly one solar year later (we crossed the line), either it will be a the same place at a different time, or it will be in a different place at the same time.

So after a year we're indeed off by about 6 hours, the next year about 12, then third one, 18, and four years later, about 24 hours. So we are off by a day. Would seem we're back on our feet (our reference meridian again points towards the sun when we reach that famous spot on the orbit). But if we didn't do anything about that, after a (long) while, winter would happen in May.

So some people determined a couple of millennia ago that we would need to make a few adjustments now and then to the duration of a year, and introduced the leap years. So as we are off by about a day after 4 years, every 4 years the year is one day longer, and we're back on our feet.

Of course some other people noticed a few centuries later that this was not exact (it's not precisely 365.25 days), and more adjustments were made (so we skip leap years on years that are multiples of 100, but not multiples of 400). Still not exact either, but a lot closer.

Guess what, the minor difference between those two actually resulted in a difference of 13 days between the two calendars, and that introduced quite a bit of confusion for quite a while as not everybody switched calendars at the same time.

As a parallel, it may be interesting to note that some cultures use calendars based on the Moon rather than the Sun. Here again, no reason for the orbit of the Earth around the Sun (a year) to be a multiple of the orbit of the Moon around the Earth (about 29.5 days -- again, not a multiple of the time it takes the Earth to rotate around its axis either). This has the consequence that such calendars either need to add some extra days to "re-align" on the year, or they just drift year after year (which explains why the month of Ramadan in the Islamic Calendar occurs in March/April this year, while back in 2011 it started in August).

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The motion of the sun in the sky doesn't exactly repeat at one-year intervals, but it's still reasonable to say that it's on a one-year cycle.

If you record the sun's position at the same time each day, the points will appear to lie on a figure-8 curve (analemma). If you parametrize the continuous curve by a real number that runs from 0 to 365.25, and your first measurement is at 0, then the points you plot will be 0, 1, 2, ..., 364, 365, 0.75, 1.75, ..., 363.75, 364.75, 0.5, 1.5, .... This repeats after four years, but of course if I'd used a more accurate estimate of the year length it wouldn't. More importantly, it doesn't repeat after one year. But it is clearly cyclic with a period of one year, nevertheless.

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We do have one planet in our solar system with a neat orbital resonance: Mercury, with an orbital period of 87.9691 Earth days, and a sidereal rotation period of 58.646 days, which works out to almost exactly 1.5 sideral days, or 0.5 solar day, in its year. Being the closest planet to the Sun, it experiences a stronger gravitational pull than the other planets, thus the most affected by tidal locking.

But Earth is farther away, and has its rotation affected by competing factors that Mercury doesn't have: a large Moon with its own gravitational pull, atmospheric pressure, and ocean currents. The length of Earth's day isn't even constant, and the rotation is gradually slowing down, so even if there ever were a whole number of days in a year, that wouldn't remain true over the course of millennia.

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Analogy

  • It takes me 3 minutes to sing Pop Goes The Weasel. I will repeat it at infinitam.
  • It takes me 20 minutes to walk around the block. I will never stop walking around the block.

Therefore, when I do a full lap of the block and reach the position I started from, I will not be starting the song at the same moment. At that point in time, I will have sung it 6 times (= 18 minutes) and be 2 minutes into the 7th time singing the song (bringing us up to 20 minutes).

The next time I reach the initial walking position, I will have sung it 13 times in total (3x13 = 39) and be 1 minute into the 14th song.

The third time I reach the initial walking position, since 3 min x 20 songs = 20 min x 3 laps around the block.

The short answer is that the time it takes to do one thing (walk around the block / orbit the sun) is not an exact multiple of the other thing (sing a song / revolve around your own axis), therefore they don't line up neatly.


Define a "year"

if the sun appears to be in the exact point in the sky as that time the previous solar year

If this were the definition of a year, it would inherently mean that a year is expressed as an integer amount of "sun seems to revolve around the sky" experiences (otherwise known as a day).

However, this is not how we define a year. The sun is in that same spot in the sky 365 times before we call it a year. What's wrong with the first 364 times it happens?

A year is not defined by the position of the sun in the sky, which is based on Earth's rotation. A year is defined by Earth's position in its orbit around the sun. Therefore the day on which the sun is in the same spot in the sky which is closest to that original orbital position is what we call a year, which is after 365-ish days.


Interesting tangent

A human defines an Earth day as seeing the sun in the same spot in the sky.

If you assume the sun does not rotate around its own axis, someone living on the surface of the sun would be able to define an Earth year as seeing the Earth in the same spot in the sky.


Some simple example math, if you're keen

I assume a year is exactly 365.25 solar days here, to make the math easy enough for an example. Finer precision is up to you to implement.

I used neat integer values in my example which means that things sometimes do line up again. In reality, however, the values are decimally complicated and it's not as easy to get things to line up.

We do still pick the nearest integer value for our calendar. The bigger loop (orbit around the sun / walk around the block) takes 365.25 times the smaller loop (Earth's rotation / singing the song), so we commonly state that a year takes 365 days.

Even though that's what I did in my example, in real life we don't "hold out" and wait for things to line up naturally. That would take 1461 years. Every year we move up .25 of a day, so we'd need 4 years to move up a single day, and it'd take 365.25 times those 4 years to catch up to an entire year.
We say that a broken clock is right twice a day, but if your calendar is only accurate once every millennium and a half, that's a worthless calendar.

Instead we continually adjust our measurement to account for the "drift" (i.e. the 1 minute "delay" in the song every time I lap the block, or the .25 day "delay" in the Earth's rotation every year).

If we adjusted this every year, which we in theory could, then we would shift the clock (= tracks the smaller loop) by 6 hours. Our day/night cycle would shift by .25 day (6 hours) every year. Imagine having daylight savings time but the jump is now 6 hours, and during the year our clocks would drift further and further from "normal" until it's 6 hours out, and then it would snap back into place.

That would be a horrible experience.

Instead, we ignore the drift until we accrue a full day, which takes 4 years (4 x 0.25 = 1). When we reach that point, we jump (or should I say we leap...) a whole day.
This means that we don't need to change our clocks, since we will move from midnight to midnight on a different day.

With this system, it only impacts our calendar, which is why we introduce this extra "catch up" day called a leap day.


As a software developer I am perpetually astounded that we humans have an innate sense of what easiest, and it's counterintuitively one of the most horribly complex and unintuitive systems that we (developers) struggle with on a daily basis.

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  • $\begingroup$ Surely this statement "The sun is in that same spot in the sky 365 times before we call it a year." is not true? Each day as we move from winter to summer, the sun rises slightly further to the east, then NE, and sets further to the west then NW. It rises higher in the middle of day, tracing an arc different to the day before. Then reverses this process as we regress back to winter. So surely it is only in the exact same position in the sky at a given time of day on 2 days of the year, but on those days, it is in infinitely many "places" that are the same as it was six months(ish) ago? $\endgroup$
    – AdamV
    Apr 13, 2023 at 17:34
  • $\begingroup$ @AdamV: You're reading that statement with much greater precision than it was intended. The context is that I'm talking about the "same spot" in reference to a day cycle. The precise location is irrelevant, only that it is the same spot in the cycle. $\endgroup$
    – Flater
    Apr 13, 2023 at 23:17
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However, if the sun appears to be in the exact point in the sky as that time the previous solar year, how does the earth reach this exact position while only being 5hrs, 48 min, 45 secs into its final rotation of the year?

It isn't in the exact same point and the earth is still 5hrs, 48 min, and 45 secs behind on its track around the sun.

You probably feel that something is amiss, because you need a nice rigorous definition of a year. Let's give a shot at making one up.

Let's stand somewhere on the equator and record the sun's maximum elevation every day. If we keep doing this for a few years, we'll find that we have recorded a function with a period of 365 days, 5 hours, 48 minutes, and 45 secs.

Now this may feel bothersome, because from one year to another we are unable to get the sun in exactly the same position that we observed it in in the previous year. But we can cheat a bit.

Suppose we record the sun's position in the sky at noon (when the sun is the highest). Then we come back to the same place on the equator exactly 365 days later. We'll see, that the sun's position is ever so slightly off. But if we travel approximately 10'000km westwards (which is very close to the rotation of the earth over a period of 5hrs 48mins, etc) and wait until noon there, we'll see a much better match of the sun's elevation with what we observed a year before.

PS For what it's worth, the definition of a year I just made up is closest to the tropical year. This is the one that mattered the most to our ancestors, since it lines up with the periodicity of the seasons. This link to wikipedia also discusses the sidereal and anomalistic years, which for various reasons are slightly longer.

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Your confusion seems to stem from there being several definitions of the word "year". If someone says "This happened exactly one year ago", they generally mean that it happened at the exact same time of day, on the same calendar date, last year. However, the term "year" can also mean the time that it takes for the Earth to make one orbit around the Sun. The former is a calendar year[1]. For the latter, there are slightly different definitions that give slightly different lengths, such as sidereal, solar/tropical, and Julian, but all of them are close enough to not matter for this question.

The solar year is about 365 days and 6 hours long. The calendar year, however, by its very definition is a whole number of days; if it's the same time of day, then there must have been an integer number of days in between. So since the calendar year and solar year are slightly different length, we need an adjustment to keep them from getting out of sync, and that adjustment is leap days. Usually the calendar year is 365 days long, but sometimes it's 366 days long. Thus, by changing the length of the calendar year, we can have a year that is always a whole number of years have an average length that's not a whole number of days.

[1] The term "calendar year" is often used to distinguish between the period from Jan 1 to Dec 31 versus a fiscal year that has other endpoints, but here I'm using it to distinguish "same date" from "one orbit".

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So far the solar day is still independent on the solar year. That's the reason not only for not-whole number of days in a year but in a quite complicated way of counting a calendar not to drift away.

By the way, there are three definitions of a "day". One defines it as period of time between same positions of sun in the sky (ignoring elevation). The second defines it as a period of time between same positions of stars in the sky. And the third defines it as 86 400 seconds.

Quite simillar system is Earth - Moon where lunar "day" is exactly one lunar year by the second definition and infinite by the first definition. It's because Moon is tidally locked to the Earth.

Earth is not (yet) tidally locked with any other planet or star so there is no reason to have any synchronisation of the day to the year except for a sheer luck.

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Perhaps an image like the following might help:

epicycloid



I am walking big circles around you. At the same time, I am also spinning on my own axis. The blue points in the graph above show the path that my nose would trace out as I spin around you.

Notice that I turn approximately eight circles on my own axis for every big circle that I walk around you. But it is not exactly eight. (Hey, no one says it has to be an exact multiple!) Therefore, on the eighth circle, I end up facing you at a slightly different place from where I started.

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@zwol made a good comment:

I feel like all of the existing answers miss the point of the question, which might be better stated "why isn't the solar year defined to be the time it takes for the sun to return to the same position in the sky, relative to the Earth's surface, as that time the previous solar year?" (Which would necessarily be a whole number of solar days.) After all, the entire reason there's a difference between a sidereal day and a solar day is that we do use a "return to the same position in the sky relative to the surface" [sort of] definition for a solar day.

The answer to that question is that defining a solar day in a way where each year is exactly 365 days would require us to do either one of the following:

  1. Change the definition of the second, so that there's still exactly 60 * 60 * 24 seconds in a single day
  2. Redefine the day as 24 hours + 59.178 seconds (to account for the fact that a solar day is slightly longer than a sidereal year)

Both of these changes would require a massive investment to change every single clock on the planet. Modern operating systems could probably be updated relatively easily, however mechanical clocks and legacy devices would all have to be replaced to account for the changes. Almost no one would agree to undertake such an investment just to get rid of February 29th, so this won't happen in the foreseeable future.

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