11
$\begingroup$

Background: I'm training to be a geography teacher. Currently I have practice lessons and I'll be discussing solar time and standard time with the class. Now I stumbled over an issue to which I could not find an answer:

We teach the students:

  • Sidereal day: In 23 h 56 min the earth rotates 360°
  • Solar day: In 24 h the earth rotates 361°

We also teach that in order to construct time zones (as a replacement of solar time), one divided 360° by 24 (to have a zone for each hour of the day), which results in 24 zones with 15° each.

Now my question is: Why does one mix the measures for sidereal day and solar day? Or put differently: Why doesn't one calculate 361°/24 or 360° / 23.933?

Improve this question
$\endgroup$
9
  • 5
    $\begingroup$ Just out of curiosity: Lets say you make time zones 361°/24. Now you start at 0° and have the first time zone to 15.0416°. And so on. The 23rd time zone is from 345.9583° to 360. And thus only 14.0416° big. Or it overlaps with the 1st again. Neither seems very practical nor desirable. How do you propose in practice time zones of 361°/24 would actually work? $\endgroup$
    – Polygnome
    Sep 27, 2020 at 7:48
  • 3
    $\begingroup$ The 361 ° are wrong anyway, it is 361.002785. But with reference to the sun the Earth rotates 360.0000 ° in 24 h. $\endgroup$
    – Uwe
    Sep 27, 2020 at 15:54
  • 3
    $\begingroup$ The time of day and time-zones are based on the apparent movement of the Sun around the Earth. (For any passing flat-earthers, note that I highlighted the word "apparent".) $\endgroup$
    – billpg
    Sep 28, 2020 at 10:47
  • 1
    $\begingroup$ Frame challenge: There are clearly way more than 24 time zones, ranging from -12:00 to +14:00 (ignoring potential DST complexities). $\endgroup$
    – Eric Towers
    Sep 29, 2020 at 4:54
  • 1
    $\begingroup$ @EricTowers: Including really weird ones. $\endgroup$
    – Eric Duminil
    Sep 29, 2020 at 11:03

7 Answers 7

Reset to default
52
$\begingroup$

The Earth takes 23 hours 56 minutes to rotate once. But that is not relevant to most people. Sure, the stars will be in the same position again after 23 hours 56 minutes, but the sun will not be in the same position.

It is far more important, for most people, to measure the time from noon to noon. And the average time from noon to noon is 24 hours. This is because the motion of the sun is a combination of both the spinning of the Earth and the orbit of the Earth around the sun. The orbital motion of the Earth adds four minutes. You should also teach the students

  • In twenty-four hours the sun advances 360 degrees. (solar day)

Time zones are based on clock time, which is based on the motion of the sun and not the motion of the stars.

Improve this answer
$\endgroup$
10
  • 19
    $\begingroup$ The motion of the sun is not the same as the motion of the stars. The sun appears to move relative to the stars. In 24 hours the stars advance 361 degrees. In 24 hours the sun advances 360 degrees. The sun has moved relative to the stars because of the orbital motion of the Earth. $\endgroup$
    – James K
    Sep 25, 2020 at 17:56
  • 1
    $\begingroup$ @mfran (or other readers) It is good to consider that the number of degrees in a circle (360) is close to the number of days ina year (365.25). Hence roughly the one degree difference. $\endgroup$
    – Vladimir F Героям слава
    Sep 26, 2020 at 19:27
  • 3
    $\begingroup$ Another way to see this motion is to look at it over larger times. Start with a point at noon. Wait six months. When the Earth/point system have the same orientation with the distant starts, it is midnight. $\endgroup$
    – Eric Towers
    Sep 26, 2020 at 20:48
  • 1
    $\begingroup$ Why 360 and not 361? Before international standards and SI units, how do you thing people defined the length of a day, and then divided it into hours and minutes? $\endgroup$
    – alephzero
    Sep 27, 2020 at 21:21
  • 1
    $\begingroup$ I think the OP is not understanding why the stars take 23h56m to return to the same position, and the sun 24h00m. Is that the fundamental problem? $\endgroup$
    – Ben Hillier
    Sep 28, 2020 at 8:33
29
$\begingroup$

We teach the students:

  • Sidereal day: In 23 h 56 min the earth rotates 360°
  • Solar day: In 24 h the earth rotates 361°

You should not teach your students that.

You should instead teach your students that it takes the Earth 23 hours and 56 minutes to rotate 360° with respect to the remote stars. So why do we use a 24 hour day? The reason we use a 24 hour day rather than a 23 hour and 56 minute day is that the remote stars do not rule our lives.

There is one star, which is not so remote, that does rule our lives. That star is the Sun. That the Earth orbits the Sun means that it takes a bit longer, about four minutes longer, for the Earth to rotate 360° with respect to the Sun. In one year, the Earth rotates one more time with respect to the remote stars than it does with respect to the Sun.

Improve this answer
$\endgroup$
3
  • 6
    $\begingroup$ "remote stars do not rule our lives" - and here I thought this were astrology.SE ;) $\endgroup$
    – Hagen von Eitzen
    Sep 28, 2020 at 21:37
  • 1
    $\begingroup$ "the remote stars do not rule our lives". I've got a very strict rule : I'm only allowed to drink a sip of rum when Polaris is above the horizon. $\endgroup$
    – Eric Duminil
    Sep 29, 2020 at 10:59
  • 3
    $\begingroup$ @EricDuminil - Don't go to Australia then. Or to New Zealand, or South Africa, or Antarctica. $\endgroup$
    – David Hammen
    Sep 29, 2020 at 14:52
14
$\begingroup$

All of the other answers are fairly technical, but a decently simple logic chain forces time zones to be 360° / 24. Consider this:

  • Since there are 24 hours in a day, it makes sense to divide the Earth into 24 time zones.
  • There are 360° in a circle, because that's how degrees are defined.
  • To completely cover the circumference of the Earth, you must account for exactly 360°. If you account for more, then you will have wrapped the Earth more than once, and end up with some overlapping time zones. If you account for less, then there will be a slice of the Earth that lies outside of all time zones.
  • Therefore, the average time zone must be exactly 360° / 24 (or 15°).
  • To be fair, it makes the most sense to just set all time zones to 15°, starting at some line that we call 0°. (Then if some nations want to mess with their own time zones, slight adjustments can be made. And that's precisely what has happened.)
Improve this answer
$\endgroup$
2
  • $\begingroup$ +1 I really like this answer, I think going into solar vs sidereal is a bit of a red herring for this question. In practice, time zones are often roughly applied and subject to all sorts of political whims and practicalities. $\endgroup$
    – eps
    Sep 28, 2020 at 17:45
  • $\begingroup$ This is really all there is to it. If you want to divide a circle into 24 equal parts, they will each be 15 degrees. The only question is whether it should be divided into 24 equal parts or 23.93 equal parts, but the sidereal day has little use in most day-to-day applications. There's a reason why the solar day is a nice round number and the sidereal day is not, rather than the other way around. $\endgroup$
    – Nuclear Hoagie
    Sep 29, 2020 at 13:51
9
$\begingroup$

Here’s a slightly different way to think about it that might be helpful to you and/or the students. To answer the question “how long does it take the Earth to rotate once on its axis?” you have to first answer “rotate with respect to what reference point?”

If you ask how long it takes to rotate enough to bring a given star back to the same position, that is 23 hours and 56 minutes. If instead you ask, how long does it take to bring the Sun back to the same position, that is 24 hours. Put differently, in 24 hours the Earth rotates 360 degrees with respect to the Sun. (And it’s equally true that in 23h56m the Earth rotates 360 degrees with respect to the stars.)

The reason the two are different is that we are moving with respect to the Sun more than we are moving with respect to the stars.

Which is more relevant for Earth-based timekeeping? Of course it’s the Sun, so we divide that 360 degrees by 24 hours.

Images like this show the perspective of an observer outside the Earth, but for an observer fixed to the Earth, the Sun really moves (on average) 360 degrees in our sky in 24 hours.

Improve this answer
$\endgroup$
3
  • $\begingroup$ The interval between times when the Sun passes the same longitude varies slightly throughout the year as a consequence of the Earth's elliptical orbit. Many globes have a figure-eight-like figure that crosses the Equator in one of the oceans, which marks what the location of the Sun would be at 24-hour intervals. It's pretty tall and skinny, but its width is definitely not zero. $\endgroup$
    – supercat
    Sep 26, 2020 at 19:46
  • 1
    $\begingroup$ @supercat - There are two factors that result in that figure-eight-like figure you mentioned, which is called an analemma. You only mentioned one. The other is axial tilt, or obliquity, whose contribution is slightly greater than that of eccentricity -- for the Earth. A Mars analemma is teardrop-shaped rather than figure-eight-like because Mars' eccentricity dominates over its axial tilt. $\endgroup$
    – David Hammen
    Sep 28, 2020 at 10:19
  • $\begingroup$ @DavidHammen: Thanks for the correction/clarification. My intended point was that the time between local noons isn't constant, though the stated reason wasn't the main one. If I'm understanding correctly, the issue is that even if the orbit were perfectly spherical, an observer on a stationary Earth-relative reference point (e.g. a plane flying around the Equator at a celestial-sphere-synchronous 1037.7mph) would perceive the Sun as tracing an elliptical path on a circular cylinder whose axis was parallel to that of Earth, but the path would be in the plane of the Earth's orbit? $\endgroup$
    – supercat
    Sep 28, 2020 at 15:56
4
$\begingroup$

Though I see the answer a few times, I feel that they are too complicated for a student. I'm 13 and my dad pointed me to this thread.

Here's how I see it. The Earth spins, and the Earth revolves around the Sun. The amount of time that it takes for a star to spin all the way around and appear again at the same spot is different than the amount of time that it takes for the Sun to spin all the way around and appear again at the same spot.

Which one is interesting to me, when some star is rising and setting? Or when the Sun is rising and setting? Because I wake up according to the Sun, and eat dinner according to the Sun, I really only care about when the Sun is setting or rising. So we measure our days by the Sun, not the stars.

Exactly how many hours and minutes each takes is not interesting.

Tell your students to read Around the World in 80 days. There is a surprise at the end due to this exactly!

Improve this answer
$\endgroup$
2
  • $\begingroup$ +1 For the teaser about Around the World in Eighty Days. I didn't know about this - now I'm intrigued and will have to read the book! $\endgroup$
    – Oscar Bravo
    Sep 29, 2020 at 8:10
  • $\begingroup$ It really is a great read, and that ending is M. Night Shyamalan-level awesome $\endgroup$
    – bobsbeenjamin
    Oct 11, 2020 at 5:30
1
$\begingroup$

This is not quite an answer but more than a simple comment. When I speak about distances I mean (angular) distances in the sky, when i speak about position I mean position in the sky. I adhere to terrestrial observer's point of view.

1, Starry sky (and everything on it, including Sun, Moon, planets) rotates around a point near Polaris (north celestial pole)

2, A given star (and everything which doesn't move relative to stars) gets to its maximum height above horizon at south (by definition). This is called culmination.

3, Interval between two consecutive culminations of the star is reasonably constant and is called sidereal day. Its length is approximately 23 hours and 56 minutes

4, Sun moves relative to stars along a well defined path, called ecliptic. This image shows starry sky along ecliptic on Apr 12, 2020 with Sun and planets. The ecliptic is the horizontal orange line in the middle.

Zodiac with Sun and planets Apr 12, 2020

5, Sun moves along ecliptic to the left and completes its journey in one year. The length of the ecliptic is 360 degrees, year has slightly more than 360 days so Sun moves by nearly one degree in one day.

At the end of May, for example, Sun every year passes between the most prominent groups of stars along the ecliptic, Pleiades and Hyades, both in Taurus. This is real photo from SOHO satellite.

SOHO photo of Sun in Taurus

6, When you measure culmination of Sun you must get more than sidereal day

7, Since, as many others noted in their comments and answers, our life is (still) controlled by daylight (Sun), we define day according to culmination of Sun, rather than stars. This is because solar (rather than sidereal) day has 24 hours.

6, If you define noon as time when real Sun is at south you get noon at different times for places at different geographical longitudes. This is serious problem when you need to coordinate say railway timetables. For this reason, time zones were introduced.

7, While sidereal day has (reasonably) constant length, solar day measured using real Sun has variable length during a year (motion of Sun along ecliptic is not uniform and different parts of ecliptic have different distance from celestial north pole). So the constant length 24 day is an abstraction - you must "observe" culmination of a fictional point called mean Sun.

Improve this answer
$\endgroup$
0
$\begingroup$

If you say that one place is one hour ahead of another, that means that the time of day at which the sun reaches its greatest apparent height for the first place is one hour ahead of the second. Time zones refer to the apparent motion of the sun, not the motion of the earth. So they are calculated as solar day divided by 360 degrees. Put another way, the earliest clocks were sundials. It takes a solar day for the shadow of a sundial to go 360 degrees (it's not visible for the night portion, but we can model it as there being an imaginary shadow that continues to move).

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .