Layperson's question regarding earth's rotational wobble

[Glenstr]

Our carpool conversation this time of year usually involves the shortening daylight hours, which are quite noticeable given that we are between the 54th and 55th parallel. After each new year arrives we start watching the horizon as we're leaving work (at 5pm) waiting for the day when we start seeing the sunset again.

Having just been supplied an iphone recently, a couple of weeks ago I put a simple sunset/sunrise app on and was looking at times for our location. When the day came when we just caught the sunset as we were leaving, I opined that on exactly this many days on the other side of the solstice we should be seeing the exact same thing, so I looked up the Xth day past Dec 21 to see what the times were. It turns out that while the (length of) daylight hours are very close, there was quite a time shift-difference in the sunrise-sunsets, almost 1/2 hour if I recall.

Our (laypersons) opinion was that this must be because of the Earth's rotational wobble, and the further towards either pole one was the bigger the time shift would be.

Are we correct in this assumption? Also, is the wobble conisistent throughout Earth's orbit?, if so then next year should be the same scenario, correct?

 

[nitsuj]

What if you consider the purpose of a Leap Year,

 

[D H]

Are we correct in this assumption?

No, you're not correct.

This effect would be present even if the Earth maintained a constant rotation rate about a fixed axis. What you have discovered is the "equation of time". It's a consequence of two things: That the Earth has a non-circular orbit, and that the Earth's rotation axis is tilted with respect to the orbital plane.

Before I go into details, I need to talk about sidereal time and solar time. One sidereal day is the time it takes for the Earth to rotate once with respect to the fixed stars. (Aside: Astronomers use quasars now rather than stars because quasars are so far away that they truly do appear to be fixed.) One solar day is the time it takes for the Earth to rotate once with respect to the Sun. High noon to high noon, for example. A solar day is a slightly longer than a sidereal day because the Earth is orbiting the Sun. There is one more sidereal day in a year than there are solar days.

Our 24 hour time scale is based on sidereal time rather than solar time, but scaled to make our clocks keep track with solar time, on average. Since there are ~365.25 solar days in a year, there are ~366.25 sidereal days in a year. Our 24 hour day is 366.25/365.25 sidereal days, aka a mean solar day. ("Mean" means average in this context.) High noon to high noon: That's a true solar day.

The reason for using sidereal time as the basis for our clocks is that the eccentricity of the Earth's orbit and the obliquity of the Earth's rotation axis make the length of a solar day vary over the course of the year. Before we had clocks, the difference between mean solar time (time kept by the fixed stars) and true solar time (time kept by a sundial) was a curiosity known to astronomers only. The invention of clocks coincided with a burgeoning industry that wanted to run things like clockwork. This resulted in a move from time as kept by the sundial to time as kept by a clock.You can google the term "equation of time" for more info. A related concept is the "analemma". The wikipedia articles on both topics are good starting points.

 

[D H]

What if you consider the purpose of a Leap Year,

That has nothing to do with the topic at hand, nitsuj. We have leap years because one year is about 365.25 days long. It's not exactly 365.25 days long, and that's why we have the goofy rules about not having leap years in years that are a multiple of 100 but not 400.

 

[Glenstr]

No, you're not correct.

This effect would be present even if the Earth maintained a constant rotation rate about a fixed axis. What you have discovered is the "equation of time". It's a consequence of two things: That the Earth has a non-circular orbit, and that the Earth's rotation axis is tilted with respect to the orbital plane.

<snip>

Thanks for the explanation, it will give us something to discuss on the ride home tonight. Although I'm still trying to figure out from your explanation, in laymans terms, what it is exactly is the reason the sun rose today at 7:57 and will set at 4:28 for 8h32m of daylight, and will rise at 8:27 & set at 4:54 for 8h27m on Jan 27, which is roughly the same distance from the winter solstice as today is.

Is it just the shifting of the mean length of day (366.25 / 365.25) over the time between now and Jan 27, or the eccentricity of Earth's orbit, or a bit of both?

 

[Ibix]

If the Earth did not spin, but just went around the Sun, the Sun would move a little bit in the sky each day. This effect is why the Sun is in a different place in the sky each day at the same time. Because the Earth doesn't circle the Sun at a constant speed, the change is different each day.

DH's explantion is covers a lot of details that I've glossed over.

 

[D H]

Thanks for the explanation, it will give us something to discuss on the ride home tonight. Although I'm still trying to figure out from your explanation, in laymans terms, what it is exactly is the reason the sun rose today at 7:57 and will set at 4:28 for 8h32m of daylight, and will rise at 8:27 & set at 4:54 for 8h27m on Jan 27, which is roughly the same distance from the winter solstice as today is.

Is it just the shifting of the mean length of day (366.25 / 365.25) over the time between now and Jan 27, or the eccentricity of Earth's orbit, or a bit of both?

It's both.

Both the eccentricity and obliquity result in quasi-sinusoidal variations in the "equation of time." The eccentricity term has an amplitude of a bit less than 8 minutes and has one cycle per anomalistic year, with zeros at the times of aphelion and perihelion passage (when the Earth is closest to and furthest from the Sun). The obliquity term has an amplitude of almost 10 minutes and has two cycles per tropical year, with zeros at the equinoxes and solstices.

Perihelion passage will occur on January 4, 2014, shortly after winter solstice, when the obliquity term crosses zero. Those two zero crossings are in the same direction. That means that eccentricity and obliquity terms add to each other at this time of year. This is why you see such a huge variance between early November and early February. The eccentricity and obliquity terms also both cross zero near the summer solstice, but there they are out of phase with respect to one another.

Another effect is that autumn and winter are collectively shorter than are spring and summer by about 7.6 days (in the northern hemisphere, that is; it's the other way around in the southern hemisphere). The obliquity term is not quite sinusoidal. Because its zero crossings are at the equinoxes and solstices, it's changing faster now than it will six months from now.

 

[Glenstr]

Thanks again - now that I can visualize it in my head I more or less get it. However I would think that the Earth's wobble must have some effect on sunsets or sunrises, but probably quite minimal. Here is another question, if the Earth's orbit direction was opposite of what it is now, yet the rotation was same, how, if at all, would this effect the solar day?

 

[Nugatory]

Thanks again - now that I can visualize it in my head I more or less get it. However I would think that the Earth's wobble must have some effect on sunsets or sunrises, but probably quite minimal. Here is another question, if the Earth's orbit direction was opposite of what it is now, yet the rotation was same, how, if at all, would this effect the solar day?

We would get one more solar day in the sidereal year, instead of the one fewer that we get.