From the University of Liverpool , something I found interesting because a few years ago, former State Climatologist Jame Goodridge said he saw correlations between length of day and other atmospheric processes.
Research reveals Earth’s core affects length of day
Research at the University of Liverpool has found that variations in the length of day over periods of between one and 10 years are caused by processes in the Earth’s core.
The Earth rotates once per day, but the length of this day varies. A yeas, 300million years ago, lasted about 450 days and a day would last about 21 hours. As a result of the slowing down of the Earth’s rotation the length of day has increased.
The rotation of the earth on its axis, however, is affected by a number of other factors – for example, the force of the wind against mountain ranges changes the length of the day by plus or minus a millisecond over a period of a year.
Professor Richard Holme, from the School of Environmental Sciences, studied the variations and fluctuations in the length of day over a one to 10 year period between 1962 and 2012. The study took account of the effects on the Earth’s rotation of atmospheric and oceanic processes to produce a model of the variations in the length of day on time scales longer than a year.
Professor Holme said: “The model shows well-known variations on decadal time scales, but importantly resolves changes over periods between one and 10 years. Previously these changes were poorly characterised; the study shows they can be explained by just two key signals, a steady 5.9 year oscillation and episodic jumps which occur at the same time as abrupt changes in the Earth’s magnetic field, generated in the Earth’s core.
He added: “This study changes fundamentally our understanding of short-period dynamics of the Earth’s fluid core. It leads us to conclude that the Earth’s lower mantle, which sits above the Earth’s outer core, is a poor conductor of electricity giving us new insight into the chemistry and mineralogy of the Earth’s deep interior.”
The research was conducted in partnership with the Université Paris Diderot and is published in Nature.
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I have some problems with the 450 days of 21 hours compared with the approximately 365.25 days of 24 hours we experience now.
Everything I’ve seen in the past indicates that the Earth is (slowly) moving into a higher — and thus longer — orbit around the Sun, just as the Moon is doing in regards to Earth. Yet these figures would have Earth experience a year of about 9450 hours 300MYA compared with our current value of around 8766. But since the length of a day has no bearing on the total number of hours in a year that seems like an exactly backward relationship to me. I could have easily believe a figure of 400 or 410 days of 21 hours, since that would be a shorter overall year but 450 I do have problems with.
Where is my understanding going wrong?
agfosterjr says:
July 12, 2013 at 8:15 pm
“Whereas tide oscillations do drive small ocean currents, some slight heat transport may turn out to be measurably correlated to tide intensity as governed by lunar orbital parameters.”
I think that the slight heat transport that you are
Are you aware of the following paper?
http://web.science.unsw.edu.au/~stevensherwood/601_02/papers/Munk_98.pdf
In this paper – i.e. Munk and Wunsch (1998) they claim that:
A. Without deep mixing, in a few thousand years, the Earth’s oceans would turn into a stagnant pool of cold salty water.
B. Hence, the Earth to Pole heat flux of 2000 Terra-watts associated with the meridional overturning circulation [of the oceans] would not exist without the comparatively minute mechanical mixing sources [i.e. deep water mixing].
C. 2.1 Terra-watts are required to maintain the global [oceanic] abyssal density distribution against 30 Servdrups of deep water formation.
D. Of the 3.7 Terra-watts of power provided by [Lunar] tidal dissipation in the Earth’s oceans, as much as 1 Terra-watt could be available for deep water mixing.
E. The winds (~ 1 TW) and [Lunar] tidal dissipation (~ 1 TW) are the only possible source of mechanical energy that can drive the deep mixing of the oceans.
These numbers have been revised (downward) since this seminal publication but the basic argument still stands.
Sorry, the top of my past post should have read:
agfosterjr says:
July 12, 2013 at 8:15 pm
“Whereas tide oscillations do drive small ocean currents, some slight heat transport may turn out to be measurably correlated to tide intensity as governed by lunar orbital parameters.”
I think that the slight heat transport that you are referring to is technically called CLIMATE on millennial to decadal time scales. .
Are you aware of the following paper?
Ian Wilson says:
July 12, 2013 at 11:23 pm
agfosterjr says:
July 12, 2013 at 8:15 pm
====================================
Munk says: “The amount [3.5TW] is tiny compared to the solar radiation of 175,000TW received by the earth, the equator-to-pole ocean heat-flow of 2000TW, and even small compared to heat flow of 30 TW from the earth’s interior.”
From this IW derives:
“Hence, the Earth to Pole heat flux of 2000 Terra-watts associated with the meridional overturning circulation [of the oceans] would not exist without the comparatively minute mechanical mixing sources [i.e. deep water mixing].”
And Munk says his study raises more questions than answers, including question #6 which he leaves unanswered: “Would the general circulation of the ocean be qualitatively different if the Earth had no Moon?”
And Munk’s paper never mentions ENSO. It seems IW is pressing wine from raisins.
So Ian, why don’t you identify this mechanism, this spigot that turns on Niños by way of tides? I find no such mechanism in Munk’s paper. He talks about tides mixing the middle and maintaining deep ocean stratification. Not a word about tides and weather. –AGF
AGF,
Here is the work of Claire Perigaud that has been done
In collaboration with her research colleagues:
Dr R. Gross, Caltech/ JPL, USA,
Dr E. Rignot, Caltech/JPL and UC Irvine, USA,
Dr D. Waliser, Caltech/JPL, USA
http://www.aviso.oceanobs.com/fileadmin/documents/OSTST/2009/poster/Perigaudabstract.pdf
ftp://ftp.cerfacs.fr/pub/globc/exchanges/GOASIS/Fermat_2009.pdf
http://www.aviso.oceanobs.com/fileadmin/documents/OSTST/2009/poster/Perigaud.pdf
http://coaps.fsu.edu/scatterometry/meeting/docs/2009_may/posters/perigaud.pdf
(http://web.gps.caltech.edu/seminars/yly_seminar/past/2011.htm
Earth-Moon-Sun alignments influencing El Niños and water/air mass momentum)
Here is some of my work:
Long-Term Lunar Atmospheric Tides in the Southern Hemisphere
Ian R. G. Wilson and Nikolay S. Sidorenkov
The Open Atmospheric Science Journal, 2013, 7, 29-54
http://www.benthamscience.com/open/toascj/articles/V007/TOASCJ130415001.pdf
El Ninos and Extreme Proxigean Spring Tides
A lecture by Ian Wilson at the Natural Climate Change
Symposium in Melbourne on June 17th 2009.
http://www.naturalclimatechange.info/?q=node/10
Ian Wilson says:
July 12, 2013 at 10:31 pm
“I let the reader decide who is actually pushing the junk science in this case.”
Don’t sweat it too much. The fellow by the screen name agfosterjr has shown many times in the past that he does not understand the dynamics of energy storage and release or resonance phenomena in general.
Ian Wilson says:
July 13, 2013 at 2:58 am
===========================
OK, that will take a while. But this I find interesting: “The most intriguing and global impact we had not anticipated is the strong biweekly peak of variability. We found it in the rain, in the winds, in the sea level, in the subsurface currents, in the Outgoing Long-wave Radiation (OLR), in the Significant Wave Height (SWH) data sets. We do not find it in SST.”
(ftp://ftp.cerfacs.fr/pub/globc/exchanges/GOASIS/Fermat_2009.pdf –pp. 5-6)
I think SST would be the most desirable link between tides and ENSO.
Tell us what you think of this paper:
http://icesjms.oxfordjournals.org/content/56/3/381.full.pdf
Regards, –AGF
Thank you agfosterjr for your openness to further inquiry.
I may have unfairly misjudged your earlier comments. Sorry.
This skat is as important as a nit on a gnats nut—relatively speaking.
tadchem says:
July 12, 2013 at 8:12 am
……………
“The tides will stop when the earth’s rotation slows so much that it always keeps one face towards the moon. That event is MANY millions of years away. By my calculations, the day and the month will both be about 1100 hours (about 45-46 days) long, as we currently measure time.”
=======
Interesting proposition! Imagine the earth-moon system viewed by an observer above the North Pole. The earth rotates anti-clockwise (all points heading east). The moon does not really rotate around the earth, we are in a “two planet in the same orbit” system. Before jumping up and down, note that the moon’s orbit, like the earth’s, is always concave to the sun. When the moon is outside the earth, it is farther from the sun and its orbital speed is less than the earth’s, so the earth overtakes the moon. Gravitational pull of the earth then attracts the moon into a lower orbit, and accelerates it, so that the moon, after passing behind the earth, then overtakes the earth. The earth’s pull then attracts the moon out again, and slows it, so the moon crosses the earth’s orbit ahead of the earth, and eventually the earth overtakes the moon again. And so on ad infinitum, or nearly. But note: to that heroic observer, the moon is in a clockwise orbit relative to the earth. If the earth then turns one face to the moon, it means that the earth will have had to change its rotation from anti-clockwise to clockwise.
Are there errors here? Perhaps, but it seems sound to me.
Note that there are two satellites of Saturn in a similar joint orbit arrangement as the Moon and the Earth but more pronounced as they are smaller and closer to each other. These are:
Epimetheus, distance to Saturn 151 422 km, diameter 113 km, orbital period 0.696 days;
Janus, distance to Saturn 151 472 km, diameter 179 km, orbital period 0.696 days.
As they are at slightly different distances, Epimetheus on the inner orbit overtakes Janus – it can’t pass it as their combined radii (146 km) is greater than the difference in orbital distances (50 km).
Their gravitational interaction attracts E[pimetheus out, and Janus in, so Epimetheus is then going slower and falls behind, while Janus goes on ahead. Som many orbits later, Janus has made one more orbiot of Saturn and catches up with Epimetheus, and they change orbits again. Source of data is:
http://jumk.de/astronomie/moons/saturn-moons.shtml
tadchem says:
July 12, 2013 at 8:12 am
uh, tad, you must be reading “modern” science books.
Don’t be fooled. The Earth cannot transfer angular momentum to the Moon.
You have been taught that the Moon rotates on its axis, I guess….
Oh, please, geran! The moon does rotate (slowly) on its axis. The moon also has angular momentum as measured from the earth (and from the sun, for that matter). Conservation of angular momentum is merely a special case of conservation of energy. You do have that in your physics books, don’t you? As the earth passes energy to the moon, to put it into a higher orbit, it can be measured as a change in angular momentum (in the earth-centered system) of both earth and moon.
ShrNfr says:
July 11, 2013 at 3:37 pm
Gosh, that almost looks like a temperature graph with the AMO…
Yes, and the length seems very similar. And when the earths spin slows down the water will be pushed mainly eastwards because of its inerta, and the top effect will be when deceleration stops and a new acceleration starts. And in combination with the regular ocean currents, hmm..
skorrent1 says:
July 14, 2013 at 5:07 am
Oh, please, geran!
>>>>>>>
Since you said “please”, I will try to help.
First, the Moon does NOT rotate on its axis. Let me demonstrate: Consider a race horse moving around an oval track. A viewer in the stands (outside the oval) sees the right side of the horse, as the horse passes nearby. But, on the far side of the track, the viewer sees the left side of the horse. So, it APPEARS the horse has rotated on it axis, But the horse in only following an oval orbit. Exactly the same as our Moon. (And, just as in the case of our Moon, a viewer INSIDE the oval always sees the same side of the horse.)
There are other errors and misconceptions in your comment, but first it is important you understand this simple point. It gets more complicated from here.
Come off it, Geran.
The moon DOES rotate. See Wikipaedia:
“With respect to the stars, the Moon takes 27 Earth days, 7 hours and 43.2 minutes to complete its orbit; but since the Earth-Moon system advances around the Sun in the meantime, the Moon must travel further to get back to the same phase. On average, this synodic period lasts 29 days, 12 hours, 44 minutes and 3 seconds. This is an average figure, since the speed of the Earth-Moon system around the Sun varies slightly over a year, due to the eccentricity of the orbit.”
If your horse starts heading east, then turns left and heads north, then left again and heads west, then turns again and heads south, then finally ends up heading east again, the horse has rotated once, even if the rotation is spread out over a full race track.
You might not like Wikipaedia, but if you posted there that the moon does not rotate you would be shot down smartish!
Ah, geran, but a viewer above the plane of the moon’s orbit notices that the moon must spin completely around once every 28 days or so to keep it’s same face toward the earth. A horse making, essentially, four left turns (or right turns in England) does wind up turning 360 degrees. Explain, please, when is a rotation not a rotation. Always open to correction of “errors and misconceptions”. Hate to think that my professional calculations of missile and satellite paths were wrong.
Dudley Horscroft says:
July 14, 2013 at 6:52 am
Exactly my point Dudley, “institutional” science is wrong in several areas, AGW, is only one such area. The Moon is so simple and straight forward that it is a perfect example.
skorrent1 says:
July 14, 2013 at 6:56 am
The difference is the horse is NOT rotating on its axis, nor is the Moon.
If you are claiming that “rotation on an axis” does not occur unless the object is stationary, then you have a weird definition of rotation. By that measure, the earth does not “rotate” because while it is “spinning” it is also moving along its orbit around the sun, just as the moon moves along its orbit around the earth.
Geran, to put it another way, if I told my buddies that their spy satellites did not have to “rotate on their axes once per revolution” to keep their camera’s pointed toward the ground, I would have been booted off the project.
skorrent1 says:
July 14, 2013 at 7:37 am
If you are claiming that “rotation on an axis” does not occur unless the object is stationary, then you have a weird definition of rotation. By that measure, the earth does not “rotate” because while it is “spinning” it is also moving along its orbit around the sun, just as the moon moves along its orbit around the earth.
>>>>>>>
If you really believe I have made such a claim, then your imagination is working REALLY well. So, try to imagine that what you see is what it is. The horse is NOT rotating on its axis, nor is the Moon.
skorrent1 says:
July 14, 2013 at 7:56 am
Geran, to put it another way, if I told my buddies that their spy satellites did not have to “rotate on their axes once per revolution” to keep their camera’s pointed toward the ground, I would have been booted off the project.
>>>>>
So, intimidation means you are not free to think for yourself?
Just curious, does anyone have the calculation for how many joules of energy is being transferred when the Earth speeds up or slows down one millisecond?
@ur momisugly Dudley Horscroft July 14, 2013 at 1:09 am
You say:-
Let’s test your statement with some observations. When the waning moon is past its last quarter we can often see it rising in the east in the pre-dawn sky ahead of sunrise. In this case, from my northern hemisphere perspective, the moon is located to the right of the sun. At new moon the moon is closer to the sun than the earth. After new moon we see the waxing crescent moon in the western sky after sunset, now located to the left of the set sun. So this series of observations tells us that the new moon passes from right to left across the daytime sunlit sky.
For this observation to be true it necessarily follows that at the time of new moon the earth passes the moon “on the outside track” and so is orbiting the sun faster than the moon, even though it is further away. Not convinced by this? Consider the example of the time in 1973 when Concorde chased the Moon Shadow across the Sahara Desert along the Path of Totality Total Solar Eclipse of 30 June 1973 from Concorde 001
At the time of the eclipse the moon shadow passed from the west over the Atlantic Ocean eastwards across the sunlit Sahara Desert towards the eastern horizon. So the direction of the shadow’s motion is from right to left for a ground based northern hemisphere eclipse observer. We know that the earth is further from the sun at the time off a solar eclipse. The earth is therefore overtaking the moon shadow. The only way this can occur is for the earth to be orbiting the sun faster than the moon when the earth is furthest from the sun.
A moment’s reflection will show that this is also true for the moon. At the time of full moon it is orbiting the sun faster than the earth even though it is further away.
geran says:
July 14, 2013 at 6:19 am
“But, on the far side of the track, the viewer sees the left side of the horse. So, it APPEARS the horse has rotated on it axis, But the horse in only following an oval orbit.”
You have a fundamental misunderstanding. We never see the “left side of the horse”, i.e., the “dark side of the Moon”. The Moon always shows us the same face.
Look at it over a month’s time. No matter where you see it, you will always see the Man in the Moon. The Moon is tidally locked. It rotates at essentially the same rate as it revolves around the Earth.
And yes, the Earth can influence both the orbital angular momentum and the spin angular momentum of the Moon because both bodies are non-rigid.
Bart says:
July 14, 2013 at 9:25 am
You have a fundamental misunderstanding. We never see the “left side of the horse”
>>>>>>
Bart, ever been to a horse race? How about a track meet? Same principle. Or, you can carry out this experiment on your living room floor. It is not a hard concept. (Use an orange for the Earth and a pencil, or other pointed object, (please be very careful with pointed objects….) to represent the Moon.
No expensive test equipment involved, but you have to use your brain.
Geran: This is getting tedious. I tried your orange/pencil experiment. I started with the marking side of the pencil facing the orange and found that, as I moved the pencil around the orange — always keeping the marking facing the orange — I had to rotate the pencil completely one time to arrive back at the starting point. Gee, what am I doing wrong ;-)? Must be my brain! LOL
geran says:
July 14, 2013 at 10:02 am
The stupid. It burns…
Try this, geran. Suppose the jockey is holding a compass. As he rounds the track, is the needle going to A) stay fixed or B) rotate 360 degrees? In your reality, you would choose (A). That is incorrect.
Anyway, sorry I got involved. I should know by now that you cannot reason someone out of something they have not been reasoned into. Hasta la vista, geran.
Philip M– dudley may be mathematically challenged, but wouldn’t it be easier to just say “At the time of a full moon the moon’s earth-orbit velocity vector is added to the earth’s orbital velocity, while at new moon, it is subtracted.”?
skorrent1 says:
July 14, 2013 at 10:42 am
Geran: This is getting tedious. I tried your orange/pencil experiment. I started with the marking side of the pencil facing the orange and found that, as I moved the pencil around the orange — always keeping the marking facing the orange — I had to rotate the pencil completely one time to arrive back at the starting point. Gee, what am I doing wrong ;-)? Must be my brain! LOL
>>>>>>
I understand, but please hang in there. I think you may be about to see it.
As you moved around the orange, you were turning the pencil, not rotating it on its axis. This is just the same as the race horse of race car.
Also, try this. Assume the orange is the middle of a clock face. Start with the pencil at the “3:00” position. Position one finger at middle of the pencil. Your finger will represent the axis of rotation. Without advancing the pencil around the clock, rotate the pencil 360 degrees. You will notice that the orange “sees” all sides of the pencil during a full rotation. If the Moon were rotating on its axis, instead of turning, we would see all sides of the Moon, which we do not.
Hang it there….
Retired Engineer John says:
July 14, 2013 at 8:38 am
Just curious, does anyone have the calculation for how many joules of energy is being transferred when the Earth speeds up or slows down one millisecond?
=============================================================
The earth’s kinetic energy of rotation is 2.138 x 10^29 J. A sidereal day is 86164 seconds.
So (2.138 x 10^29J x 2) / (86164 x 1000) = 4.96 x 10^21 J.
LOD loses 2.3ms/century irreversibly, less a temporary .6ms/century reversibly due to GIA–which doesn’t enter into the equation, so it takes 43 years to lose a ms worth of energy, enough to heat the whole ocean almost 1/1000 of a degree and enough to heat your kitchen with some left over. Plenty of energy but hard to harness. –AGF