# Canute Ponders The Tides

Guest Post by Willis Eschenbach

Short Post.  You can skip this if you understand the tidal force. Some folks seem confused about the nature of tidal forces. Today I saw this gem: “The tide raising force acts in both directions (bulge on each side in the simplistic model)” … the author of that statement may understand the tides, but that isn’t how tidal forces work, they don’t act in different directions.

To clarify the situation, let me give an example. Imagine three one-kilogram masses M1, M2, and M3, connected by a blue rope, that are in free-fall directly into the sun. Figure 1 shows the situation.

Figure 1. Forces on three one-kilogram masses M1—M3, which are in free-fall straight towards the sun, and are connected by the blue rope. The equations at the bottom show the force of gravity on each of the three masses. The “tidal force” is the difference between two gravitational forces on two objects. Solar tidal forces on a 1-kg mass at the earth’s surface average about half a micronewton (µN, or 10-6 newtons), and the lunar tidal forces are about twice that, about a micronewton. By comparison, Earth’s gravity exerts a force of about 10 newtons on a 1-kg mass at the surface …

Clearly, the gravitational force on M1 is greater than the force on M2, which in turn is greater than the force on M3. The difference between those two gravitational forces is the tidal force. There are two different tidal forces of interest in the diagram, which are GF1 – GF2, and GF2 – GF3. As a result of these tidal forces, as the three masses fall towards the sun. the near mass M1 moves the furthest, the far mass M3 moves the least, and the blue rope is always under tension.

For tidal forces on a theoretical planet, just imagine that the surface of the planet is the gray dotted line, and that “r” is the radius of the planet. The tidal force is the difference between the force of gravity on the unit mass M2 at the center of the planet, and the forces of gravity on the unit masses M1 and M3 at the points nearest and furthest from the sun.

Calculating the Tidal Forces

As shown in Figure 1, to calculate the tidal force you can just subtract the gravitational force on the unit mass at the center of the planet from the force on the other unit mass, for example GF1 – GF2. So we can use the equations in Figure 1 to calculate the force.

However, when the distance D is much, much larger than the radius r, we can make some simplifying assumptions. First, we can assume that the force between M1 and M2 is the same as the force between M2 and M3. It’s not, as you can see from the equations … but it is not a large error. For tides from the moon, the individual tidal forces are about ±5% different from their average value.

The next simplifying assumption we can make if D is much larger than r is that the average tidal force can be calculated as

Tidal Force = 2 * G * sunmass * r / D^3

This leads to tiny errors with respect to the sun’s average tidal force on the earth, and slightly larger errors with respect to the moon. For the sun we have

Near side true tidal force on earth (GF1 – GF2) = 0.50590 µN

Far side true force (GF2 – GF3) = 0.50583

True average = 0.50587

Approximation = 0.50587

So the approximation of the average tidal force is good to five significant digits …

And for the moon we have:

Near side true tidal force on earth (GF1 – GF2) = 1.1350 µN

Far side true force (GF2 – GF3) = 1.0796

True average = 1.1073

Approximation = 1.1067

For the moon, the approximation of the average is good to four significant digits. For most results, of course, those differences are far too small to be meaningful, and we can use the approximation without concern.

Why Are There Two Tidal Bulges?

Moving on, let’s look at why there are tidal bulges on both the near and far sides. It’s not because the tidal force acts in different directions, because tidal force is always directed towards the sun. So lets look at Figure 1 again, and this time we’ll include a planet and an ocean. As in Figure 1, the planet is simply free-falling into the sun. That result is shown in Figure 2:

Figure 2. Tidal forces elongating the planet and the ocean. Note that the planet is elongated as well, but this is not shown in the diagram because obviously, tides in the solid planet are much smaller than tides in the ocean. NOTE THAT THIS PLANET IS NOT THE EARTH.

As the planet free-falls into the sun, the near side of the ocean is pulled the most. So it moves the most, pulling away from the planet. And the planet, in turn is pulled more than the far side, so it pulls away from the ocean. These relative motions create the “bulges”

In other words, the “bulges” on the two sides of the planet are simply a result of the tidal forces stretching the entire system. Indeed, the atmosphere is subject to the same phenomenon, so there are atmospheric tides as well. In prior centuries, these atmospheric tides were said to affect the weather, although in a cyclical rather than a secular manner. However, such connections, while certainly possible, have proven to be elusive and difficult to establish. One complicating factor is that the atmospheric tides that are caused by the sun occur in combinations with solar heating, which makes the atmosphere swell. Also, the variations in pressure caused by the atmospheric tides are small, on the order of a tenth of a millibar (or of a hectopascal), or a variation of only 0.01% …

Finally, I’ve said that the situation in Figure 2 shows that planet free-falling into the sun. As it gets closer and closer, the tidal forces stretching the system get greater and greater, stretching the system more and more. Eventually, at a distance called the “Roche limit”, the tidal forces can reach such strength that they rip the planet into pieces … dang, to have a ring-side seat right on the Roche limit line for that spectacle would be awesome, even if it were perhaps a bit warm …

My best to all,

w.

PS—As with gravity, every mass in the universe exerts a tidal force on every other mass, so you can personally take credit for raising a tide on the surface of the sun … yeah, it’s a small tide, but it’s real. Just sayin’, I had someone claim the opposite. Lots of tidal misinformation out there.

THE USUAL: If you disagree with me or anyone, please quote exactly what it is you disagree with, and tell us precisely why you disagree. That way, we can understand exactly what it is you object to.

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February 14, 2014 6:43 pm

millinewton (mN, or 10-6 newtons
1-^(-3) newton. ‘Milli’ means 1/1000 th

matthew
February 14, 2014 6:57 pm

Figure 2 and it’s accompanying explanation seem to imply that the major axis of the simplified ellipse of the ocean (in cross section), is always pointed towards the sun. But the major axis (of the simplified cross sectional view of the oceans) should surely be oriented towards the moon, which is why the tides change from day to day.

February 14, 2014 6:58 pm

Those Greek letters can be a real pain! Teachers may have a question with micro newtons, but micro is represented by the Greek letter mu, however if the symbol font is not available, then mu becomes m and the answer is off by a factor of 1000 and students get confused.

earwig42
February 14, 2014 6:59 pm

Willis, I learn something from each of your posts.
Thank you!

William Sears
February 14, 2014 7:06 pm

Willis, not sure about your numbers but also you seem to be calculating a force per unit mass (N/kg) and not a force in newtons. The tidal force (per unit mass) formula should also have D-cubed and not D-squared.

February 14, 2014 7:11 pm

Nicely done Willis, a very clear explanation. Thank you.

February 14, 2014 7:16 pm

Werner – alt+230 = mu = µ
I also like alt+0176 = °, alt+0153 = ™, alt+0177 = ± alt+0174 = ®

Baa Humbug
February 14, 2014 7:41 pm

For tidal forces on the Earth, just imagine that the surface of the planet is the gray dotted line, and that “r” is the radius of the planet. The tidal force is the difference between the force of gravity on the unit mass M2 at the center of the planet, and the forces of gravity on the unit masses M1 and M2 at the points nearest and furthest from the sun

Possible typo, I believe the farthest unit mass is M3.
[Thanks, fixed. -w]

Richard D
February 14, 2014 7:55 pm

For me the math is not fine with paper, pen, calculator…..but entering on the compute really sucks. Thanks Willis.

noaaprogrammer
February 14, 2014 8:01 pm

With 3 bodies in play here, (earth, moon, and sun), are there not more than two bulges on opposite sides of the earth? In other words, if the earth was an all-water sphere with no land, what would be the length of the radius as measured from the earth’s center to the surface of the water as a function of its angles in a spherical coordinate system? With no continents to set any boundaries, it seems to me that it would be possible for more than one mode of a standing wave (bulge).

eyesonu
February 14, 2014 8:11 pm

Willis,
I must admit that I have never given much thought about the tides. I knew the sun was a factor and thought the moon was the other important factor but never really thought much deeper.
From your explanation, could the bulge opposite the sun be explained as a resultant force created by the centripetal force resulting from the earth’s orbit around the sun?
You’ve got me thinking and now I can’t rest until I fully understand the tides. Ohh, that need to know thingy.

William Sears
February 14, 2014 8:16 pm

Yes Willis, this would be true except for the fact that you do not have an m1 and an m2 in your formulae. You have only the sunmass with no earthmass. Also the numbers that you have calculated appear to have used only one mass, that of the sun.

William Sears
February 14, 2014 8:30 pm

Willis,
Also in your figure 1 caption you say “Solar tidal forces on a 1-kg mass at the earth’s surface average about half a micronewton (µN, or 10-6 newtons)” which means that you actually agree with me and were just a little sloppy elsewhere. Although I don’t think this is quite the right way of saying this as the tidal force that you have calculated is applied over the radius of the earth (per unit mass) and is not the force on a kilogram mass laying on its surface, tidal or otherwise. I’m in a pedantic mood this evening but will soon fall asleep. I think that I’m in a later time zone than you are.

eyesonu
February 14, 2014 8:33 pm

OK Willis, now my computer has refreshed after making the above comment and I see what you are expressing in a free falling earth. But even if outside the scope of your discussion, for my need to know, in reality would the bulge opposite the sun be from centripal force?

Barry L.
February 14, 2014 8:33 pm

Willis,
Ever about the tides and if they could induce some kind of spherical harmonics within the ocean?
http://en.wikipedia.org/wiki/File:Spherical_harmonics.png

Steve Reddish
February 14, 2014 8:33 pm

Thanks for presenting a clear explanation of tidal forces and thus why there is a tidal bulge on the backside of the Earth for both the Sun and the Moon, Willis.
I would like to add 2 points:
1. Atmospheric contraction and expansion contribute to its tidal bulges, but not so for the oceans. Sea tides rise and fall solely due to the lateral flow of sea water.
2. While the tidal bulge on the “backside” of the Earth is due to the Earth being pulled toward the Sun and the Moon, and NOT due to any force directly acting on the seas, to an observer it appears as though the waters are flowing away from the Sun and the Moon. It is the same type of effect that makes the Sun appear to rise or set when it is actually the horizon that is moving. Just as it is normal to speak of the Sun rising, it is normal to speak of the seas flowing towards the backside bulge.
SR

February 14, 2014 8:36 pm

Thanks Willis, very good explanation, but I think this one is also very clear:
Ocean Tides:
The tides that we see in the oceans are due to the pull of the Moon and the Sun. The simplest explanation is that the water on the side of the Earth closest to the Moon is pulled, by the Moon’s gravitational force, more strongly than is the bulk of the Earth, whereas the water on the side furthest from the Moon is pulled less strongly than the Earth. The effect is to make bulges in the water on opposite sides of the Earth. The effect of the Sun’s pull is similar, and the tides that we see are the net effect of both pulls.
When the pull from the Sun adds to that of the Moon, the tides are large and we call them Spring tides, whereas when the pulls are at 90 degrees, the tides are small and we call them Neap tides. The heights of spring tides are governed by the distance of the Moon from the Earth, being largest at Perigee (when the Moon is closest to the Earth) and smallest at Apogee (when the Moon is at its furthest).
Because the Sun’s pull is aligned with that of the Moon at New Moon and Full Moon, these are the times when Spring Tides occur. The pull of the Sun is less than half that of the Moon, and so the frequency of the tides is determined by the apparent passage of the Moon around the Earth, which takes just over a day. We, therefore, in most places on the Earth have two tides a day, with the time of each becoming later from one day to the next by just under an hour a day. (The actual period is, of course, determined by the rotation of the Earth and the orbit of the Moon).
Produced by the Information Services Department of the Royal Greenwich Observatory (1996).
From http://www.oarval.org/tides.htm

February 14, 2014 8:42 pm

Problems
One the earth is not falling into the sun it is semi steady in orbit around it.
Therefore the gravity would be attracting each piece of the earth and sea at the right amount to keep it rotating around the sun. Hence nothing should be moving in respect to anything else.
Imagine 4 balls going around in orbit next to each other. There is no reason for them to move out of their orbits relative to each other.
Hence how do tides form?
Know I am wrong but look forward to explanation .
Second the moon is actually the greater attractor than the sun by a small percentage (WIki) and both together give king tides and both apart 45 degrees give neap (low) tides.
Not that this detracts from your explanation of tides, just rounding out the edges.
Third is the rotation of the earth important in tide formation, not mentioned here.

William Sears
February 14, 2014 8:57 pm

Angech,
Consider that the four balls are next to each other but at different distances from the sun. They will have different orbital periods and thus will separate from their initial alignment. Connect them with a string as in Willis’ analogy and a tensile force is required to keep the alignment. This is the tidal force. The rotating earth will have an effect both because of the centrifugal and Coriolis forces of a non-inertial frame and the rotation of the tidal bulge. Willis has discussed the latter elsewhere.

Jerry Gustafson
February 14, 2014 9:10 pm

For a great explanation of the tides as caused by the moon on Earth watch the series The Mechanical Universe And Beyond. It is a physics course created at Cal Tech for high school or introductory physics. Look at Program 25 Kepler to Einstein. http://www.learner.org/resources/series42.html#
Great explanation and computer graphics illustrating what they are talking about. This whole series is really worth watching. It is available in the US and Canada through Annenberg Learner, I think it’s on Youtube as well.

Richard D
February 14, 2014 9:11 pm

in reality would the bulge opposite the sun be from centripal force?
++++++++++++++++++++==
Nope.

eyesonu
February 14, 2014 9:15 pm

Before I get called on it I should have said centrifugal force instead of centripetal in my comments above.
Long day, time for bed. 🙂

Richard D
February 14, 2014 9:34 pm

Before I get called on it I should have said centrifugal force instead of centripetal in my comments above.
========
Nope.

Richard D
February 14, 2014 9:47 pm

Its circular motion not apparent forces and inertia

Richard D
February 14, 2014 10:33 pm

Willis Eschenbach says: Not so, Richard. In the example shown in Figure 2, there is no circular motion, and inertia is not an issue.
++++++++++++++++++
You’re right re:figure 2, thanks……. I was looking at centripetal/centrifugal which got me off tract.

Richard D
February 14, 2014 10:44 pm

As with gravity, every mass in the universe exerts a tidal force on every other mass, so you can personally take credit for raising a tide on the surface of the sun … yeah, it’s a small tide, but it’s real.
+++++++++++++++++++++++++
Thanks for that Willis. Takes me back to the physics classroom.

FightingScallion
February 14, 2014 10:48 pm

Problems
One the earth is not falling into the sun it is semi steady in orbit around it.
Therefore the gravity would be attracting each piece of the earth and sea at the right amount to keep it rotating around the sun. Hence nothing should be moving in respect to anything else.
Imagine 4 balls going around in orbit next to each other. There is no reason for them to move out of their orbits relative to each other.
Hence how do tides form?
Know I am wrong but look forward to explanation .
———————————————————————————————————–
So, this will be hard to explain without drawing a diagram, but….
For simplicity, consider a two body system, with a fixed, Sol-centered reference frame and Earth as a moving body, however not yet in an orbit about Sol. As the Earth passes, the mutual gravitational attraction of Sol and Earth modifies the trajectory of Earth, which we call gravitational force. The gravitational force is equal to the change in momentum of earth with respect to time (F = dp/dt = d(mv)/dt = v*(dm/dt) + m*(dv/dt)). We assume both Sol and Earth are fixed mass, reducing the first term to zero and recognize dv/dt to be acceleration. Thus, the gravitational force is F = ma. Because of how the gravitational force equation is defined, the induced acceleration on Earth is towards Sol (the sign convention used in the derivations skipped in this and above). From this definition, you can see that Earth “falls” towards Sol, in the same way that an object dropped from a height “falls” to the ground.
You must remember the fact that momentum, velocity, and acceleration are all vectors. Even in a constant elliptical orbit, Earth will always be accelerating towards Sol. Were you to remove Sol (and, therefore, the force of gravity), Earth’s inertia will carry Earth along the same vector it was traveling at the moment of the disappearance of Sol. The continuing presence of the gravity force continuously accelerates the velocity of Earth about Sol. Even in a truly circular orbit (i.e. radial velocity is constant; rectilinear velocity magnitude is constant), the rectilinear velocity must be constantly accelerated to maintain the orbit. This is done by “pulling” Earth towards Sol. This is why we say it is constantly falling.
Now, for all of that, we traditionally think in terms of point masses, because it’s easier to think about (remember the physics student joke: a horse is a sphere, if it makes the equations easier). In reality, we should treat every body as a continuum of mass elements.
Properly speaking, Willis’s diagram would have a near infinity of mass elements, each working a different amount on each other, with a bunch of other interactions drawn on there. For ease, though, let’s leave the “solid” Earth as a single body. This is an easy idealization to do, since we assume that the frictional forces between the elements are such that they overcome the gravitational force differential between the elements and that the Earth is incompressible. Thus, we assume that the solids would move as one (i.e. the definition of a solid being that it holds its shape and volume in any container). In fact, there is a certain amount of deformation that takes place, as with all solids, in something called “creep”, but we deem this negligible and cancelled out by the rotation (think of it as a rotisserie chicken; we cook it evenly on all sides, so we neglect the temporal differences).
The fluid portions, however, do not have the same amount of friction between them, and therefore “slosh” more easily. The friction between the elements is insufficient to overcome the differential. As a bulk, it will still “fall” about the same as the solid Earth towards Sol (consider the magnitudes of each value input). The different individual elements, however, have are just different enough for us to perceive it.
In truth a fully solid Earth has the same “tidal force” as our blue and green Earth. It just manifests differently, due to the internal forces. The atmosphere “sloshes” far more than the oceans, but it doesn’t really matter to us (the pressure difference being small due to the low densities involved). Were the oceans made of molasses, the same tidal force would again be generated, but the oceans would deform far less (more viscosity = more internal friction to resist deformation).
I hope that helps a bit.

Coldlynx
February 14, 2014 10:53 pm

Still is the tide calculation based on Laplace’s tidal equations which states:
from http://en.wikipedia.org/wiki/Tide
“Ocean depths are much smaller than their horizontal extent. Thus, the response to tidal forcing can be modelled using the Laplace tidal equations which incorporate the following features:
1 The vertical (or radial) velocity is negligible, and there is no vertical shear—this is a sheet flow. 2 The forcing is only horizontal (tangential).
3 The Coriolis effect appears as an inertial force (fictitious) acting laterally to the direction of flow and proportional to velocity.
4 The surface height’s rate of change is proportional to the negative divergence of velocity multiplied by the depth. As the horizontal velocity stretches or compresses the ocean as a sheet, the volume thins or thickens, respectively.”
I repeat: The forcing is only horizontal (tangential).
More here:
http://en.wikipedia.org/wiki/Theory_of_tides

February 14, 2014 11:40 pm

“One complicating factor is that the atmospheric tides that are caused by the sun occur in combinations with solar heating, which makes the atmosphere swell. Also, the variations in pressure caused by the atmospheric tides are small, on the order of a tenth of a millibar (or of a hectopascal), or a variation of only 0.01% …”
I fear there may be a misconception here, or perhaps 2. Take a column of air, or water, and measure the pressure at the bottom. Now heat the air, or the water, so it expands upwards. Unless any can leak out sideways the same mass of air or water is till there in the column so the pressure at the bottom must be constant. So whether the atmosphere swells or not, the pressure remains constant.
Now for atmospheric tides. As the earth rotates so the bulge of water tries to remain directly on the line connecting the centre of the earth to the attractor. But friction, inertia, and the continents prevent this happening. So water piles up on the edges, and we see the phenomenon of tides. The atmospheric bulge also tries to stay in the same place, but still there is friction, and hills and mountains get in the way. So the effect of the atmospheric tide is amplified in some places and is minimized in others. I believe it true to say that were the earth a perfect sphere, with a uniform layer of water, the actual rise and fall of the sea level due to tides would be as little as a foot (the surface of the solid earth is also reputed to distort about a foot as well). As all well know, in certain places the tidal effect is greatly magnified, eg, Bay of Fundy with tides up to 40 ft. Just so with the atmospheric tide, and a barograph will record the daily rise and fall of air pressure of up to 3 mb in particular locations. So while the theoretical variation in pressure doe to the atmospheric tide may be 0.1 mb, the observed variation can be 20 or 30 times as great.

Coldlynx
February 15, 2014 1:13 am

The tidal force is affecting every part of the planet.
The force that actually move fluid as water and air is according to Laplace the horizontal (tangential) force.
The vertical force can not move the fluid.
The result of the tangential force is the bulges. It is not the vertical force that create the bulges since water does not expand.
The reason the vertical force are at maximum at bulge maximum is causing the common misinterpretation that the vertical force is the reason for the bulges,
The horisontal forcing is the reason why the earth rotation and tilt and moon inclination have an impact on the direction of tidal induced movement of earths fluids.
Moon inclination is 5.145° to the ecliptic which is between 18.29° and 28.58° to Earth’s equator.
That is then also the direction of the tidal induced fluid movement direction. Fluid is our ocean water and atmosphere are depending on how high the moon are above (or below) the horizon in mornings and evenings when vertical tidal force are at maximum.
Same for solar tide.

Greg
February 15, 2014 1:33 am

Quite a nice explanation Willis. Since there is a huge amount of confusion about this subject, it’s good to have this over-view.
However, one thing is inconsistent in your description.
“There are two different tidal forces of interest in the diagram, which are GF1 – GF2, and GF2 – GF3. ”
That is correct and you will see that are roughly equal and of opposite sign.
” It’s not because the tidal force acts in different directions, because tidal force is always directed towards the sun.”
There you are confusing the tidal raising force (a difference) which you had correctly explained up to that point and the gravitational attraction which is indeed in the same direction, just slightly different in magnitude.
Your piece of blue string is of course the earth’s own gravitational field that is many orders larger than the tiny tidal forces. That’s why tides are only a few metres on a planet thousand of kilometres across.
“string” is not a particularly good idea, a strong steel cable may be better. The cable would hold M1 and M3 anchored in place, but would stretch a little, allowing the ‘tides’ to rise.
So as the proud author of the gem you opened the article with, I note your explanation is in agreement with what I wrote, if you don’t confuse tidal force, the difference in gravitational attraction with gravitational attraction itself.
“The tide raising force acts in both directions (bulge on each side in the simplistic model)”
( GF1 – GF2 ) is negative ( GF2 – GF3. ) is positive. They are roughly equal in size.
Otherwise this nice and clear and easily digestible to a more general readership. Nice work.
Here is my attempt at presenting it. It’s somewhat more technical in language
http://climategrog.wordpress.com/?attachment_id=776
Clive Best has also written a good description with some fairly detailed equations rather then just considering the straight line forces along the earth-moon axis.
http://clivebest.com/blog/

Greg
February 15, 2014 1:38 am

( GF1 – GF2 ) is negative ( GF2 – GF3. ) is positive.
No, that badly put. The point is M3 is accelerated faster towards the sun , faster than the solid earth , M1 less fast than the solid earth. So _when viewed from the point of view of the solid earth_ the forces act in opposite directions.

wayne
February 15, 2014 1:53 am

“I think the units are correct. Consider gravitational force. The units of G are N m2 / kg2.
As a result, G * m1 * m2 /D^2 (gravitational force) has units of
G ( N m2 / kg2) * m1 (kg) * m2 (kg) / D^2 (m2)
The kg and the meters cancel out leaving newtons for the unit of gravitational force. Since tidal forces are the difference of graviational forces, they also are measured in newtons.”
Yes Willis but drop your logic and just look at your tidal force equation that need to be explained:
Tidal Force (N/kg) = 2 * G ( N m2 / kg2) * sunmass (kg) * r (m) / D^3 (m3)
Cancel away. m2 is cancelled by the m/m3 and only one of the kg’s cancel.
Are the units not N/kg?
Two other commenters above seem to have come up with this answer.
You are familiar with these tidal force equations and not myself, yet, so maybe you can explain the discrepancy in either your logic or the tidal force equation itself.

Greg
February 15, 2014 1:55 am

This is all about what scientists refer to as “frame of reference” , it depends up on where you’re sitting as to how you view things.
Willis adopts a frame of reference outside the earth which is rotation about the sun but he does not discuss the centrifugal and Coriolis “fictitious forces” that are necessary if you do that. The other way is regard it as an instantaneous snapshot of a system in rotation where gravity provides the centripetal force that causes the roughly circular orbital motion.
That is the approach I adopted because you can avoid the complication of fictitious forces.
http://climategrog.wordpress.com/?attachment_id=776
Sorry folks, but bits of string and billiard ball mechanics don’t work correctly if the billiard table is on a rotating platform. All this “frame of reference” stuff is not a simple idea but you can’t just stop the earth going round the and forget about it.
I like the simplicity of Wilis’ diagram but he needs to add the centrifugal “fictitious force” acting outwards (to the left). When you do that you find the net force on M1 is to the left and the net force on M3 is to the sun . We then see the equivalence with my statement about opposite forces.

Greg
February 15, 2014 2:13 am

Tidal Force (N/kg) = 2 * G ( N m2 / kg2) * sunmass (kg) * r (m) / D^3 (m3)
“Are the units not N/kg?”
Well, no. It’s because Willis had written his equations for a unit mass of 1kg. and thus left out one of the m’s
The force should be of the form GmM/d^2 , he leaves out the m which is 1kg. It would be clearer to leave it in.

February 15, 2014 2:30 am

Music for physicists, outside of my benchtop chemistry and lately jewelry comfort zone. Thanks to Willis, WUWT is becoming Khan Acadamy, an online college, so here I can at least at times study my old man eyes out, as an empiricist, delving into new ways to prove those rat bastard theorists of Nature wrong again. Sometimes correctly.
I still think there’s a better groking of Maxwell’s equations out there somewhere in rebel controlled space.
Willis? Torqued space? Wither the aether?
Lubos won’t go there. He’s not a classicist, even in spirit, ever.
The three body problem alone, but add to it the severe inability of overheated supercomputers to even fold simple Tinkertoy proteins right, the stuff of life, and theory becomes just an adjunct to real lab work. Benchtop chemists always bust limits, lately the diffraction limit of traditional optical microscopes.
Oh, they’ll catch up, dragged kicking and screaming, into Reality.

February 15, 2014 3:05 am

Willis,
I usually enjoy reading your articles because I think you have a gift for explaining complex matters in a correct and simple way, but unfortunately I think this is an exception.
You say:

Finally, I’ve said that the situation in Figure 2 shows that planet free-falling into the sun. As it gets closer and closer, the tidal forces stretching the system get greater and greater

From this one could think that the distance between the Sun and the Earth is decreasing, but that is not the case, it is increasing.
The tidal force interacts with the rotation of the Sun and causes the Earth to be pushed slightly outwards every year. The energy for the push is taken from the Sun’s rotational energy.
We have the same effect between the Earth and the Moon. The earth is rotating slower and is pushing the moon away.
See:
Sun – Earth distance increasing one micrometer per year:
http://curious.astro.cornell.edu/question.php?number=317
Earth – Moon distance increasing 3 cm per year
/ Jan

February 15, 2014 3:26 am

Angech ponders the tides.
There is a minute change in the value of g which leads to a minute bulge .
The g value is the gravity of the moon not the sun.
It is actually twice as strong as the gravity of the sun on the earth tides as it is much closer to the earth.
The moons gravity is a lot weaker than the earth’s so any effect on bulges (tides) is a lot less than I originally contemplated and the effect of the sun is even less!
Why are the tides longer than 12 hours
The moon orbits the earth every 28 days so the tides are 24 hours divided by 28 means about 50 minutes later every day.
Are their minor tides due to the effect of the sun alone separate to the main 2 we realise.
Once actual movement occurs rotational effects and shallowness of sea beds and Coriolanus forces may make the waves much bigger

February 15, 2014 3:55 am

Willis
Hi
not strictly on topic but I saw an interesting post on Bishop Hill by a commenter called Paul K which I set out below. The connection to your piece is the gravitational effect on the ‘solid earth’.
Regards
“Personally, I think that it wouldn’t do to underestimate the importance of the England paper – even if it is founded on poor data.
As Nic correctly points out, from the observed data, the total global ocean heat flux shows a peak around 2001-2005 depending on which dataset one takes. TOA radiative measurements show a peak in net radiative incoming flux somewhere around 1997-2000, driven largely by SW changes in net albedo. Modern MSL data from satellite altimetry (or indeed from tide gauge data) shows a peak in its derivative function around 2001-2003, which should also be a proxy for net heat flux going into the ocean. (Using gravimetric data from GRACE, we can rule out the possibility that the peak in MSL derivative was caused by mass addition – it is a peak clearly driven by thermosteric expansion. There is a useful presentation here by Nerem: http://conference2011.wcrp-climate.org/orals/B3/Nerem_B3.pdf)  So there is a consistent story from three data sources which says that the net incoming flux hit a peak and has since been decreasing overall for about a decade. This is not compatible with increasing forcing from GHGs and flat or declining tropospheric temperature – a mini paradox, if you will.
The question it leaves is: what then controls the equatorial trade winds? The answer was actually known more than 40 years ago when science was still relatively unsullied, but it will not be accepted easily by mainstream climate science today, since the answer makes not one but two major breaches in fundamental assumptions of climate science.
The first part of the answer is that the climate oscillations are triggered by gravitationally forced changes in the angular velocity of the solid Earth. These changes transmit a (non-radiative) momentum flux into the hydrosphere and atmosphere via frictional torque and conservation of angular momentum. These changes explain the fluctuations in trade winds and, just as importantly, the latitudinal meanderings of the jet streams. Before anyone starts calling for the men in white coats, I would suggest that you have a look at this 1976 paper: http://gji.oxfordjournals.org/content/46/3/555.full.pdf and this: http://gji.oxfordjournals.org/content/64/1/67.full.pdf . For the excellent correlation apparent in the higher frequency data between Earth’s rotation velocity, atmospheric angular momentum and ENSO events, you might also try this paper: http://trs-new.jpl.nasa.gov/dspace/bitstream/2014/17186/1/99-0613.pdf .
So it seems that England has probably confirmed that the multidecadal oscillations are driven by atmospheric tides which are driven by a non-radiative orbital forcing. He just hasn’t realised yet that what he has done is to demonstrate that the GCMs are all missing a massively important piece of physics which was considered small enough to be neglected on energetic grounds.   The story doesn’t end there. The orbital forcing is a triggering and control mechanism, but it is “energetically deficient” to explain the full amplitude of the climate oscillations. On my sums the trough-to-peak transfer of energy via momentum flux and friction amounts to something less than 2*10^22 joules during the 60-year cycles. The amplification factor comes from the cloud response to the change in phase of the orbital forcing, which is why we note the dominant effect of SW changes in the radiative signature. This is a feedback mechanism of sorts, but it is not a “temperature dependent” feedback mechanism; it does not correlate simply with global surface temperature, but rather with the phase of orbital forcing. This post is already too long for me to try to explain how that works. I am hoping if I live long enough to try to get some of this stuff down in more detail in an article for Lucia, but I do keep getting distracted, not to mention beaten up by my wife for wasting time on that climate change rubbish instead of doing something useful.
Feb 14, 2014 at 2:35 PM | Paul_K”

February 15, 2014 3:55 am

Let’s test this theory. How does one go about measuring the “tidal force”?
The logical consequences of the tidal theory (such as elongation and Roche limit) have been falsified by observations. Metis and Pan orbit inside the Roche limit. Pan seems to be accreting. Far from being elongated, the Moon has a deficit of material on the near side.
Other than that, can any 2- or 3-body theory (plus rope) explain the real ocean tide? As understand, tide prediction is not physical. It uses harmonic curve-fitting.
http://web.vims.edu/physical/research/TCTutorial/tidepredict.htm
The idea that there are two bulges on the planet is also too ideal. It is true that there are places where one sees the passage of two maxima every day, but that does not mean there are two bulges on the planet.
I would love to see a similar animation with the position of the Moon plotted on it.

Mike M
February 15, 2014 6:11 am

For some reason my instinct says that this explanation cannot be completely correct because I ask that if we take away the moon altogether, (leaving only the sun and earth), would there be an ocean tide at all?
If the earth wasn’t rotating then the answer is easily no. If there was any bulge at all it would be very slight, at the same spot and never vary. But earth is rotating and rotating so that the outermost material at midnight is going the fastest above earth’s center’s orbital velocity resulting in it bulging outward away from the sun and the material at noon time going the slowest below earth’s center’s orbital velocity resulting in it bulging toward the sun thus resulting in a double bulge – without any moon at all.
Is this rotational factor significant? I’ll have run the numbers or wait for someone else to do it.

Mike M
February 15, 2014 6:18 am

I really meant, instead of “If the earth wasn’t rotating”, – – “If earth rotated once per year”

catweazle666
February 15, 2014 6:26 am

Fascinating, Willis.
Isn’t it amazing how such things are often (always?) more complex than they seem at first glance.
No wonder the “Post-Normal” science lot suffer from physics envy.

Mike M
February 15, 2014 7:13 am

A quick look gave me a result for a “no moon” variation of “midnight fling” acceleration over average center V^2/R of 3%. For now I’d say it looks significant.
Earth’s average orbit velocity V1 as 29885 m/sec. Crust WRT center V2 as 463 m/sec. So combined velocity in orbit around the sun at midnight V1+V2=V3 as 30349 m/sec V3^2 / V1^2 = 1.03

February 15, 2014 7:36 am

More on the Roche limit.
Images for the Roche limit
NASA on the Roche limit: Tidal Forces: Let ‘er Rip

Mike M
February 15, 2014 7:51 am

Boiling it down, I’ve produced a two bulge tide model with no moon at all. Earth to Sun acceleration due to gravity is .006 M/sec^2 Plus 3% at midnight, minus 3% at noon makes it a change plus or minus .0002 m/sec^2 against our earth gravity of 10M/sec^2
Certainly tiny WRT to earth’s gravity at sea level but how much does that contribute to the formation of two lobes?

Ulric Lyons
February 15, 2014 7:54 am

Willis, could you show the combined tidal forces on Earth from the Sun and Moon for both new and full moon positions.

Mike M
February 15, 2014 8:12 am

In other words I’m stating that, simply by virtue of earth’s rotation, everything on the surface of earth weighs .002% less at midnight and noon than it does at 6am and 6pm.

February 15, 2014 8:15 am

Mike M says:
February 15, 2014 at 8:12 am
In other words I’m stating that, simply by virtue of earth’s rotation, everything on the surface of earth weighs .002% less at midnight and noon than it does at 6am and 6pm.
When is it noon at the North Pole?

February 15, 2014 8:19 am

Willis you say : “… because tidal force is always directed towards the sun.”
That sentence end could use a little correction/clarification I think.
Mike

spen
February 15, 2014 8:26 am

A slight digression. Canute is mentioned in the title. May I put a word in for this king. When he said that he was commanding the tides he was not doing this out of hubris but to show his people that he had his limitations. Now there is degree of modesty that some warmists could benefit from.

Mike M
February 15, 2014 8:35 am

lsvalgaard says: When is it noon at the North Pole?
Maximum kinetically induced differential is obviously at the equator so times the cosine of the latitude.

February 15, 2014 8:37 am

As the planet free-falls into the sun, the near side of the ocean is pulled the most. So it moves the most, pulling away from the planet. And the planet, in turn is pulled more than the far side, so it pulls away from the ocean. These relative motions create the “bulges”

By all due respect, this is a rather misleading statement
The planet does not “fall into the Sun”. The gravitational pull is countered by the centripetal acceleration of the Earth in its orbit around the Sun.
The centripetal acceleration is the phenomenon one can feel in a carousel. One can feel an outward force away from the center of the spinning point.
The centripetal acceleration is given by the formula:

a = v*v/r
Where:
a is the centripetal acceleration
v is the Earth’s orbital velocity
r is the orbital radius, i.e, the distance from the Sun

The planet stay in the orbit because the centripetal acceleration, a, is equal to the gravitational pull from the Sun
But as we see from the formula, a increases with the distance from the Sun. On the other hand, the gravitational pull from the Sun decreases with the distance from the Sun.
The centripetal acceleration is therefore greater than the gravitational pull in point M3 and weaker than the gravitational pull in point A1. In point A2 they are equal.
This means that the greater “carousel force” felt by the centripetal acceleration in M3 combined with the less gravitational pull, makes the oceans bulge out and away from the Sun.
In M1 we have the opposite situation. The smaller carousel effect and larger gravitational pull from the Sun makes the oceans bulge out from the Earth, but toward to the Sun.
/Jan

February 15, 2014 8:39 am

Mike M says:
February 15, 2014 at 8:35 am
Maximum kinetically induced differential is obviously at the equator so times the cosine of the latitude.
everything on the surface of earth …
And you statement is not correct anyway, as the rotation of the Earth causes things to weigh less at all times at the equator.

Mike M
February 15, 2014 8:44 am

lsvalgaard says: “everything on the surface of earth …”
Okay, you got me, everything BUT an infinitesimal sized point at each pole – happy?
“And you statement is not correct anyway, as the rotation of the Earth causes things to weigh less at all times at the equator.”
That’s a CONSTANT, (also per cosine of latitude)… I’m looking at variations in acceleration.

February 15, 2014 8:45 am

Jan Kjetil Andersen says:
February 15, 2014 at 8:37 am
By all due respect, this is a rather misleading statement
The planet does not “fall into the Sun”. The gravitational pull is countered by the centripetal acceleration of the Earth in its orbit around the Sun.

Amazing how many experts we have around here. Willis is completely correct [apart from a few typos in the beginning]. This tidal effect has nothing to do with rotation, centripetal forces, orbital movement or any of that.

February 15, 2014 8:46 am

Mike M says:
February 15, 2014 at 8:44 am
I’m looking at variations in acceleration.
Variations caused by what?

Mike M
February 15, 2014 8:49 am

Willis, the idea that a person free falling into a black hole feet or head first will be at some point ripped apart lengthwise is a common one and consistent with your assertion.

Gerg
February 15, 2014 8:52 am

“DO THE FREAKING MATH, Greg. The two have the same sign. ”
“Not true at all. I have a heliocentric (sun centered) frame of reference, which does NOT rotate about the sun. Neither does the planet in my example. The clue is where I say it is free-falling into the sun …”
OK, I didn’t realise you intended not to rotate anything. So you want to call be a fool for saying something about how planets moons and tides work by comparing to a situation where nothing rotates.
Should still be an interesting exercise.
It’s not clear what the mechanical properiteis of your piece of “string” are so I’ll stick with a steel cable.
As I think we are agreed GF1>GF2>GF3 so the outer mass would accelerate less and get left behind unless it was tethered. Inversely M1 would fall faster. So both cables are under (rougly equal) tension.
You show that we can ignore the asymmetry:
True average = 0.50587
Approximation = 0.50587
So the net force accelerating the three masses is 3GF2 acting 3kg, They experience a common acceleration (since they are tethered) equal to that of the centre mass alone.
But M3 is experiencing less gravitational attraction and has to be pulled by the tether to the tune of GF2-GF3. The tension in the tether is thus that value. Similary the other side which should accelerate more but is held back by the tether.
So at M2 the tensions in the tethers are equal and in opposite directions.
The cables will extent a little due to the tension and both masses will end up a little further away from M2 and continute accelerating as an ensemble. This is the equivalent of the two tideal bugles. The tidal force and the ‘tide’ (stretch in cable) will remain constant until they get significantly closer to the sun and the gravity gradient gets steeper.
All you have done is substituted free-fall acceleration to the sun , without rotation, for a centripetal acceleration towards the sun with rotation.
The difference terms remain the same and act in opposite directions. If they didn’t one of the cables would be lack.
Even a “fool” like me can see that.

Gerg
February 15, 2014 8:56 am

slack.

Gerg
February 15, 2014 8:56 am

The difference terms remain the same and act in opposite directions. If they didn’t one of the cables would be slack.

Mike M
February 15, 2014 8:58 am

lsvalgaard says: “Variations caused by what?”
Read BACK further to my calculations of difference in heliocentric orbital velocity of a point (on the equator) due to earth’s rotation; fastest at midnight, slowest at noon. The kinematic difference compared to the acceleration due to Sun’s gravity, (at earth’s distance), compared to 6am or 6pm, is 3%.

February 15, 2014 9:01 am

Mike M says:
February 15, 2014 at 8:58 am
Read BACK further to my calculations of difference in heliocentric orbital velocity of a point (on the equator) due to earth’s rotation; fastest at midnight, slowest at noon. The kinematic difference compared to the acceleration due to Sun’s gravity, (at earth’s distance), compared to 6am or 6pm, is 3%.
Not so. The Earth is is free fall and knows nothing about orbital velocity.

Gerg
February 15, 2014 9:05 am

So my comments are on moderation now. ?? Is “slack” a moderation tripwise or what?
REPLY: dunno, some triggers are wordpress controlled. May be some spam circulating that uses “slack”…maybe Viagra like – A

Mike M
February 15, 2014 9:05 am

lsvalgaard says: Not so. The Earth is is free fall and knows nothing about orbital velocity.
I’m done with ya…

February 15, 2014 9:09 am

Jan (and others), the planet shown free-falling into the sun IS NOT THE EARTH.

Ok, I see
I thought it was something totally wrong here.
I then see that my other comment on misleading description is also misplaced.
/ Jan

February 15, 2014 9:12 am

Mike M says:
February 15, 2014 at 9:05 am
I’m done with ya…
Good that you stop digging your hole deeper and deeper. According to your ‘theory’ a pendulum clock should vary its ticking rate during the day and it does not.

Greg
February 15, 2014 9:29 am

Just noticed I’d inverted my name to Gerg . LOL, was that what tripped moderation

Greg
February 15, 2014 9:30 am

“DO THE FREAKING MATH, Greg. The two have the same sign. ”
“Not true at all. I have a heliocentric (sun centered) frame of reference, which does NOT rotate about the sun. Neither does the planet in my example. The clue is where I say it is free-falling into the sun …”
OK, I didn’t realise you intended not to rotate anything. So you want to call be a fool for saying something about how planets moons and tides work by comparing to a situation where nothing rotates.
Should still be an interesting exercise.
It’s not clear what the mechanical properiteis of your piece of “string” are so I’ll stick with a steel cable.
As I think we are agreed GF1>GF2>GF3 so the outer mass would accelerate less and get left behind unless it was tethered. Inversely M1 would fall faster. So both cables are under (rougly equal) tension.
You show that we can ignore the asymmetry:
True average = 0.50587
Approximation = 0.50587
So the net force accelerating the three masses is 3GF2 acting 3kg, They experience a common acceleration (since they are tethered) equal to that of the centre mass alone.
But M3 is experiencing less gravitational attraction and has to be pulled by the tether to the tune of GF2-GF3. The tension in the tether is thus that value. Similary the other side which should accelerate more but is held back by the tether.
So at M2 the tensions in the tethers are equal and in opposite directions.
The cables will extent a little due to the tension and both masses will end up a little further away from M2 and continute accelerating as an ensemble. This is the equivalent of the two tideal bugles. The tidal force and the ‘tide’ (stretch in cable) will remain constant until they get significantly closer to the sun and the gravity gradient gets steeper.
All you have done is substituted free-fall acceleration to the sun , without rotation, for a centripetal acceleration towards the sun with rotation.
The difference terms remain the same and act in opposite directions. If they didn’t one of the cables would be lack.
Even a “fool” like me can see that.

Greg
February 15, 2014 9:38 am

The difference terms remain the same and act in opposite directions. If they didn’t one of the cables would be _slack_.
[Mods , you can delete the three posts by “Gerg” 😉 Must have been taken as posting a pseudo form the same email. ]

Kevin Kilty
February 15, 2014 10:05 am

Short Post. You can skip this if you understand the tidal force. Some folks seem confused about the nature of tidal forces. Today I saw this gem: “The tide raising force acts in both directions (bulge on each side in the simplistic model)” … the author of that statement may understand the tides, but that isn’t how tidal forces work, they don’t act in different directions.

Willis, I like your rope model, it is very elegant. However, with all due respect, we often formulate dynamical problems from a convenient frame of reference. Pretending that the Earth is an inertial frame of reference is one such convenience. Within this frame of reference the gradient of tide-raising potential appears to act in opposite directions from the center of mass of the Earth-Moon system. The reason for this is the manner in which centrifugal acceleration gets included on the left side of Newton’s equation (The F of F=ma) in the translation from one frame of reference to the other. I don’t see that the view of the matter from this alternative frame of reference is wrong–it does provide correct predictions.

Kevin Kilty
February 15, 2014 10:06 am

I mean centripetal not centrifugal in my last post.

February 15, 2014 10:08 am

Kevin Kilty says:
February 15, 2014 at 10:06 am
I mean centripetal not centrifugal in my last post
It doesn’t matter as the tidal effect has nothing to do with rotation or orbital movements.

February 15, 2014 10:43 am

Willis,
What you write is clear and simple. However the earth is a non-inertial accelerating frame of reference because it is in orbit around the moon-earth barycenter. The mystery of gravity is why the ‘m’ in gravity is the same as the inertial ‘m’ in F=ma. Einstein used this equivalence principal to derive general relativity. The bulge on the opposite side to the moon is not because the centre of the earth feels a stronger gravity than the far side ocean. That only applies to fee falling bodies. It is instead due to the centrifugal acceleration caused by its orbit around the joint barycenter.
The tidal force on the ocean is 10 million times smaller than the earth’s gravity. What moves enormous quantities of water is the tidal component parallel to the surface which is approximately $= \frac{2GmR \sin \theta}{r^3}$
PS—As with gravity, every mass in the universe exerts a tidal force on every other mass.
No this is not true – An electron does not exert a tidal force on a quark.

Mike M
February 15, 2014 11:04 am

lsvalgaard says: “Good that you stop digging your hole deeper and deeper. According to your ‘theory’ a pendulum clock should vary its ticking rate during the day and it does not. ”
Yes, by my theory it does vary! It’s just too small to notice and averages out, cycling every 12 hours. (Like I said, if my numbers are right, the factor is only .002% )
Given that you will not deny that a particle of mass M at the equator is moving faster than earth’s orbital velocity around the sun at midnight and slower than earth’s orbital velocity around the sun at noon – it’s inertial V^2/R reaction (“centrifugal force”) will be greater than its gravitational attraction to the sun (centripetal force) at midnight and less than its gravitational attraction to the sun at noon.

February 15, 2014 11:09 am

Mike M says:
February 15, 2014 at 11:04 am
Yes, by my theory it does vary! It’s just too small to notice and averages out, cycling every 12 hours. (Like I said, if my numbers are right, the factor is only .002% )
Then your theory is wrong as the orbit around the Sun and around the center of the Milky Way and around the center of the Local Group of Galaxies, and around the… have no measurable effect at all of the weight of ‘everything’ on Earth.

Jeff D.
February 15, 2014 11:19 am

Gravity by far my most favorite subject to read about and ponder. Semi quantified by Newton and refined by my personal hero [Einstein]. The one thing I can take away from the above comments as well as the bulk of my amateur research is that nobody really has a clue. Experimental data results says the math is very close that we currently use to calculate it but the actual mechanism that produces it still eludes us.
For me the simple fact that something so seemingly mundane as a coin falling to the ground is beyond our understanding is captivating. When i contrast this to the state of ” Climate Science ” which does not even get freaking close to meeting model projections makes me laugh and cry at the same time….

Mike M
February 15, 2014 11:25 am

lsvalgaard says: “Then your theory is wrong as the orbit around the Sun and around the center of the Milky Way and around the center of the Local Group of Galaxies, and around the… have no measurable effect at all of the weight of ‘everything’ on Earth.”
True or false – They ALL have an affect?
You said I was digging a hole and now look who is crawling into one with “measurable”?

Expat
February 15, 2014 11:28 am

The greatest tidal effect if from the moon. The near bulge on earth is from gravitational force. The far side bulge is created by centripetal force. The moon does not rotate around the earth. They both orbit around a common center of mass which is not the center of the earth.

February 15, 2014 11:32 am

Mike M says:
February 15, 2014 at 11:25 am
lsvalgaard says: “Then your theory is wrong as the orbit around the Sun and around the center of the Milky Way and around the center of the Local Group of Galaxies, and around the… have no measurable effect at all of the weight of ‘everything’ on Earth.”
True or false – They ALL have an affect?

They have no effect due to rotations or orbital movements. The do have [mostly unmeasurable] tidal effects due to gravity.

Bart
February 15, 2014 11:34 am

FTA: “the author of that statement may understand the tides, but that isn’t how tidal forces work, they don’t act in different directions.”
They do. It’s conceptually a little tricky, but you have made a simple math mistake.
You calculate GF1 – GF2 on the near side. What is this? It is the specific force acting in the direction of the Sun relative to the Earth, i.e., the arrow is starting from the Earth and pointing toward the Sun.
You then calculate GF2 – GF3. This is the specific force acting opposite of the direction of the Sun. The arrow is pointing from the Sun to the Earth.
To get them both in the same direction, you would need to calculate both with respect to the same reference point, GF2.
You see, independently, M1 is experiencing GF1 in the direction of the Sun. M3 is experiencing GF3 with respect to the Sun. With respect to one another, they are experiencing |GF1 – GF3|. This is non-zero, which means the objects are not being pushed in the same direction.
It may help to note that GF2 is actually zero, when you include the centripetal acceleration. Let’s assume we have a circular orbit with angular rate omega = sqrt(mu/D^3). The total specific force acting at the center of the Earth is
GF2 = -mu/D^2 + D*omega^2 = 0
i.e., we are in free fall at the center of the Earth.
The specific force acting on the near side is
GF1 = -mu/(D-r)^2 + (D-r)*omega^2 := (-2*mu/D^3)*r – r*omega^2 = (-3*mu/D^3)*r
The specific force on the far side acting in the direction of the Sun is
GF3 = -mu/(D+r)^2 + (D+r)*omega^2 := (+3*mu/D^3)*r
These specific forces are relative to the Sun. The near one is radially inward toward the Sun, and the far one is radially outward from the Sun. Note that |GF1 – GF3| = (6*mu/D^3)*r, i.e., they are pushing away from each other.

February 15, 2014 11:35 am

Expat says:
February 15, 2014 at 11:28 am
The greatest tidal effect if from the moon. The near bulge on earth is from gravitational force. The far side bulge is created by centripetal force.
No, very wrong. Both are gravitational. Willis is correct. Rotation and orbital movements have nothing whatsoever to do with the tides.

Mike M
February 15, 2014 11:36 am

lsvalgaard says “They have no effect due to rotations or orbital movements.”
FALSE!
Let’s try this again, step by step…
DO YOU DENY that a particle of mass M at the equator is moving faster than earth’s solar orbital velocity around the sun at midnight and slower than earth’s solar orbital velocity around the sun at noon?

Greg
February 15, 2014 11:41 am

“In other words, the “bulges” on the two sides of the Earth are simply a result of the tidal forces stretching the entire system.”
That is a fair desciption. It is the difference in two differenct forces, applied at two different points in the body that exert a tensile stress on the body. You cannot stretch a body with one force. So in the case of two co-linear forces, there will be a net force (the average of the two) actiing to accelerate the body and two opposing components (each half the difference) acting in opposite directions, at the same two points, to stretch the body.
In the case above (GF3+GF1)/2 = GF2 will accelerate each of the three tethered weights at the common rate towards the sun.
Two opposing forces (GF3-GF1)/2 will act at each end mass acting to stretch the tethers applying a tensional stress to the ensemble.
In the case of an orbitting planet, most of the gravitational force is excerted to supply the centrpetal force that produces the centripetal acceleration which ensures the circular orbit. In this experiment it goes to linear acceleration towards the sun.
In both cases there is a residual of two equal and opposing forces that stretch the masses/planat creating the distortions we call “tides”. These are the tide raising forces.
The statement I made yesterday remains valid.
“The tide raising force acts in both directions (bulge on each side in the simplistic model)”
All Willis has done here is to add a large additional force to everything in one direction and then conclude that they all act in the same direction (but with different magnitudes. ). Again, its a question of frame of reference.
It’s a bit like measuring tides as haveing extremes of +/- 2m . Then someone else comes a long , moves the measuring stick and says “no you’re wrong! Here tides are between 8m and 12m , they’re always well above zero”. They then start calling everyone who does not agree with them a “fool”.

February 15, 2014 11:42 am

Bart says:
February 15, 2014 at 11:34 am
It may help to note that GF2 is actually zero, when you include the centripetal acceleration. Let’s assume we have a circular orbit with angular rate …
Please, Bart. The tidal effects have nothing to do with centripetal accelerations, rotation, orbital movements, etc. Three bodies lined up in the direction of the Sun and falling towards the Sun [[think] of pieces of comet] all are falling under the gravitational force of the Sun directed towards to Sun for all three bodies.

February 15, 2014 11:46 am

Mike M says:
February 15, 2014 at 11:36 am
DO YOU DENY that a particle of mass M at the equator is moving faster than earth’s solar orbital velocity around the sun at midnight and slower than earth’s solar orbital velocity around the sun at noon?
Whatever you may think of this, such movements have nothing to do with the tidal effects.

Bart
February 15, 2014 11:52 am

lsvalgaard says:
February 15, 2014 at 11:42 am
“The tidal effects have nothing to do with centripetal accelerations…”
Not so. Perhaps you are confusing rotations of the Earth, versus revolution about the Sun. The former has little effect on tides, because it is equally distributed about the Earth. But, the near side of the Earth is indubitably instantaneously accelerating slower with respect to inertial space than the far side.

February 15, 2014 11:57 am

Bart says:
February 15, 2014 at 11:52 am
But, the near side of the Earth is indubitably instantaneously accelerating slower with respect to inertial space than the far side.
That has nothing to do with the tidal effects. Perhaps the easiest way to see this is that the equations of the tidal heights and for the locations of the Roche Limit make no reference to rotation or orbital movements, only to the dimensions, masses, and distances of the bodies involved.

Mike M
February 15, 2014 12:00 pm

lsvalgaard says: Whatever you may think of this, such movements have nothing to do with the tidal effects.
Oh, so now you clam up refusing to recognize even a simple statement of relative motion?
I never said it was THE explanation of ocean tides but I stand by my assertion the force differential is REAL and acts to reduce the net gravitational force at the equator at both midnight and noon which would cause a perfect 12 hour tide period if there was no moon present.
The moon injects another component of variation to screw up the timing but then there is the consistence of highest tides occurring at periods of new and full moon which only amplifies my contention.

Bart
February 15, 2014 12:01 pm

lsvalgaard says:
February 15, 2014 at 11:57 am
“Perhaps the easiest way to see this is that the equations of the tidal heights and for the locations of the Roche Limit…”
You are wrong. Please do not bother me with any more silly stuff like this.

Greg
February 15, 2014 12:02 pm

CliveBert
” The bulge on the opposite side to the moon is not because the centre of the earth feels a stronger gravity than the far side ocean. That only applies to fee falling bodies. It is instead due to the centrifugal acceleration caused by its orbit around the joint barycenter.
The tidal force on the ocean is 10 million times smaller than the earth’s gravity. What moves enormous quantities of water is the tidal component parallel to the surface ”
Yes, I think that is what primarily moves the water volume. However, I think it is gravity that sets a limit on how high it can get. As you say its by far the stronger force.
This is the approach the I took the other day in equating the centripetial force to the local value of gravitational attraction. That provides an instananeous equilibrium level to which it would tend.
http://climategrog.wordpress.com/?attachment_id=776
As we all know , the dynamics is a world away from two (or four) bulges.

February 15, 2014 12:15 pm

Mike M says:
February 15, 2014 at 12:00 pm
I never said it was THE explanation of ocean tides but I stand by my assertion the force differential is REAL and acts to reduce the net gravitational force at the equator at both midnight and noon which would cause a perfect 12 hour tide period if there was no moon present.
A force differential is due to gravity, but is not due to rotation or orbital movement.
Bart says:
February 15, 2014 at 12:01 pm
“Perhaps the easiest way to see this is that the equations of the tidal heights and for the locations of the Roche Limit…”
You are wrong.

The roche limit depends only on the cube root of the ratio of the densities of the bodies [and on the radius of the central body]. As your link says: “we can ignore the centrifugal potential VC and consider only the tidal potential VT”. The centrifugal force being ignored is that of the rotation of a nonspherical body.
Please do not bother me with any more silly stuff like this.
There you go again. A little thought on your part would do you good.
A planet with the dimension and mass as the Earth falling into the Sun from infinity with no centripetal acceleration at all would at the time it distance reached 1 AU experience precisely the same tidal force as the Earth.

Mike M
February 15, 2014 12:15 pm

Let’s not forget that earth’s crust is extremely thin WRT earth’s diameter and is going to flex to “bulging forces” mostly at the equator along with the ocean. Could that explain why ocean tide varies the least at the equator and greatest at the higher latitudes by virtue that those regions of crust are not being bulged as much making the determination of which way is “downhill” at various times of the day rather complex.. ??

February 15, 2014 12:16 pm

Willis, you *do* have a problem with your signs. You yourself state:
“As shown in Figure 1, to calculate the tidal force you can just subtract the gravitational force on the unit mass at the center of the earth from the force on the other unit mass, for example GF1 – GF2.”
At the other side, your definition would give “GF3 – GF2” (not “G2 – G3”), which is indeed negative. Using the center of mass of the the earth as the frame of reference, the tidal force is away from the center., This is positive toward the moon (as tyou defined your directions), so the far force is indeed negative.

Bart
February 15, 2014 12:18 pm

lsvalgaard says:
February 15, 2014 at 12:15 pm
Look farther down. Do you see the equation immediately prior to and following the statement “let us consider how the centrifugal force from rotation will affect the result”? They then add in the centrifugal effects and, as you see, it is the same boost from a factor of 2 to a factor of 3 that I gave in my equations.
Quit making a fool of yourself, Leif. You are wrong. Let it go.

Mike M
February 15, 2014 12:21 pm

lsvalgaard says “A force differential is due to gravity, but is not due to rotation or orbital movement.”
Oh really, only gravity huh? So if earth’s orbital velocity around the sun slows down it will not fall closer towards the sun and if it speeds up it won’t go further way from the sun? Do you understand orbital mechanics at all?

Bart
February 15, 2014 12:21 pm

Tim Folkerts says:
February 15, 2014 at 12:16 pm
Yes, that is me point, before Leif sidetracked me with an irrelevant tangent.

yirgach
February 15, 2014 12:22 pm

Although we are not talking about earth-moon tidal dynamics, that seems to be shading some of these comments. A review of the detail Noaa Technical Tide description may help…
http://www.co-ops.nos.noaa.gov/restles3.html

February 15, 2014 12:26 pm

Bart says:
February 15, 2014 at 12:18 pm
They then add in the centrifugal effects and, as you see, it is the same boost from a factor of 2 to a factor of 3 that I gave in my equations. Quit making a fool of yourself, Leif. You are wrong. Let it go.
As I said, a little more thought on your part would do you good:
From the link: This is the Roche limit,
d \approx 2{.}423 \cdot R \cdot \sqrt[3]{ \frac {\rho_M} {\rho_m} } \,.
Surprisingly, including the centrifugal potential makes remarkably little difference, though the object becomes a Roche ellipsoid, a general triaxial ellipsoid with all axes having different lengths. The potential becomes a much more complicated function of the axis lengths, requiring elliptic functions. However, the solution proceeds much as in the tidal-only case, and we find
d \approx 2{.}455 \cdot R \cdot \sqrt[3]{ \frac {\rho_M} {\rho_m} } \,.
the change of the numeral factor is from 2.423 to 2.455 thus negligible, and nil for tides on an almost spherical body like the Earth.

February 15, 2014 12:30 pm

Mike M says:
February 15, 2014 at 12:21 pm
Oh really, only gravity huh? So if earth’s orbital velocity around the sun slows down it will not fall closer towards the sun and if it speeds up it won’t go further way from the sun? Do you understand orbital mechanics at all?
The closer the Earth would be to the Sun, the faster it would go, but the orbital speed has nothing to do with the tides.

February 15, 2014 12:33 pm

Einstein and others sought a mechanism to explain how a rotating body might experience a centrifugal force as a result of some sort of interaction with all the distant matter in the universe. A clue came from the theory of gravitation: after all, centrifugal force is sometimes even called artificial gravity.

The equivalence principal is the fundamental assumption of general relativity.
So gravity and inertia are two different ways of looking at the same thing. Anything falling into a black hole will get stretched so that a unit of length increases but time then will also go slower ! It takes an infinite amount of time to pass through the event horizon.
I prefer to use rotational dynamics to derive tides on earth because they better explain the origin of the opposite tide and give quantitative answers – see my calculations here

Mike M
February 15, 2014 12:40 pm

lsvalgaard says: “The closer the Earth would be to the Sun, the faster it would go….”
(In Lief’s universe the slowest rockets reach the highest orbits!)

Bart
February 15, 2014 12:45 pm

lsvalgaard says:
February 15, 2014 at 12:26 pm
Those are the formulas for fluid bodies. And, even though the Roche limit is relatively insensitive there, the shape the fluid assumes is quite sensitive. So, you’re no longer dealing with a spherical body. But, my equations for a spherical body are still correct. And, there is a significant effect.
Look higher at the rigid body calculation. You will see the factor of 2 change to the factor of 3 in the formula as I explained above.
It has an effect, just as I outlined. You are wrong.

Steve Fitzpatrick
February 15, 2014 12:48 pm

Leif,
I’m pretty sure Clive Best (February 15, 2014 at 10:43 am) has it correct for tides on Earth. Yes, tidal forces on objects in free fall is exactly as Willis describes, but I think that explanation is not a very good one for Earth’s ocean tides. Please see Clive’s web page (http://clivebest.com/blog/?p=5572) for his complete derivation. The magnitude of the ocean tides on Earth is too large to be accounted for by Willis’s classic free-falling-object explanation. Clive describes his background as follows:
“I have a Bsc in Physics and a PhD in High Energy Physics and have worked as a research fellow at CERN for 3 years, Rutherford Lab for 2 years and the JET Nuclear Fusion experiment for 5 years. Thereafter I worked at the Joint Research Centre in Italy until April 2008 being seconded to the African Union in Addis Adaba Nov 2007 until March 2008.”
He does know a bit of physics; I would not discount what he says without looking at his derivation, which looks correct to me.

February 15, 2014 12:55 pm

Mike M says:
February 15, 2014 at 12:40 pm
lsvalgaard says: “The closer the Earth would be to the Sun, the faster it would go….”
(In Lief’s universe the slowest rockets reach the highest orbits!)

Here you can learn about the orbital speeds of the planets and see that the closer they are to sun the faster they go: http://www.sjsu.edu/faculty/watkins/orbital.htm
Bart says:
February 15, 2014 at 12:45 pm
Look higher at the rigid body calculation. You will see the factor of 2 change to the factor of 3 in the formula as I explained above.
The factor increases because of the assumed rotation of the body [if you rotate the body fast enough it will break up just because of rotation], not because of orbital centripetal acceleration.
again: the orbital speed has no bearing on the tidal effects. From your link:
“As the orbit has been assumed circular, the total gravitational force and orbital centrifugal force acting on the main body cancel. That leaves two forces: the tidal force and the rotational centrifugal force.”

February 15, 2014 1:01 pm

Steve Fitzpatrick says:
February 15, 2014 at 12:48 pm
Please see Clive’s web page (http://clivebest.com/blog/?p=5572) for his complete derivation.
If you go there, you will see that the orbital centripetal force omega actually falls out of the equation and it is never used for anything.
The magnitude of the ocean tides on Earth is too large to be accounted for by Willis’s classic free-falling-object explanation.
I don’t think so. As far as I know they are in the right ballpark.

February 15, 2014 1:07 pm

Steve Fitzpatrick says:
February 15, 2014 at 12:48 pm
Please see Clive’s web page (http://clivebest.com/blog/?p=5572) for his complete derivation.
Tides are very important in astrophysics and is something we know a lot about. Here is my derivation of the tidal equation [the actual height of the tidal bulge], slide 21 of
http://www.leif.org/research/AGU%20Fall%202011%20SH34B-08.pdf
On slide 4 you can see the formula used to calculate the tidal effects of the planets on the Sun.

Bart
February 15, 2014 1:13 pm

lsvalgaard says:
February 15, 2014 at 12:55 pm
“That leaves two forces: the tidal force and the rotational centrifugal force.”
And, the “rotational centrifugal force” to which they refer is embodied in the equation below which, as you see, depends on the orbit rate. Why? Because their body is tidally locked.
This treatment is, in fact, precisely equivalent to what I have written. It is just another way of writing the same thing for a specific scenario. You do not understand it, so you jump to conclusions. But, the bottom line is that the orbital rate does have an effect.
Really, Leif, this is elementary. You aren’t even close to being right here. Give it up.

Bart
February 15, 2014 1:17 pm

lsvalgaard says:
February 15, 2014 at 1:07 pm
“On slide 4 you can see the formula used to calculate the tidal effects of the planets on the Sun.”
A very different situation than the one we are exploring. The effect of centripetal acceleration is, indeed, negligible here. Why? Because the Sun is massive, and the barycenter is close to the center of the Sun. The Sun does not orbit the planets, but the planets can very nearly be said to orbit the Sun.
Similarly, centripetal effects are very small on lunar tides – the Earth does not orbit the Moon. But, we are talking here of solar induced Earth tides.

February 15, 2014 1:27 pm

Lief says

If you go there, you will see that the orbital centripetal force omega actually falls out of the equation and it is never used for anything.

It drops out because inertial mass = gravitational mass. This mystery was eventually explained by Einstein and it was gravity that turned out to be in some sense the fictional force!
Centrifuges separate U256 from U238. Nobody told the machines that what they were doing was fictitious or impossible. The engineers observing them working from the lab – an inertial external frame – didn’t stand around scratching their heads wondering how the hell it all works !

stevefitzpatrick
February 15, 2014 1:27 pm

Lief,
So do you disagree also with the Wikipedia description of lunar driven tides on Earth?
“The plane of the Moon’s orbit around Earth lies close to the plane of Earth’s orbit around the Sun (the ecliptic), rather than in the plane perpendicular to the axis of rotation of Earth (the equator) as is usually the case with planetary satellites. The mass of the Moon is sufficiently large, and it is sufficiently close, to raise tides in the matter of Earth. In particular, the water of the oceans bulges out towards the Moon. There is a roughly opposing bulge on the other side of Earth that is caused by the centrifugal force of Earth rotating about the Earth–Moon barycenter. The average tidal bulge is sychronized with the Moon’s orbit, and Earth rotates under this tidal bulge in just over a day. However, the rotation drags the position of the tidal bulge ahead of the position directly under the Moon. ”
Which leads to gradual (geologic periods) increase in the Moon’s distance from Earth and gradual slowing of Earth’s rotational rate. Like I said, Willis’s explanation is perfectly accurate for a free falling body, It just doesn’t seem to me a very useful description for the orbiting Earth-Moon (and Sun) system that generates Earth’s ocean tides.

stevefitzpatrick
February 15, 2014 1:32 pm

clivebest says:
February 15, 2014 at 1:27 pm
I think you mean U235 from U238. (The only possibilities for natural Uranium are 234, 235, and 238).

February 15, 2014 1:43 pm

Steve,
Yes indeed it should have been U235
Although God knows we could do with another energy source !

Greg
February 15, 2014 1:40 pm

Oh dear !
http://www.leif.org/research/AGU%20Fall%202011%20SH34B-08.pdf
Triple Peak Periods of 9.91 yr [C], 10.78 yr, and 11.87 yr [J]
because in general we have:
cos α + cos β = 2 cos [(α + β)/2] cos [(α − β)/2]
So, the sum of two
cosines can be written
as the product of two
cosines [‘amplitude
modulation’].
====
Hopefully Willis will be along soon to explain how Leif is making a fool of himself , how dare he make up all this “bullshit” and he’d better damn well explain it, or else……
I used to firmly believe all that crap too until Willis put me straight. Good job he’s around.

Bart
February 15, 2014 1:40 pm

stevefitzpatrick says:
February 15, 2014 at 1:27 pm
“There is a roughly opposing bulge on the other side of Earth that is caused by the centrifugal force of Earth rotating about the Earth–Moon barycenter.”
I disagree with that. The bulge would exist regardless of centrifugal “force”. Centrifugal force simply adds to the effect.
Gravity is stronger on the near side than it is at the center, so it draws material on that side toward the focus of the orbit. It is weaker on the far side than it is at the center, so the material on that side is drawn in the other direction.
Similarly, centrifugal “force” on the near side is weaker than it is at the center, so material on the near side is drawn away towards the focus of the orbit. Mutatis mutandis on the far side.
The Wikipedia description appeals to intuition, because we thing of gravity as pulling into the center, and centrifugal force as pulling away. But, that is erroneous, because all the pushing and pulling is with respect to the central of gravity of the orbiting object.

stevefitzpatrick
February 15, 2014 1:41 pm

Leif,
I think a quick though experiment will settle it. Imagine for a moment not an orbiting Earth-Moon system, but Earth and Moon fixed relative to each other, but held apart at a fixed distance by a (very strong!) mass-less rod. What would happen to the ocean? It would bulge (become deeper) on the Moon side of the Earth, and at the same time become more shallow on the side opposite the Moon. Not two bulges, only one. It is the joint rotation about the Earth-Moon barycenter that generates the opposing bulge away from the Moon.

stevefitzpatrick
February 15, 2014 1:47 pm

Bart says:
February 15, 2014 at 1:40 pm
Please consider my thought experiment described above. Centrifugal forces around the barycenter seem to me key to the actual tidal behavior of the Earth’s oceans.

Mike M
February 15, 2014 1:52 pm

.Mike M says: February 15, 2014 at 12:21 pm …. So if earth’s orbital velocity around the sun slows down it will not fall closer towards the sun and if it speeds up it won’t go further way from the sun? …
—————————
lsvalgaard says: …The closer the Earth would be to the Sun, the faster it would go …
—————————
My ORIGINAL question: If earth’s orbital velocity around the sun slows down it will:
A) Go closer to the Sun
B) Go further away from the Sun
Answer the question, A or B?

Greg
February 15, 2014 1:56 pm

CliveB: Centrifuges separate U256 from U238. Nobody told the machines that what they were doing was fictitious or impossible.
If you climb in with the gas and start spinning you will need to invoke a _fictitious_ force to explain why you are being thrown against the wall of the centrifuge.
If you watch from the lab you will explain that the walls of vessel are providing the centripetal force needed to constrain the gases in a circular motion. Nothing fictitious needs to be explained to the machines or the lab techs.

Bart
February 15, 2014 1:57 pm

stevefitzpatrick says:
February 15, 2014 at 1:41 pm
“What would happen to the ocean? It would bulge (become deeper) on the Moon side of the Earth, and at the same time become more shallow on the side opposite the Moon. Not two bulges, only one. “
That is an inappropriately constrained analogy. The the rod is canceling the attractive force on the Earth via compressive forces.
In the case of pure gravity, the Earth is being pulled at one level. The oceans on the near side are being pulled harder, and those on the far side are being pulled less.
The centrifugal “force” does not create the symmetry in the bulge, it only adds to it.

February 15, 2014 1:58 pm

Bart says:
February 15, 2014 at 1:13 pm
And, the “rotational centrifugal force” to which they refer is embodied in the equation below which, as you see, depends on the orbit rate. Why? Because their body is tidally locked.
Fast rotation can also disrupt a body [nothing to do with tides]. One way of estimating the rotation period of a close-in body is to assume that it is tidally locked and then equal to the orbital period.
The combined effect of fast rotation and tidal forces makes the break up happen before it would without rotation. But the tides themselves have nothing to do with rotation or orbital motions. I have you a simple example of a non-rotating body falling straight into the Sun from infinity [no orbital ‘sideways’ movement. Calculate for us what the tidal height would be for that body and you will see where you go wrong [you get the same answer as the ‘standard’ tidal formula gives].

Bart
February 15, 2014 2:05 pm

I know it is a difficult logical hurdle when you first approach these kinds of problems, but intuition largely fails in accelerated reference frames.
The Earth is, itself, being accelerated by gravity. Stress occurs not when objects are accelerated, but when they are differentially accelerated.
The amount by which materials on the near side are being accelerated faster than the solid Earth is the same as the amount by which objects on the far side are being accelerated slower. As a result, you get a symmetric bulge.

February 15, 2014 2:06 pm

Steve Reddish says:
February 14, 2014 at 8:33 pm
1. Atmospheric contraction and expansion contribute to its tidal bulges, but not so for the oceans. Sea tides rise and fall solely due to the lateral flow of sea water.
===========
good point. water is largely incompressible. Thus its cannot expand or contract to create a tidal bulge. Thus, the water must flow in response to the tidal force.
This will affect the mixing rate. And due to the large thermal capacity of the oceans, even a small change in the mixing rate will have significant affects on climate.
So, if there is any long term cyclical behavior in the tides, and it is found to coincide with climate cycles, then tidal methods could provide a reliable mechanism to predict climate change.

February 15, 2014 2:08 pm

Willis:
A very nice and clear explanation of the governing tidal force and how it is correctly derived. There are many poor examples out there which your example puts to shame.
Thank you.
I hope that you will agree with me though that this is just the tip of the iceberg.
How that force plays out on the real ocean surface and internally below that surface is where the complexity and interest really lies.

stevefitzpatrick
February 15, 2014 2:10 pm

Bart,
“The the rod is canceling the attractive force on the Earth via compressive forces.”
Nonsense. The water on the surface of the Earth does not know anything about the existence of my imaginary rod. The water only reacts to net acceleration forces acting on it. What makes the second tidal bulge in an orbiting system is indeed centrifugal acceleration around the barycenter.

Bart
February 15, 2014 2:12 pm

lsvalgaard says:
February 15, 2014 at 1:58 pm
Are you laboring under the presumption that I am claiming something along the lines of stevefitzpatrick, that the bulge is only symmetric due to the centripetal acceleration? If so, you can see that, that is incorrect from the recent posts.

February 15, 2014 2:13 pm

Mike M says:
February 15, 2014 at 1:52 pm
“My ORIGINAL question: If earth’s orbital velocity around the sun slows down it will:
A) Go closer to the Sun
B) Go further away from the Sun
Answer the question, A or B?”
http://lmgtfy.com/?q=Speeding+up+and+slowing+down+in+orbit
https://howthingsfly.si.edu/media/speeding-and-slowing-down-orbit

Greg
February 15, 2014 2:14 pm

“So, if there is any long term cyclical behavior in the tides, and it is found to coincide with climate cycles, then tidal methods could provide a reliable mechanism to predict climate change.”
what like lunar declination affecting Indian Ocean you mean?
http://climategrog.wordpress.com/?attachment_id=774
this comes from a movement of water in and out of tropical Indian.O.
http://climategrog.wordpress.com/?attachment_id=777
That implies non linear feedbacks in tropics compensate when warm water is moved in and out.

Bart
February 15, 2014 2:16 pm

stevefitzpatrick says:
February 15, 2014 at 2:10 pm
“The water on the surface of the Earth does not know anything about the existence of my imaginary rod.”
Then, it is an alien entity which destroys the analogy. The water on the surface of the Earth clearly experiences the gravity field pulling on the Earth.
“The water only reacts to net acceleration forces acting on it.”
Acceleration relative to what? To the acceleration of the Earth. If your Earth is not accelerating, then your analogy is off the mark.
“What makes the second tidal bulge in an orbiting system is indeed centrifugal acceleration around the barycenter.”
Sorry, no.

Mike M
February 15, 2014 2:19 pm

RichardLH says: ,….
Reducing velocity AT A GIVEN ORBITAL RADIUS, (the whole basis of my contention) = less energy = go closer to the sun, (yeah, it will speed up tangentially because gravity is stronger but it’s overall orbital energy will be LOWER; it has to be.

Mike M
February 15, 2014 2:20 pm

RichardLH says: ,….
What do astronauts do to come back to earth? Speed up?

February 15, 2014 2:22 pm

Bart says:
February 15, 2014 at 2:12 pm
Are you laboring under the presumption that I am claiming something along the lines of stevefitzpatrick,
I’m not sure what misconception you are laboring under, but to clarify things “consider a non-rotating body falling straight into the Sun from infinity [no orbital ‘sideways’ movement]. Calculate for us what the tidal height would be for that body according to what you are laboring under.

February 15, 2014 2:24 pm

lsvalgaard says:
February 15, 2014 at 12:30 pm
“The closer the Earth would be to the Sun, the faster it would go, but the orbital speed has nothing to do with the tides.”
The closer the Moon to the Earth, the faster its orbit, and therefore the faster the change in its tides I am sure you would agree.
So nothing is a bit short of the true explanations. Sure the 24 hour – 0/1/2 tides rate does not change as that is Earth rotational but the rest does.

stevefitzpatrick
February 15, 2014 2:26 pm

Bart,
OK you have convinced me; you will never understand this relatively simple problem. A deus.

February 15, 2014 2:26 pm

Mike M says:
February 15, 2014 at 2:20 pm
“What do astronauts do to come back to earth? Speed up?”
Ar but now you ask a different question. How do I change my orbit so as to not stay in orbit but instead move into a elliptical path that crosses the Erath’s surface.

February 15, 2014 2:29 pm

Mike M says:
February 15, 2014 at 2:19 pm
“Reducing velocity AT A GIVEN ORBITAL RADIUS, (the whole basis of my contention) = less energy = go closer to the sun, (yeah, it will speed up tangentially because gravity is stronger but it’s overall orbital energy will be LOWER; it has to be.”
Shall I just quote from the link I provided.
“Speeding up and slowing down in orbit works just opposite to what you might expect. The larger a spacecraft’s orbit, the slower the spacecraft travels. So if you wanted to pass a spacecraft just ahead of you, you would have to fire a thruster in a forward direction. This would decrease your orbital energy and drop you into a lower orbit, where you will travel faster! The “passing lane” in orbit is always lower.”

Mike M
February 15, 2014 2:31 pm

RichardLH says: February 15, 2014 at 2:26 pm “Ar but now you ask a different question. How do I change my orbit so as to not stay in orbit but instead move into a elliptical path that crosses the Erath’s surface.”
Nice dodge, I want come back DOWN! I fire RETRO rocket engines to SLOW my velocity and LOWER my orbit. Less energy = LOWER orbit. You want to deny that – go ahead.

February 15, 2014 2:33 pm

Mike M says:
February 15, 2014 at 2:19 pm
Think of it like this. You have acquired some potential energy to get to a given orbit. If you fire a forward thruster you lose some of that energy. To stay in orbit you must therefore be closer to the surface. i.e. less potential energy from that previously given you by the rocket that got you there.

Mike M
February 15, 2014 2:35 pm

RichardLH says: February 15, 2014 at 2:29 pm
Shall I just quote from the link I provided.
“Speeding up and slowing down in orbit works just opposite to what you might expect. The larger a spacecraft’s orbit, the slower the spacecraft travels. So if you wanted to pass a spacecraft just ahead of you, you would have to fire a thruster in a forward direction. ”
—–
“In a FORWARD direction” = RETRO ROCKET = OPPOSITE the direction of travel = OPPOSITE direction of thrust that got you into orbit!

Mike M
February 15, 2014 2:40 pm

RichardLH says: …
“My ORIGINAL question: If earth’s orbital velocity around the sun slows down it will:
A) Go closer to the Sun
B) Go further away from the Sun
Hmmmm? A or B ?

February 15, 2014 2:45 pm

Here is Feynman on the basic tidal force:
http://www.feynmanlectures.caltech.edu/I_07.html#Ch7-S4
There is of course no force on the mass at the planet centre:
http://www.feynmanlectures.caltech.edu/I_13.html#Ch13-S4

February 15, 2014 2:46 pm

Mike M says:
February 15, 2014 at 2:40 pm
“My ORIGINAL question: If earth’s orbital velocity around the sun slows down it will:
A) Go closer to the Sun
B) Go further away from the Sun

Perhaps it will be clearer if you think about the Moon instead. The Moon is moving away from us to a larger orbit where it moves slower because it is gaining angular momentum [stealing it from the Earth’s rotation which is slowing down]. So if you slow down the Moon, the answer is A. But your question is ill-posed in the first place as you did specify what mechanism you had in mind.

February 15, 2014 2:47 pm

Mike M says:
February 15, 2014 at 2:31 pm
“Nice dodge, I want come back DOWN! I fire RETRO rocket engines to SLOW my velocity and LOWER my orbit. Less energy = LOWER orbit. You want to deny that – go ahead.”
Indeed you do as I explained above. Less energy = LOWER orbit = higher speed. Strange isn’t it?

Bart
February 15, 2014 2:49 pm

stevefitzpatrick says:
February 15, 2014 at 2:26 pm
“OK you have convinced me; you will never understand this relatively simple problem.”
It is too simple. Don’t you realize that? Do you really think I would not understand your very simple setup?
But, it bears no relationship to reality. This is a more subtle problem than you yet apprehend. A lot of newbies make the same mistake as you do here.

February 15, 2014 2:56 pm

Mike M says:
February 15, 2014 at 1:52 pm

My ORIGINAL question: If earth’s orbital velocity around the sun slows down it will:
A) Go closer to the Sun
B) Go further away from the Sun
Answer the question, A or B?

This is trickier to answer than you may think
Imagine that you could have a big rocket engine mounted in the sky with the exhaust upwards from the Earth. Imagine that you started this engine when the exhaust pointed in the orbital “forward” direction. You should then expect the Earth was slowing down, right?
Well, it will not slow down, it will fall closer too the Sun and the orbital speed will increase. So if you try to brake the orbital velocity of the Earth the speed will increase.
The only way to slow down the orbital velocity of the Earth is to push it forward. The Earth will then move further away from the Sun and the velocity will counter-intuitively slow down.
This is not only a theoretical possibility, it actually happens each time we send a rocket away from the Earth. If for instance a rocket is sent away with destination Mars, it will be sent in the orbital “forward” direction, and the recoil will then make the Earth fall a little bit closer to the Sun and the speed will increase a little bit
To sum up: Closer orbits means higher velocity, but we have to brake the orbital objects to speed them up.
/ Jan

Tom in Florida
February 15, 2014 2:57 pm

So why is the orbital speed of Earth faster at perihelion? Or is that something different due to an elliptical orbit?

Mike M
February 15, 2014 2:58 pm

RichardLH says: Indeed you do as I explained above. Less energy = LOWER orbit = higher speed. Strange isn’t it?
It is NOT a higher speed AT the same orbital radius – that’s the WHOLE point!

February 15, 2014 2:58 pm

Tom in Florida says:
February 15, 2014 at 2:57 pm
So why is the orbital speed of Earth faster at perihelion? Or is that something different due to an elliptical orbit?
Because the closer you are to the Sun, the faster you have to move [‘sideways’] to stay in orbit.

February 15, 2014 3:04 pm

stevefitzpatrick says:
February 15, 2014 at 2:26 pm
“OK you have convinced me; you will never understand this relatively simple problem.”
Try these for an explanation which is very similar to Wills’s work.
http://co-ops.nos.noaa.gov/restles3.html
http://www.lhup.edu/~dsimanek/scenario/tides.htm
“It is important to note that the centrifugal force produced by the daily rotation of the earth on it axis must be completely disregarded in tidal theory. This element plays no part in the establishment of the differential tide-producing forces.”
The forces that arise from moving around the barycentre are tiny compared to those. In there as described above but swamped by all the rest.

Mike M
February 15, 2014 3:08 pm

Jan Kjetil Andersen says February 15, 2014 at 2:56 pm ” This is trickier to answer than you may think”

Mike M says:
February 15, 2014 at 1:52 pm
My ORIGINAL question: If earth’s orbital velocity around the sun slows down it will:
A) Go closer to the Sun
B) Go further away from the Sun
Answer the question, A or B?

Not ‘tricky’ at all …in the context of my explanation of a mass M on the equator going faster at midnight or slower at noon with respect to earth’s orbital velocity – the answer is A…. PERIOD.
The mass experiences more net force toward the sun at noon and less at midnight by virtue of the difference in its orbital velocity as imparted by the rotation of the earth. (The difference in gravity itself by difference of one earth’s diameter versus 93 million miles is miniscule as WIllis already calculated.)

February 15, 2014 3:10 pm

Mike M says:
February 15, 2014 at 2:58 pm
“It is NOT a higher speed AT the same orbital radius – that’s the WHOLE point!”
But it is impossible to stay at the same radius with lower radial velocity. In this universe anyway.
As you move closer and closer to the Earth’s surface you will get faster and faster as you do. Like a skater bringing in their arms in a spin.
As the Moon gains energy from the Earth by the friction on the tides, it is moving further and further away as Lief says. More potential energy = less radial velocity = larger orbital distances.

February 15, 2014 3:16 pm

Mike M says:
February 15, 2014 at 3:08 pm
“My ORIGINAL question: If earth’s orbital velocity around the sun slows down it will:
A) Go closer to the Sun
B) Go further away from the Sun
Answer the question, A or B?”
If it has a lower orbital radial velocity (and therefore higher potential energy to be stable) it will be B.
If it has a higher orbital radial velocity (and therefore lower potential energy to be stable) it will be A.
As noted in the link above, you fire a forward thruster to loose height but gain orbital speed and vice versa.

don penman
February 15, 2014 3:20 pm

We can think of the of the Sun and earth or the earth and moon being stationary at small point in time would the earth not have a bulge nearest the moon and furthest from the moon at this small instant of time.I think it would.

eyesonu
February 15, 2014 3:24 pm

I just dropped in to see what condition my condition was in.
I considered the gravitational force between two bodies and the centrifugal force of one rotating around another with a fluid surface of the smaller. This is where the discussion seemed to move to even though Willis’ original post was the attraction of a non orbiting body to another.
The previous discussion has me concerned that I may not know sh*t from apple butter. I will not cook breakfast at home tomorrow but will instead eat @ a fast food restaurant w/ no apple butter.
Therefore my condition is unknown by me at this moment in time.
LOL
————-
Sorry mods, I post this comment on another thread by accident. My condition seems to be deteriorating.

Mike M
February 15, 2014 3:32 pm

RichardLH says: As noted in the link above, you fire a forward thruster to loose height but gain orbital speed and vice versa.
No.. not INITIALLY! Initially your velocity HAS TO slow down because you are DEcelerating! Yes, it speeds back up as you get lower and gravity increases BUT per my tidal assertion, you DO NOT get lower – you are staying within an earth radius of earth’s orbit around the sun.

February 15, 2014 3:35 pm

The question about what will happen if a satellite “slows down” or “fires retro-rockets” is ill-posed. There are two different interpretations with two very different answers. Let’s assume the orbit is circular of radius “r1” to start with.
1) If on-board rockets are fired backwards briefly, the rocket will slow at that point in the orbit. It will stop orbiting in the circular orbit and orbit in an elliptical orbit with a semimajor axis r2 < r1. The time it takes to orbit will become less. It will have less overall energy, less average gravitational potential energy, and more average kinetic energy.
2) If on-board rockets are fired forwards briefly, the rocket will speed at that point in the orbit. If on-board rockets are fired forwards AGAIN briefly when the satellite reaches apogee so that it takes on a new circular orbit with r2 > r1, then the new orbit will have more average gravitational potential energy, and less average kinetic energy. The average speed over the whole orbit will decrease. (Rather counter-intuitive that your can fire the rockets forward TWICE and still end up with less KE!)
Basically, you have to know if you are forcing the satellite back into a circular orbit or not.
Interesting bit of trivia — for circular orbits, the potential energy is always twice the magnitude and opposite in sign to the kinetic energy.

Mike M
February 15, 2014 3:36 pm

RichardLH says: But it is impossible to stay at the same radius with lower radial velocity. In this universe anyway.
No – not when you’re stuck to earth by 10 m/sec^2 gravity it isn’t impossible.

u.k.(us)
February 15, 2014 3:37 pm

It is fun to try to wrap (or would it be warp) my mind around all these ideas.
But something keeps nagging me…
Why is the rope blue ?

Mike M
February 15, 2014 3:38 pm

Tim Folkerts says: February 15, 2014 at 3:35 pm
The question about what will happen if a satellite “slows down” or “fires retro-rockets” is ill-posed. There are two different interpretations with two very different answers. Let’s assume the orbit is circular of radius “r1″ to start with.
1) If on-board rockets are fired backwards briefly, the rocket will slow at that point in the orbit.
…………………
STOP! Right there! … that’s all I was ever talking about!

February 15, 2014 3:41 pm

Mike M says:
February 15, 2014 at 3:38 pm
STOP! Right there! … that’s all I was ever talking about!
and what has that to do with the tides that are around all the time.

February 15, 2014 3:44 pm

Mike M says:
February 15, 2014 at 3:32 pm
“No.. not INITIALLY! Initially your velocity HAS TO slow down because you are DEcelerating! Yes, it speeds back up as you get lower and gravity increases BUT per my tidal assertion, you DO NOT get lower – you are staying within an earth radius of earth’s orbit around the sun.”
You seem to have a very strange view of the universe. It takes a while to catch up with what you do.
Unfortunately in the real universe, if you fire the thruster for a millisecond or an hour. the rate of change will happen instantaneously and so will velocity/height/etc.
If you can’t manage going down. Think the other way instead. As your rocket engine – now pointing along the orbit – pushes you forwards, you will slow down in forward radial velocity as you rise in height. The energy has gone into potential energy for height above the surface and thus lost from radial forward velocity.
Below stable orbital radial velocity you will of course get faster as you rise to it. It is what happens from then we are talking about.

February 15, 2014 3:47 pm

lsvalgaard says:
February 15, 2014 at 3:41 pm
“and what has that to do with the tides that are around all the time.”
Nothing. But people don’t get how to manoeuvre in space at all. Most of the games have it wrong as well.

Mike M
February 15, 2014 3:51 pm

RichardLH says: “…will happen instantaneously and so will velocity/height/etc.”
Nope. You fire the rocket, your velocity increases as the rocket fires and THEN you start going up. As you go up, OVER TIME, your velocity then decreases until your reach a higher orbit at a lower velocity than that at which you started. You do NOT “instantaneously” change orbit altitude!

February 15, 2014 3:56 pm

Mike M says:
February 15, 2014 at 3:51 pm
“Nope. You fire the rocket, your velocity increases as the rocket fires and THEN you start going up. As you go up, OVER TIME, your velocity then decreases until your reach a higher orbit at a lower velocity than that at which you started. You do NOT “instantaneously” change orbit altitude!”
Fire it for a millisecond in an hour and you will still see a change in orbit. Or are all the guys who fly in orbit wrong as well?

William Larson
February 15, 2014 3:56 pm

Mr. Eschenbach–
” In prior centuries, these atmospheric tides were said to affect the weather, although in a cyclical rather than a secular manner.” I am confused (or, Kung-Fu-sed, as my son’s karate teacher likes to say): I have never used “secular” in such a context as you have here. What does it mean here? (If you respond to this, thank you.)

February 15, 2014 4:00 pm

William Larson says:
February 15, 2014 at 3:56 pm
I have never used “secular” in such a context as you have here. What does it mean here?
From the prayers IN OMNIA SAECULA SAECULORUM meaning [roughly] ‘for ever and ever’, thus ‘long-term’

February 15, 2014 4:02 pm

Mike M says:
February 15, 2014 at 3:36 pm
“No – not when you’re stuck to earth by 10 m/sec^2 gravity it isn’t impossible.”
But you’re not in free orbit at that point. If you were to try and stay in free orbit just above the surface (1cm) you will need to have an airless planet and a very flat planet to achieve it. The required orbital velocity is just too great.

February 15, 2014 4:05 pm

u.k.(us) says:
February 15, 2014 at 3:37 pm
“It is fun to try to wrap (or would it be warp) my mind around all these ideas.
But something keeps nagging me…
Why is the rope blue ?”
Cheapest on offer at K-Mart?

Mike M
February 15, 2014 4:06 pm

Mike M says:February 15, 2014 at 3:38 pm
STOP! Right there! … that’s all I was ever talking about!
——————
lsvalgaard says: February 15, 2014 at 3:41 pm and what has that to do with the tides that are around all the time.
————–
So you forgot? I didn’t Let’s go over it shall we? – you never answered the question:
My question: If earth’s orbital velocity, (AT IT’S CURRENT ORBIT), around the sun slows down it will:
A) Go closer to the Sun
B) Go further away from the Sun
Answer the question, A or B?
Still waiting for an answer FROM YOU. (And when you can give a straight answer.. THEN we’ll move on to explaining more to you .)

February 15, 2014 4:08 pm

Mike M says:
February 15, 2014 at 4:06 pm
(And when you can give a straight answer.. THEN we’ll move on to explaining more to you .)
That is very generous of you. But, as I said, your question is ill-posed and borders on nonsense, so you perhaps see my problem…

Mike M
February 15, 2014 4:09 pm

RichardLH says: …airless planet and a very flat …”
You’re missing the whole point. The orbit is earth around the sun, the thing stuck by earth’s gravity is NOT in orbit around earth, IT IS EARTH! Specifically a drop of water at the equator.

Mike M
February 15, 2014 4:15 pm

lsvalgaard says: “..perhaps see my problem…”
Yeas I do – you’re obstinate.
The answer is A and it happens to a mass M sitting on earth’s equator at noon – it’s going SLOWER than earth’s overall orbital velocity.
On the other side, at midnight, it is going FASTER than earth’s overall orbital velocity.
The slower particle gets dragged in more by the sun and the particle going faster is being flung away from the sun.
Now disprove it directly instead of trying to weasel around it. Did I state something that is not true?

February 15, 2014 4:20 pm

Mike M says:
February 15, 2014 at 4:15 pm
Now disprove it directly instead of trying to weasel around it. Did I state something that is not true?
You have discovered the amazing fact that the Earth’s rotation [measured on a mass sitting firmly and fast on the equator] goes in the same direction as the orbital movement as midnight and in the opposite direction at noon. Fascinating! Now what has that to do with tides?

Mike M
February 15, 2014 4:28 pm

lsvalgaard says: Fascinating! Now what has that to do with tides?
What I already explained above but I guess you have a short attention span.
February 15, 2014 at 7:13 am: A quick look gave me a result for a “no moon” variation of “midnight fling” acceleration over average center V^2/R of 3%. For now I’d say it looks significant.
Earth’s average orbit velocity V1 as 29885 m/sec. Crust WRT center V2 as 463 m/sec. So combined velocity in orbit around the sun at midnight V1+V2=V3 as 30349 m/sec V3^2 / V1^2 = 1.03

February 15, 2014 at 7:51 amBoiling it down, I’ve produced a two bulge tide model with no moon at all. Earth to Sun acceleration due to gravity is .006 M/sec^2 Plus 3% at midnight, minus 3% at noon makes it a change plus or minus .0002 m/sec^2 against our earth gravity of 10M/sec^2
Certainly tiny WRT to earth’s gravity at sea level but how much does that contribute to the formation of two lobes?

I’m stating that the difference EXISTS. Are you stating that it does not exist? Are my numbers way off?

February 15, 2014 4:29 pm

Mike M says:
February 15, 2014 at 4:09 pm
“You’re missing the whole point. The orbit is earth around the sun, the thing stuck by earth’s gravity is NOT in orbit around earth, IT IS EARTH! Specifically a drop of water at the equator.”
Then the majority of the forces operating (to the first quite a lot of decimal places) will be the frictional, compressive, etc. components from the bits on water on either side of it.
Then will come the various forces that come from the Earth’s rotation alone, centripetal, coriolis, etc.
Then will come the 3d Tidal vectors.
That should cover the top 10 decimal places or so.

February 15, 2014 4:30 pm

Mike M says:
February 15, 2014 at 4:09 pm
Oops. Sorry – left out tidal currents etc.

February 15, 2014 4:34 pm

Mike M says:
February 15, 2014 at 4:28 pm
I’m stating that the difference EXISTS. Are you stating that it does not exist? Are my numbers way off?
As Bart has said many times, I am a dummy, embarrassing myself constantly, so please explain to this dummy how your differences produce tides. And how you would explain tides if the Earth were not rotating. I guess you would say that on a non-rotating Earth there would no tides ‘flung’ around, but explain that in some more detail, please.

Mike M
February 15, 2014 4:51 pm

RichardLH says: February 15, 2014 at 4:29 pm “Then the majority of the forces operating (to the first quite a lot of decimal places) will be the frictional, compressive….”
If you say so… I’m only pointing out the change in net acceleration force that goes up and down twice a day at the equator without any moon at all and simply because of earths’ rotation slowing/speeding earth’s crust slower/faster than earth’s solar orbital velocity.
lsvalgaard to his credit compartmentalized it best – you will weigh .002 less at midnight and noon than you will at 6am or 6pm. (Females using spring loaded bathroom scales take note.)
I would think seismographs register this effect but it isn’t noticed because it’s a much longer period than an earthquake and the amount of change is so small it’s probably ignored.

February 15, 2014 4:56 pm

Mike M says:
February 15, 2014 at 4:51 pm
I’m only pointing out the change in net acceleration force that goes up and down twice a day at the equator without any moon at all and simply because of earths’ rotation slowing/speeding earth’s crust slower/faster than earth’s solar orbital velocity.
So you are claiming that if the Earth were not rotation there would be no tidal effects…
Good news for the proverbial astronaut falling into a black hole: just be sure that you don’t rotate and you’ll not be shredded by tidal forces…

February 15, 2014 5:04 pm

Mike M says:
February 15, 2014 at 4:51 pm
“I would think seismographs register this effect but it isn’t noticed because it’s a much longer period than an earthquake and the amount of change is so small it’s probably ignored.”
I think you will find that the Earth Tide, the bulge in the rock itself from the Moon and Sun is an order to two higher but….
http://en.wikipedia.org/wiki/Earth_tide

February 15, 2014 5:04 pm

Mike M says:
February 15, 2014 at 4:51 pm
you will weigh .002 less at midnight and noon than you will at 6am or 6pm.
Here is get-rich scheme using your insight: buy gold at noon, e.g. 100 kg, file off 0.2 kg [=6 oz @ US$1300/oz =$7600], then sell the gold at 6 pm as it at that time will weigh 100 kg and recover what you paid for the 100 kg at noon, pocketing $7600/day. In a year that is almost$3 million [tax free].

Mike M
February 15, 2014 5:05 pm

lsvalgaard says: “So you are claiming …”
I did NOT discount or preclude other factors! Stop putting words into other people’s mouths!

Mike M
February 15, 2014 5:07 pm

lsvalgaard says: Here is get-rich scheme
Ummm, only an idiot would use anything but a MASS BALANCE scale for stuff like gold – works on the moon.

AmusedObserver
February 15, 2014 5:11 pm

Wow. Leif never stops amazing me with his infinite patience here. I’d be into the whiskey by now.

February 15, 2014 5:14 pm

Mike M says:
February 15, 2014 at 5:05 pm
I did NOT discount or preclude other factors! Stop putting words into other people’s mouths!
So, answer the question: if the Earth did not rotate would there be tidal effects? If so, how large would they be compared to the ones caused by rotation? Huge or tiny? Or don’t you know?
Mike M says:
February 15, 2014 at 5:07 pm
Ummm, only an idiot would use anything but a MASS BALANCE scale for stuff like gold
And Mother Nature does not do that when it comes to create tides. How do you know?

Mike M
February 15, 2014 5:25 pm

RichardLH says: “I think you will find that the Earth Tide, the bulge in the rock itself …”
Oddly, that’s what I generally guessed at above. ( February 15, 2014 at 12:15 pm)

Mike M
February 15, 2014 5:31 pm

lsvalgaard says: February 15, 2014 at 5:14 pm “So, answer the question: if the Earth did not rotate would there be tidal effects? ”
From the moon, of course.

Kirk c
February 15, 2014 5:35 pm

I think there is a gross misunderstanding of “free fall”. Willis is correct in his analogy and may even underestimate his own clarity.. the planet is actually “free falling” and is plunging toward the sun as described in his picture. There is also a sideways motion (orbital velocity) so we (earth) manage to keep missing the sun as we fall into it. Since the sideways motion is perpendicular it has no effect on tidal force so we can ignore it and are left with only the free fall condition. So the bodies M1, M2 and M3 are each different distances from the main attractor so each has a slightly decreasing pull. Lets call the forces magnitudes of 3,2, and 1 .All are positive pulls toward the sun ( or moon) in this frame of reference. The closest is hardest, the middle less and the extreme even less again. so you end up with tension in the theoretical string (local gravity replaces the string) and this results in a front-side and back-side bulge relative to the middle . Rather simply presented I think.
Good one Willis!
Trying to bring in orbital and rotational centripetal force is not required and has insignificant affects.

Mike M
February 15, 2014 5:37 pm

lsvalgaard says: “And Mother Nature does not do that when it comes to create tides. How do you know?”
Ho do I know what exactly? How to spot non-sequitur blather?

February 15, 2014 5:37 pm

Mike M says:
February 15, 2014 at 5:25 pm
“Oddly, that’s what I generally guessed at above. ( February 15, 2014 at 12:15 pm)”
I Googled it with much less effort (to get the value – I already knew it existed). As I said., an order of magnitude or two greater than the discovery you are looking at.

Mike M
February 15, 2014 5:42 pm

RichardLH says: “I Googled it with much less effort ”
Keep up the good work!

Mike M
February 15, 2014 5:55 pm

Kirk c says: “Trying to bring in orbital and rotational centripetal force is not required and has insignificant affects.”
Willis’ is pointing out a gravitational differential producing a difference of ~1.1 MICRO Newtons, (.0000011 Newtons) between two 1KG masses.
I’m pointing to a .0002 m/sec^2 difference in acceleration from a rotational induced difference in sun’s gravitational field which, for a 1 KG mass, is .0002 Newtons. Insignificant?

February 15, 2014 5:56 pm

Mike M says:
February 15, 2014 at 5:42 pm
“Keep up the good work!”
Just to point out that it existed and was significantly greater than you were looking for……
I can’t see why you didn’t try that route first anyway.
And looked at the internal tide which is much more significant than all the stuff going on at the surface anyway.

Steve Richards
February 15, 2014 5:57 pm

It is amusing to see bright people communicating with such difficulty.

Mike M
February 15, 2014 6:12 pm

RichardLH says: I can’t see why you didn’t try that route first anyway.
Well I guess that’s just a difference between us. I always try to come up with my own answer first and then look to the answers by others to see if/how my answer compares. Just looking up stuff doesn’t put my brain in gear.

February 15, 2014 6:16 pm

Mike M says:
February 15, 2014 at 6:12 pm
“Well I guess that’s just a difference between us. I always try to come up with my own answer first and then look to the answers by others to see if/how my answer compares. Just looking up stuff doesn’t put my brain in gear.”
I just like to order the stuff I look at I quantitate order first. Tend not to get lost in the minutia then 🙂

Ulric Lyons
February 15, 2014 6:18 pm

RichardLH says:
February 15, 2014 at 3:04 pm
“At about 54.7° from the Earth-moon line, the vector difference in the forces happens to be parallel to the surface of the Earth.” http://www.lhup.edu/~dsimanek/scenario/Field_tidal.jpg
That’s an interesting geometry, focused at sqrt of 3 Earth radius away from the center.

Kirk c
February 15, 2014 6:34 pm

Mike M says:
February 15, 2014 at 5:55 pm
“Trying to bring in orbital and rotational centripetal force is not required and has insignificant affects.”
I’m pointing to a .0002 m/sec^2 difference in acceleration from a rotational induced difference in sun’s gravitational field which, for a 1 KG mass, is .0002 Newtons. Insignificant?
Ok. but rotational is not tidal. the rotational is uniform and causes an average equatorial bulge in 360 degrees 24/7. without moon or sun then, the net sea level rise is constant . I agree there is probably some small Coriolis effect set up as well… but this is not really the point of the thread. “Pondering tides” was the title.. not, “Would there be ocean currents on a rotating non-orbiting body”
cheers

February 15, 2014 6:35 pm

Mike M says:
February 15, 2014 at 5:31 pm
From the moon, of course.
So if the Earth did not rotate there would be solar tides?

February 15, 2014 6:35 pm

Ulric Lyons says:
February 15, 2014 at 6:18 pm
““At about 54.7° from the Earth-moon line, the vector difference in the forces happens to be parallel to the surface of the Earth.” http://www.lhup.edu/~dsimanek/scenario/Field_tidal.jpg
That’s an interesting geometry, focused at sqrt of 3 Earth radius away from the center.”
That’s actually from Wiki as the url notes (and yes it is deliberately exaggerated).
http://en.wikipedia.org/wiki/Tide

February 15, 2014 6:36 pm

Mike M says:
February 15, 2014 at 5:31 pm
From the moon, of course.
So if the Earth did not rotate there would be no solar tides?

February 15, 2014 6:39 pm

Ulric Lyons says:
February 15, 2014 at 6:18 pm
““At about 54.7° from the Earth-moon line, the vector difference in the forces happens to be parallel to the surface of the Earth.”
You weren’t suggesting the geometry is wrong are you? The arrows are too long but that is all.

Ulric Lyons
February 15, 2014 6:54 pm

RichardLH says:
February 15, 2014 at 6:39 pm
“You weren’t suggesting the geometry is wrong are you? The arrows are too long but that is all.”
I was remarking upon the described 54.7° angle rather than the diagrammatic representation of it.
It makes a sqrt 2, sqrt 3, sqrt 5 sided triangle.

Ulric Lyons
February 15, 2014 7:08 pm

Ulric Lyons says:
February 15, 2014 at 6:54 pm
correction 1, 2, sqrt 3.

Ulric Lyons
February 15, 2014 7:14 pm

Ulric Lyons says:
February 15, 2014 at 7:08 pm
Third time lucky lol, 1, sqrt 2, sqrt 3.

February 15, 2014 7:18 pm

Mike M says:
February 15, 2014 at 5:55 pm
I’m pointing to a .0002 m/sec^2 difference in acceleration from a rotational induced difference in sun’s gravitational field which, for a 1 KG mass, is .0002 Newtons. Insignificant?
Irrelevant as there is not tidal effect stemming from that as the rotation does not make any difference to the Sun’s gravitational field.

Mike McMillan
February 15, 2014 8:32 pm

Thought I’d jump in here in case there’s any confusion about Mike M.
Not me.

February 15, 2014 11:04 pm

lsvalgaard says:
February 15, 2014 at 5:04 pm

Mike M says:
February 15, 2014 at 4:51 pm
you will weigh .002 less at midnight and noon than you will at 6am or 6pm.
Here is get-rich scheme using your insight: buy gold at noon, e.g. 100 kg, file off 0.2 kg [=6 oz @ US$1300/oz =$7600], then sell the gold at 6 pm as it at that time will weigh 100 kg and recover what you paid for the 100 kg at noon, pocketing $7600/day. In a year that is almost$3 million [tax free].

I know a better one:
Why not buy gold in Mexico City where the gravity acceleration is 9.776 m/s*s and sell it in Anchorage where the gravity acceleration is 9.826 m/s*s.
The difference is 0.5% so you can pocket 0.5 kg each time, not a tiny 0.2 kg.
Alas, the trick will not work for any of us with balance scales, as I think is the preferred method in gold trading.
Perhaps we should concentrate on answering Mike’s concrete question rather than evading to something that some may perceive as quite arrogant.
/ Jan

Mike M
February 15, 2014 11:16 pm

Kirk c says: February 15, 2014 at 6:34 pm “Ok. but rotational is not tidal. ….”
Nothing of what you wrote had anything to do with what I’m asserting. You simply don’t understand. For any given particle orbiting the sun at a given radius from the sun there is an orbital velocity V with an inertial reaction of V^2/R acceleration EXACTLY EQUALING the gravitational pull it experiences from the sun at that orbital radius R. If it suddenly “finds” itself going slower it will be pulled toward the sun. If faster it will move away from the sun.
Such is what a particle sitting on earth’s equator experiences every day. The speed, S, the particle is traveling WRT to earth’s center is simply earth’s equatorial radius times earth’s angular velocity. From ~6am it begins slowing down from V (its component of solar orbital speed, being perpendicular to it rotational velocity spinning around earth at that time of day ), until noon when it reaches the point of maximum decrease, V- S, from solar orbit velocity and then speeds up again until ~6pm when its solar orbital speed is again back to V. Then it continues, now speeding up to V+S at midnight, the amount S more than solar orbital speed V.
When it is going slower than V it is pulled toward the sun by the amount of sun’s gravity which is, at noon, greater than the particle’s now reduced solar orbital inertial reaction (V-S)^2/R . When it is going faster by S at midnight than V it is being flung away from the sun by the increase of solar orbital inertial imbalance (V+S)^2/R which is greater than sun’s gravity.
I calculated the maximum differences in acceleration (V+S)^2/R or (V-S)^2/R versus V^2/R (which are actually slightly different in magnitude but beyond 5 places), as being -.00018 meters/sec^2 offset countering earth’s gravity occurring twice per day, at noon and midnight – two bulges.
For simplification I ignored the difference of R for each by the amount of earth’s radius as well as the sun gravitational field variation that WiIlis addressed which is a much smaller effect.
Is there an seismic instrument, appropriately damped to ignore earthquakes etc. and look only at the difference in gravity occurring over hours? If so it ought to record this effect and see a maximum gravitation pull being reached at 6am and 6pm with a minimum reached at noon and midnight with a difference by about .0002 meters/sec^s.

February 16, 2014 12:57 am

Willis,
Yes you’re right
The correct equation is more complicated. Taking x as the direction from the centre of the earth to the moon and taking y the vertical axis then the net tidal force has 2 components.
$Fx = Gm( \frac{ \cos \phi}{R^2 + r^2 - 2rR \cos \theta} - \frac{1}{r^2})$
$Fy = \frac{ -Gm \sin \phi}{R^2 + r^2 - 2rR \cos \theta}$
where
$\cos \phi = \frac{r-R \cos \theta}{\sqrt{R^2 + r^2 - 2rR \cos \theta}}$
and
$\sin \phi = \frac{R \sin \theta}{\sqrt{R^2 + r^2 - 2rR \cos \theta}}$
The tractional force acting along the surface and therefore unaffected by the earth’s gravity is
Fy cos(theta) + Fx sin(theta) . The tractional force is zero at the central bulge and equal and opposite from either side.

Coldlynx
February 16, 2014 12:48 am

Willis Eschenbach says:
February 15, 2014 at 11:30 pm
“This is because the tidal force depends on the difference in distance to the sun from a given point on the planet, and the distance to the sun from center of the planet. And at the poles, that difference is zero, so the tidal force is zero.”
That is valid only 2 days a year for the sun due to earth tilt. All other days of the year is the “Sun tidal zero poles” not at earth geographics poles. Moon inclination will also have large impact on where on earth the “moon tidal force zero poles” are located. And also where on earth the maximum horisontal tide forcing are located. The horisontal forcing that actually move air and water according to Laplace.
That latitude change between 45°N and 45°S +/- 28.36°
In a cyclic way. 18.6 years the most obvious.

Greg Goodman
February 16, 2014 1:19 am

Some of the motivation for this article seems to be to dismiss my comment of a couple of days ago: “The tide raising force acts in both directions (bulge on each side in the simplistic model)”
The context of that remark was to explain why Willis’ plot of variation in the magnitude of the tidal force was not correct. He was subtracting the lunar and solar forces at full moon.
His headline graph in that article only peaks once per year , while the tides peak twice per year.
http://wattsupwiththat.com/2014/02/09/time-and-the-tides-wait-for-godot/#comment-1569175
I pointed out the mistakes and provided a modification for his R-code to give the correct addition of forces and a graph that produces the twice per year variations.
So does Willis correct his code and update the graph?
No, he writes another article to insist that he’s right, ridicules my comment and calls me a fool.
The fact that his tidal forces only peak once a year should be a clue to him, but I’m sure there’s a good reason to ignore that too.
As I said above, “You cannot stretch a body with one force” You can accelerate a body with one force. You need to separate the two effects.
http://wattsupwiththat.com/2014/02/14/canute-ponders-the-tides/#comment-1568736
You cannot stretch a body with one force.
In the case above (GF3+GF1)/2 = GF2 will accelerate each of the three tethered weights at the common rate towards the sun.
Two opposing forces (GF3-GF1)/2 will act at each end mass acting to stretch the tethers applying a tensional stress to the ensemble.
The lay reader may wish to decide whether to be guided by Willis or Feynman on this question.
http://www.feynmanlectures.caltech.edu/I_07.html#Ch7-S4
“The result of this imbalance is that the water rises up, away from the center of the earth. On the near side, the attraction from the moon is stronger, and the imbalance is in the opposite direction in space, but again away from the center of the earth. The net result is that we get two tidal bulges.”

February 16, 2014 1:22 am

Both sides in this argument are correct. You don’t need to use the centrifugal force to derive the formula for tides. This is because there is a perfect balance between the centrifugal force and the gravitational force at the center of the earth when in orbit around the earth-moon barycenter. This balance also determines the strength of tides on earth. When in doubt see what Feynman says.

What do we mean by “balanced”? What balances? If the moon pulls the whole earth toward it, why doesn’t the earth fall right “up” to the moon? Because the earth does the same trick as the moon, it goes in a circle around a point which is inside the earth but not at its center. The moon does not just go around the earth, the earth and the moon both go around a central position, each falling toward this common position. This motion around the common center is what balances the fall of each. So the earth is not going in a straight line either; it travels in a circle. The water on the far side is “unbalanced” because the moon’s attraction there is weaker than it is at the center of the earth, where it just balances the “centrifugal force.” The result of this imbalance is that the water rises up, away from the center of the earth. On the near side, the attraction from the moon is stronger, and the imbalance is in the opposite direction in space, but again away from the center of the earth. The net result is that we get two tidal bulges.

So a body in free fall into the sun experiences ever increasing tidal forces until it is torn apart. A body in orbit however experiences regular varying tidal forces depending on the eccentricity of the orbit.
So on a purely logical basis Willis is correct. But in order to calculate the variations of tides on earth you need to include orbital dynamics because they change the earth-moon distance.

Carl Brannen
February 16, 2014 2:10 am

The formula: “Tidal Force = 2 * G * sunmass * r / D^3” is wrong. This is the formula for the “Tidal acceleration”. To make it a tidal force, you need to have another mass. The usual way it’s written is with m for the little mass (you used 1kg masses before) and M for the big mass (the sun):
Tidal Force = 2G m M r/D^3
The sentence “It’s not because the tidal force acts in different directions, because tidal force is always directed towards the sun.” is incorrect. On the side nearest the sun, the tidal force is directed towards the sun and moves the water towards the sun. On the side away from the sun, the tidal force is directed away from the sun and moves the water away from the sun. It should be clear that these are opposite directions. On the other hand, it is true that the “gravitational force” is always directed towards the sun. Maybe that was the source of some confusion. “Tidal forces” are defined with respect to the center of mass of the body (earth in this case), not the gravitating object.
On “Eventually, at a distance called the “Roche limit”, the tidal forces can reach such strength that they rip the planet into pieces …” it might be useful to note that the Roche limit is achieved when the tidal forces exceed the gravitational force of the planet. This makes it clear that “tidal forces” have to act oppositely on opposite sides of the planet — as does the planet’s gravitational force. In fact, I got the impression that someone reading the article might conclude that tidal forces should cause people on the sunny side of the planet to fall off of the planet. The reason this doesn’t happen is that the tidal force is smaller than the planet’s gravitational force.
By the way, I teach this subject.

Asmilwho
February 16, 2014 2:17 am

there is a correct explanation of tides, written by some scientists involved in that kind of thing, here:
http://tidesandcurrents.noaa.gov/restles1.html

February 16, 2014 3:05 am

Ulric Lyons says:
February 15, 2014 at 6:54 pm
“I was remarking upon the described 54.7° angle rather than the diagrammatic representation of it.
It makes a sqrt 2, sqrt 3, sqrt 5 sided triangle.”

February 16, 2014 3:10 am

Asmilwho says:
February 16, 2014 at 2:17 am
“there is a correct explanation of tides, written by some scientists involved in that kind of thing, here:
http://tidesandcurrents.noaa.gov/restles1.html
I know, I provided it to Willis on his previous thread about Tides. I don’t think it got through then.
And all this
“We must concentrate on the Tide Raising Force”
In an abstract way it IS interesting. But in a minutia sort of way really. How the planet reacts to that tiny force and the massive outcomes that occur here on the surface and below the surface is the real source of interest.
http://i29.photobucket.com/albums/c274/richardlinsleyhood/TheInternalTideatHawaii_zps7c7d5dbf.png
http://i29.photobucket.com/albums/c274/richardlinsleyhood/M2Tides_zps758f7faa.png
http://i29.photobucket.com/albums/c274/richardlinsleyhood/K1Tides_zps453d8381.png

February 16, 2014 3:16 am

clivebest says:
February 16, 2014 at 12:57 am
“The tractional force acting along the surface and therefore unaffected by the earth’s gravity is
Fy cos(theta) + Fx sin(theta) . The tractional force is zero at the central bulge and equal and opposite from either side.”
I have tried to point this out to Willis already.
The important thing here is that this tractional force is not a delta on a larger force, Gravity. That is what the vertical component Willis is so keen on concentrating on is. A delta, a small difference. One capable of affecting the planet in a way that is difficult to miss but a delta none the less.
The tractional or tangential to the surface force is NOT a delta. There is nothing opposing it. If operates in its full magnitude which makes it much more interesting IMHO.

February 16, 2014 3:22 am

Coldlynx says:
February 16, 2014 at 12:48 am
“That is valid only 2 days a year for the sun due to earth tilt. All other days of the year is the “Sun tidal zero poles” not at earth geographics poles. Moon inclination will also have large impact on where on earth the “moon tidal force zero poles” are located. And also where on earth the maximum horisontal tide forcing are located. The horisontal forcing that actually move air and water according to Laplace.
That latitude change between 45°N and 45°S +/- 28.36°
In a cyclic way. 18.6 years the most obvious.”
Again, something that has been pointed out to Willis before on another thread, by me and Greg.
On each occasion he has seen the force pattern, drawn on a vertically oriented globe and looked at it as though that was the true picture.
What he has never done is drawn the North Pole to South Pole line through that globe and considered how that changes the picture.

February 16, 2014 3:24 am

Willis Eschenbach says:
February 15, 2014 at 11:41 pm
Steve Fitzpatrick says:
February 15, 2014 at 12:48 pm
“Steve, neither Leif or I are talking about actual tides on earth. We’re talking about how the tidal forces work on a planet. So whether Clive has it correct or not for the earth is material for a discussion of the actual Earthly tides … it means nothing to this discussion.”
Concentrate on what I’m talking about – not on how this is all relevant to the planet. We are off in la la land.

February 16, 2014 3:27 am

Willis Eschenbach says:
February 15, 2014 at 11:30 pm
“And at the poles, that difference is zero, so the tidal force is zero.”
If, and only if, the Earth’s rotational axis and orbit were not tilted to the line of both of the items in question, the Moon and the Sun.
If everything was vertical, then things would be trivially simple and not nearly as complicate as they really are.

February 16, 2014 3:32 am

I have given this to Willis before
“And this for explaining why it is not due to centrifugal forces :-)”
http://www.lhup.edu/~dsimanek/scenario/tides.htm

February 16, 2014 3:45 am

And this provided by Greg, on the other thread about tides, about how the heat (missing or otherwise) dances to a tidal pattern down deep in the oceans when no-one is looking 🙂

February 16, 2014 3:47 am

EDIT – damn ” got in there by mistake.
And this provided by Greg, on the other thread about tides, about how the heat (missing or otherwise) dances to a tidal pattern down deep in the oceans when no-one is looking 🙂

Coldlynx
February 16, 2014 4:03 am

Add to that the earth moon barycenter is only average 1707 km below earth surface. Compare with earth radius of 6,371 km.

February 16, 2014 4:19 am

Coldlynx says:
February 16, 2014 at 4:03 am
“Add to that the earth moon barycenter is only average 1707 km below earth surface. Compare with earth radius of 6,371 km.”
You do have to keep a sense of proportion here though when talking about the barycentre. Sure the whole thing does revolve around that point. But at a rate that is 28 times slower than it is running round the Earth central spin axis.
So the effect is spread out over that much faster rate and the higher rate centripetal force is correspondingly that much higher as well. 28 times as much in fact.

February 16, 2014 4:32 am

clivebest says:
February 16, 2014 at 12:57 am
“Yes you’re right
The correct equation is more complicated. Taking x as the direction from the centre of the earth to the moon and taking y the vertical axis then the net tidal force has 2 components.
Fx = Gm( \frac{ \cos \phi}{R^2 + r^2 – 2rR \cos \theta} – \frac{1}{r^2})
Fy = \frac{ -Gm \sin \phi}{R^2 + r^2 – 2rR \cos \theta}
where
\cos \phi = \frac{r-R \cos \theta}{\sqrt{R^2 + r^2 – 2rR \cos \theta}}
and
\sin \phi = \frac{R \sin \theta}{\sqrt{R^2 + r^2 – 2rR \cos \theta}}
The tractional force acting along the surface and therefore unaffected by the earth’s gravity is
Fy cos(theta) + Fx sin(theta) . The tractional force is zero at the central bulge and equal and opposite from either side.”
There are three components to the tidal vector space as oriented to the Earth’s surface at any single point on that surface.
There is x, a vertical up and down vector that varies from 1 day to 1 year depending on Latitude.
There is y, a North/South vector that varies likewise
and z, an East-West vector that also varies likewise.
Now that will turn into a 3 channel, 3 colour normalised Mollweide projection movie of how this all varies over the 4 * 18.6 year solar/lunar cycle.
Any one with movie making skills out there?

February 16, 2014 5:07 am

Willis Eschenbach says:
February 16, 2014 at 4:42 am
“RichardLH says:
February 16, 2014 at 3:16 am
The important thing here is that this tractional force is not a delta on a larger force, Gravity. That is what the vertical component Willis is so keen on concentrating on is. A delta, a small difference. One capable of affecting the planet in a way that is difficult to miss but a delta none the less.
The tractional or tangential to the surface force is NOT a delta. There is nothing opposing it. If operates in its full magnitude which makes it much more interesting IMHO.
I fear I don’t understand what you mean by “a delta on a larger force”. Usually, a “delta” means a change in something, that is to say the value at time 2 minus the value at time 1. Obviously, you are using “delta” in some non-conventional way.”
No in its normal usage. That is a small relative change in a larger force. Sort of like ‘Normals’ rather than ‘Absolutes’!

February 16, 2014 5:08 am

EDIT: Make that No in its normal usage. That is a small relative change in a larger force. Sort of like ‘Anomalies’ rather than ‘Absolutes’!

February 16, 2014 5:15 am

Willis Eschenbach says:
February 16, 2014 at 4:42 am
“It seems that what you are calling a “tractional force” is the horizontal component of the tidal force. It is not a separate force. A force can be divided into a vertical and a horizontal components, but this is a vectorial representation, not separate forces. The horizontal and vertical components of the tidal force are real, but they are not separate forces.”
I was re-using the terms used by others so as to not add further complications….but you’re getting there.
“At any instant, at two locations on the earth (points A and B in the graphic above), the tidal force is pointed directly vertically. Everywhere else, the tidal force is pulling at some angle to the surface. Using vectors we can divide this angled tidal force into a vertical and a horizontal component.”
As I have been trying to point out to you for the last week or so.
“Very near the spots A and B in the graphic above, the tidal force is almost directly vertical, and the horizontal forces are smallest. Now given that, you’d think that near the north and south poles (top and bottom of the earth in the above diagram), the horizontal component of the angled tidal forces would be the largest. And you’d be right.
The problem is that as you move towards the poles, the horizontal forces indeed are a larger and larger component of the tidal force … but at the same time the tidal forces themselves become smaller and smaller as you move towards the poles.”
But now we are not adding or subtracting a small delta from a larger constant force. We are balanced on the centre of a see-saw. The slightest change around the centre will have a large effect on the position of the plank.
That is why you need to consider the 45-60 degree band as relative to the two orbital planes and its changes.
The force is not opposed by any other force.

February 16, 2014 5:23 am

Willis Eschenbach says:
February 16, 2014 at 4:42 am
“Very near the spots A and B in the graphic above, the tidal force is almost directly vertical, and the horizontal forces are smallest. Now given that, you’d think that near the north and south poles (top and bottom of the earth in the above diagram),”
WRONG. This is only true if the North/South Earth axis line is at the same angle as the orbital planes. That is only true for 2 days a cycle, one each for the orbital planes on Sun and Moon. That damn Saros cycle again!

February 16, 2014 5:32 am

Willis: See
http://www.nfo.edu/tilt.jpg
But you knew that right?

February 16, 2014 5:54 am

Willis:
Let me try to come up with a presentation that you can see what I mean.
Take some heavy planks (like your buckets)
Align them vertically on the Earth’s surface suspended/pivoted in a gimballed/bungee cord sense so that they float just above the surface.
Assume that the Earth, Sun and Moon are orbit/spin axis wise aligned to start with.
Now watch the planks as things rotate.
At the Equator they bounce up and down on a yearly, monthly and daily pattern.
At the Poles they wave left to right (say) and bounce up/down on a yearly and monthly pattern.
At 54.7° angle to the orbital plane they only wave left/right on a yearly, monthly and daily pattern, no vertical bounce at all.
Now make the orbital planes and the spin axis how they are in reality and…..

February 16, 2014 5:58 am

EDIT: Damn it.
At the Poles they bounce up/down on a yearly and monthly pattern.

Mike M
February 16, 2014 6:00 am

Jan Kjetil Andersen says: February 15, 2014 at 11:04 pm ” Perhaps we should concentrate on answering Mike’s concrete question rather than evading to something that some may perceive as quite arrogant.”
Thank you!!

eyesonu
February 16, 2014 7:17 am

Guest Post by Willis Eschenbach
“Short Post. You can skip this if you understand the tidal force. ……”
==============
You made it sound as if it were going to be a simple discussion. There have probably been a lot of lurkers on this one.

February 16, 2014 7:56 am

Willis: See Fig903d and its accompanying text of how the Moon’s declination causes variation in your simple plots above.
From the NGA (so as you don’t think this is some random site I have plucked from Google)
msi.nga.mil/MSISiteContent/StaticFiles/NAV_PUBS/APN/Chapt-09.pdf
http://msi.nga.mil

Coldlynx
February 16, 2014 8:19 am

“As the planet free-falls into the sun, the near side of the ocean is pulled the most. So it moves the most, pulling away from the planet.”
No, water will not move vertically. Only horisontally. The bulges are result of water flow by the horisontal tidal forcing. Mr. Laplace got that right already in the 18th century.
http://en.wikipedia.org/wiki/Theory_of_tides#Laplace.27s_tidal_equations
In fact is the near side the place where it moves the least.

wayne
February 16, 2014 8:27 am

Greg, of course, the missing unit mass factor. Thanks for the reply, should have caught that myself.

Bart
February 16, 2014 9:26 am

clivebest says:
February 16, 2014 at 1:22 am
“You don’t need to use the centrifugal force to derive the formula for tides.”
As I have demonstrated in my previous comment, you do, and you don’t.
You do not need centrifugal “force” to get a symmetric bulge. The bulge occurs on both sides in equal measure because gravity is stronger on the near side than it is at the center, and it is weaker on the far side than it is at the center. Relative to the center, this produces stress inducing acceleration relative to the Earth center which is towards the gravitating body on the near side, and away from it on the far side.
You do need centrifugal “force” to get the precise formula, though the effect could be said generally to be second order. It is more important for solar induced tides on the Earth’s oceans than it is for lunar induced tides on the Earth’s oceans, or for other cases where the distance from the barycenter of the orbiting system, of the heavenly object experiencing tidal forces, is relatively small. The centrifugal “force” adds to the stresses which produce the tides because it is weaker on the near side than at the center, and it stronger on the far side than at the center. Because the centrifugal “force” is directed outward rather than inward, this has similar impact to the gravitational effect: stress inducing acceleration relative to the Earth center which is towards the gravitating body on the near side, and away from it on the far side.

February 16, 2014 9:33 am

Bart says:
February 16, 2014 at 9:26 am
“You don’t need to use the centrifugal force to derive the formula for tides.”
As I have demonstrated in my previous comment, you do, and you don’t.

No, Bart, the centrifugal force has nothing to do with it. You went very quiet when asked about calculating the tidal effect on a planet falling straight into the Sun [Willis’ example]. Try again.

Kevin Kilty
February 16, 2014 9:56 am

lsvalgaard says:
February 15, 2014 at 10:08 am
Kevin Kilty says:
February 15, 2014 at 10:06 am
I mean centripetal not centrifugal in my last post
It doesn’t matter as the tidal effect has nothing to do with rotation or orbital movements.

Isvalgaard, you always appear to me more deliberative than you seem here. If one treats the rotating Earth-Moon system as an inertial frame, then the rotation of the frame does indeed have to do with the tides because of pseudoforces (centrifugal force) than appear on the left side of the equation of motion. You aren’t arguing with me on this point, but rather with Lagrange, Laplace, D’Lambert, and Geo. Darwin. Good luck with that.

February 16, 2014 9:58 am

Bart says:
February 16, 2014 at 9:26 am
For a full explanation of why centripetal is the wrong way to go see
http://www.lhup.edu/~dsimanek/scenario/tides.htm
“So what about those centrifugal forces so many books make such a fuss about? You’ll notice we never mentioned them in our simple explanation. Should we have?
Many misleading accounts of the tides result from a common confusion about centrifugal effects due to rotation. “

February 16, 2014 10:00 am

EDIT: Sorry should have been “why centripetal/centrifugal is the wrong way to go see”

yirgach
February 16, 2014 10:37 am

Interesting 2005 monograph on Tides and centrifugal force by the astronomer Paolo Sirtoli.
Analyzed from several different Reference Frames, all with the same result.
At many websites, but also in some textbooks, one finds plainly wrong explanations of the lunar and solar tides. In particular much confusion arises when authors try to explain the existence of the second tidal bulge, the one opposite to the Moon. Often they invoke the centrifugal force as an “explanation”. But centrifugal force is a fictitious force, and we can’t justify a real effect with a fictitious force, can we?
The necessary and sufficient reason for the tides is the fact that over the volume of the earth, the gravitational force (from Moon or Sun) has different size and direction over the earth’s volume as a result of the varying distance from the external body. This is the “differential” force, sometimes called the “tidal force” and is the gradient of the force due to the external body. The centrifugal force is required only when the gravitational problem is done in a rotating reference frame. Often it is incorrectly used, even when its use is appropriate to such a rotating frame. It is a quantity necessary for accounting purposes, without any real effect on tides.
We will demonstrate this by considering different reference frames, rotating and non-rotating. We will calculate the direction and the intensity of the total acceleration for just two trial points, the well known A and B in figure 2. They are, respectively, the sublunar point and the anti-lunar point. If the results are identical to the those emerging from the case in wich there is only the gravitational force, we may rule out the centrifugal force as necessary to explain tides.
Analysing the commonest incorrect explanation, we find that that confusion arises from the incorrect choice of the reference frame (RF). In physics, the first step to successfully deal with a problem, is the choice of the RF. If mistakes are made in this starting phase, one will surely arrive at plainly wrong results. In physics, as in many other systems, the “GIGO” rule applies: Garbage In, Garbage Out.
Another misleading thing is that to get the tidal field, we must subtract from the moon’s gravity a uniform force field. But we learned in college that when we deal with rotation, the fictitious forces are proportional to the distance from the rotation’s axis. So where does this uniform field arise? All this will be revealed in the examples which follow.

http://vialattea.net/maree/eng/index.htm