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.

tidal force on three bodiesFigure 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:

tidal force is stretchingFigure 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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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.. ??

Tim Folkerts
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….”
That answers my last question…
(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….”
That answers my last question…
(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).

Reply to  stevefitzpatrick
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 …
—————————
You avoided answering the question.
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?

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