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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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.

ferdberple
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.

RichardLH
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.

RichardLH
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?”
The answer you seek is in Google.
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.

RichardLH
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.

RichardLH
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.

RichardLH
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.

RichardLH
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: …
So with all your links and education why can’t you just ANSWER 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
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.

RichardLH
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?

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