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.
RichardLH says: February 18, 2014 at 8:50 am
Earth’s average orbit = 149,597,870.00 kilometres
Earth’s equatorial radius = 6,378.14 kilometres
Remind me again about how much difference this makes as to the forces involved.
************************
I said 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
****************************
I don’t have the other numbers in front me at my current location but as stated above, the difference in earth orbit centripetal acceleration was about 3% which yielded an acceleration difference of .00018 M/sec^s at noon and midnight on the equator.
(Bart did it with rotational vectors and I did it with velocities – same concept.)
I just now realized that how I earlier described how this effect drops off as you move from the equator toward either pole was in error. I think I stated per cosine(latitude) but now see it drops off per [cosine(latitude)]^2 where Willis’ effect drops off more gradually as cosine(latitude).
lsvalgaard says:
February 18, 2014 at 3:44 am
“Is an irrelevant straw man, only used to give an example of a possible rotation rate.”
It can hardly be irrelevant when it’s the case I’ve been arguing.
Mike M. says:
February 18, 2014 at 8:22 am
We may be arguing along the same lines. Will have to review. More later.
Bart says:
February 18, 2014 at 9:51 am
“Is an irrelevant straw man, only used to give an example of a possible rotation rate.”
It can hardly be irrelevant when it’s the case I’ve been arguing.
You are laboring under the false assumption that your argument is relevant. Let me explain: If a rock pile [the moon approaching the planet] is rotating fast enough it will fly apart regardless of the tidal effects. If it does not rotate at all, then only the tidal effect plays a role. In the intermediate case both the tidal effect and the rotation work in concert to make the moon break up. Under the assumption that the rotation rate is that of a tidally locked moon [not a bad assumption], the rotational role of the breakup is to raise the factor from 2 to 3. This has nothing to do with the orbital centripetal acceleration in causing tides, and is thus irrelevant.
I am amazed that this thread broke out in full argument last night again. Mike M, Bart, and I have been arguing that there are alternative ways to look at the tides, and that the pseudoforces like “centrifugal” can be used in a consistent explanation. Willis has correctly explained the tides with his rope model, but seems insistent that other models making use of centrifugal force are just plainly wrong. Edwin Smith has stated that “gravity” is the entire story to the tides. It is not. This would imply that the tides are a problem in engineering statics, but in fact they need an acceleration of some sort to complete the explanation. They are dynamical phenomena.
The tides do not just slosh across the ocean basins, but actually circulate around amphidromic points in the various ocean basins, and even in lakes. You see, the “coriolis force” another of the pseudoforces that arise from treating the Earth as an inertial frame, is important to the complete picture as well.
Willis, if I could make a suggestion that could become the basis of an interesting thread, numerous textbooks, popular science books, and even articles from Scientific American make the claim that enhanced tides in places like the Bay of Fundy, result from resonance. Not so. A far more correct explanation is available if one simply examines the shape that most estuaries and bays obtain. Are you game to tackle this?
Kevin Kilty says:
February 18, 2014 at 10:22 am
This would imply that the tides are a problem in engineering statics, but in fact they need an acceleration of some sort to complete the explanation. They are dynamical phenomena.
There seems to be some confusion about this. The ‘static’ part has to do with the tidal forcing, and the ‘dynamical’ part has to do with the response of the oceans to that forcing. This was understood and explained by Laplace way back in the middle of the 18th century and is thoroughly discussed in http://www.leif.org/EOS/Munk-Tides-1966.pdf
Mike M. says:
February 18, 2014 at 9:40 am
“I don’t have the other numbers in front me at my current location but as stated above, the difference in earth orbit centripetal acceleration was about 3% which yielded an acceleration difference of .00018 M/sec^s at noon and midnight on the equator.”
And that compares to
Angular velocity of Earth’s axial rotation in radians per second = 7.29 x 10^-5 rad/s
Angular velocity of Earth around the Sun in radians per second = 1.99 × 10^-7 rad/s
I am not sure how you got to that figure.
Remember the Earth rotates against the orbit on the near side. (looking from above as conventional).
That axial rotation just HAS to be the largest component in all this. Then you can add/subtract the two orbital components. You may be right in that the differential dark/light is bigger than the orbital but I am not sure that you are right, I think it is the other way round.
lsvalgaard says: February 18, 2014 at 9:59 am
If I was able to prove to you using a sensitive (and appropriately highly damped to take out fast seismic signals) accelerometer or other such apparatus that things weigh less at the equator by a factor of .00018 Newtons per Kg maximum at noon and midnight versus what they weigh at 6am and 6pm …
Would not such a measurement provide a valid reason for having two bulges at both noon and midnight and it’s magnitude exclude it’s existence being caused by the much smaller effect per WIllis?
In your link, http://www.gps.caltech.edu/classes/ge167/file/agnew_treat_tide.pdf on pub page 171, PDF page 9,
Doodson’s harmonic analysis clearly shows a fairly strong semidiurnal tide factor with a period of exactly 2.000000 that he associated with the sun. Something is causing it and part of that is WIllis’ effect but if measured to be a lot stronger then ~something else~ is much more responsible.
Mike M. says:
February 18, 2014 at 9:40 am
Centripetal (outwards) acceleration at Earth’s surface due to Earth’s axial rotation = 0.034 m/s^2
Centripetal (outwards) acceleration at Earth’s surface due to Earth’s orbit around the Sun ~ 0.0000952 m/s^2
How you get the tiny differential between orbit + radius and orbit – radius centred around orbital to be bigger than the orbital itself is a mystery to me.
Mike M. says:
February 18, 2014 at 10:35 am
Doodson’s harmonic analysis clearly shows a fairly strong semidiurnal tide factor with a period of exactly 2.000000 that he associated with the sun. Something is causing it and part of that is WIllis’ effect but if measured to be a lot stronger then ~something else~ is much more responsible.
Your question is sufficiently obscure that I’m not sure what you are hinting at. Perhaps our resident mathematical genius can interpret what you said and provide a compelling answer.
RichardLH says: February 18, 2014 at 10:33 am
“You may be right in that the differential dark/light is bigger than the orbital but I am not sure that you are right, I think it is the other way round.”
Bigger than orbital!?! I never stated anything of the sort. I stated: Earth’s average orbit velocity V1 as 29885 m/sec. Crust WRT center V2 as 463 m/sec. (WRT EARTH’S center)
Solar orbital velocity DIFFERENCE is 463m/sec (minus at noon, plus at midnight) – so either subtract it or add it to/from 29885 m/sec.
RichardLH says: February 18, 2014 at 10:40 am “Centripetal (outwards) acceleration at Earth’s surface ….” BLAH BLAH BLAH!!!!
This is the 2nd time I’ve told you that has NOTHING to do with this (and I think so has Bart).
I already took it away from you anyway! Go back and read my free body description February 17, 2014 at 5:25 am I describe the difference in the centripetal required for a particle traveling along earth’s orbital path at a slower or faster speed but WITHOUT any earth at all let alone a spinning one.
Stop bringing it up, it is TOTALLY non sequitur to the description.
Mike M. says:
February 18, 2014 at 10:56 am
So we have two particles – non rotating – in orbit around the Sun.
One at 149,597,870.00 + 6,378.14 kilometres
One at 149,597,870.00 – 6,378.14 kilometres
Centripetal (outwards) acceleration at Earth’s centre due to Earth’s orbit around the Sun ~ 0.0000952 m/s^2 at the mid point between them.
You really think that the tiny differences from that mid point for the far and the near particle given their differences in orbital values are significant?
When compared to G 9.8 m/s^2 (holding particles onto the surface) and that Centripetal (outwards) acceleration at Earth’s surface due to Earth’s axial rotation = 0.034 m/s^2
I tell you what – do the vector sums and come back with the figures. If they are anything other than up/down to Earth centre on almost any scale of graph you can construct…..
Dudley Horscroft says: February 16, 2014 at 8:33 pm “BTW, the earth and the moon do not rotate around each other, or around the barycentre. ”
Quick, tell NASA not to launch Apollo 8! … oh… too late ….
http://upload.wikimedia.org/wikipedia/commons/5/5c/Apollo-8-mission-profile.png
RichardLH says: February 18, 2014 at 11:32 am
So we have two particles – non rotating – in orbit around the Sun.
One at 149,597,870.00 + 6,378.14 kilometres
One at 149,597,870.00 – 6,378.14 kilometres
**********************
Wrong. They are BOTH at 149,597,870.00 km
Try again…
Mike M. says:
“We’ve already been over this. Gravity is pulling you down and the inertial reactive force is the ground pushing you up – equal and opposite. Take the latter one away and down you go!”
It would be helpful if you used standard names. Above you are referring to the normal force or reactive force. Using the vague term “inertial reaction” makes it very hard to know what you are talking about.
“How about I construct a trap door over a vertical mine shaft, have you stand on it and then you scream to me about non-inertial reference frames after I pull the lever?”
This sentence is absolutely devoid of meaning. The same phenomenon can be described in either inertial or non-inertial frames of reference. One or the other is chosen for mathematical convenience, but you can not mix the two as you seem to do.
“If you are on rusty roller skates and I’m pushing 50 pounds against you to keep you moving at 2 feet/sec I feel that same 50 pounds, as a reaction, pushing BACK on me thus I am doing REAL work = F*V REAL heat comes from your rusty roller skate bearings in ANY frame of reference.”
You are right, it does apply in all reference frames, although the method of approach will vary. As hard as I try I do not see what it is that you are getting at. By the way, please note that F*V is power not work.
I’m beginning to suspect that the problem is to be found in this quote from Wikipedia:
“Centrifugal force (from Latin centrum, meaning “center”, and fugere, meaning “to flee”[1][2]) is the apparent force that draws a rotating body away from the center of rotation. It is caused by the inertia of the body as the body’s path is continually redirected. In Newtonian mechanics, the term centrifugal force is used to refer to one of two distinct concepts: an inertial force (also called a “fictitious” force) observed in a non-inertial reference frame, and a reaction force corresponding to a centripetal force.”
You may be using the second definition when others are using the first, and more common, definition. I don’t think that this is the only problem though, as you do not seem to like non-inertial frames at all.
William Sears says: February 18, 2014 at 11:55 am ” You may be using the second definition when others are using the first, and more common, definition.”
Guess no longer, I am using the second one because it is the correct one and being “uncommon” is of no concern to someone like me.
William Sears:
I wrote, “I am doing REAL work” Being in the course of DOING work infers the expenditure of power. Sorry I wasn’t scientifically precise enough with the terminology to prevent you from using it for a cheap shot….
For folks frustrated by Google giant URLs: Don’t just “copy link” from a google search, click the link, then copy it from the browser bar. Or use a more polite search engine that doesn’t track tag and bag you, like this one: https://duckduckgo.com/?q=duckduckgo that does not feel compelled to drive you through their site with obscure URL tricks…
Mike M. says:
February 18, 2014 at 11:54 am
“Wrong. They are BOTH at 149,597,870.00 km”
So following each other in the same orbit?
Mike M,
Power vs Work: It wasn’t a cheap shot since there was nothing critical in what I said. I’m sorry that you took it that way.
“Guess no longer, I am using the second one because it is the correct one and being “uncommon” is of no concern to someone like me.” The correct one, really? You have to be able to communicate with other people and if the language is physics you have to know the terms. There was enough mention of fictitious forces and reference frames for you to understand the context. If all this was just a misunderstanding of terms then I expect that you will now withdraw your objections to my previous posts. Although I feel obliged to point out that the initial objection had to do with asteroids falling into the sun and there wasn’t any centrifugal force there, by either definition.
William Sears says: February 18, 2014 at 12:41 pm “There was enough mention of fictitious forces…”
I’m an engineer why should I give a rat’s rear end about “fictitious forces” when the real ones are plenty enough to keep me busy?
RichardLH says: February 18, 2014 at 12:27 pm “So following each other in the same orbit?”
Yeah… okay. It’s a circular orbit for the purpose of illustration so they can be on opposite sides of the sun for all I care.
lsvalgaard says: February 18, 2014 at 10:43 am “Perhaps our resident mathematical genius can interpret what you said and provide a compelling answer.”
Color me consigned to wait for such an eventuality. (My wife doesn’t understand me half the time either so I’m used to it.)
William Sears says: February 18, 2014 at 12:41 pm “I’m sorry that you took it that way. ”
Don’t be, I’m the one who was joking so I should be apologizing to you. Sorry.
We may be talking past one another here, but I don’t see that anyone on this thread denies that gravity is the “cause” (forcing) of the tides, but you, yourself, used the term “tidal effect” which suggests to me what we ultimately observe as the tides–i.e. things beyond the cause. There is no means of applying nothing more than gravity to explain what one observes of the tides. Willis included an acceleration of the three masses in his explanation, not just gravity, and as a linear acceleration, or as the centripetal acceleration in an orbit, these can be represented by a “fictitious force” (translational in the linear case, or centrifugal in the case of orbits). Every textbook I have ever consulted on dynamics, and I have taught the subject in both physics departments and in engineering programs for 22 years, has a chapter regarding non-inertial coordinate systems, and their conversion to an inertial approximation using fictitious forces–all of these forces have names–translational, azimuthal, centrifugal, coriolis. I don’t see that explanations invoking these fictitious forces are just plain wrong, and that is the one and only criticism I had of Willis’ essay.
If nothing else this essay has certainly set off a busy thread.