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.
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!!
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.
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
“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.
Greg, of course, the missing unit mass factor. Thanks for the reply, should have caught that myself.
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.
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.
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.
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. “
EDIT: Sorry should have been “why centripetal/centrifugal is the wrong way to go see”
RichardLH says:
February 16, 2014 at 5:23 am (Edit)
Richard, I have put several simplified situations up for discussion. As I have said over and over, these are NOT THE REAL EARTH. Please stop telling me ad nauseum that the real Earth is more complicated than the simplified systems under discussion in my post. I know that, Richard, truly I do. That’s why humans use simplified examples, to understand what’s going on in the simplified system BEFORE thinking about all the real-world complications.
Can we return to the subject under discussion, which are the horizontal and vertical components of the tidal forces? At the points nearest and furthest from the sun IN THE SIMPLIFIED FREAKIN’ SYSTEM SHOWN IN FIGURE 2, the horizontal component of the tidal force is approximately zero. In addition, this is where the tidal force is the strongest.
At the points nearest to the north and south poles IN THE SIMPLIFIED FREAKIN’ SYSTEM, the vertical component of the tidal force is approximately zero, what force exists is almost all horizontal. However, that’s also where the tidal forces are approximately zero.
Since the amount of the total tidal force is proportional to cos(latitude), and the horizontal component of that tidal force is proportional to sin(latitude), the horizontal tidal force is equal to the tidal force time sin(latitude) cos(latitude). This peaks at 45°N/S, at 0.5 times the tidal force. At both the equator and the poles IN THE SIMPLIFIED SYSTEM, the horizontal tidal force is zero.
w.
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
RichardLH says:
February 16, 2014 at 7:56 am
Richard, the point of my simple plots is to assist in understanding the basics of the system BEFORE we get to such complications as declination. Keep your mind fixed on Figure 2. I’m talking about the tidal forces on a non-rotating planet and ocean free-falling into the sun … patience, my friend. We may get to declination some day, but not today.
w.
yirgach says:
February 16, 2014 at 10:37 am
I can’t thank you enough for that, yirgach.
The link that you gave, to this discussion of the tides, is stupendous. It should be required reading for anyone interested in the subject. It examines the tides from a variety of frames of reference, complete with moving graphics, and it shows that all reference frames come to the same conclusion. For me the money quote was in the conclusion (emphasis mine):
Note that the fact that the earth is rotating on its own axis, or the fact that the Earth and Moon both rotate around their common center of mass, has absolutely nothing to do with tidal forces.
w.
PS—Curiously, while writing this it strikes me that one place where rotational forces would come into play is where a planet in free-fall into the sun is approaching the Roche limit. If it is spinning very fast, it will break apart sooner than a non-rotating planet … but other than that unusual situation, the rotation is immaterial to the tidal force.
Larry Niven wrote a short story based on this phenomena called “Neutron Star”. Of course, mentioning that in this context is something of a spoiler.
The tide opposite the Moon is from centripetal force as the Earth and moon revolve around their combined centre of mass.
Willis Eschenbach says:
February 16, 2014 at 11:22 am
The article essentially points out the fact that intuitive reasoning may be blindsided by reality.
You have to know enough to ask the right questions.
Thank you for providing that knowledge.
Servius (@Servius) says:
February 16, 2014 at 11:28 am
I would also recommend “Dragon’s Egg” by Robert L. Forward for a hard science look at life on a neutron star.
https://en.wikipedia.org/wiki/Dragon%27s_Egg
Willis Eschenbach says:
February 16, 2014 at 10:33 am
“Richard, I have put several simplified situations up for discussion. As I have said over and over, these are NOT THE REAL EARTH. ”
Fine. Except that you (and others) then try to draw direct conclusions from them. I am just trying to point out that you can draw no conclusions from them at all.
They are useful only to the point where they explain to others what they do may not properly understand.
Can you try and keep it to just that then and not reach further? Perhaps put a proviso up that this is NOT relevant directly to the Earth? An abstract talking point.
And not try in head posts to dismiss how others see that the complications are what make this relevant and interesting?
lsvalgaard says:
February 16, 2014 at 9:33 am
“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].”
No, Leif. You are trivially wrong. The equations are right up in my post above. The factor I get is shown in the Wiki page on the Roche limit.
I didn’t answer because it is already answered. The tidal effect for falling straight in is missing the additional component due to centripetal acceleration, so its a factor of 2 instead of 3.
Stop being stupid. Read the equations I provided. They are correct.
RichardLH says:
February 16, 2014 at 9:58 am
No, you are arguing against stevefitzpatrick’s position as detailed in his previous posts, and with which I specifically disagreed over a number of posts in reply. I think this is Leif’s misconception as well, and he refuses to read my previous post to see my position, and just assumes he knows my argument.
I do not know how many times I can say this with nobody paying any attention, but let me state quite clearly for the umpteenth time: THE CENTRIPETAL ACCELERATION IS NOT RESPONSIBLE FOR THE SYMMETRY OF THE TIDAL BULGE. IT WOULD BE SYMMETRIC REGARDLESS. THE CENTRIPETAL ACCELERATION MERELY ADDS TO THE BULGE.
This is very annoying. Read my explanation of what is happening before assuming you know it.
Willis Eschenbach says:
February 16, 2014 at 2:30 am
“That’s big news, Carl, that there exists a repulsive gravitational force, that instead of attracting the water towards the sun, actually moves the water away from the sun…”
No, no, no, no, no. This is an ACCELERATED coordinate system. The center of the Earth is accelerating toward the Sun. The differential acceleration of the ocean on the far side away from the Sun is, indeed, away from the Sun.
If does not mean that the ocean on the far side is being pushed away from the Sun. It means it is being pulled toward the Sun LESS than the Earth is being pulled toward the Sun. The net effect is indistinguishable from an incremental repulsive gravitational force, or rather, a less attractive overall gravitational force.
Willis Eschenbach says:
February 16, 2014 at 10:33 am
As in
“The 54+ year cycle gets a lot of airtime, because people claim it is reflected in a sinusoidal approximately 54-year cycle in the for example the HadCRUT temperature records.”
where you most definitely tried to link on of your toy, abstract, (over?) simplified discussions to draw very definite conclusion that others were WRONG, WRONG, WRONG (and called me lots of names into the bargain).
I am quite prepared to forgive and forget but please don’t patronise me.
I have to leave, so I’m going to leave a final note. My equations are right. My interpretation of what is going on is right. If anyone thinks I am wrong, then they are either themselves wrong, or are addressing an argument I am not making. So, read what I have written before you imagine you think you know what I am claiming.
what make this relevant and interesting?
Bart says:
February 16, 2014 at 12:27 pm
“RichardLH says:
February 16, 2014 at 9:58 am
No, you are arguing against stevefitzpatrick’s position as detailed in his previous posts, and with which I specifically disagreed over a number of posts in reply. I think this is Leif’s misconception as well, and he refuses to read my previous post to see my position, and just assumes he knows my argument.
I do not know how many times I can say this with nobody paying any attention, but let me state quite clearly for the umpteenth time: THE CENTRIPETAL ACCELERATION IS NOT RESPONSIBLE FOR THE SYMMETRY OF THE TIDAL BULGE. IT WOULD BE SYMMETRIC REGARDLESS. THE CENTRIPETAL ACCELERATION MERELY ADDS TO THE BULGE.”
As I have never, ever claimed otherwise you’ve lost me.
I have provided countless links to other web sites that show precisely just that outcome. centripetal/centrifugal forces are almost totally irrelevant in tidal calculations.
Do you read anything I post or do you just assume I wrong regardless?
Willis,
“Since the amount of the total tidal force is proportional to cos(latitude), and the horizontal component of that tidal force is proportional to sin(latitude), the horizontal tidal force is equal to the tidal force time sin(latitude) cos(latitude). This peaks at 45°N/S, at 0.5 times the tidal force. At both the equator and the poles IN THE SIMPLIFIED SYSTEM, the horizontal tidal force is zero.”
Have read about seventeen pages thoroughly in my trusty astronomy book so I’m closer to trust my views on this topic. You seem to be limiting this entire discussion as if drawn on a 2d graph and when you say “… the tidal force time sin(latitude) cos(latitude)” I have to stop right there. Where is the longitude in that equation? Because your “SIMPLIFIED SYSTEM” is strictly 2d and not a water-only Earth? Seems kind of right but in 3d it seems to fail. At the equator 45° either side of the zenith point would not the tangential be also 0.5 that way? And at 45°N 45°E would it not be 0.25. That doesn’t seem correct at all.
I seem to get this… tangentially, as “the tidal force time √(sin²(long.azimuth°) + sin²(latitude))”, a circle on the Earth’s surface 45° any direction from the zenith point (just assume 0°N 0°E to use longitudes) and there would be 1.0 tangentially, at the zenith point zero and at 90° any direction from zenith point once again, zero.
If so I’m now following what you are saying. If not, what do you disagree with? Didn’t see anyone else addressing that.
RichardLH says:
February 16, 2014 at 12:32 pm
“As I have never, ever claimed otherwise you’ve lost me.”
I was stating my position, not imputing yours. We are in effing agreement on this. Why are you arguing?
“centripetal/centrifugal forces are almost totally irrelevant in tidal calculations.”
No, they are not. There are not the largest contributor, in fact are insignificant for any but solar induced tides on the Earth. And, they are NOT the cause of the symmetry in the bulge.
But they are not insignificant. Read what I have written. The centrifugal “force” is stronger on the far side than on the near side. This causes additional tidal deformation.