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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I really do not know what point you are trying to make.
Earth orbital round Sun = 149.59787e9*(1.99e-7)^2 = 0.0059 m/s^2
Bart:
So you have the orbital round the Sun centripetal force greater than the size of centripetal force at Earth’s surface due to Earth’s circular motion around the Earth-moon barycenter at 3.3 x 10-5 m/s2?
How do you manage that?
Given that Earth axis rotational centripetal force is 0.034 m/s^2, are you seriously suggesting that the orbital figure is that close to that rotational figure and larger than the barycenter figure? I suspect you have the maths wrong somewhere.
These are deltas against the 9.8 m/s^2 in any case which will triumph in all cases. You can barely detect the difference between being weighed at the poles and the equator. You need a really good set of scales. To detect the differences then added by any barycentre movement and Earth orbital values gets to an even higher degree of precision.
Also you seem to have failed to notice that any such values that do exist will be expressed almost entirely in the rock, the oceans response to all this is even tinier still. 21-23 km (depending on sources) of Earth rotational is in the rock after all. We would still have that oblate spheroid even if there were no oceans at all.
Bart: I apologise – the value you give is correct – the point about it being expressed in the rock stands though. Any water on top of the rock will see very little.
Define “very little”. It is 1/3 of the total solar tidal differential force at the sub-solar point and its antipode.
21-23 km of the Earth’s rotational budget in the rock as against .3m or so in the oceans from the same source. You do the maths for the rest but it is very tiny.
Sorry that should be 0.03 m, I missed a decimal place.
i.e. Earth rotation of 0.034 m/s^2 = 21-23 km in the rock, about 0.03m in the ocean above the rock..
I think you are getting lost in the forest. It is 1/3 of the solar tidal forcing. Not particularly small.
There are certainly effects on ocean dynamics due to the spin of the Earth beyond the orbit rate, but that is another subject entirely.
Look, the only thing I care about here is my statement above. If you have a problem with that, state so and why. Otherwise, you are engaging in discussion with which I am not particularly interested at this time.
Bart:
You obviously have decided you have lost and are walking away.
The majority of the centripetal is in the rock (from what ever source). The Earth Tide is much larger than the Ocean tide (for the same reason).
As I said up thread, there are 6,378.14 km of rock and around 3.4 km of water. Where do you expect the majority of the forces to show up and why?
Please come back with figures that show that to be wrong.
RichardLH says:
February 27, 2014 at 1:55 am
Lost what? I made a correct argument which you and others have in no way impeached. That is all I ever cared about. I’m not interested in further details. I made a statement. The statement was correct. The statement was challenged. The challengers were wrong. That’s it.
Bart:
I too have made an observation which you have very carefully failed to address. The majority of ALL forces show up in distortions in the rock, not the water.
You have not and cannot address that as it throws your whole case into failure.
9.8 m/s^2 triumphs and the rest is barely able to raise a ripple. The main centripetal rotational figure causes the largest outcome and that is most definitely in the rock. Then we have the two tides, again mainly in the rock. Then somewhere down in the dust we have whatever other forces you believe are important as very tiny modifiers of those. And, again, most of those show up in the rock.
Until you come to terms with that state of affairs you are most definitely lost.
Bart,
You are obviously very intelligent as you have made quite detailed arguments trying to support your view. If I may suggest, go back to the beginning of this thread and follow my comments and the responses/comments to those comments by others. Watch my learning curve however vaguely presented.
Now try my simple thought experiment as noted in comment ” eyesonu says: February 21, 2014 at 10:59 am”.
Follow the remaining comments by me. A lot of warts in my presentation (to say the least) but It was quite an eye-opener when I grasped what Willis and Leif had presented.
To follow/interpret what I offered would take much less of your time than many of your arguments. I understand your reason for your arguments as that point of view is where I started out.
Note that I am here on WUWT as my only motivation is to understand things presented here. If I had more time and motivation I could have been much better with the written composition of my comments. But then keep in mind that I’m just a country bumpkin.
Just trying to help as it is clear that you have an interest. This thread is now two weeks old and almost 500 comments. No one would still be here if not interested. Hope this helps.
RichardLH says:
February 28, 2014 at 2:26 am
“I too have made an observation which you have very carefully failed to address.”
Don’t be paranoid. I have very uncarefully failed to address it. I do not know if you are wrong or right. I do not even know specifically what you are arguing, and have no real desire to know. I have been focused on showing that my argument above is correct. It is. That is all I care about. If you want to engage on that topic, I can oblige. If not, you should probably find someone else to argue with. This entire thread is going to expire any day or moment now, and there is no time to get involved in a protracted additional discussion anyway.
eyesonu says:
February 28, 2014 at 7:22 am
Thank you for your concern, but please take my word for it that I have actual, real world experience with these things, and know very well of what I speak. Better than these other guys do. If you find yourself disagreeing with me, then you are either A) correct, and there exists a 1:1 mapping with what I am saying so that we are actually both correct or B) wrong.
The fundamental thing people have got wrong, it seems, is in assuming the entire Earth is in free fall about the Sun. It isn’t. Only the center of gravity is in free fall. The rest of the body is being pulled along for the ride, and this creates internal stresses.
It is the same with your three-mass problem. The center of gravity of the three masses together is in free fall, but individually, they are not. Each mass is carving out its own trajectory in space and time relative to that center of gravity. They are not being equally pulled by gravity, and they are not uniformly accelerating through space.
Bart says:
February 28, 2014 at 9:35 am
“…. please take my word for it that I have actual, real world experience with these things, and know very well of what I speak. Better than these other guys do ….”
=================
I believe I understand clearly now.
eyesonu says:
February 28, 2014 at 11:17 am
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Great thread. I see that this is a save for future study thread. I,m glad I noticed your last post. By the way, Bart,s last comment is very interesting.
On another note, when I moved into the mountains in the early 70s, I became very good at making hasenpfeffer.