# General relativity and tidal forces

#### altergnostic

That's not possible, at least not if the orbits are circular
You are correct, maybe I oversimplified the problem. When I said "started to orbit" I wasn't implying the bodies would complete the orbit, but then I negligently stated that the inner body would complete one revolution before the outer body, which deserves your criticism. What I should have said is that, since the outer body has to travel a longer distance in order to stay aligned with the inner body towards the central mass (conjunction configuration), it would get behind the inner body because they have the same linear velocities.

Now look at the torque around the center of mass of the body-spring system when the outer object is leading and when it is lagging; we know that there will be some torque because both the direction and the strength of the forces on the two bodies is different.

When the outer body is leading, that torque acts against the rotation, and when the inner body is leading that torque tends to act with the rotation. That is, when the body-spring system is rotating around its center of mass at less than one rotation per orbit, the torque from the different forces on the two ends acts to increase the rotation rate; and when the body-spring system is rotating at more than that rate, the torque acts to reduce it.
Following your logic, when the bodies are supposedly stabilized in tidal lock, we will always have a greater force from gravity on the inner body than on the outer body, so there's a tendency to increase the distance from one to the other. Also, the force that would cause the torque has to travel through the spring until it reaches the outer body, at a later time. As a result, the outer body will always be behind, and this is precisely the shearing I was talking about in the beginning of this thread. When you have torque in the real world, you can be certain that you'll get some shearing.

Thus, we expect the system to stabilize at exactly one revolution per orbit, and that's tidal lock.
Ok, plus shearing.

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#### altergnostic

I believe we have struck the heart of the problem. If tidal lock was really achieved by these mechanisms, there should be measurable shearing, but we have no evidence of that on the moon. I'd also like to point out that we've been mostly using Newton, but if you take GR, where there are no forces (only geodesics), you get to shearing much faster and with a much simpler analysis. The outer body has to travel a larger geodesic – a longer distance – than the inner body. Both have the same constant linear velocity and cross the same distance over the same time, so logically the outer body gets behind. Connect them with a spring and you get shearing, just like Newton.

I think we are facing a real problem here. It is clear that the rigidity of the body (the spring we are using to simplify the problem) has no possible mechanism to produce enough torque to keep the far side aligned towards the central mass without shearing – that would mean infinite rigidity, and transmission of forces at infinite speed! Something else must be accelerating the far side so it stays aligned. In other words, tidal locked bodies have some mechanism that adjusts the tangential components of the orbital velocities on the near and far sides so they match precisely the magnitudes needed to maintain alignment towards the central mass, and it can't be the torque due to rigidity, and it can't be gravity directly.

Any thoughts?

#### Nugatory

Mentor
Following your logic, when the bodies are supposedly stabilized in tidal lock, we will always have a greater force from gravity on the inner body than on the outer body, so there's a tendency to increase the distance from one to the other
Yes, that is exactly what happens. As long as that outward pull doesn't exceed the strength of the forces holding the orbiting body together, the resulting situation is stable. The earth's tidally locked moon is an example - the near and the far side are being pulled apart, but the lunar rock is strong enough that the moon doesn't fall apart. If the tidal forces were stronger or the forces holding the moon together were weaker the earth wouldn't have a moon, it would have a ring like Saturn. But as long as the moon doesn't disintegrate the tidal locked situation is stable (Try googling for "Roche limit").
Also, the force that would cause the torque has to travel through the spring until it reaches the outer body, at a later time.
Also quite true, but for most of the weak-field situations that we're considering here not all that important. For example, a force applied to one side of the moon will propagate to the other side in a thousand seconds or so while the moon rotates about its axis once every million seconds or so... If we treat the propagation as instantaneous, we'll be close enough.

#### russ_watters

Mentor
I believe we have struck the heart of the problem. If tidal lock was really achieved by these mechanisms, there should be measurable shearing, but we have no evidence of that on the moon.

Any thoughts?
This may be the heart of your misunderstanding, but there is no such "problem" with science's understanding of tidal force.

In the diagram you posted on the first page, the two examples show exactly why shearing occurrs when a body is not in tidal lock (the first diagram) and why there is no shearing when the body is in tidal lock (the second diagram).

It almost looks from your posts like you are saying that both diagrams show the body in tidal lock. Is that the issue? If the body isn't rotating, then it isn't in tidal lock. Tidal lock means the body is rotating at the same rate as it is revolving.

From that post:
My question is what causes the changes in these tangential velocities. Or alternatively, why is there no shearing.
Did you read the wiki link I posted? It has a nice diagram showing the forces involved in causing a tidal lock.

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#### Nugatory

Mentor
I think we are facing a real problem here. It is clear that the rigidity of the body (the spring we are using to simplify the problem) has no possible mechanism to produce enough torque to keep the far side aligned towards the central mass without shearing – that would mean infinite rigidity, and transmission of forces at infinite speed! Something else must be accelerating the far side so it stays aligned. In other words, tidal locked bodies have some mechanism that adjusts the tangential components of the orbital velocities on the near and far sides so they match precisely the magnitudes needed to maintain alignment towards the central mass, and it can't be the torque due to rigidity, and it can't be gravity directly.
I'm sorry, but I don't see the problem you see - which probably means that we're differing on some underlying assumption that hasn't come to the surface yet.

Tidal lock develops very slowly over a period of many orbits as the modest torque generated by the imbalanced forces on the leading and lagging parts slowly pushes the system towards one rotation per orbit. There's no need for infinite rigidity or instantaneous transmission of force. All that's necessary is that the forces propagate quickly compared to the orbital and rotational periods, and that the the tidal forces not be so strong that they tear the system apart instead of locking it.

A digression: You've mentioned "shear" several times. We'd probably be better off speaking in terms of tension and compression - shear is a combination of tension and compression working in different directions, and it's much easier to analyze tidal forces in terms of tension and compression than the combination. The compressive forces act tangentially (the horizontal rod is compressed) while the tensile ones act radially (the vertical rod is stretched). Of course the people building bridges and buildings and airplane wings think in terms of shear as a force in its own right... But they're working with a different class of problems.

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#### Nugatory

Mentor
the two examples show exactly why shearing occurrs when a body is not in tidal lock (the first diagram) and why there is no shearing when the body is in tidal lock (the second diagram).
Are there not internal shearing forces at work even when the body is in tidal lock (unless it's a ideal rod, in which case it it will be oriented radially and only under tension)? The body is being stretched in the radial direction and compressed tangentially.

However, these are internal forces trying to change the shape of the body and resisted by the rigidity of the body - they produce neither net force at the center of mass nor torque around the center of mass, so do not disturb the one revolution per orbit equilibrium of tidal lock.

#### pervect

Staff Emeritus
This may be the heart of your misunderstanding, but there is no such "problem" with science's understanding of tidal force.
Yes, apparently the OP's intuition is leading him astray somehow. I'm not quite sure how, exactly. Currently, all I"m hoping for is that he'll realize from the number of SA's and other posters that are disagreeing with his conclusions (about Newtonian gravity, nevermind general relativity) that he probably made a mistake somewhere.

It's hard to tell what the OP's background is in order to help him figure out where he's gone astray. From the lack of calculus in any of his arguments, I' suspect the OP's background probably doesn't include calculus :-(.

#### altergnostic

This may be the heart of your misunderstanding, but there is no such "problem" with science's understanding of tidal force.
Probably science's understanding sees no problem with tidal forces, otherwise I would have found this discussion somewhere, but I urge you to notice that my problem is not with tidal forces, it is with the result shape of the body.

In the diagram you posted on the first page, the two examples show exactly why shearing occurrs when a body is not in tidal lock (the first diagram) and why there is no shearing when the body is in tidal lock (the second diagram).
What I intended to point out in that diagram is the difference in tangential velocities only, later we entered the discussion regarding torque which is the main cause of tidal lock, so it is clear that torque must speed up the far side so it stays aligned with the near side towards the central mass. My problem is that this torque should cause shearing stress and strain, and the far side should be skewed behind the near side, but we don't see that in any of the bodies in tidal lock (at least I haven't found any evidence of this).

It almost looks from your posts like you are saying that both diagrams show the body in tidal lock. Is that the issue? If the body isn't rotating, then it isn't in tidal lock. Tidal lock means the body is rotating at the same rate as it is revolving.
From that post:
Did you read the wiki link I posted? It has a nice diagram showing the forces involved in causing a tidal lock.
No, that's not the issue, as I said, I was only trying to point out the difference in tangential velocities. I actually, the diagram has a mistake, now that you mention it: those difference in tangential velocities would never cause shearing as I drew it, it would cause rotation, the shearing I'm talking about here is due to the torque in a body in tidal lock. And yes, I read the post, and it makes sense, I just think the forces would necessarily cause skewing, just like any torque causes stress, strain, shearing, distortion, yada yada.

#### altergnostic

I'm sorry, but I don't see the problem you see - which probably means that we're differing on some underlying assumption that hasn't come to the surface yet.

Tidal lock develops very slowly over a period of many orbits as the modest torque generated by the imbalanced forces on the leading and lagging parts slowly pushes the system towards one rotation per orbit. There's no need for infinite rigidity or instantaneous transmission of force. All that's necessary is that the forces propagate quickly compared to the orbital and rotational periods, and that the the tidal forces not be so strong that they tear the system apart instead of locking it.

A digression: You've mentioned "shear" several times. We'd probably be better off speaking in terms of tension and compression - shear is a combination of tension and compression working in different directions, and it's much easier to analyze tidal forces in terms of tension and compression than the combination. The compressive forces act tangentially (the horizontal rod is compressed) while the tensile ones act radially (the vertical rod is stretched). Of course the people building bridges and buildings and airplane wings think in terms of shear as a force in its own right... But they're working with a different class of problems.

#### altergnostic

Are there not internal shearing forces at work even when the body is in tidal lock (unless it's a ideal rod, in which case it it will be oriented radially and only under tension)? The body is being stretched in the radial direction and compressed tangentially.

However, these are internal forces trying to change the shape of the body and resisted by the rigidity of the body - they produce neither net force at the center of mass nor torque around the center of mass, so do not disturb the one revolution per orbit equilibrium of tidal lock.
Perfect.

#### Nugatory

Mentor
The problem is that if tidal lock occur due to torque, there should be skewing, meaning the far side would lag behind the near side.
Why so? When the far side lags, the torque acts to increase the rotation of the body, reducing the lag; when the far side leads the torque acts to reduce the rotation of the body and hence to reduce the lead. It's only when the body is rotating once per orbit so that the far side is neither lagging nor leading, the situation that we call tidal lock, that there is no net torque to further change the rotation.

#### Nugatory

Mentor
Are there not internal shearing forces at work even when the body is in tidal lock (unless it's a ideal rod, in which case it it will be oriented radially and only under tension)? The body is being stretched in the radial direction and compressed tangentially.

However, these are internal forces trying to change the shape of the body and resisted by the rigidity of the body - they produce neither net force at the center of mass nor torque around the center of mass, so do not disturb the one revolution per orbit equilibrium of tidal lock.
Hmmm... I hope I have not inadvertently contributed to the confusion here. To be clear:

These forces (radial stretching and tangential compression) are there no matter what the rotation of the body is; they arise from gravity acting in slightly different directions and with slightly different strength on the various parts of the body.

Only when the body is rotating exactly once per orbit (so that the same side always faces in, the situation that we call gravitational lock) do these forces balance and produce no net torque at the center of mass.

#### D H

Staff Emeritus
You point out one thing I figured out yesterday before I read your post just now: this is closer to an engineering problem that happens to be inside celestial mechanics, so it's no surprise we haven't given it a good amount of thought.
That is utter nonsense.

Any thoughts?
Yes. You need to stop posting nonsense and you need to start using math. You need to start doing that right now.

You posted nonsense in your other thread, which you apparently are trying to reinvoke. You repeatedly posted that there are no forces in general relativity, that scientists don't know / don't study the shapes of objects. Regarding the former, physicists talks about proper acceleration, model the interiors of stars with relativistic hydrodynamics. They wouldn't do either if there are no forces in general relativity. Regarding the latter bit of nonsense, which you have now repeated in this thread (see above), I suggest you google the phrase "lunar k2 love number" as a start.

Regarding your use (misuse) of logic: Start using math. If you don't know the math, you need to learn it before you start saying what physicists and astronomers know / don't know.

#### russ_watters

Mentor
Are there not internal shearing forces at work even when the body is in tidal lock (unless it's a ideal rod, in which case it it will be oriented radially and only under tension)?
You are correct, but the OP's diagram appears to me to be analyzing just three points on that circular body in his diagram, so I was indeed treating it as a rod.

#### russ_watters

Mentor
...but I urge you to notice that my problem is not with tidal forces, it is with the result shape of the body.
Ok...you haven't really explained why that would be though.
What I intended to point out in that diagram is the difference in tangential velocities only, later we entered the discussion regarding torque which is the main cause of tidal lock, so it is clear that torque must speed up the far side so it stays aligned with the near side towards the central mass. My problem is that this torque should cause shearing stress and strain, and the far side should be skewed behind the near side, but we don't see that in any of the bodies in tidal lock (at least I haven't found any evidence of this).
The torque exists until the body reaches tidal lock. Once in tidal lock, the bulges are directly aligned with the object being orbited, so there can't be any more torque.

On Earth, for example, our oceans move due to tidal forces, but since there is friction with the land, they aren't exactly aligned with the moon. Hence, a torque exists which is slowing our rate of rotation. Once the rate of rotation slows to equal the moon's rate of revolution, two sides of earth will have permanent high tides.
And yes, I read the post, and it makes sense, I just think the forces would necessarily cause skewing, just like any torque causes stress, strain, shearing, distortion, yada yada.
What is caused, internally, by the tidal force is the tidal bulges.

#### altergnostic

Yes, apparently the OP's intuition is leading him astray somehow. I'm not quite sure how, exactly. Currently, all I"m hoping for is that he'll realize from the number of SA's and other posters that are disagreeing with his conclusions (about Newtonian gravity, nevermind general relativity) that he probably made a mistake somewhere.

It's hard to tell what the OP's background is in order to help him figure out where he's gone astray. From the lack of calculus in any of his arguments, I' suspect the OP's background probably doesn't include calculus :-(.
I haven't seen calculus in any posts, so I don't know why you are taking me to task for this. But regardless, I was trying some calculus solutions with my father yesterday (who is a mechanical engineer). One thing you will notice if you try it is that this is a huge, complex problem. Really, you can't calculate it with high precision overnight unless you are very obstinate, there are so many things to consider. But there's a way to solve it very roughly with algebra:

- Position the moon vertically with the axis aligned towards the earth.
- Divide the mass of the moon equatorially in two halves so that each half has the length of one moon radius and diagram each center of mass m1 and m2. Consider each half as a cube with half the volume of the moon - this will give you a block shape)
- Find the lengths of the sides of the cubes.
- Put the center of this system Mm at a distance D from a point Me, with D=moon-earth distance.
- Move the system (the moon) a bit in the direction of the constant innate tangential velocity as if there was no gravity - I moved it half the moon's radius). – You have to so this because if you don't, there's no orbit, no linear velocity and the moon would fall straight to earth, I hope you can see that.
- Turn on earth's gravity at point Me and draw both gravity vectors from Me to point m1 and m2 and give them the right magnitudes (the magnitudes are not the same, you have to consider the distance between them - the correct value would be one moon radius, but since we're looking for the forces on the near and far sides I use the magnitudes of gravity on the near and far sides).
- Project the gravity vectors in the opposite direction of the linear innate velocity of the moon at right angles from the vertical axis of the moon (the line connecting m1, Mm and M2) and find their magnitudes.
- Subtract them and find the net force F.
- You can ditch the earth now, we won't need it anymore. Apply F on either point m1 or m2 (it doesn't matter because all we have to do now is find out how much a force on one point skews the whole system).
- Consider a moon of iron (or choose the material you find closer to what's the main composition of the moon)
- Find the elasticity modulus of the material E (you can find it easily on the internet).
- The displacement of the point where F is applied is Δb.
- Calculate Δb:

Δb = FL3//3EI

where:
I = bh3//12

h=b= the length of the sides of the cubes = 1 moon radius
L= the diameter of the moon

If you do this you will see there's no possible way whatsoever that the moon could ever maintain it's shape with no shearing. The forces are too great and the moon is not nearly strong enough to overcome the shear stress.
The value I found for Δb was so large I can't believe it, either I've done something wrong, or... I don't even want to think about it. I'll take it back to my father to analyze, but meanwhile, why don't you guys try to calculate it and then we can compare the numbers?

The numbers I used are:
3474km - Diameter of the moon
3.84x109 - Distance from the center of the earth to the near side of the moon
3.85x109 - Distance from the center of the earth to the far side of the moon
211 GPa (Giga Pascal) - elasticity module of iron - I took the moon as a ball of iron for simplicity

Notice this gives us a very rough estimate and it's already a bit dense. If you want to do the complete analysis, you have to take the moon as a sphere and do a lot of calculus indeed, but it seems pointless to try and get a more precise number since just for this simple estimate the work is already a bit tiring, and to consider it fully we'd have to know the full composition of the moon, where each element is located, know their densities on the moon, test our numbers against each element (and do it in a laboratory since you won't find these on the net), know the strength of the bonds between the various elements, consider pressure from the inside out (mantle pressure, etc) and outside in (moon's own gravity), plug in the eccentricity of the moon and of it's orbit and on and on and on. Since all our numbers would be rough estimates from the start, I can live with the above analysis. Also see that final number is probably rounded up, since the moon has mainly 60% the density of the earth (and so much less pressure and much less rigidity).

I conclude that a body in tidal lock should always be skewed, with the inner side ahead of the far side by a very clear amount, or even breaking up entirely. Either in GR or Newton. If you have forces of torsion, torques and anything like that, the body must suffer distortion. Always. Engineers know this very well and take very much care to consider the forces involved and how much of it a material can take without breaking up.

The point is that, if tidal lock is the result of a torque, the near side of the body should travel ahead of the far side in the direction of orbit, dragging the far side with it, skewing the shape of the body by a large amount. Since there's no absolute rigidity and instantaneous transmission of forces and gravity drops with the square of the distance, I'm starting to think that the only plausible scenario shearing wouldn't happen without reconsidering current theory is if the rotation is innate and just happens to match the orbit period by coincidence, but that can't be, there are too many bodies in tidal lock, too much to be just an accident. There must be a mechanism... but if it is due to torque, the shape of the bodies in tidal lock don't show any sign of it (namely, shearing).

I wasn't expecting this. I thought there should be shearing, but I was open to consider that it was too small to notice, that's why I went and did the calculations. The results just blew me away. I'm starting to believe that no one has ever thought of applying "engineering" equations to check if a body like the moon could hold itself together with the forces predicted/assumed by current theory. We see that it does, so we just assumed the forces were not strong enough to cause any significant stress, but I urge you to try the math yourselves.

Something is wrong.
Or I am wrong and desperately in need a careful consideration of the variables and calculations and a good explanation on why there's no shearing at all in tidal locked bodies.

#### zonde

Gold Member
I conclude that a body in tidal lock should always be skewed, with the inner side ahead of the far side by a very clear amount, or even breaking up entirely.
There are no forces in tangential direction when body is in tidal lock. All parts of body are moving inertially in tangential direction. The only forces are radial.

So please explain why do you think there should be forces that cause shear?

#### altergnostic

There are no forces in tangential direction when body is in tidal lock. All parts of body are moving inertially in tangential direction. The only forces are radial.

So please explain why do you think there should be forces that cause shear?
The problem is that you are considering the forces on a body after it has achieved tidal lock. But the forces that work to put a body in tidal lock are the same forces that cause shearing. Torque causes shearing every day in the real world. And if the force is constant, so is the shearing. Take a long plastic stick and spin it around, if you keep the force constant, you keep the stick bent by the same amount all the way around.

#### russ_watters

Mentor
The problem is that you are considering the forces on a body after it has achieved tidal lock. But the forces that work to put a body in tidal lock are the same forces that cause shearing. Torque causes shearing every day in the real world. And if the force is constant, so is the shearing. Take a long plastic stick and spin it around, if you keep the force constant, you keep the stick bent by the same amount all the way around.
That's all fine. So what's the problem?
I conclude that a body in tidal lock should always be skewed, with the inner side ahead of the far side by a very clear amount...
Why do you believe that a body can be "skewed" and in tidal lock at the same time? Why do you not believe that if the body is "skewed", there will be a torque on it that is trying to "unskew" it? That's the entire point of tidal locking in a nutshell.

#### altergnostic

That is utter nonsense.
If it is nonsense please show me where these calculations have been done, where has shear strain have ever been calculated for a body in tidal lock? I looked for it and found none, I'd be glad if you post it here, though.

Yes. You need to stop posting nonsense and you need to start using math. You need to start doing that right now.
Done - previous to your "warning" (I've been preparing my last post since yesterday).

You posted nonsense in your other thread, which you apparently are trying to reinvoke. You repeatedly posted that there are no forces in general relativity, that scientists don't know / don't study the shapes of objects. Regarding the former, physicists talks about proper acceleration, model the interiors of stars with relativistic hydrodynamics. They wouldn't do either if there are no forces in general relativity. Regarding the latter bit of nonsense, which you have now repeated in this thread (see above), I suggest you google the phrase "lunar k2 love number" as a start.
Regarding the first bit of nonsense, gravity is not a force in GR, and I'm quoting Einstein, what's the problem? Other forces are still forces, just gravity isn't, if that wasn't clear, i hope it's cleared up now.
And I think love numbers are related to tidal compression and stretching, not to how the body holds itself against shear stress and strain, and this was already on your first wiki link on tidal locking where I clearly responded further that I've seen no consideration of shearing. Using your own words, I suggest you google "elasticity modulus" as a start.

Regarding your use (misuse) of logic: Start using math. If you don't know the math, you need to learn it before you start saying what physicists and astronomers know / don't know.
Regarding your use (misuse) of courtesy: start using math. If you don't know the math, you need to learn it before you start saying what I know / don't know. I'm not judging what you know/don't know, we don't even know each other, so drop it.

I think you're tone is starting to be a little disrespectful here. Have I been rude to you or anyone? I have been asking questions nicely and I just did post the math. If you are convinced I'm mistaken and don't feel like addressing my questions, that's fine, let others try to explain what am I missing. This has been a nice discussion so far. Maybe you can post some math or draw some diagrams as well instead of shushing and scoffing me off to wikipedia before you even consider my questions respectfully.

I don't want to take this unscientific debate any further, I will continue to stay focused on my questions and on physics (if you allow me). And I'd be happy if you did the same. The only "content" of your post was "google love number".

#### altergnostic

That's all fine. So what's the problem? Why do you believe that a body can be "skewed" and in tidal lock at the same time? Why do you not believe that if the body is "skewed", there will be a torque on it that is trying to "unskew" it? That's the entire point of tidal locking in a nutshell.
The torque is what causes skewing. If the body wasn't feeling any torques, there would be no reason for skewing! And I don't "believe" it, it is both logical and I have calculated it. The plastic stick is in tidal lock: the period of one revolution matches the period of rotation. If you are at the center turning around holding this stick, you will always be facing the near side and the far side will always be facing away. That's the entire point of tidal locking in a nutshell.

#### russ_watters

Mentor
The torque is what causes skewing. If the body wasn't feeling any torques, there would be no reason for skewing!
Maybe it isn't clear what you mean by "skewing", since that isn't a scientific term. It sounded like you meant the moon would be symmetrically oval-shaped (like the theory predicts), but the oval would not aligned with the planet.

Now what bothers me here is that you're saying there is a net torque. But a net torque would cause a continuous acceleration -- the moon would start spinning faster and faster if there was a net torque: it wouldn't stay in your "skewed" alignment.
And I don't "believe" it, it is both logical and I have calculated it.
Er, no. One of the things that's getting on DH's nerves (and mine, frankly) is that you're making statements against established theory, while simultaneously displaying misunderstandings of how the theories work. I'm not sure who you think you are, but these theories have been around for one hundred and several hundred years, respectively and have been examined by millions of scientists in that time -- scientists who actually understand what the theories are saying. They aren't wrong. You are. You need to start dealing with that or this thread will be locked too. We teach established science here, we don't argue with crackpots about whether the science is right or wrong.

Second, your calculation - rather, the description of the setup - was incomprehensible. A diagram would be a big help, for us to know what you are calculating. Odds are, you just caluclated something nonsensical.
The plastic stick is in tidal lock: the period of one revolution matches the period of rotation. If you are at the center turning around holding this stick, you will always be facing the near side and the far side will always be facing away. That's the entire point of tidal locking in a nutshell.
If there is a net torque on it, then it isn't rotating at a constant rate.
...gravity is not a force in GR, and I'm quoting Einstein, what's the problem?
The problem is that you don't understand what that means and won't accept explanations of it.

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#### altergnostic

Maybe it isn't clear what you mean by "skewing", since that isn't a scientific term. It sounded like you meant the moon would be symmetrically oval-shaped (like the theory predicts), but the oval would not aligned with the planet.
Yes, that's what I meant, it would be elliptical, but the far side would be a little behind in the direction of orbit.

Now what bothers me here is that you're saying there is a net torque. But a net torque would cause a continuous acceleration -- the moon would start spinning faster and faster if there was a net torque: it wouldn't stay in your "skewed" alignment. Er, no. One of the things that's getting on DH's nerves (and mine, frankly) is that you're making statements against established theory, while simultaneously displaying misunderstandings of how the theories work. I'm not sure who you think you are, but these theories have been around for one hundred and several hundred years, respectively and have been examined by millions of scientists in that time -- scientists who actually understand what the theories are saying. They aren't wrong. You are. You need to start dealing with that or this thread will be locked too. We teach established science here, we don't argue with crackpots about whether the science is right or wrong.
The constant torque here is the difference between the centripetal forces on the near and far side. There's a constant force on the near side that is stronger than the one on the far side. If there's a constant acceleration, there's a constant force, and a constant torque. And I am not even saying that current theory is wrong, what I am saying is that following the forces the theory predicts a body suffers to achieve tidal lock, the torque will cause shearing and the far side will not be aligned with the near side (although will be trying to). It isn't, and I am asking why. All I hear is that there should be no shearing, that somehow the torque causes no deformation on the orbiter and the far side reaches the point where it's aligned with the near side towards the central mass, but the only way I see this a possible is if the forces are too small compared to the rigidity of the body to notice, and that's the calculation I am seeking. I did it roughly, if the set up has flaws I would like you to point them out so I can correct them.

Second, your calculation - rather, the description of the setup - was incomprehensible. A diagram would be a big help, for us to know what you are calculating. Odds are, you just caluclated something nonsensical.
If there is a net torque on it, then it isn't rotating at a constant rate. The problem is that you don't understand what that means and won't accept explanations of it.
You are saying there shouldn't be any shearing expected, but I can't see that. The example of the plastic stick is clear, apply a constant torque on the near side, the far side will try to catch up but will never be able to, a constant centripetal force is a constant acceleration, but does not necessarily cause a varying period of rotation, in this problem it doesn't. This is clear from the plastic stick example.
I will try to diagram each step and post it since it is a complex calculation, but it will take some time. You are welcome to ask what is it that you don't understand from the description (I took it straight from a book on torsion, shearing, stress, strain, pressure and all sort of effects different forces have on materials, it's my father's so I don't know the name or the author, I can try to check it for you). I am not asking anyone to believe anything, the problem I need you to address is:

Before a body achieves tidal lock, there is a torque applied on the body. This torque emerges from the differing forces on the near and far sides, the near side experiencing a stronger force. This torque tends to push/pull the far side in line with it towards the central mass, but due to the elasticity/rigidity of the orbiting body, it should suffer shearing – the far side would not catch up to be perfectly aligned with the near side towards the central mass. Depending on the relations between the forces and the rigidity, the shearing will achieve a state of equilibrium at some point or, if the forces are strong enough, the body will tear apart. If the applied force is constant (which it is, gravity is a constant force/acceleration), the far side will always stay behind, it will never catch up. The body will achieve a state of equilibrium where the period of rotation matches the period of orbit, but the far side will not be aligned, the body will be elliptical with the far side behind the near side.

It is just like holding a plastic stick horizontally from one end and start spinning around at a constant rate. You have to apply a constant force, which will result in a constant torque. The period of "orbit" will match the period of rotation and the stick will behave just like if it was in tidal lock, but the far side will be behind, not perfectly aligned with the near side towards you. The plastic stick will be bent. The torque could only make the far side align perfectly with the near side if the stick was perfectly rigid. This is the shearing I am talking about.

#### zonde

Gold Member
You are saying there shouldn't be any shearing expected, but I can't see that. The example of the plastic stick is clear, apply a constant torque on the near side, the far side will try to catch up but will never be able to, a constant centripetal force is a constant acceleration, but does not necessarily cause a varying period of rotation, in this problem it doesn't. This is clear from the plastic stick example.
Before a body achieves tidal lock, there is a torque applied on the body. This torque emerges from the differing forces on the near and far sides, the near side experiencing a stronger force. This torque tends to push/pull the far side in line with it towards the central mass, but due to the elasticity/rigidity of the orbiting body, it should suffer shearing – the far side would not catch up to be perfectly aligned with the near side towards the central mass. Depending on the relations between the forces and the rigidity, the shearing will achieve a state of equilibrium at some point or, if the forces are strong enough, the body will tear apart. If the applied force is constant (which it is, gravity is a constant force/acceleration), the far side will always stay behind, it will never catch up. The body will achieve a state of equilibrium where the period of rotation matches the period of orbit, but the far side will not be aligned, the body will be elliptical with the far side behind the near side.
The far side will catch up with the near side and overcome it. Perfectly elastic body would be oscillating around middle point with far side sometimes being ahead and sometimes behind the near side.

The example of the plastic stick is clear, apply a constant torque on the near side, the far side will try to catch up but will never be able to, a constant centripetal force is a constant acceleration, but does not necessarily cause a varying period of rotation, in this problem it doesn't. This is clear from the plastic stick example.
A constant torque on the near side means that you are spinning faster and faster.
And then - yes, the far side will try to catch up but will never be able to.
And then - no, it does cause a varying period of rotation.
And centripetal force is not constant either in that case.

#### Austin0

It might help if you considered the mechanics of planetary condensation from an orbiting cloud.
If the forces and effects you are proposing as inferred from Keplerian mechanics were operating as you surmise, then how could condensation ever occur?
The differing orbital velocities at the near and far sides of the diffuse mass would prevent anything like spherical symmetry from ever forming wouldn't they? Would keep the matter smeared out around the orbit.
But it seems evident that the combined Ricci and Weyl effects bring it about as there are a bunch of pretty round planets

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