[rmolnav]

Quote
But today, I´ll only tackle what I consider ESSENTIAL for any possible understanding: NOT to have completely opposite ideas regarding INERTIA, that can manifest itself in different ways, one of them as “centrifugal force", real force (or, at least, causing quite real effects ...)
I suppose it is a matter of interpretation if it is real or not.  I kind of go with force being the F=ma in Newton’s 2nd law, whereas inertial effects (centrifugal being among them) are the ‘m’ in that equation, not the F.  It isn’t causing any real acceleration of anything in the direction of the supposed force.

Come on !! Many forces can be real without "causing any real acceleration of anything in the direction of the supposed force" ... if the affected object is not free to move ! In that case what they cause is deformations, of the object and/or contiguous objects (or parts of an object) !!
That´s the essence of internal stresses and strains, and even the actual manifestation of tides on solid parts of earth. And on our oceans they can produce water pressure changes, and subsequently the transportation of water towards where the bulges build !!

You mean the CM can be used to track Earth’s motion as if it were a point mass.  Indeed, it doesn’t work when the discussion is about more local effects like tidal forces.  It also doesn’t work in general.  I can have two pairs of all identical mass objects with identical difference in separation by CM, but one is a pair of spheres and the other is a pair of long barbells arranged perpendicularly.  The spheres will have more gravitational attraction than the crossed barbells despite the same mass and distanace between their respective centers of mass, because most of the mass is nowhere near that center like it is with the spheres.  So point taken (??).  You can’t treat objects as point masses when the difference matters.

Sorry, but that is utterly absurd ...
A not uniforme distribution of masses, even two o more separated objects, can be treated (for certain purposes) as a whole, with a common center of mass, useful for its dynamic analysis ...
E.g., the very stuff we are talking about: earth-moon system.
It has a common CM (the barycenter), which is what follows the elliptical orbit around the sun. It´s quite reasonable to analyze the complex dynamic of moon-related tide formation in different stages: mentioned orbiting of common CM, earth-moon revolving/rotation around the barycenter, gravitational and inertial forces on the earth as a whole (where earth CM has an important roll), and then the detail of forces acting on each area of our planet (internal force changes due to earth-moon dynamic included) ...
Those last ones are mainly in the moon´s CM-barycenter-earth´s CM direction, both "moonwards" and outwards (or "moonfugal" or "centrifugal"...)
And they are the direct cause of tides ... certainly generated by moon varying pull, but also the inertial effects on all and each particle of the earth.
As I´ve said many times, earth particles only can "feel" gravity at their respective location, and pushes and/or pulls from contiguous matter. And they always have their tendency not to change their velocity vector (inertia), what causes inertial effects ...
What they can´t "feel" is either the so called "differential gravity" or the gravity on locations many km away, let alone they can "subtract" different gravitational pulls to react to that difference !!

[Halc]

That´s the essence of internal stresses and strains, and even the actual manifestation of tides on solid parts of earth. And on our oceans they can produce water pressure changes, and subsequently the transportation of water towards where the bulges build !!

Agree with this.  Its about stress and strain.

You mean the CM can be used to track Earth’s motion as if it were a point mass.  Indeed, it doesn’t work when the discussion is about more local effects like tidal forces.  It also doesn’t work in general.  I can have two pairs of all identical mass objects with identical difference in separation by CM, but one is a pair of spheres and the other is a pair of long barbells arranged perpendicularly.  The spheres will have more gravitational attraction than the crossed barbells despite the same mass and distanace between their respective centers of mass, because most of the mass is nowhere near that center like it is with the spheres.  So point taken (??).  You can’t treat objects as point masses when the difference matters.

Sorry, but that is utterly absurd ...
A not uniforme distribution of masses, even two o more separated objects, can be treated (for certain purposes) as a whole, with a common center of mass, useful for its dynamic analysis ...
E.g., the very stuff we are talking about: earth-moon system.

Earth and moon are hardly shaped like barbells.  If all objects can be treated as point masses, then I should weigh more and more as I dig myself into a hole.

It has a common CM (the barycenter), which is what follows the elliptical orbit around the sun.

Well, to be fully anal about it, and in an ideal situation with no other planets, we would orbit the CM of the solar system, not the sun itself.  The two points are almost the same thing if we discard the other planets.

It´s quite reasonable to analyze the complex dynamic of moon-related tide formation in different stages: mentioned orbiting of common CM, earth-moon revolving/rotation around the barycenter, gravitational and inertial forces on the earth as a whole (where earth CM has an important roll), and then the detail of forces acting on each area of our planet (internal force changes due to earth-moon dynamic included) ...
Those last ones are mainly in the moon´s CM-barycenter-earth´s CM direction, both "moonwards" and outwards (or "moonfugal" or "centrifugal"...)

Complicated but reasonable to analyze the dynamics this way, sure.  Some of them assume a rotating frame of reference, and it is reasonable to analyze the dynamics in that frame.

And they are the direct cause of tides ... certainly generated by moon varying pull, but also the inertial effects on all and each particle of the earth.

The tides are directly proportional to the varying pull, and not directly proportional to the inertial effects.  So the latter is a related cause, but not a direct cause.  As I said, it is reasonable to analyze it in such terms, but it is more complicated due to this non-proportional relationship.  The inertial effects from the sun are far greater (about 175x) than those from the moon, yet produces a significantly smaller tidal effect.

As I´ve said many times, earth particles only can "feel" gravity at their respective location, and pushes and/or pulls from contiguous matter. And they always have their tendency not to change their velocity vector (inertia), what causes inertial effects ...

Yep

What they can´t "feel" is either the so called "differential gravity" or the gravity on locations many km away, let alone they can "subtract" different gravitational pulls to react to that difference !![/b]

Right.  No individual particle can detect a gravity gradient.  It requires pairs of separated particles.  What the particles might 'feel' is the changing forces put on them by their immediate neighbors, which in turn is effectively them 'feeling' the differential gravity on locations many km away.  This is similar to the way you feel a train go by 100m away.  The stresses are communicated between particles at different locations.

[rmolnav]

Quote
If that happens as an inertial reaction of the water (and the man´s mass) when "forced" to follow the equatorial curved path, why wouldn´t happen something similar when the water has to revolve, following during some 28 days a circumference (considering only earth-moon "dance") of a radius equal to the distance between earth CM and earth-moon barycenter ??
Inertial effects are indeed involved, but not directly like that.  The far tide has far more acceleration about the barycenter (which represents the inertial frame of a two-body model), but both tides are the same magnitude, which they would be even if the moon was further away putting the barycenter above the surface.  The direct inertial explanation of the tide would push the water down, not upward, since it is now rotating on the same side of the barycenter.  Inertial of the water on the sides would tend to carry it to the far side.
So no, I cannot agree with this wording.  In a uniform gravitational field (say at a sufficient distance from the galactic black hole to have a gravitational force on earth similar to that of our moon (2e20N), you have the same inertial forces at work, but undetectable tides.  It is the non-uniformity of the gravitational field, not inertia, that causes the stress, and  tides which are strain resulting from that stress.

I´m afraid you haven´t fully grasp the difference between rotating and revolving (what earth does around the barycenter), not in their layman acceptation but in physicist´s …
Disregarding earth daily spinning (which has nothing to do with moon-earth dynamics), only earth´s CM follows a circular path around the barycenter. Any other earth point follows identical circumferences, but with their centers at its actual distance from the CM, and keeping its orientation relative to far distant stars. Perhaps best analogy is the movement of a child  waist making a hulla-hoop rotate …
As said on one of the NOAA previously linked sites:
" The Effect of Centrifugal Force. It is this little known aspect of the moon's orbital motion which is responsible for one of the two force components creating the tides. As the earth and moon whirl around this common center-of-mass, the centrifugal force produced is always directed away from the center of revolution. All points in or on the surface of the earth acting as a coherent body acquire this component of centrifugal force. And, since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”
At sublunar area, direct moon´s pull is bigger than above mentioned centrifugal force, it prevails, and we have there the more obvious bulge ...
And at antipodal area, being moon´s pull there smaller, centrifugal force (or “outward” inertial force if you like) prevails, and we also have a bulge ...

Not sure if all such accelerated motion can be considered centrifugal since the acceleration is more typically inward instead of outward, and 'centrifugal' is typically used only for the tendency towards outward.

I still have to post here my ideas about the generally badly used issue of inertial and non-inertial reference frames ...
I´m procrastinating in that realm, because I know it is tough to convey.
But please note that the term "centrifugal acceleration" doesn´t exist ... "Centripetal force" is the name we give to any force that is bending a moving object trajectory. And "centripetal acceleration" the one that object gets due to mentioned force.
Centrifugal force is a controversial term, but it is always an inertial reaction of the affected body to the fact that it is being obliged to change the direction of its speed ...
But if the object moved outward, that would actually mean that it´s got somehow free, because both centripetal and centrifugal forces had ceased to exist, at least partially.
And regarding:
"...In a uniform gravitational field (say at a sufficient distance from the galactic black hole ... you have the same inertial forces at work, but undetectable tides ..."
 please kindly note that in that case centrifugal forces would also be very, very tiny, proportional to v²/r, being "r" enormous !! 

[Halc]

Oh, hey, I am a human now, instead of a modal robot.  The edit window didn't get larger.  Still have to do it offline

I´m afraid you haven´t fully grasp the difference between rotating and revolving (what earth does around the barycenter), not in their layman acceptation but in physicist´s …

Thanks for that correction.  Rotate is what something does around its own axis, and revolve is around a different one, even if off-center within the thing revolving.  Orbital motion is a form of revolving.

As said on one of the NOAA previously linked sites:

Which admitted to being for children and laymen, and just wrong after the 50’s….   Hardly an argument from authority.

”since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”

This contradicts with the actual situation since the barycenter is within Earth and thus points on the lunar side of that get centrifugal force directed towards the moon.  They're being sloppy, something they'd not get away with in a publication meant for a higher audience.
Yes, I understand this reasoning.  You are now just repeating it instead of addressing my objections to it in prior posts.  It isn’t completely wrong (the centrifugal force needs to be described in the frame in which it exists), just not proportional to the tides, and most importantly, it doesn’t explain tides in situations where there are no centrifugal forces, such as all these hypothetical situations with multiple moons and such.  That lack of a general explanation makes it wrong I think

At sublunar area, direct moon´s pull is bigger than above mentioned centrifugal force, it prevails, and we have there the more obvious bulge ...

This is the actual cause, and it works without computing the difference with centrifugal forces to see which is prevailing.  It is a general solution that works even in non-revolving situations.

Not sure if all such accelerated motion can be considered centrifugal since the acceleration is more typically inward instead of outward, and 'centrifugal' is typically used only for the tendency towards outward.

I still have to post here my ideas about the generally badly used issue of inertial and non-inertial reference frames ...

Other frames are valid, but the laws of physics are not the same as in inertial frames.  Centrifugal force is a valid force in rotating frames, which is contrary to what I first said.  Your use of it in descriptions of the scenario in inertial frames seems like bad use of frames.
The moon and Earth are stationary in the rotating frame of the barycenter, and gravity prevents the natural tendency for both objects to accelerate away due to centrifugal force.

I´m procrastinating in that realm, because I know it is tough to convey.
But please note that the term "centrifugal acceleration" doesn´t exist

It does.  An object at rest begins to move if not impeded.  Sounds like acceleration to me.  The acceleration is initially outward, but quick bends until it is mostly inward, but always accelerating.

... "Centripetal force" is the name we give to any force that is bending a moving object trajectory.

In an inertial frame yes, and then only sometimes. The thrust of a rocket turning left isn't usually described as centripetal. In a rotating frame, that centripetal force might be holding an object stationary against its unimpeded tendency to start moving.

Centrifugal force is a controversial term, but it is always an inertial reaction of the affected body to the fact that it is being obliged to change the direction of its speed ...
But if the object moved outward, that would actually mean that it´s got somehow free, because both centripetal and centrifugal forces had ceased to exist, at least partially.

That bolded part I didn’t understand.  Maybe if you specify a frame it would make sense, but it sounds like you’re just describing an object moving forever in a line in an inertial frame, which yes, it does if no forces act upon it, per Newton’s first law.  I hadn’t seen that law described as something “getting free”.

And regarding:
"...In a uniform gravitational field (say at a sufficient distance from the galactic black hole ... you have the same inertial forces at work, but undetectable tides ..."
 please kindly note that in that case centrifugal forces would also be very, very tiny, proportional to v²/r, being "r" enormous !! 

No, the centrifugal forces would be identical to what the moon generates since they have to counter the gravitational pull of that black hole at the exact distance where the force was the same.  The acceleration of Earth in such a situation would be the same as in a pure Earth/moon system, countered by the same inertial resistance that you’ve been labeling centrifugal force.  I chose that scenario specifically for its identical gravitational forces.

[rmolnav]

Quote
What they can´t "feel" is either the so called "differential gravity" or the gravity on locations many km away, let alone they can "subtract" different gravitational pulls to react to that difference !![/b]
Right.  No individual particle can detect a gravity gradient.  It requires pairs of separated particles.  What the particles might 'feel' is the changing forces put on them by their immediate neighbors, which in turn is effectively them 'feeling' the differential gravity on locations many km away ... The stresses are communicated between particles at different locations.

Right ... But that "chain of transmission" of information what actually transmits is not "differential gravity", but the part of total forces acting directly on each particle (moon´s pull, gravitational pull from the rest of the planet, and pushes or pulls from contiguous particles, in your own words "the changing forces put on them by their immediate neighbors"), not used to cause the centripetal acceleration required for the revolving of the considered particle !!.
Those "communicated" stresses, according to 3rd Newton´s Motion Law, happen in pairs of opposite forces exerted on each other (between two considered "neighbors") ... Most of them are parallel to global moon´s pull, half of them centripetal ("moonwards") and half of them outward, or in the broad sense of the term, centrifugal ("moonfugal").

[rmolnav]

By the way, what you said:

I took issue with his blocky diagram above in the 'stress' section, which shows two stacked blocks in a gravitational field and a weight force preventing the whole setup from accelerating from the gravity.  It labels the spring as tension, despite the arrows depicting the spring pushing on the two blocks, which would compress the spring.  The arrows are correct, but text incorrectly labels this as tension.  If the arrows represent tension T, they need to point the other way, and the tension would be negative.  So while I understand the concept of stress and strain (I took civil engineering courses), the description of it there needs a bit of review.

I have to check the details of that figure and explanation, and I´ll come back to that. But please keep in mind that, as said on recent post in relation to earth internal stresses, the concept "tension" (of a spring in this case) always includes the inherent two opposite forces ... Just an "arrow" where two contiguous objects A and B should always include the information if it is the force exerted by A on B, or by B on A !!   

[Colin2B]

Orbital motions are not considered as rest frames

I think you mean ‘inertial frames’

I haven’t had time to follow this thread and so am not sure what the point of contention is about, so best I keep out particularly as there is a lot of antagonism going on. However, skimming through I did notice your comment below and as it doesn’t appear to have been picked up I will comment.

I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast. If we accelerate a satellite which is ,  already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon.

This is a valid analysis. I thought about it and decided it’s easier to think of the moon where you don’t have the complication of the barycentre being inside the earth.
The moon will orbit with its centre of mass at an equilibrium position based on the balance between centrifugal force and gravitational force (rotating, noninertial frame). As you say, the inner part will want to move to a lower orbit and the outer to a higher orbit. Because the moon keeps its same face to the earth it will be elongated along the earth-moon line by the difference in force.
When looking at the earth, the barycentre is within the earth and, as you say, the far side is going to fast and the near side too slow. You comment on the earth’s daily rotation, but this is an added complication we can remove. Despite what some books or websites show, the moon does not orbit around the equator so the earth’s spin doesn’t align with it. Fortunately, as with most motion/force scenarios, we can deal with these as separate issues and say that the spinning creates a symmetrical equatorial bulge and take that out of the equation. In this case we are left with the earth’s centre of mass orbiting the barycentre at it’s equilibrium position and, as you correctly say, the far side will go too fast and the nearside too slow, hence each will try to seek its own equilibrium point.

     BY THE WAY, on a link posted by PmbPhy some days ago it is said:
..the CENTRIFUGAL FORCE and the moon´s gravitational attraction are equal and opposite.

If the two forces where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not. 

@PmbPhy wouldn’t make that mistake. I think by now you will have realised that he was talking about non inertial, rotating frames.
I think the whole subject of centrifugal and what are called fictitious forces is badly taught. NASA tends to refer to all 3 as inertial forces and includes G force in linear acceleration as a 4th. I think it’s better terminology.
Certainly these inertial forces are only seen in a non-inertial, rotating frame (or a close approximation), so they aren’t prime movers, but they do have real, measurable effects.

As said on one of the NOAA previously linked sites:

Which admitted to being for children and laymen, and just wrong after the 50’s….   Hardly an argument from authority.

”since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”

This contradicts with the actual situation since the barycenter is within Earth and thus points on the lunar side of that get centrifugal force directed towards the moon.  They're being sloppy, something they'd not get away with in a publication meant for a higher audience.

I think they are considering the earth to be a rigid body, so any force at the centre of mass will be transferred to near and far sides. Although the water isn’t rigid it is incompressible and would tend to find it’s own level so might be considered to act as the earth does. Whether it’s a reasonable assumption is up for debate, but I don’t think the article was intended to be completely rigorous.

[Halc]

If the two forces where equal and opposite, the particles at earth’s center would be moving in a straight line, but they’re not.

@PmbPhy wouldn’t make that mistake. I think by now you will have realised that he was talking about non inertial, rotating frames.

I have since corrected my statement above.  Yes, in a rotating frame reference, the two forces exist and can cancel, holding the object stationary.

I think the whole subject of centrifugal and what are called fictitious forces is badly taught. NASA tends to refer to all 3 as inertial forces and includes G force in linear acceleration as a 4th.

What are the other two?  Centrifugal is an inertial force in a rotating frame.  G force is an inertial force in an accelerating reference frame.  There are inertial effects involved anytime some mass is accelerated in an inertial frame.  That doesn't mean that inertial effects are all centrifugal.

Tides happen or not in both cases, so tides are not a function of either.  That was my point.  Tides are a function of the stresses and strain on an object due to non-uniform gravitational field, and are a function of only that.

”since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”

This contradicts with the actual situation since the barycenter is within Earth and thus points on the lunar side of that get centrifugal force directed towards the moon.  They're being sloppy, something they'd not get away with in a publication meant for a higher audience.

I think they are considering the earth to be a rigid body, so any force at the centre of mass will be transferred to near and far sides.

They seem to be making an explicit reference to forces on specific points, not about where those forces get transferred due to the rigidity of the body. It was a nit, not really detracting from the point being made by NOAA.

Although the water isn’t rigid it is incompressible and would tend to find it’s own level so might be considered to act as the earth does. Whether it’s a reasonable assumption is up for debate, but I don’t think the article was intended to be completely rigorous.

Sure.  For one, the tides would be higher if the day were longer since it would give the water more time to move to where gravity is shoving it.

[rmolnav]

Thanks for that correction.  Rotate is what something does around its own axis, and revolve is around a different one, even if off-center within the thing revolving.  Orbital motion is a form of revolving.

There also can be “rotations” around other axis … E.g., the “hammer" to be thrown by the athlete: it simultaneously follows an “orbit”, and "spin" around its own axis (hammer hook always points towards the athlete ...).
Curiously it is the same case of our moon, because it is tidal locked to the earth.
But earth, in its dance with the moon, keeps its orientation relative to distant stars, and that is "revolving" (orbiting without rotation).
That´s why I prefer to reserve the term “spinning” for when around own axis, “revolving” for orbiting without any spinning, and “rotation” for orbiting with simultaneous spinning (when due to the dynamics of same pair of objects “dance”, such as moon´s rotation around barycenter) ...

Quote
”since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”
This contradicts with the actual situation since the barycenter is within Earth and thus points on the lunar side of that get centrifugal force directed towards the moon.

Sorry, but that is erroneous. You seem not to have fully grasped important nuances of actual earth revolving …
I suggest you put your open hand horizontally on a table. Spot the table point app. under the middle of the first bone of your middle finger (B, from barycenter). And try slowly to make your hand “revolve”, with the first knuckle of same finger (C, app. at hand center) following a circle around B .
If initially you had imagined a “moon” a couple of feet away in line with your middle finger, and that “moon” rotated around B simultaneously to hand revolving, ALWAYS in line with B and C and at side opposite to C, you would have a scenario quite similar to actual earth-moon dance.
If you carefully observe the end of any of your fingers (or any other point of your hand), you will find that THEY ALL follow equal circumferences, ALWAYS keeping their location farthest from the moon (logically, within each own path).
THAT´S, I think, A GOOD WAY to see what quoted from NOAA´s scientits work IS QUITE CORRECT:
" since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”

[Colin2B]

What are the other two?  Centrifugal is an inertial force in a rotating frame.  G force is an inertial force in an accelerating reference frame.  There are inertial effects involved anytime some mass is accelerated in an inertial frame.  That doesn't mean that inertial effects are all centrifugal.

They are Coriolis force, which we all recognise, and Euler force -  tangential force caused by angular acceleration/deceleration - which happens during any pitch, roll, yaw manoeuvre.

[Halc]

They are Coriolis force, which we all recognise, and Euler force -  tangential force caused by angular acceleration/deceleration - which happens during any pitch, roll, yaw manoeuvre.

OK.
I thought of Coriolis force after asking the question.  I never heard of Euler force, which seems to be the rotating equivalent of G-force in an accelerating rotating frame.  I wasn't considering frames with variable rotation.

My guess was a different one: In a rotating frame, a stationary object accelerates to move away due to centrifugal force.  Coriolis curves it to the side once it picks up speed and the acceleration vector swings in a direction against the rotation.  After a bit more time, that acceleration vector points nearly back to the axis.  What force accelerates objects back to the origin like that?

The example is stars in the sky, which, in the rotating frame of Earth's surface, go around once a day, all faster than light.  Their acceleration vector is very high, proportional to their distance, and directed straight at us.  What force is responsible for this apparent acceleration in the opposite direction of the centrifugal force?

[rmolnav]

Quote from: Le Repteux on 14/09/2018 15:11:18
I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast . If we accelerate a satellite which is ,  already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon. [/quote]
As “Colin2B” correctly says, it is much more logical to disregard the daily spinning of earth when analyzing moon (or sun) related tides … It causes the circular and permanent so called “equatorial bulge”, which has nothing to do with earth-moon dynamics …
Quite another thing is that due to that spinning, we usually perceive the tide cycle as if it had two high and two low tides a day, when the actual cycle has two high and two low tides in a little less than a month.
Apart from that, what you say:
"... while the earth's C.G. was going at the right orbital speed around the moon... ”
is erroneous: the earth´s C.G. doesn´t orbit around the moon whatsoever !!
The moon pull certainly “supplies” the forces necessary for the circular movement of the earth, which is a revolving around earth-moon barycenter, actually very, very far from the moon !!
That´s why I consider applying proper orbiting dynamics features to our case is an erroneous approach …

[rmolnav]

This is a valid analysis (NOAA scientist´s). I thought about it and decided it’s easier to think of the moon where you don’t have the complication of the barycentre being inside the earth.
The moon will orbit with its centre of mass at an equilibrium position based on the balance between centrifugal force and gravitational force (rotating, noninertial frame). As you say, the inner part will want to move to a lower orbit and the outer to a higher orbit.

Though you are quite right on your reference to NOAA´s analysis, you aren´t on what in bold ...
If you read carefully second half of my post « on: Yesterday at 18:52:07 » (and recent one) I do hope you´ll see why I say so ...

[rmolnav]

... the tides would be higher if the day were longer since it would give the water more time to move to where gravity is shoving it.

Again: you continue to mix daily earth spinning with moon related tides … Gravity and inertial effects (centrifugal forces) don´t "need" longer days to build full-size tides, because they have plenty of time to deform ocean surface: a little more than a week from low to high tide !!
If the day were longer, what would imply a smaller angular and tangential speed, what we would have is a smaller gap between sublunar meridian and actual bulge location (the same at antipodes), apart from some more time between our perception of low and following high tide ...

[Le Repteux]

Quote from: Le Repteux on 14/09/2018 10:11:18

I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast. If we accelerate a satellite which is already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon.

This is a valid analysis. I thought about it and decided it’s easier to think of the moon where you don’t have the complication of the barycentre being inside the earth.
The moon will orbit with its centre of mass at an equilibrium position based on the balance between centrifugal force and gravitational force (rotating, noninertial frame). As you say, the inner part will want to move to a lower orbit and the outer to a higher orbit. Because the moon keeps its same face to the earth it will be elongated along the earth-moon line by the difference in force.
When looking at the earth, the barycentre is within the earth and, as you say, the far side is going to fast and the near side too slow. You comment on the earth’s daily rotation, but this is an added complication we can remove. Despite what some books or websites show, the moon does not orbit around the equator so the earth’s spin doesn’t align with it. Fortunately, as with most motion/force scenarios, we can deal with these as separate issues and say that the spinning creates a symmetrical equatorial bulge and take that out of the equation. In this case we are left with the earth’s centre of mass orbiting the barycentre at it’s equilibrium position and, as you correctly say, the far side will go too fast and the nearside too slow, hence each will try to seek its own equilibrium point.

Hi Collin, and thanks for noticing that my idea looks right.

As you said, it might be easier to use the moon instead of the earth to analyze it, so let's do that. I said that the proper rotation of the planet had something to do with the fact that the far and the near sides are not going at the right orbital speed, and I must defend that feature otherwise my idea wouldn't work. Of course, that kind of rotation also produces an equatorial bulge, but even if the rotation is as tiny as the moon's one, that's what makes its far and near sides not going at the same orbital speed as its center of gravity. These sides could very well be going at the right speed if they weren't tidally locked for instance, and in this case, if my explanation is right, there would be no tides. What would be producing the tides if the speed was wrong would also produce another kind of motion that we attribute to tidal friction: the slow deceleration of rotations. In the earth's case, when its far side starts going too fast around the moon, it also starts decelerating and getting away from it as if it was at its perigee, while its near side does the inverse, and they both go on doing that all along their transit. This way, no need for friction to explain the observation, the principle of orbital motion does it all, and it can also be applied directly to orbital rotations.

For instance we can consider that only the barycenter of the earth/moon system is going at the right orbital speed around the sun, and it explains why the two bodies are constantly moving away from one another while their orbital speed is constantly decelerating without needing to use the less evident tidal friction explanation, which is so unclear to me that I decided to start a thread about it. Naturally, it also applies to the rotation of the solar system around the galaxy core, which should also be recessing. The only data that I found about that is the slow deceleration of the probes that we send throughout the solar system, for which the NASA is still looking for an explanation. If the orbital motion of the probes is constantly recessing, and if the calculations don't account for that, they are certainly not at the place where they are expected. They should actually be farther from the sun than expected, and moving at a slower orbital speed around it than expected. No need to change the gravitational theory to explain the phenomenon though, just the tidal one.

Let me now try to test my principle while applying it to the orbital rotation of the stars belonging to two different orbiting galaxies. If my proposition is right, while two galaxies would be orbiting around one another, the stars belonging to one of the galaxies would not be orbiting at the right orbital speed with regard to the other galaxy during their transit, so they should also be recessing slowly from the galaxy core with time, and the outer ones would necessarily be recessing faster than the inner ones for all their orbital distances to stay proportional. Now, the more the galaxy we are observing is far from us, the less it would have time to recess, so we see it as it was when the orbits of its stars were much closer to one another, thus also when their respective orbital speeds were much faster than they actually are, but without the shape of the galaxy looking any different from the one of the closer galaxies that we observe. Now the big question: if our estimate of the distant galaxies' dimension is wrong, could that mechanism replace the dark matter one?

[rmolnav]

I said that the proper rotation of the planet had something to do with the fact that the far and the near sides are not going at the right orbital speed, and I must defend that feature otherwise my idea wouldn't work. Of course, that kind of rotation also produces an equatorial bulge, but even if the rotation is as tiny as the moon's one, that's what makes its far and near sides not going at the same orbital speed as its center of gravity. These sides could very well be going at the right speed if they weren't tidally locked for instance, and in this case, if my explanation is right, there would be no tides.

I have to go more slowly through your post, but for now I have to say, as I´ve already done many times, that the two cases are quite different ...
As all earth particles follow identical circular paths (of a radius equal to the distance between earth CM and barycenter), "the fact that the far and the near sides are not going at the right orbital speed" hasn´t actually any reasonable meaning ... There is no a "proper" orbiting around the moon !
But being the moon tidal locked to the earth, and orbiting around the barycenter (to say around the earth is sufficiently right too...), that would allow you to apply your idea.
With my approach (and NOAA´s scientists, "French Tides" ...), further not visible side, apart from having a tangential speed bigger than the moon CM and still bigger than closer side, centrifugal forces are also the further the bigger (ω²r per unit of mass). That increases tidal effects, because not only the so called "differential gravity" counts for tides.
That rotation could be considered as the addition of a revolving without rotation (as earth moves), and a spinning around its own axis over same some 28 days of the orbiting.
But the effect of that "spinning" shouldn´t be considered like an "equatorial bulge" as what happens on earth ... Moon´s one is part of earth-moon global dynamics, but earths´s one (with much, much higher angular speed) is not whatsoever ...   

[Halc]

... the tides would be higher if the day were longer since it would give the water more time to move to where gravity is shoving it.

Again: you continue to mix daily earth spinning with moon related tides … Gravity and inertial effects (centrifugal forces) don´t "need" longer days to build full-size tides, because they have plenty of time to deform ocean surface: a little more than a week from low to high tide !!

A little more than 6 hours actually.  A week between spring tide and whatever they call the low between them.  The shoving I spoke of happens every 6 hours.  The river by me runs backwards twice a day due to that shoving.

[Halc]

I know at least another way to analyze the tides: tangential speed. I used it before in this thread but it didn't seemed to interest anybody either. I said that while the earth's C.G. was going at the right orbital speed around the moon, neither of its near or far sides were going at the right speed: due to the daily rotation of the earth, the near side is going too slow and the far side is going too fast. If we accelerate a satellite which is ,  already on a circular orbit, it will get away from the earth, and if we slow it down, it will get closer, so the same thing should happen to the two halves of the earth while it orbits around the moon.

This is a valid analysis. I thought about it and decided it’s easier to think of the moon where you don’t have the complication of the barycentre being inside the earth.
The moon will orbit with its centre of mass at an equilibrium position based on the balance between centrifugal force and gravitational force (rotating, noninertial frame). As you say, the inner part will want to move to a lower orbit and the outer to a higher orbit.

My bold.  I disagree.  This explanation predicts that tides would be negative if the Earth spun the other way since the inner part has the greatest tangential speed and wants to move outward, and the outer part has the least tangential speed and wants to move inward.  Venus is such a case (it rotates backwards), at it has normal solar tides towards and away from the sun just like Earth.

[rmolnav]

... the tides would be higher if the day were longer since it would give the water more time to move to where gravity is shoving it.

Again: you continue to mix daily earth spinning with moon related tides … Gravity and inertial effects (centrifugal forces) don´t "need" longer days to build full-size tides, because they have plenty of time to deform ocean surface: a little more than a week from low to high tide !!

A little more than 6 hours actually.  A week between spring tide and whatever they call the low between them.  The shoving I spoke of happens every 6 hours.  The river by me runs backwards twice a day due to that shoving.

Your basic error comes from same misconception as if somebody said moon (or sun) rotates around earth once every 24 hours ... That is only an apparent movement !
"Sublunar" bulge is always almost in line with the moon, and its actual cycle is, as moon´s, 28/29 days ... A little more than 7 (rather than 6) days  ahead of next low tide !!
Solid earth below (and partially ocean deep water) experiences a daily spin (which has nothing to do with earth-moon dynamics), and  as we also move with solid earth,  what you say apparently happens !!

[rmolnav]

Quote from: Colin2B on 23/09/2018 14:51:12
This is a valid analysis. I thought about it and decided it’s easier to think of the moon where you don’t have the complication of the barycentre being inside the earth.
The moon will orbit with its centre of mass at an equilibrium position based on the balance between centrifugal force and gravitational force (rotating, noninertial frame). As you say, the inner part will want to move to a lower orbit and the outer to a higher orbit.
My bold.  I disagree.  This explanation predicts that tides would be negative if the Earth spun the other way since the inner part has the greatest tangential speed and wants to move outward, and the outer part has the least tangential speed and wants to move inward. 

You keep insisting on your error, probably because you haven´t read (or disagree with) some of my recent posts, one of them #328:

... what quoted from NOAA´s scientits work IS QUITE CORRECT:
" since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon”

ALL material earth points have the same tangential speed (considering only moon-earth "dance"), because their paths are identical circles, logically with same cycle period (28/29 days) ...