I Why the Moon always shows the same face

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The Moon always shows the same face to Earth due to tidal locking, a phenomenon explained by classical Newtonian mechanics. The discussion critiques alternative theories suggesting uneven density in the Moon's core as a reason for this, emphasizing that such views lack scientific basis. Tidal forces create a torque that eventually synchronizes the Moon's rotation with its orbit, leading to the observed stability. The uneven mass distribution on the Moon, particularly the concentration of denser materials on the near side, facilitates this process. Ultimately, the consensus explanation is supported by extensive scientific research and modeling, leaving little room for speculation.
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From Earth, we always see precisely the same half of the moon.
Isn't the simplest explanation for this odd fact that the Moon's core is not uniformly dense, and it's centre of gravity is situated some distance further way from us than its geometric centre but on a direct line extended beyond the centre of its face from the centre of gravity of the Earth? The heaviest bit will always be the furthest away.
If you stick a weight on one side of a tennis ball, put it in a bucket of water on a rope and swing it round fast, the weighted side of the floating ball will stay furthest way from you.
The common consensus explanation has to do with our tides but I don't buy that.
 
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Tony Hambro said:
The common consensus explanation has to do with our tides but I don't buy that.
The consensus explanation builds on a basic understanding of classical Newtonian mechanics and can be shown mathematically from Newton's laws of motion. Your "explanation" is, forgive the expression, word sallad with no basis in Newtonian physics. Gravitation does not work like your example with the bucket and the tennis ball. That is based on a pressure gradient and potential energy in a rotating frame of reference with dissipation of energy to the surroundings.

In the end, it does not matter if you "buy" it or not. It is the correct explanation. It is up to you if you want to learn actual physics or base your understanding of the world upon your own imagination.
 
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Tony Hambro said:
From Earth, we always see precisely the same half of the moon.
Isn't the simplest explanation for this odd fact that the Moon's core is not uniformly dense, and it's centre of gravity is situated some distance further way from us than its geometric centre but on a direct line extended beyond the centre of its face from the centre of gravity of the Earth? The heaviest bit will always be the furthest away.
If you stick a weight on one side of a tennis ball, put it in a bucket of water on a rope and swing it round fast, the weighted side of the floating ball will stay furthest way from you.
The common consensus explanation has to do with our tides but I don't buy that.
That's not the reason for the water in the bucket. There is a rope pulling the handle towards the person from an off-centre position on the bucket. The Moon has no rope and, left to itself, it would rotate at some other rate for ever. The 'real' reason is that parts of the Earth and Moon are free to flow a small amount as the other body's gravity pulls at it and acts as a Drag force between the two. That is a mechanism for loss of rotational energy and tends to dissipate differential rotational energy. Eventually the two will end up facing each other all the time. Because there has to be conservation of angular momentum, the actual separation will increase, eventually. But we are talking in terms of a verrrry long time. That link from @Orodruin will tell you the details. His explanation is perfectly correct but may need a bit of translation. I have given a Noddy Version for you. :smile: No matter if you didn't need it; someone else may find it useful.
PS the Moon rocks a bit from side to side so we actually see a bit more than half of it over a period of time.
 
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sophiecentaur said:
PS the Moon rocks a bit from side to side so we actually see a bit more than half of it over a period of time.
It also "nods" up and down as its axis of rotation is perpendicular to the plane of its orbit around the Earth.

This video shows this libration.
 
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Tony Hambro said:
From Earth, we always see precisely the same half of the moon.
Yes, look up - see the dark features that make up the Man in the Moon? When Soviet satellites first imaged the far side of the moon in the early 60's, they discovered that there were no similar features, only a heavily cratered surface. It turns out that those dark features were more than just a difference in coloration. After the Apollo missions it was learned that those dark areas, called Maria, consisted of a very dense mineral called basalt, while the rest of the crust was made of much lighter silica.
Tony Hambro said:
If you stick a weight on one side of a tennis ball, put it in a bucket of water on a rope and swing it round fast, the weighted side of the floating ball will stay furthest away from you. The common consensus explanation has to do with our tides but I don't buy that.
Yes, I can see why you'd be skeptical - if the gravity of the Earth acted on the Moon the same way that the rope acted on the weighted tennis ball, then the heavier side of the Moon should be facing away from the Earth, not towards it. Well, in the tennis ball example, it's not the rope that's exerting a force on the ball, it's a fictitious force from the curved motion called centrifugal force, which is directed outward, not inward. The weighted side of the ball, being heavier, having more inertia, is affected more by this outward directed force. The near side of the Moon however, is affected more by the inward force of gravity, and so faces towards the Earth, rather than away.
 
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Orodruin said:
The consensus explanation builds on a basic understanding of classical Newtonian mechanics and can be shown mathematically from Newton's laws of motion. Your "explanation" is, forgive the expression, word sallad with no basis in Newtonian physics. Gravitation does not work like your example with the bucket and the tennis ball. That is based on a pressure gradient and potential energy in a rotating frame of reference with dissipation of energy to the surroundings.

In the end, it does not matter if you "buy" it or not. It is the correct explanation. It is up to you if you want to learn actual physics or base your understanding of the world upon your own imagination.
 
The explanation given in Wikipedia, that the bulge in the shape of the moon from the stronger gravity on the nearer side facing the Earth produces torque that eventually cancels rotation, may be true if the moon really does bulge enough. But if the moon’s centre of gravity is not in its geometric centre (who knows?), as with my weighted tennis ball in a bucket of water swung round on a rope, it would produce a similar effect, the face with the weight always being furthest away from the centre of rotation. So both views could be true. Incidentally neither explanation tells us why the moon/ball has no rotation around the axis linking the two centres of gravity. Any ideas?
 
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Tony Hambro said:
The explanation given in Wikipedia, that the bulge in the shape of the moon from the stronger gravity on the nearer side facing the Earth produces torque that eventually cancels rotation, may be true if the moon really does bulge enough. But if the moon’s centre of gravity is not in its geometric centre (who knows?)...
Scientists know. The shape and mass distribution of the moon is very accurately known, and scientists have mathematically modeled its behavior. There are no open questions warranting speculation here.
...as with my weighted tennis ball in a bucket of water swung round on a rope, it would produce a similar effect, the face with the weight always being furthest away from the centre of rotation.
I'm sorry, but repeating it does not help. Moreover you've made your scenario overly complicated; you don't need to swing the bucket, just set it down on the floor and the tennis ball assumes the same orientation; center of gravity low due to gravity and buoyancy. The moon does not have buyoyancy.
So both views could be true.
Sorry, but no. You're going to have a lot of trouble learning science if you want to develop the answers yourself. It is much better to learn from the thousands of scientists who have spent hundreds of years figuring this stuff out.
Incidentally neither explanation tells us why the moon/ball has no rotation around the axis linking the two centres of gravity. Any ideas?
If the Moon was formed in an impact, the impact would impart a rotation in the direction of a motion...like bouncing a tennis ball at an angle does.
 
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  • #10
alantheastronomer said:
the heavier side of the Moon should be facing away from the Earth, not towards it.

alantheastronomer said:
The near side of the Moon however, is affected more by the inward force of gravity, and so faces towards the Earth, rather than away.

You appear to be claiming that the near side of the Moon is heavier than the far side. That is not correct. And it is not necessary to account for the Moon keeping the same face to the Earth. Tidal locking accounts for that, as has already been pointed out several times.
 
  • #11
PeterDonis said:
This is not correct. The rope does exert a force on the ball.

The rope is not connected to the ball, it is connected to the bucket. The centripetal force on the ball is that from the pressure gradient of the water in the bucket.
 
  • #12
Orodruin said:
The rope is not connected to the ball, it is connected to the bucket.

Oops, you're right, I misread that in the OP.
 
  • #13
PeterDonis said:
You appear to be claiming that the near side of the Moon is heavier than the far side. That is not correct. And it is not necessary to account for the Moon keeping the same face to the Earth. Tidal locking accounts for that, as has already been pointed out several times.
The GRAIL spacecraft provides a map of lunar surface gravity that shows a clear asymmetry between the near and far sides. An uneven mass distribution is a facilitator of tidal locking, not meant as a substitute for it.
 
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  • #14
alantheastronomer said:
The GRAIL spacecraft provides a map of lunar surface gravity that shows a clear asymmetry between the near and far sides.

Ah, that's right, I was forgetting about the mass concentrations beneath the maria, which are concentrated on the near side.
 
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  • #15
alantheastronomer said:
An uneven mass distribution is a facilitator of tidal locking, not meant as a substitute for it.
Yes. That's the point. If there is no internal friction, the Moon could rotate or swing back and forth like a pendulum for ever. To get a stable situation, you need damping.
 
  • #16
sophiecentaur said:
If there is no internal friction, the Moon could rotate or swing back and forth like a pendulum for ever. To get a stable situation, you need damping.

I don't see how an unequal mass distribution (a better term might be "unsymmetrical" since nobody expects the density to be the same everywhere) is required to get frictional damping. Even a perfectly symmetrical Moon won't stay that way under tidal forces; it will be distorted, and the distortion will vary with time as the Moon swings back and forth about the equilibrium position. This time-varying distortion will cause friction in the Moon's material and hence will damp the vibrations.
 
  • #17
PeterDonis said:
I don't see how an unequal mass distribution (a better term might be "unsymmetrical" since nobody expects the density to be the same everywhere) is required to get frictional damping. Even a perfectly symmetrical Moon won't stay that way under tidal forces; it will be distorted, and the distortion will vary with time as the Moon swings back and forth about the equilibrium position. This time-varying distortion will cause friction in the Moon's material and hence will damp the vibrations.
It's a matter of Gravitational Potential of an asymmetrical body* in different orientations and not just the separation of the CMs. PE is added and taken away and the KE changes accordingly - it's not uniform rotation. Tidal effects are due to the forces on the body as Energy changes cyclically. Where there's any movement (water or molten core) there can be losses.
In your response I think there's a hidden a truth(?) and that is there can be movement and losses, for any shape if there is any non-uniformity and some liquidity (if that's the right word.) A lop-sided distribution will presumably make the tidal losses faster.

*Do the sums for a dog bone shape and a nearby sphere. The GP when the bone lies along a radius and when at right angles to the radius are easy to calculate and the difference is apparent. (Bed time for me though!:wink:)
 
  • #18
sophiecentaur said:
It's a matter of Gravitational Potential of an asymmetrical body

No asymmetry of gravitational potential is required for tidal gravity to be present. A perfectly spherically symmetric, non-rotating Earth would have a perfectly spherically symmetric gravitational potential, but would still cause tidal gravity in the space around it. Of course the non-sphericity and rotation of the actual Earth add additional effects.

sophiecentaur said:
Tidal effects are due to the forces on the body as Energy changes cyclically.

No, they're due to tidal gravity, i.e., an extended body like the Moon, moving in the tidal gravity of the Earth, will be stretched in the radial direction and squeezed in the tangential directions. This will happen even if the Moon's orbit is perfectly circular and its kinetic and potential energies never change. Of course the non-circular orbit of the actual Moon means the tidal gravity it feels won't be exactly the same as this idealized example.

sophiecentaur said:
there can be movement and losses, for any shape if there is any non-uniformity and some liquidity (if that's the right word.)

"Non-uniformity" ("asymmetry" would be a better word for what I think you mean here) is required only in the sense that the tidal gravity will stretch the object in some directions and squeeze it in others. It is not required for either the Earth or the Moon themselves to have any asymmetry.

sophiecentaur said:
A lop-sided distribution will presumably make the tidal losses faster.

I haven't done the math but intuitively this seems right to me, yes.
 
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  • #19
How would tidal gravity operate? Whet mechanism would be at work?
 
  • #20
Tidal forces are due to the gravitational acceleration changing from point to point, leading to a different acceleration of different parts of the same object (leading to internal stress and thereby deformation if the object is deformable). As long as the separation ##d\vec x## between two points is small, the difference in gravitational acceleration at the points is given by
$$
\delta \vec g = \vec g(\vec x + d\vec x) - \vec g (\vec x) \simeq d\vec x \cdot \nabla \vec g.
$$
In terms of the gravitational potential ##\Phi## such that ##\vec g = - \nabla\Phi##, you would have
$$
\delta \vec g \simeq - \vec e_j dx^i \partial_i \partial_j\Phi = - (d\vec x \cdot \nabla) \nabla \Phi.
$$
Thus, tidal forces are best described by the rank 2 tensor that is the second derivative of the gravitational potential. You do not need any asymmetric objects to generate a potential with non-zero second derivatives. A point mass works just fine (although the object affected needs to be extended for the tidal forces to be noticeable).
 
  • #21
Try rotating a well-balanced bicycle wheel vs. rotating a poorly balanced one.
A well-balanced bicycle wheel set in motion would still slow down because of friction against hub and air, and eventually stop. But it would slow down and stop without any rocking back or forth in the end, and stop in a random position each time. Whereas a poorly balanced bicycle wheel would slow down and stop at the same specific position each time, and oscillate before stopping.
Further, since a well-balanced wheel stops at random position, even a small nudge will make a small but permanent movement, accumulating over time.
Moon acts like an ill-balanced wheel, not like a well-balanced one.
 
  • #22
sophiecentaur said:
How would tidal gravity operate? Whet mechanism would be at work?

What mechanism makes gravity in general work? In either Newtonian gravity or General Relativity, tidal gravity is no more mysterious than gravity in general.
 
  • #23
snorkack said:
Try rotating a well-balanced bicycle wheel vs. rotating a poorly balanced one.
A well-balanced bicycle wheel set in motion would still slow down because of friction against hub and air, and eventually stop. But it would slow down and stop without any rocking back or forth in the end, and stop in a random position each time. Whereas a poorly balanced bicycle wheel would slow down and stop at the same specific position each time, and oscillate before stopping.
Further, since a well-balanced wheel stops at random position, even a small nudge will make a small but permanent movement, accumulating over time.
Moon acts like an ill-balanced wheel, not like a well-balanced one.
The problem with this description (the OP's take) is that you have two separate forces: one holding the wheel up at its geometric center and another pulling down at the center of mass. The moon in this model (where we ignore tides) only has one force, pulling "down" (toward Earth) on the center of mass. There's no second force to provide the torque (couple).

There is no geometric bias with the tidal force. The fact that the moon is oriented this way is a coincidence or result of a different process such as biased meteor bombardment.
 
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  • #24
snorkack said:
Try rotating a well-balanced bicycle wheel vs. rotating a poorly balanced one.
For fairness, you need to either do this in a weightless environment (difficult) or with a rigidly mounted axle (easier). Do not hold it in your hands or use an upside-down bicycle. You do not want to compromise the experiment by having an energy dump into a partially damped axle.

You should find that a small motion imparted to an ill balanced wheel persists for a long time, just like a well balanced wheel. It is just that it is a pendulum motion rather than a rotary one.
 
  • #25
russ_watters said:
The problem with this description (the OP's take) is that you have two separate forces: one holding the wheel up at its geometric center and another pulling down at the center of mass. The moon in this model (where we ignore tides) only has one force, pulling "down" (toward Earth) on the center of mass. There's no second force to provide the torque (couple).

There is no geometric bias with the tidal force. The fact that the moon is oriented this way is a coincidence or result of a different process such as biased meteor bombardment.
Are you sure there is no "geometric bias"?

Tidal force is principally a difference between two forces. Centrifugal repulsion and gravitational attraction. They add to zero over the whole Moon, but since the repulsion increases with distance (proportional to it) and attraction decreases (with inverse square), the far side of Moon experiences more repulsion and near side more attraction. Both are thus away from centre of Moon.

Now what kind of asymmetry can cause Moon to librate?
Take the example of bicycle wheel again.
A well balanced wheel is a perfect circle with hub in its exact centre.
An ill-balanced wheel might be imbalanced in several ways.
It might be a perfect circle but with hub off circle. Then it would, on a road, bump once each turn - and when rotating on suspended hub, swing off the hub as a pendulum.
It might be elliptical, with hub at centre. A wheel like this would on the road bump twice each turn - but rotating on suspended hub, the two long ends of the elliptic wheel would balance each other exactly, so it would rotate freely then.
Or it might have higher order deviations. A square wheel would not roll on a road, but its corners would balance each other when hanging in the air.
In case of Moon, elliptical Moon would experience a torque. Two bulges would be attracted to near and far sides respectively.
But higher order asymmetry? Would a cubic moon turn one face (or edge or corner) to Earth, or would the perfect balance of pairs of opposing corners leave Moon freely rotating?
 
  • #26
snorkack said:
Tidal force is principally a difference between two forces. Centrifugal repulsion and gravitational attraction.

No, it isn't. The difference between those two forces is the net "acceleration due to gravity" on a body in a rotating frame, not the tidal force. In the language you are using here, tidal force would be the "difference in the difference" from one point to another.

Also, tidal gravity is present in an inertial frame, where there is no centrifugal force at all, so thinking of it in terms of centrifugal force is not the best way to think of it. A much better way was given by @Orodruin in post #20.

snorkack said:
what kind of asymmetry can cause Moon to librate?

No asymmetry in the Moon itself is necessary for the Moon to librate. The effects of the asymmetries due to the eccentricity and inclination of the Moon's orbit, plus the asymmetry between the Moon's period of rotation and the Earth's, are sufficient.
 
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  • #27
russ_watters said:
There is no geometric bias with the tidal force. The fact that the moon is oriented this way is a coincidence or result of a different process such as biased meteor bombardment.
I think it would be a good idea if someone specified what they actually mean by the terms "tidal gravity" and "tidal locking". Is it just the forces involved or is it to do with distortion of a body (and friction losses)? Clearly, it is the result (can be described in terms) of Newtonian Gravity. The forces on any distributed mass will be different in all parts of the mass (CM is just an approximation) and the geometry of the object will have an influence on those forces.
I am sure that any form of "locking" must be due to the loss of Energy due to distortion and friction and the tendency towards a minimum of Potential Energy in the system. It's not clear which parts of the argument are accepted or rejected by contributors to the thread so could I have a response to that last sentence?
 
  • #28
PeterDonis said:
No asymmetry in the Moon itself is necessary for the Moon to librate.
I haven't understood why a symmetrical, rigid body would have any potential minimum dependent on its orientation. I think we need to be defining terms more precisely as we may well not be disagreeing.
 
  • #29
sophiecentaur said:
I haven't understood why a symmetrical, rigid body would have any potential minimum dependent on its orientation. I think we need to be defining terms more precisely as we may well not be disagreeing.
In a gravitational field with a uniform gradient, it would not. However, the Earth's gravitational field is neither uniform nor does it have a uniform gradient.

Edit: I have to correct myself. A stick is symmetrical, but nonetheless has an orientation-dependent potential in a field with uniform gradient.
 
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  • #30
sophiecentaur said:
I think it would be a good idea if someone specified what they actually mean by the terms "tidal gravity" and "tidal locking".

@Orodruin gave a good explanation of tidal gravity in post #20.

To understand tidal locking, consider the Earth-Moon system. Not only does each body exert tidal gravity on the other, the tidal gravity of each on the other changes with time, for a number of reasons:

The Earth is rotating much faster than the Moon revolves around it;

The Moon's orbit about the Earth is inclined to the Earth's equator;

The Moon's orbit about the Earth is not perfectly circular.

All three of these things cause the deformation of both the Earth and the Moon due to tidal gravity to change with time, which causes frictional damping and torques on both bodies. These effects act to slow the Earth's rotation, to make the Moon's orbit less inclined to the Earth's equator, and to make the Moon's orbit more circular. Also, conservation of angular momentum means that as the Earth's rotation slows, the Moon's average distance from the Earth increases.

At some point in the future, the Earth's rotation will have slowed to exactly match the Moon's rate of revolution about the Earth; the Moon's orbit will be in the Earth's equatorial plane; and the Moon's orbit will be perfectly circular. At that point, the deformation of each body due to the tidal gravity of the other will no longer be time-dependent; it will be "locked" in a certain configuration on each body. Once that point is reached, the orbital parameters will stop changing, and the system is said to be tidally locked.
 
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  • #31
PeterDonis said:
No asymmetry in the Moon itself is necessary for the Moon to librate. The effects of the asymmetries due to the eccentricity and inclination of the Moon's orbit, plus the asymmetry between the Moon's period of rotation and the Earth's, are sufficient.
That´´ s forced libration, not free. And has nothing to do with Moon showing the same face to Earth.
Suppose that Moon were a perfectly elastic perfect sphere of cheese, with craters and seas painted on with weightless paint, exactly equal density black and white, for markers.
Then Moon would have forced libration due to eccentricity - the orbital movement would be nonuniform due to eccentricity, but rotation uniform. Craters and seas would rock back and forth over the edge of Moon... but there would be no free libration. Because no restoring force. Nothing would stop Moon´´ s rotation period from being just slightly different from orbital period, causing the craters and seas to slowly drift from near side to edge, vanish to far side after several times of rocking back and forth and eventually cross far side and return from the other edge.

Some asymmetry of Moon must allow free libration, and therefore near side staying near side.
 
  • #32
snorkack said:
That´´ s forced libration, not free.

Forced or free, the causes I gave are the ones that account for the observed libration of the Moon. So if there is any other kind, its magnitude must be comparatively very small.

snorkack said:
And has nothing to do with Moon showing the same face to Earth.

The causes I gave certainly do. See below.

snorkack said:
Because no restoring force.

Yes, there is: tidal gravity. See my explanation of tidal locking in post #30. That explanation applies perfectly well to a Moon that would be spherically symmetric in the absence of tidal gravity, because in the presence of tidal gravity, it is not spherically symmetric: there are tidal bulges on it, and if the Moon's period of rotation is different from its period of revolution around the Earth, those tidal bulges will move, which will create a restoring force tending to make the two periods the same.

It is true that, with a Moon that would be slightly asymmetric in the absence of tidal gravity, there will be an additional force tending to make the mass asymmetry radial (whereas with a Moon that was perfectly symmetric in the absence of tidal gravity, there would be no preference for any particular orientation once it was tidally locked--but that would not stop some orientation from being tidally locked). However, there are two local equilibria for this: more mass towards the Earth, and more mass away from the Earth. The latter equilibrium is only local; the former is the global one. But if the Moon were stuck in the latter local equilibrium, it's quite possible that it could stay there, since the forces acting on it might be insufficient to kick it out of the local equilibrium hard enough to overcome the potential barrier between that local equilibrium and the global one.
 
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  • #33
PeterDonis said:
@Orodruin gave a good explanation of tidal gravity in post #20.
Yes. That post describes well how the force / acceleration on each part of the object varies over the object and with its orientation. However, the term "tidal gravity" is confusing because it implies more than that simple description. It can cause 'tidal' movement when the object is not rigid so why use the term "tidal" when what is really meant is "tide-causing" or even "anisotropic", which would be precise? I guess my problem here is that 'tidal' can refer to both a cause and an effect. Many of the red herrings that have appeared in the thread seem to be based on just that confusion.
The rest of your post is fine by me - even if the 'details' of the Earth - Moon motions are more involved than necessary for a general discussion.
I still think that these things are better dealt with in terms of Energy, rather than forces and torque. Every such system will eventually reach the condition of least potential energy but the forces approach doesn't make that as clear as it should, imo.
On the whole, this thread has been pretty useful at getting to the right conclusion; I got a lot from it.
 
  • #34
sophiecentaur said:
I guess my problem here is that 'tidal' can refer to both a cause and an effect.

In both Newtonian gravity and General Relativity, the term "tidal gravity" is pretty standardly used to refer to the cause: in Newtonian terms, the difference in the "acceleration due to gravity" from place to place; in GR terms, spacetime curvature (or sometimes specifically Weyl curvature). One of the advantages of GR is that it gives a separate term, "spacetime curvature", which unambiguously refers to the cause and doesn't give rise to any confusion with various effects, as you say the word "tidal" does.
 
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  • #35
sophiecentaur said:
Every such system will eventually reach the condition of least potential energy

I'm not sure I would specify potential energy here. Consider the Earth-Moon system: as tidal gravity slows the Earth's rotation, it moves the Moon further away (to conserve angular momentum), so the potential energy of the system increases. (This is balanced by a decrease in kinetic energy due to slowing the Earth's rotation and the Moon's orbital speed.)
 
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  • #36
PeterDonis said:
I'm not sure I would specify potential energy here.
You are right, of course - the total orbital Energy needs to be considered and it will go to a minimum when locking is reached.
 
  • #37
PeterDonis said:
The causes I gave certainly do. See below.

Yes, there is: tidal gravity. See my explanation of tidal locking in post #30. That explanation applies perfectly well to a Moon that would be spherically symmetric in the absence of tidal gravity, because in the presence of tidal gravity, it is not spherically symmetric: there are tidal bulges on it, and if the Moon's period of rotation is different from its period of revolution around the Earth, those tidal bulges will move, which will create a restoring force tending to make the two periods the same.
Does not follow.
If Moon were perfectly elastic, the tidal bulges would move - but without creating any force.
Even if a force existed, it need not be a restoring force.
PeterDonis said:
It is true that, with a Moon that would be slightly asymmetric in the absence of tidal gravity, there will be an additional force tending to make the mass asymmetry radial (whereas with a Moon that was perfectly symmetric in the absence of tidal gravity, there would be no preference for any particular orientation once it was tidally locked--but that would not stop some orientation from being tidally locked).
It would.
A rubber tire is not axisymmetric when supported on road. Because the weight on the tire presses into the point of support.
But if you turn it then it rolls without bumps. Any part of the tire will be equally compressed when loaded. Therefore there will be no restoring force.
A rubber tire does exert a force against movement. But it is a retarding force, not restoring one. A rolling wheel comes to a stop, but at a random new position - it will not return to its previous position.
So, a wheel without a parking brake on will move slightly but permanently on any small nudge. Over time and many small nudges, it adds up.
Whereas a wheel with a parking brake on does have a restoring force and returns to a fixed position when displaced by a small nudge.

Moon shows one side because Moon possesses a permanent asymmetry - Moon acts like a wheel with parking brake on, not like a wheel without a parking brake on.
 
  • #38
snorkack said:
- but without creating any force.
No net force at equilibrium? That always has to be true but surely what counts is not Force but Work done on the system, with or without Hysteresis.
PeterDonis said:
no preference for any particular orientation once it was tidally locked--but that would not stop some orientation from being tidally locked)
snorkack said:
It would.
If there was perfect symmetry, the orientation at which it would lock would be indeterminate. It would just go slower and slower, as with any exponential damping. Any slight Energy Minimum / dip would eventually be the orientation where locking would occur. (See where Energy considerations can be useful?)
snorkack said:
A rubber tire does exert a force against movement. But it is a retarding force, not restoring one.
Why make the distinction? Both forces are present; if you cut the drive, the tyre will spring back to another position (restoring force) which will change the slip angle. The tyre analogy doesn't really add to this argument, imo; we are past analogies at this point.
 
  • #39
snorkack said:
If Moon were perfectly elastic, the tidal bulges would move - but without creating any force.

So what? The actual Moon is not perfectly elastic. But a perfectly symmetric Moon (i.e., symmetric in the absence of tidal gravity) would still experience tidal locking, which is what I've been saying.

snorkack said:
Even if a force existed, it need not be a restoring force.

If you don't like the term "restoring force", then just say "force". The point is that even if the Moon were perfectly spherically symmetric in the absence of tidal gravity, there will still be a force tending to make the Moon's orbit circular and in the Earth's equatorial plane, and to make the Earth's period of rotation the same as the Moon's period of revolution around the Earth. You don't need a asymmetric Moon for that.

snorkack said:
Moon shows one side because Moon possesses a permanent asymmetry

No, it shows one side because that part of tidal locking in the Earth-Moon system has had enough time to run to completion. Whereas the parts where the Moon's orbit is circular in the equatorial plane, and the Earth's rotation has the same period as the Moon's revolution, have not.

What is true is that, if the Moon did not have an asymmetry (meaning an asymmetry that would be there in the absence of tidal gravity), which side of the Moon permanently faced the Earth as a result of tidal locking would have been a random result, since no side would have been preferred. But since the Moon does have such an asymmetry, if some other side had ended up originally facing the Earth as a result of tidal locking, i.e., if the asymmetry were not radial, there would have been an additional force that gradually changed things until the asymmetry was radial.
 
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  • #40
snorkack said:
If Moon were perfectly elastic, the tidal bulges would move - but without creating any force.

Actually, even this is not true. There would still be torques on the tidal bulge even with a perfectly elastic Moon.
 
  • #41
sophiecentaur said:
the total orbital Energy needs to be considered and it will go to a minimum when locking is reached.

Actually, the main mechanism of tidal locking is torques on the tidal bulges, for which no dissipation is required, so the total energy of the system actually remains constant.

In actual objects there is also dissipation going on, but its effects are very small compared to the effects of the torques. Even in the absence of dissipation, the tidally locked configuration is an equilibrium because any small perturbation from it will set up torques that tend to restore it.

The Wikipedia page gives a good overview:

https://en.wikipedia.org/wiki/Tidal_locking
 
  • #42
PeterDonis said:
for which no dissipation is required, so the total energy of the system actually remains constant.
No problem there but. without losses there will always be oscillatory motion about some Energy minimum. The Moon would go past its maximum orbit size and then return ad infinitum.
Tell me why there is such bad reception for the Energy - based argument. Somewhere along the line, the Force approach will need to consider Work in and Work out and a 'hysteresis' style of diagram. A good answer is available, based on Energy which needs no complicated calculations; it's based on fundamentals and needs no details.
 
  • #43
sophiecentaur said:
without losses there will always be oscillatory motion about some Energy minimum
The Moon would go past its maximum orbit size and then return ad infinitum.

That's not quite true either as you state it. There will indeed be oscillations about an equilibrium, but these are not oscillations about an energy minimum, because there is no energy minimum; all the relevant states of the system have the same total energy. The equilibrium is an equilibrium of zero torque, but it's only meta-stable (if that's the right word); it's not an energy minimum.

The effect of dissipation is to slowly reduce the total energy of the system; what that does is slowly move the equilibrium of zero torque to a smaller average distance between the two bodies (i.e., make the system more tightly bound). This will not damp oscillations about that equilibrium, because the oscillations are not due to the system having more energy than some minimum at equilibrium.

sophiecentaur said:
Tell me why there is such bad reception for the Energy - based argument.

I don't think it's a matter of "bad reception". I just think that for this particular problem, since the primary process of tidal locking conserves total energy, thinking in terms of energy doesn't help to understand that primary process.
 
  • #44
Isn’t a stable equilibrium situation at an energy minimum? I realize my assumptions are not based on orbits but where is the difference in principle? How can friction forces not lose energy and bring the system to a limiting low level of energy?
An alternative view can be valid also even if it’s not conventional.
 
  • #45
sophiecentaur said:
Isn’t a stable equilibrium situation at an energy minimum?

It depends on what you mean by "stable equilibrium". In the case under discussion, the equilibrium of zero torque for a system subject to tidal locking, like the Earth-Moon system, is not an energy minimum, since the forces involved are conservative and so all states of the system accessible with those forces have the same energy. So if "stable equilibrium" requires an energy minimum, then this equilibrium is not one. That's a matter of terminology, not physics.

sophiecentaur said:
How can friction forces not lose energy and bring the system to a limiting low level of energy?

Friction forces do. But friction forces aren't what make tidal locking happen. Please read the Wikipedia article I linked to in post #41. Friction forces do what I said the "effects of dissipation" are in post #43; that's not the same as tidal locking. Briefly, friction forces change the equilibrium of zero torque; but they are not what drives the system towards whatever that equilibrium is.
 
  • #46
PeterDonis said:
But friction forces aren't what make tidal locking happen.
Have I misunderstood the actual meaning of Tidal Locking then? Is it correct to say that he Moon faces the Earth due to Tidal Locking? Is it not true that it is losses that caused that basically stable situation? Without losses (purely elastic deformations) why would the situation be as it is?
 
  • #47
sophiecentaur said:
Have I misunderstood the actual meaning of Tidal Locking then?

I don't know. What do you think tidal locking is? Did you read the Wikipedia article?

sophiecentaur said:
Is it correct to say that he Moon faces the Earth due to Tidal Locking?

Yes, with a correct understanding of what "tidal locking" means.

sophiecentaur said:
Is it not true that it is losses that caused that basically stable situation?

No, it's not true. Tidal locking would happen even with a perfectly elastic Earth and Moon.

sophiecentaur said:
Without losses (purely elastic deformations) why would the situation be as it is?

Because tidal locking is due to torques exerted by the gravity of each body on the tidal bulges in the other. Please read the Wikipedia article I linked to; it gives a good explanation of the mechanism. Feel free to ask questions if something there isn't clear to you.
 
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  • #48
PeterDonis said:
The Wikipedia page gives a good overview:
It's full of different instances of the effect and it has some good graphics but it doesn't actually include the words 'energy' or 'loss' and the word 'friction' only occurs in the bibliography. That, to my mind does not constitute a full explanation of what happens between Earth and Moon. The article makes mention of the rapid drop-off of the effect (inverse cube law), which explains why the effect is only seen in bodies with close separation.
If you are using the Wiki article as your source of information for me then there is no wonder you are not including the Energetic principles used. Wiki should always be read with care and using it as a slope source can be risky. Physics (and Science in general) advances by starting with simple systems and trying to understand the actual mechanism at work.
PeterDonis said:
with a correct understanding of what "tidal locking" means.
Wiki does not supply that. It would be nice if you were to address my issue, rather than to ignore it and refer me only to that article. If you don't understand what I am getting at then, fair enough, say so but don't just dismiss it.
 
  • #49
PeterDonis said:
Actually, even this is not true. There would still be torques on the tidal bulge even with a perfectly elastic Moon.
No. If Moon were perfectly elastic, there would be tidal bulges, but there would be no torques whatsoever on the tidal bulges, because the tidal bulges would move freely with rotation of Moon and be always exactly aligned with the tidal forces.
 
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  • #50
snorkack said:
No. If Moon were perfectly elastic, there would be tidal bulges, but there would be no torques whatsoever on the tidal bulges, because the tidal bulges would move freely with rotation of Moon and be always exactly aligned with the tidal forces.
+1
No loss element would mean no phase shift and, hence, no net torque.
 
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