Tidally locked planets/moons

[Paul Wally]

Are tidally locked moons or planets really tidally locked over very long time periods? I.e. is there something analogous to precession of the Earth's axis where the planet or moon slowly rotates over a period of say 10 000 years, or does "tidally locked" really mean that such relative rotation will never happen, not in a million years?

[grapes]

Are tidally locked moons or planets really tidally locked over very long time periods? I.e. is there something analogous to precession of the Earth's axis where the planet or moon slowly rotates over a period of say 10 000 years, or does "tidally locked" really mean that such relative rotation will never happen, not in a million years?Pretty much. But I can imagine other scenarios.

[Hornblower]

Are tidally locked moons or planets really tidally locked over very long time periods? I.e. is there something analogous to precession of the Earth's axis where the planet or moon slowly rotates over a period of say 10 000 years, or does "tidally locked" really mean that such relative rotation will never happen, not in a million years?

I cannot imagine what you mean by "something analogous to precession".

As a close-in body approaches a synchronous spin rate, it will take on a slightly oblong shape from the tidal action, and that should lock it into a stable truly synchronous rotation.

[lone77star]

I remember reading somewhere about a relativistic effect on Mercury which forces that "tidally locked" body to rotate. The rotation relative to revolution is slight, but not based on angular inertia.

[Paul Wally]

I cannot imagine what you mean by "something analogous to precession".

As a close-in body approaches a synchronous spin rate, it will take on a slightly oblong shape from the tidal action, and that should lock it into a stable truly synchronous rotation.

With "analogous" I mean that the relative rotation is just as slow relative to the orbital period as precession is relative to the Earth's orbital period. So I was wondering whether it is not also possible for a body to be slightly asynchronous spin rate, so that the effect is only noticeable after many orbital periods. I don't know what to call it, what about asynchronous drift?

We say that the moon is tidally locked to the Earth, so the same side will always face the Earth, but what if we go back a million years, did the same side still face the Earth then?

[Jeff Root]

A tidally-locked moon or planet can only get that way if it is
at least slightly asymmetrical. The Earth's Moon, for example,
is quite asymmetrical, with a much thicker crust on the far
side than on the Earth-facing side. It could probably have
ended up exactly the other way around, but it was destined
to become tidally locked one way or the other.

For a brief second or two I was thinking of saying something
about the Moon's libration, then remembered that what I was
thinking is not correct, but it is worth pointing out because
others could make the same mistake. The Moon rotates at
a very constant rate. It doesn't speed up and slow down
over short time periods. Its orbital speed around the Earth
does change continuously throughout the month, though, as
its distance changes. The combination of constant rotation
speed and varying orbital speed result in the majority of the
side-to-side libration seen every month. The libration does
mean that the "center" of the Moon's near side is not aimed
exactly at Earth more than occasionally, but on average, it
is very constant.

-- Jeff, in Minneapolis

[Paul Wally]

A tidally-locked moon or planet can only get that way if it is
at least slightly asymmetrical. The Earth's Moon, for example,
is quite asymmetrical, with a much thicker crust on the far
side than on the Earth-facing side. It could probably have
ended up exactly the other way around, but it was destined
to become tidally locked one way or the other.

-- Jeff, in Minneapolis

So is original asymmetry the cause of tidal locking or can it still happen with a perfectly spherical body? How does tidal locking happen, is it a gradual process or is it sudden. If it is gradual then it must approach the tidally locked state asymptotically such that it looks locked when it is in fact not quite yet so.

[Hornblower]

So is original asymmetry the cause of tidal locking or can it still happen with a perfectly spherical body? How does tidal locking happen, is it a gradual process or is it sudden. If it is gradual then it must approach the tidally locked state asymptotically such that it looks locked when it is in fact not quite yet so.

If the spinning body were absolutely spherical when undistorted by tidal stress, and perfectly uniform in composition, elasticity, etc., I would expect it to asymptotically approach a true lock, but not quite get there.

[Jeff Root]

Original asymmetry enables tidal locking and makes it happen
more readily. As Hornblower said, tidal forces elongate the bodies.
Tides raised in a solid generate a lot of friction, so they tend to
retain their shape to some degree. When a body approaches the
synchronous spin rate, the tidal bulge will be in one place for a
long time, so we can easily imagine it getting stuck there -- a
permanent bulge in one location, enabling the tidal lock.

-- Jeff, in Minneapolis

[grapes]

A tidally-locked moon or planet can only get that way if it is
at least slightly asymmetrical.Tides alone make the body spherically asymmetrical, which is enough to cause tidal braking, which would start the process of tidal locking.

[Paul Wally]

Now when such a body finally approaches the tidally locked state, does it make one last very long rotation or does it kind of go back and forth like a pendulum until it finally settles down. The Wikipedia articles says that tidal locking is caused by gravitational gradient and torque.

[grapes]

It's complicated. :)

As Jeff points out, the librations of the moon mean that it is not absolutely locked onto the earth even now, it rocks back and forth--even though its rotation rate matches its revolution rate, essentially "locked."

[George]

So is original asymmetry the cause of tidal locking or can it still happen with a perfectly spherical body? If both bodies were magically perfectly spherical and homogenous all the time then there would be nothing to grab hold of one another to alter either bodies' rotation or orbital change. I think this is correct. The other answers are the more appropriate real universe answers, but the magical approach is certainly helpful.

[grapes]

If both bodies were magically perfectly spherical and homogenous all the time then there would be nothing to grab hold of one another to alter either bodies' rotation or orbital change. I think this is correct. The other answers are the more appropriate real universe answers, but the magical approach is certainly helpful.You mean, perfectly rigid so that there is no tide? Even in a perfect sphere, the tidal distortion produces the handles that can cause tidal braking.

[Paul Wally]

I'm trying to understand how the tidal braking actually works, in terms of Newtonian mechanics. Does the rotation slow down because of energy dissipated in the tidal action friction or does it still happen if we ignore friction. Wikipedia says that angular momentum is conserved and that the slow down in rotation is balanced by a change to a higher orbit, but doesn't this suggest that friction is not a big player in the slowdown process?

Say, if we take an exaggerated case, a long rigid rod for example and its rotating around a central axis and its moving in an orbit around a planet, and there is a significant difference in gravity between the two ends when they're aligned with the center of the planet. Why must the rotation slow down if there is no internal friction and no friction in space?

[George]

You mean, perfectly rigid so that there is no tide? Even in a perfect sphere, the tidal distortion produces the handles that can cause tidal braking. Yes, I should have said perfectly rigid, though, I suppose, only one body needs to be pefectly rigid (ie inealstic). [Of course, I could argue that my "perfectly spherical" implies perfectly rigid. But you know how I like to eschew obfuscation. :)] Any bulging by both would produce the necessary handles. I'm not 100% sure I'm right on this answer, however, because I've raised this question before and don't recall ever getting a clear answer since the real universe hasn't any inealstic bodies, or homogeneous, or spherical, or... :)

[Oops, I think both bodies must be inealstic.]

[chornedsnorkack]

If a body were fluid (not magical or unrealistic) then tides would elongate the sphere, plastically.

There would be no tidal lock and no libration. For if the distorted fluid sphere were slightly turned relative to the tide, it would deform so that the new elongation would be aligned to the tide exactly, and there would be no restoring force.

Of course fluid flow causes friction; but as the rotation relative to the tide slows down, the resistance force decreases proportional to the square of the flow speed, so there would be slow remaining rotation.

Now consider perfectly elastic solid perfect sphere. (this case is unrealistic.)

Tidal force would elastically elongate it. But since the deviation from sphere is caused solely by elastic response to force, if the sphere were turned relative to the tide then the elastic bulge would be turned to exactly the new position of tidal force. So again absolutely no restoring force, no libration and no tidal lock.

Now consider a perfectly rigid solid nonspherical body. (again unrealistic)

When it is slightly out of alignment with tidal force, there is a restoring force. It will librate. Since it is perfectly rigid, it will have no damping, and the amplitude of the libration is constant until changed. If the amplitude of damping were big enough, the body could be unlocked - it could rotate with rotation significantly slowed down as it approaches 90 degrees, but after crossing 90 degrees accelerate again, and then slow down to the same speed at 270 degrees.

Again consider a perfectly elastic solid nonspherical body (again unrealistic).

When the permanent tidal bulge is turned out of alignment with the tidal force, elastic deformation would cause the body to deform, so that the temporary bulge would be somewhere else than the permanent bulge. But it would also be somewhere else than the direction of tides - so again a restoring force and libration. But perfectly elastic body also has no damping, so again constant amplitude libration, or constant average speed rotation.

A solid but not perfectly elastic body (which is realistic) will respond to a small disturbance with damped free libration, and when it dissipates it returns precisely to its previous position.

A fluid body will have no restoring force and no free libration, so a small disturbance will cause it to rotate and the rotation will slow down but last forever.

Will the accumulated rotation angle, over all eternity, of a fluid body damped from a slow initial disturbance, converge to a finite value or go to infinity?

I am pretty sure about the case of a ship pushed off from rest. Work is the product of distance and force. The total kinetic energy is proportional to square of speed - and so is resistance force. Therefore the ship will cover infinite distance.

Unless it encounters a bigger resistance. Such as resistance force proportional to speed rather than its square, which would bring it to halt in a finite distance.

[grapes]

A solid but not perfectly elastic body (which is realistic) will respond to a small disturbance with damped free libration, and when it dissipates it returns precisely to its previous position.

A fluid body will have no restoring force and no free libration, so a small disturbance will cause it to rotate and the rotation will slow down but last forever.

Will the accumulated rotation angle, over all eternity, of a fluid body damped from a slow initial disturbance, converge to a finite value or go to infinity?Damping is typically from friction, and that is key in tidal braking, in getting things off line.


I am pretty sure about the case of a ship pushed off from rest. Work is the product of distance and force. The total kinetic energy is proportional to square of speed - and so is resistance force. Therefore the ship will cover infinite distance.

Unless it encounters a bigger resistance. Such as resistance force proportional to speed rather than its square, which would bring it to halt in a finite distance.You seem to be saying that a real ship in real water, given a push, will go forever. We know that's not true, right?

[chornedsnorkack]

You seem to be saying that a real ship in real water, given a push, will go forever. We know that's not true, right?

Do we?

Real trains on real wheels come to stop because solid friction is a nonzero force when approaching and reaching rest. Not so with fluid resistance. Fluids take the shape of vessel, exactly, because a force no matter how small will cause them to flow.

A real ship in real water, when not anchored or moored to any solid, will drift albeit slowly under no matter how feeble force. (And in the absence of any force at all, it undergoes Brownian motion).

Will a real ship also keep drifting under inertia, or will it come to a complete stop in a finite time?

[grapes]

Do we? Yes, I think so. :)

Isn't that what we mean by viscosity?

[Paul Wally]

Now consider a perfectly rigid solid nonspherical body. (again unrealistic)

When it is slightly out of alignment with tidal force, there is a restoring force. It will librate. Since it is perfectly rigid, it will have no damping, and the amplitude of the libration is constant until changed. If the amplitude of damping were big enough, the body could be unlocked - it could rotate with rotation significantly slowed down as it approaches 90 degrees, but after crossing 90 degrees accelerate again, and then slow down to the same speed at 270 degrees.



So this suggests that there can be no tidal locking without some form of dissipation, usually internal friction, caused by tidal action, and damping is a function of the mechanical properties of the material.

I'm just a bit perplex about conservation of angular momentum in the Wikipedia article on tidal locking. If this is a dissipative system how can angular momentum be conserved? It must be that the slow down in rotation is at most only partially balanced by movement to a higher orbit. On the other hand, the principle of conservation of angular momentum, suggests that tidal locking could be possible even when there is no dissipation, but what would cause this to happen in the case of perfect conservation of angular momentum?

[grapes]

So this suggests that there can be no tidal locking without some form of dissipation, usually internal friction, caused by tidal action, and damping is a function of the mechanical properties of the material. Friction doesn't have to be dissipative. The friction of a conveyor belt surface allows the belt to do work, with virtually no induced dissipation, from that friction.

Basically, what is necessary is hysteresis--the lag of the tidal bulge, in "catching up" with the line between the bodies. The fast rotation of the earth moves this bulge ahead of the moon, and it is pulled back (the farside bulge has the opposite, but slightly (though significantly) weaker effect).

[swampyankee]

I'm sure that everybody here knows that for a body to become tidally locked, it must have some level of internal damping; without this mechanism, the rotational kinetic energy of the body will not be dissipated as heat. Most solids have fairly low levels of internal damping, at least when stressed within the elastic limit.

A perfectly elastic or perfectly rigid body would never become tidally locked, regardless of its shape.

[grapes]

I'm sure that everybody here knows that for a body to become tidally locked, it must have some level of internal damping; without this mechanism, the rotational kinetic energy of the body will not be dissipated as heat. Why must it be dissipated as heat?

[Githyanki]

If you run Celeste and increase the time, eventually, you'll see the Dark-Side of the Moon facing Earth; could be an error in programing but still...

[swampyankee]

Why must it be dissipated as heat?

What else would it be dissipated as in vacuo?

[grapes]

What else would it be dissipated as in vacuo?Good point. :)

[publius]

What else would it be dissipated as in vacuo?

Gravitational radiation. :) But that's only significant in highly relativistic systems like binary pulsars and colliding black holes. IIRC, the earth-sun system is radiating about 300W as gravitational radiation, which is nothing.

[Jeff Root]

I don't understand the physics here, but I would think that much
of the kinetic energy of one body would simply be transferred to
the other body, or become gravitational potential energy, rather
than heat.

-- Jeff, in Minneapolis

[swampyankee]

Gravitational radiation. :) But that's only significant in highly relativistic systems like binary pulsars and colliding black holes. IIRC, the earth-sun system is radiating about 300W as gravitational radiation, which is nothing.

Well, that answers my question. :doh: My excuse is that I'm an engineer, and we rarely (as in never) deal with relativistic systems, so gravitational radiation is negligible.

[Paul Wally]

I don't understand the physics here, but I would think that much
of the kinetic energy of one body would simply be transferred to
the other body, or become gravitational potential energy, rather
than heat.

-- Jeff, in Minneapolis

This is what I'm trying to understand. Can tidal locking be understood as such a process from e.g. potential to kinetic, without any dissipation as heat or what have you. Or perhaps from an unstable to stable state without any dissipation, like soap bubbles settle at a stable configuration? Basically, can tidal locking (theoretically) happen with angular momentum perfectly conserved (macroscopically of course)? I think it's about minimization of some variable.

[George]

This is what I'm trying to understand. Can tidal locking be understood as such a process from e.g. potential to kinetic, without any dissipation as heat or what have you. Or perhaps from an unstable to stable state without any dissipation, like soap bubbles settle at a stable configuration? Basically, can tidal locking (theoretically) happen with angular momentum perfectly conserved (macroscopically of course)? I think it's about minimization of some variable.
Conversion from PE to KE happens all the time and I am confident it applies in the Earth-Moon case. The rotational drag upon Earth due to the bulge or other mass variation is "felt" by the Moon in a way that causes the Moon to be pulled faster (KE) along in its orbit. This causes the Moon to constantly move to a larger orbit.

The friction issue is a side affect that the 2nd law of thermo addresses. When you drop a rock onto the ground you convert PE to KE and heat is generated due to friction, so the process is never quite reversible.

[Hornblower]

Let us consider an isolated 2-body system. The total angular momentum is conserved, period. The total amount of kinetic energy is not. As tidal interaction slows down the rotation of a primary and forces the satellite into a higher orbit, part of the kinetic energy is transformed into increased gravitational potential energy. If the primary has significant internal friction or hysteresis, some additional kinetic energy is transformed into heat. If the body is perfectly insulated so the heat cannot escape, the total energy of the system is conserved. The final state of the system as it approaches a synchronous spin/orbit state depends on the amount of internal friction. In a thought experiment we could end up with a hot primary spinning relatively fast, and the satellite in a relatively low and short period synchronous orbit, or a cold primary spinning somewhat more slowly with the satellite in a higher, longer period synchronous orbit.

If I am not mistaken, internal friction is not necessary for tidal interaction to slow down the rotation of the primary. I expect the tidal bulges to be carried forward of the point of conjunction with the satellite by the momentum of the material. This creates the necessary retrograde gravity vector on the bulge, which is balanced by the prograde component acting on the satellite. The presence of friction would increase the angle between the bulge and the satellite, thus hastening the spindown.

Even with the friction, we may not get a true lock if the body is perfectly uniform, since the amount of drag will diminish as the the rotation rate diminishes. The spin will decelerate more quickly, but it still could asymptotically approach a lock without quite getting there.

[chornedsnorkack]

If I am not mistaken, internal friction is not necessary for tidal interaction to slow down the rotation of the primary. I expect the tidal bulges to be carried forward of the point of conjunction with the satellite by the momentum of the material. This creates the necessary retrograde gravity vector on the bulge, which is balanced by the prograde component acting on the satellite. The presence of friction would increase the angle between the bulge and the satellite, thus hastening the spindown.


If there were no friction, the tidal bulge might be carried by rotation all around the primary, and be subject to prograde gravity vector when it is behind the point of conjunction.

[Jeff Root]

If there were no friction, the tidal bulge might be carried by
rotation all around the primary, and be subject to prograde
gravity vector when it is behind the point of conjunction.
I think you have it backwards. If there were no friction,
the tidal bulge would not be carried forward at all, but
would be directly under the satellite.

If the tidal bulge could be carried forward a large distance,
it would mean that it was not collapsing under its own weight
plus the backward pull from the satellite. Unless it was a local
feature like the uplifted Tharsis region on Mars or a continent
on Earth, the bulge would be an addition to the centrifugal
bulge, making the entire body more oblate.

-- Jeff, in Minneapolis

[Jeff Root]

If the body is perfectly insulated so the heat cannot escape,
the total energy of the system is conserved.
I think it would be simpler just to include escaping heat
as part of the system.

-- Jeff, in Minneapolis

[George]

As tidal interaction slows down the rotation of a primary and forces the satellite into a higher orbit, part of the kinetic energy is transformed into increased gravitational potential energy. Yes, but how?

I can imagine something like the Moon being pulled a bit closer to the Earth when a more masive side of the Earth rotates towards the Moon. When this higher density area of the Earth rotates passed the Earth Moon line it will pull the Moon along with it slightly, which gives more KE to the Moon advancing its orbital distance. [I'm ignoring a moon's rotation and any mascons to see if this element of the total transfer effect is correct in your view. I'm also ignoring, for the moment, the tidal bulge affect on the angular momentum transfer.]

[Jeff Root]

As tidal interaction slows down the rotation of a primary and
forces the satellite into a higher orbit, part of the kinetic energy
is transformed into increased gravitational potential energy.
Yes, but how?
This seems pretty straightforward to me and you essentially
have it.



I can imagine something like the Moon being pulled a bit
closer to the Earth when a more massive side of the Earth
rotates towards the Moon.
Well, not that bit. The Moon doesn't get pulled closer.



When this higher density area of the Earth rotates past the
Earth Moon line it will pull the Moon along with it slightly, ...
Let's just consider the tidal bulge raised by the Moon. It is
always ahead of the Moon because Earth's rotation carries
it there continuously.



... which gives more KE to the Moon, advancing its orbital
distance.
That's about it. Plus slowing Earth's rotation, of course,
because the Moon is always pulling back on that bulge.

-- Jeff, in Minneapolis

[publius]

You know, this is a rather complex subject. What we have with tidal locking is a transfer of energy via the gravitational field from one body to another. This is called "gravitational induction" or inductive energy transfer or variations thereof. What is happening in the earth-moon system is some of the earth's rotational kinetic energy is transfered to the moon's orbital energy, and rotational angular momentum is converted to orbital angular momentum, a type of spin-orbit coupling.

With EM, all field energy flow is handled by the Poynting vector, E x H. This handles all energy flow, radiative as well as near field transfers from one system to another. The trouble with Newtonian gravity is we have no H, no gravitomagnetic component, and no wave equation. When we go to GR, of course we get a (much more complex) wave equation, and a H_g/B_g notion, which figures prominently in the so-called GEM, gravito-electromagnetic linear approximation.

But what of Newtonian gravity itself, with no GEM framework? Turns out you can define a sort of Poynting vector, and this works with instananeous Newtonian gravity. Bondi did a lot of work on this, and this energy flow thing is usually called Bondi's Newtonian Poynting vector or similiar. I forget how it goes, but if Phi is the gravitational potential, Bondi's vector goes something like

P_bondi ~ d(Phi)/dt*del(Phi) - Phi*d(del Phi)/dt, where del is the del operator. There is constant involving G and pi in front, I forget exactly what.

You should be able to come up with something like that for EM, by letting mu_0 go zero, and thus c to infinity. This would mean that B would vanish and we'd just have an electric field that updated instantly.

In this one gets no radiation, only near field transfers of energy, and you can show with Bondi's vector that there must a receiver for any transmitter of energy via the field.

And in GR, it can be shown that Bondi's vector falls out of a weak field approximation and all that good stuff.


-Richard

[publius]

Yeah, I was just able to sastify myself than the Poynting expression will become Bondi's vector in the limit as mu --> 0 and thus
c --> infinity up to the minus sign of gravitational attraction for like charges rather than repulsion. And I'm almost certain, though I can't prove it and can't easily find such a proof, that the process of gravitational inductive transfer requires some friction dissipation mechanism.

One thing Bondi was able to do was demonstrate a system where such an inductive transfer took place, but there was no net change in gravitational potential energy after the transfer. Thus you had a "pure" inductive transfer of energy and momentum from one body to another via the gravitational field.

The fascinating thing is that this Newtonian "induction", mediated by an instantaneous field is conceptually a very different thing than GR (and EM's) radiative transfers of energy, the latter two having wave equations and thus an intutive notion of energy and momentum being moved around by the field. What Bondi and others have done is shown that the two are mathematically equivalent, with GR reducing to the Newtonian "induction" in the appropiate weak field limits.

The thing is Newtonian gravity can have no radiation. Radiation is "throwing energy away to infinity" and is a consequence of the finite propagation speed. An instantaneous field can have no radiation. But it can still transfer energy.

In the GR model, the earth-moon system is radiating. But there is a near-field component that accomplishes the transfer of earth's rotational energy to the moon's orbital energy. A tiny, tiny little bit is carried away to infinity, but most of it just goes from earth to moon. In Newtonian induction, there is no far field radiation, only the near field transfer.

-Richard

[George]

This seems pretty straightforward to me and you essentially
have it. It is straightforward, but show me the devil. How does the transfer take place?


Well, not that bit. The Moon doesn't get pulled closer. If a more massive element of the Earth rotates into a position that is closer to the Moon then why wouldn't the mutual graviational effect bring them a tiny bit closer to one another?

The bigger factor that applies, I suspect, is the frictional story for the bulge, so perhaps my point is not that big a deal, though maybe it is.

If I imagine the Earth and Moon are fixed and held in place, I can easily imagine a bulge for both. But this bulge is symmetric. If I put the Earth in rotation then I can now see the bulge's symmetry become offset as the water bulge trys to slide over the rotating hard stuff beneath it. Friction will drag some of the bulge along with it. Also, land masses swinging around will push on the water bugle. These effects, perhaps more, will cause the bulge to be ahead of the Earth-Moon line and will cause the Moon to feel a tugging vector that pulls it forward in its orbit, once we restore the Moon from a held position to a normal orbit.


Let's just consider the tidal bulge raised by the Moon. It is
always ahead of the Moon because Earth's rotation carries
it there continuously. Yes, but I had to imagine why it isn't symmetric first. The assymetry seems to be critcal, though perhaps the mountains and lunar mascons are not so trivial either.

It takes all of this to explain the broader statements.

[George]

In the GR model, the earth-moon system is radiating. But there is a near-field component that accomplishes the transfer of earth's rotational energy to the moon's orbital energy. A tiny, tiny little bit is carried away to infinity, but most of it just goes from earth to moon. In Newtonian induction, there is no far field radiation, only the near field transfer. Always nice to see your take on these things, though GR is only something I can spell.

Induction seems to be the perfect word to describe the energy transfer but I don't see the GR view here. Is the warpage of spacetime and, perhaps frame dragging, the heart of the energy transfer? Would a leading bulge on Earth produce a slightly steeper "grade" along the spacetime lines (geodesics?)?

[Hornblower]

To review the gravitational mechanism of tidal interaction, in Newtonian terms:

The Moon's gravity gradient raises tidal bulges on the Earth, whose rotation drags them off center from the Moon's angular position. The bulge on the near side causes a slight prograde component in the gravity acting on the Moon, while the bulge on the far side, slightly farther away, causes a somewhat weaker retrograde component. Thus there is a net prograde action on the Moon which forces it to spiral out slowly as if pushed by a steady thruster directed tangent to its orbital path. At any moment in time the Moon is moving slightly faster than it would if unperturbed at the same distance from the Earth, but it slows down as it spirals out. Both bodies lose kinetic energy, but the gravitational potential energy increases and the total energy is conserved.

At these low energy levels a GR calculation gives virtually the same results as the Newtonian version.

[George]

To review the gravitational mechanism of tidal interaction, in Newtonian terms:

The Moon's gravity gradient raises tidal bulges on the Earth, whose rotation drags them off center from the Moon's angular position. The bulge on the near side causes a slight prograde component in the gravity acting on the Moon, while the bulge on the far side, slightly farther away, causes a somewhat weaker retrograde component. Thus there is a net prograde action on the Moon which forces it to spiral out slowly as if pushed by a steady thruster directed tangent to its orbital path. At any moment in time the Moon is moving slightly faster than it would if unperturbed at the same distance from the Earth, but it slows down as it spirals out. Nicely stated.

What about density asymetries? Any idea how much they contribute and what about the gain from eastward moving shorelines that add to the bulge, I think?

[publius]

Induction seems to be the perfect word to describe the energy transfer but I don't see the GR view here. Is the warpage of spacetime and, perhaps frame dragging, the heart of the energy transfer? Would a leading bulge on Earth produce a slightly steeper "grade" along the spacetime lines (geodesics?)?

Danged if I know. :lol: To produce some descriptive language of what's going in the GR picture, one would need to understand that picture and the math of it all very well. And even then, some descriptive language might not easily be constructed. But frame dragging, which can be called gravitomagnetism, the gravitational analog of the B field, doesnt' have anything much to do with it. The terms are very small, and the only significant thing is the g field, the E field analog. The induction goes all through g. The Newtonian picture works well enough, and there's absolutely no B_g there.

But I'll give you an EM analogy. Imagine a radio transmitter, some big radio station thing. If you get close enough with a system of your own, you can couple to that antenna, and pull energy out of it, say heating water with a resistor driven by currents induced by the field or something fancier. Or you can imagine some power line coupling, a fun topic I once went through, where one steals energy from a long transmission line by setting up a clever coupling mechanism.

The thing about this is it is "near field" coupling. You change the system via the close proximity of your parasitic system. Thus things look different to the source when the near field device is in place. In the antenna case, however the system is radiating on its own, in the power line case, only a vanishingly small amount.

So inductive transfer is like the near field coupling, and gravitational radiation is the throwing away of energy to the wind on its own. In Newtonian gravity you can have the former but not the latter. In EM and GR gravity, you can have both.

Some crazy analogy you might play around with in your mind's eye that comes to mind -- the trouble is gravity is so darned weak that something like would require incredible masses, but here goes:

Imagine some driver mass. You've got a piece of nuetron star material and you wiggle it up and down. As you do that, the g field all around changes. No imagine a little mass off to the side. You restrain it so it can't get any closer horizontally to the driver mass, but it can slide up and down. As you move the driver up and down, the other mass will try to wiggle up and down with it, and you have friction and thus energy being transfered.

The restraint and the friction is going to change how much force you must exert and work you have to do to keep wiggling the driver mass up and down. Stop wiggling the driver mass, and the system will settle back in the same configuration as when you started, same gravitational potential energy. Thus you accomplish a clean transfer of energy via the gravitational field.

-Richard

[George]

Danged if I know. :lol: Yeah, translating to georgeeze can require some real contortions. Few probably realize just how sympathetic I am to their efforts. My contribution is to be a catalyst to pedagogory. :)


To produce some descriptive language of what's going in the GR picture, one would need to understand that picture and the math of it all very well. And even then, some descriptive language might not easily be constructed. But frame dragging, which can be called gravitomagnetism, the gravitational analog of the B field, doesnt' have anything much to do with it. The terms are very small, and the only significant thing is the g field, the E field analog. The induction goes all through g. The Newtonian picture works well enough, and there's absolutely no B_g there.

But I'll give you an EM analogy. Imagine a radio transmitter, some big radio station thing. If you get close enough with a system of your own, you can couple to that antenna, and pull energy out of it, say heating water with a resistor driven by currents induced by the field or something fancier. Or you can imagine some power line coupling, a fun topic I once went through, where one steals energy from a long transmission line by setting up a clever coupling mechanism. Yes, that sounds like induction, all right, but where's the gravitational devil?


So inductive transfer is like the near field coupling, and gravitational radiation is the throwing away of energy to the wind on its own. In Newtonian gravity you can have the former but not the latter. In EM and GR gravity, you can have both. I keep thinking that GR is about spacetime gradients altered by mass. Is there a radiation component or is it just the flip side of the GR coin, where, say gravitons, are tossed out of mass whereby they act directly upon space to alter the pathways (ie warpage). In that imaginative sense, I suppose I can find more touchyness.


Imagine some driver mass. You've got a piece of nuetron star material and you wiggle it up and down. As you do that, the g field all around changes. No imagine a little mass off to the side. You restrain it so it can't get any closer horizontally to the driver mass, but it can slide up and down. As you move the driver up and down, the other mass will try to wiggle up and down with it, and you have friction and thus energy being transfered.

The restraint and the friction is going to change how much force you must exert and work you have to do to keep wiggling the driver mass up and down. Stop wiggling the driver mass, and the system will settle back in the same configuration as when you started, same gravitational potential energy. Thus you accomplish a clean transfer of energy via the gravitational field. Hmmm. I see the inductive view whenever the other mass moves, but not how friction contributes anything but irreversibility.

[Jeff Root]

This seems pretty straightforward to me and you essentially
have it.
It is straightforward, but show me the devil. How does
the transfer take place?
You essentially have it.

The gravitational gradient of each body makes the other
somewhat prolate. A bulge on each body aimed at the other.
Earth's rotation carries its bulge ahead. It takes time for the
bulge to rise and to slump back down after the Moon goes
over any part of the Earth, so it is always ahead of the Moon.
The Moon pulls back on that bulge. That part of the Earth
has a force on it pulling it back, slowing its motion. And the
bulge pulls the Moon forward, speeding its motion, which
results in an increase in the Moon's orbital distance, which in
turn slows its motion to less than it was to begin with.

Since the speeding, rising, and slowing of the Moon all take
place simultaneously, the Moon never actually speeds up -- it
just moves away and slows down.



If a more massive element of the Earth rotates into a position
that is closer to the Moon then why wouldn't the mutual
graviational effect bring them a tiny bit closer to one another?
If one side of the Earth had greater density than the rest, then
Earth's center of mass (center of gravity) would be on that side.
The planet rotates about that center of mass/gravity, not the
geometric center. The Moon's distance from Earth's center of
gravity would be constant even if its distance from the dense
side varies.



If I imagine the Earth and Moon are fixed and held in place, I
can easily imagine a bulge for both. But this bulge is symmetric.
If I put the Earth in rotation then I can now see the bulge's
symmetry becomes offset as the water bulge tries to slide over
the rotating hard stuff beneath it. Friction will drag some of the
bulge along with it. Also, land masses swinging around will push
on the water bulge. These effects, perhaps more, will cause the
bulge to be ahead of the Earth-Moon line and will cause the Moon
to feel a tugging vector that pulls it forward in its orbit, once we
restore the Moon from a held position to a normal orbit.
I'm now thinking that friction isn't absolutely required. It helps,
but I think the bulge would move forward even without friction.
In any case, sliding isn't required.

I think the friction isn't needed because it takes some time for
the material of the Earth to move from one position to another
in response to the Moon's gravity, just because mass has to
accelerate -- it doesn't move about instantaneously. So the
bulge raised by the Moon should move ahead of the Moon even
if the Earth were entirely zero-viscosity fluid.

Fluid isn't required, either. As long as the Earth can deform from
the Moon's gravity, a bulge will be raised that will move ahead
of the Moon's position. Imagine the Earth as a big ball of Jello.
It is essentially a solid (there's a technical name for what Jello
actually is, a colloid), and can easily be deformed, but no part
slips over any other part. Every part is securely connected to
the parts around it. So the Moon's gravity raises a bulge in the
Jello Earth, and the rotation of the Earth moves the bulge ahead
of the Moon. The bulge pulls on the Moon and the Moon pulls
on the bulge. No need for water and land.

I don't know what that kind of deformation is called, but it might
be called "elastic deformation", in parallel to "plastic deformation"
such as you get with soft clay.

I think lunar mountains and mascons are pretty small factors.
They can help to produce tidal locking, but don't contribute to
changing Earth's rotation and the Moon's orbit.

-- Jeff, in Minneapolis

[Jeff Root]

I agree with that dastardly fast-typing hornblower who
said the Moon moves in its orbit faster than it would
without the bulge on Earth pulling it along, for the
distance it is from Earth's center of gravity. Ignoring
the fact that its orbit is actually elliptical, the Moon's
speed would be reducing constantly as it spirals out,
but the speed is always slightly higher than it would
be without the bulge.

-- Jeff, in Minneapolis

[Hornblower]

Nicely stated.

What about density asymetries? Any idea how much they contribute and what about the gain from eastward moving shorelines that add to the bulge, I think?

The shorelines drag the sea water part of the tidal bulge farther off center than it would be otherwise, and thus increase the magnitude of the tidal interaction.

I would expect density asymmetries to put a small ripple into the otherwise steady effect. My educated guess is that this effect would be slight in comparison to the aforementioned coastal water effects.

[chornedsnorkack]

A solid, frictionless pendulum off exact resonance will not be drawing energy from exciting force. When the exciting force is in exact resonance, the amplitude of the pendulum shall diverge without bound; but out if resonance the amplitude should acquire a definite value.

Likewise a perfect fluid sloshing without friction should not take any energy from tidal force, unless it is in exact resonance.

[George]

If one side of the Earth had greater density than the rest, then
Earth's center of mass (center of gravity) would be on that side. Right, and that would be true if a planet only had one mass offset, but their are entire ranges of mountains and other density anomalies. Imagine a perfectly rigid sphere where centroid and c.g. are the same point. Now add a gaint mountain range that is a North-South ring around the planet. This would have the effect I described due to the inverse square law. The approaching mountain range would pull the moon closer without the wobble factor you mentioned. Of course, I think we are both right so that Earth's behavior is a combination of both, yet the net would contribute to the Moon's orbital drift. [I'm just thinkin' out loud on this, as usual.]

[George]

The shorelines drag the sea water part of the tidal bulge farther off center than it would be otherwise, and thus increase the magnitude of the tidal interaction. Thanks, that's how I see it, too.


I would expect density asymmetries to put a small ripple into the otherwise steady effect. My educated guess is that this effect would be slight in comparison to the aforementioned coastal water effects. If the tidal bulge asymmetry were greater I would agree, but a mountain range seems to have more potential impact on the process, though more of a pulse rather than a steady tug that tides produce. I assume a perfectly symmetric tidal bulge, though imaginary only, would cause no momentum transfer, or am I wrong?

[Jeff Root]

Imagine a perfectly rigid sphere where centroid and c.g. are
the same point. Now add a gaint mountain range that is a
North-South ring around the planet. This would have the effect
I described due to the inverse square law. The approaching
mountain range would pull the moon closer without the wobble
factor you mentioned.
I don't think I mentioned a wobble. Were you thinking of the
"ripple" that hornblower suggested in the post after mine?
I'm probably the one who is confused.

It seems to me that the mountain ring can be considered as
just a ring, without the planet. I'm not sure that helps any,
except that it might be something that can be looked up:
The gravity of a ring.

-- Jeff, in Minneapolis

[HenrikOlsen]

What else would it be dissipated as in vacuo?
Not dissipated but exchanged for kinetic and potential energy as the distance between the orbiting bodies is increased.

[grapes]

If the tidal bulge asymmetry were greater I would agree, but a mountain range seems to have more potential impact on the process, though more of a pulse rather than a steady tug that tides produce. I assume a perfectly symmetric tidal bulge, though imaginary only, would cause no momentum transfer, or am I wrong?You mean farside/nearside bulge symmetry? The bulges are very nearly symmetric, and are, for purposes of calculating the tidal slowing.

You can exaggerate this effect to the point where it does make a difference, though, like in the gravitational gradient stabilization of satellites (http://en.wikipedia.org/wiki/Space_tether#Gravitational_Gradient_Stabilization) .

[Hornblower]

Thanks, that's how I see it, too.

If the tidal bulge asymmetry were greater I would agree, but a mountain range seems to have more potential impact on the process, though more of a pulse rather than a steady tug that tides produce. I assume a perfectly symmetric tidal bulge, though imaginary only, would cause no momentum transfer, or am I wrong?

I think you are mistaken. See my explanation in post #43.

I would expect any lateral perturbation from fixed mascons such as mountains to be alternating between prograde and retrograde, while the effect of the tidal bulge is prograde all the time.

[George]

I think you are mistaken. See my explanation in post #43. Am I? [It wouldn't be a first.]

Note that I said "perfectly symmetric". How could a perfecly symmetric bulge, which implies zero friction, act to prograde the Moon's orbit? If this hypothetical case can do so, or even if it can't, then it should prove effective in helping us understand what seems to be taking place, at least in a Newtonian sense.


I would expect any lateral perturbation from fixed mascons such as mountains to be alternating between prograde and retrograde, while the effect of the tidal bulge is prograde all the time. Their contribution would be the same as the Earth's density asymmetry -- whatever these regions are called. The greater the proximity of the mass concentrations, the greater the gravitational attraction (their product), the greater will be the effect on both bodies.

[Hornblower]

Am I? [It wouldn't be a first.]

Note that I said "perfectly symmetric". How could a perfecly symmetric bulge, which implies zero friction, act to prograde the Moon's orbit? If this hypothetical case can do so, or even if it can't, then it should prove effective in helping us understand what seems to be taking place, at least in a Newtonian sense.I did not realize that by "perfectly symmetric" you meant perfectly in conjunction with the Moon. I don't think it could be that way even with no friction. As an increment of Earth's body comes around toward conjunction with the Moon, it will be rising away from the center as a result of the gravitational gradient and will have momentum in that direction. My hunch is that it will continue to rise a bit longer after passing the conjunction point, for the same reason that a rocket that is lifting off vertically will rise a while longer after the engine cuts off. This would put the bulge in the prograde position. Here I yield to those with some real expertise in this analysis.



Their contribution would be the same as the Earth's density asymmetry -- whatever these regions are called. The greater the proximity of the mass concentrations, the greater the gravitational attraction (their product), the greater will be the effect on both bodies.

These mascons would cause small fluctuations in the magnitude and direction of the gravity vector acting on the Moon, which could cause the orbit to precess in ways that it would not with a perfectly uniform Earth. My father observed just such perturbations of the orbit of Vanguard I while tracking it back around 1960, as part of a quest to firm up the orbital elements of beacon satellites. That is not the same as a net transfer of energy from Earth's spin to the satellite's orbit and forcing the orbit higher. The Vanguard most certainly did not go higher. It descended a bit from atmospheric drag, as expected. These perturbations would be so much weaker at the Moon's distance that I would expect them to be virtually negligible.

[George]

It seems to me that the mountain ring can be considered as
just a ring, without the planet. Yes, that is a great way to see it, though the calculus might be a little rough. We could simplify it even more and use a rotating barbell set of exceeding mass. As one side of the "weights" approach our moon, the moon will accelrate toward it and the greater proximity will allow a greater pull as the weight receeds away from it.

Hmmm, the phase angle concerns me, though. The moon's greatest proximity won't come when the weight is in alignment with it, but perhaps @ 90 deg., since it is a 2-weight barbell. It may be opposite of what I thought. The moon would be furthest away when the weight was aligned with it, which means the strongest pull will be a retrograde vector. That's interesting. Is this right?

[Jeff Root]

George,

Why would the perfect symmetry of the bulges imply zero friction?
By "symmetry", you mean that the bulge on Earth's far side is the
same as the bulge on the side facing the Moon, don't you? The
entire Earth becomes symmetrically prolate, on top of its existing
oblateness.

-- Jeff, in Minneapolis

[George]

I did not realize that by "perfectly symmetric" you meant perfectly in conjunction with the Moon. I don't think it could be that way even with no friction. Right. These fictional approaches help me isolate the appropriate vectors at work. There are several nuances to this that make it more interesting. Often something simple is stated about conservation of angular momentum and the broad logic is fairly easy to see, but the details are not once the deep digging takes place.


My hunch is that it will continue to rise a bit longer after passing the conjunction point, for the same reason that a rocket that is lifting off vertically will rise a while longer after the engine cuts off. This would put the bulge in the prograde position. I suspect you're right, though friction seems to me to be the big dog in a prograde vector creation.


These mascons would cause small fluctuations in the magnitude and direction of the gravity vector acting on the Moon, which could cause the orbit to precess in ways that it would not with a perfectly uniform Earth. My father observed just such perturbations of the orbit of Vanguard I while tracking it back around 1960, as part of a quest to firm up the orbital elements of beacon satellites. Interesting. Do such precessions imply greater eccentricity? [Greater eccentricity would accelerate the orbital advance, I think.]


That is not the same as a net transfer of energy from Earth's spin to the satellite's orbit and forcing the orbit higher. The Vanguard most certainly did not go higher. It descended a bit from atmospheric drag, as expected. These perturbations would be so much weaker at the Moon's distance that I would expect them to be virtually negligible. True, but hundreds of billions of orbits can add up after a while. :)

[Jeff Root]

George,

With the barbell, you are now describing a situation which
is identical to a binary star orbited in the same plane at a
short distance by a third star.

-- Jeff, in Minneapolis

[George]

George,

Why would the perfect symmetry of the bulges imply zero friction?
By "symmetry", you mean that the bulge on Earth's far side is the
same as the bulge on the side facing the Moon, don't you? The
entire Earth becomes symmetrically prolate, on top of its existing
oblateness. Sorry for the confussion. I mean complete symmetry where there is no variation in the shape and density of either bulge regardless of axis. Friction would alter the symmetry, I'm fairly sure, or would it simply shift the top of the bulge off the center-line between the two bodies?

[Jeff Root]

George,

I said in post #47 much the same as Hornblower says about
matter raised in a tidal bulge in the Earth necessarily moving
ahead of the Moon, with or without friction. The bulge cannot
be directly under the Moon if Earth is rotating, even if there
is no friction.

-- Jeff, in Minneapolis

[George]

George,

With the barbell, you are now describing a situation which
is identical to a binary star orbited in the same plane at a
short distance by a third star. I know and thought about tieing two neutron stars of equal mass together and spinning them, but Im not as strong as I used to be. :) Since you mentioned it, though, it would be the same, so is there an easy way to picture the prograde action they would have on an orbiting body just outside their orbit?

[Jeff Root]

George,

It is looking to me like the central problem is that you are
imagining the tidal bulge as being directly under the Moon.
Yes? No?

-- Jeff, in Minneapolis

[George]

George,

I said in post #47 much the same as Hornblower says about
matter raised in a tidal bulge in the Earth necessarily moving
ahead of the Moon, with or without friction. The bulge cannot
be directly under the Moon if Earth is rotating, even if there
is no friction. Even here I'd like to divide and go hypothetical again.
Instead of Earth, let's use a water-only planet. Without friction the surface water, let's say, isn't rotating with the planetary our perfectly spherical and inelastic interior. Would there really be asymmetry as the Moon orbits?

Now let's have the water rotate, yet w/o friction. Are you both saying then that the inertial motion of the water will be effectual? If so, after restoring friction, will the inertial motion be wimpy relative to the friction caused bulge shift? [I'm just trying to find all those little cute vectors at work.]

What about my view that I was bass akwards on the proximity of the Moon during the dumbell's -- suddenly a more appropriate name -- position closest to the Moon? I'm fairly sure the 90 deg. phase angle is right.

[George]

George,

It is looking to me like the central problem is that you are
imagining the tidal bulge as being directly under the Moon.
Yes? No? Yes, I see the squeeze as symmetric. Hmmm. Ok, the rotational vector would not allow that. I understand now what you're saying. It can't be in symmetric with the line between the bodies. That helps.

[Jeff Root]

I said in post #47 much the same as Hornblower says about
matter raised in a tidal bulge in the Earth necessarily moving
ahead of the Moon, with or without friction. The bulge cannot
be directly under the Moon if Earth is rotating, even if there
is no friction.
Instead of Earth, let's use a water-only planet.
I was already doing better than that. "Earth" was an ideal
gas. (Okay, I only have one cylinder of ideal gas, not enough
for an entire planet, but extrapolate a bit...) The rotating
ball of gas is oblate because of its rotation and inertia, and
prolate because of the Moon's gravity. The prolate bulges
move ahead of the Earth-Moon line because of their inertia,
as part of the rotation.

Adding friction would reduce the size of the bulge, so Earth
would not be as prolate. It would also reduce the speed at
which the surface rose and fell due to the Moon's gravity, so
the bulge would be farther ahead of the Earth-Moon line.

I just found myself humming a bit of Bach's Little Fugue in
G, obviously as a result of thinking, several minutes ago,
how this conversation resembles a canon.

-- Jeff, in Minneapolis