[Le Repteux (OP)]

Here is wiki on tidal acceleration:

Tidal acceleration is an effect of the tidal forces between an orbiting natural satellite (e.g. the Moon), and the primary planet that it orbits (e.g. Earth). The acceleration causes a gradual recession of a satellite in a prograde orbit away from the primary, and a corresponding slowdown of the primary's rotation.

The similar process of tidal deceleration occurs for satellites that have an orbital period that is shorter than the primary's rotational period, or that orbit in a retrograde direction.

The naming is somewhat confusing, because the speed of the satellite relative to the body it orbits is decreased as a result of tidal acceleration, and increased as a result of tidal deceleration.

Here is the problem: if the orbital motion of the earth around the moon is able to slow down the earth's rotation speed, then shouldn't the orbital motion of the earth-moon system around the sun be able to slow down the earth/moon's orbital speed, or on a larger scale, shouldn't the orbital motion of the solar system around the galaxy core be able to slow down the solar system's orbital speed?

Of course, there is a difference between orbital rotations and proper rotations of planets, but as it emerges from this actual discussion on tides, the tidal bulges are not affected by that difference since they depend only on gravitational forces, so if it is true, I think that the two kinds of rotation should produce the same tidal effects. Unfortunately, the tidal theory says that the deceleration of the proper rotational speed of the earth is due to tidal friction opposing it, and that the deceleration of the orbital speed of the earth/moon system is due to the tidal bulges being dragged sideways by that friction. Since the proper rotation speed of the earth is working against the bulges, I can imagine how it would decelerate, but I can't imagine how the resultant offset of the bulges would decelerate the orbital speed of the earth/moon system. As they are presented on the wiki page, those offset bulges would on the contrary accelerate the orbital speed of the moon, which is probably why it has been confusedly named "Tidal acceleration".

I looked for a more precise explanation of the mechanism but I didn't find any. Does anybody know how it works exactly?

 

[Janus]

. Since the proper rotation speed of the earth is working against the bulges, I can imagine how it would decelerate, but I can't imagine how the resultant offset of the bulges would decelerate the orbital speed of the earth/moon system. As they are presented on the wiki page, those offset bulges would on the contrary accelerate the orbital speed of the moon, which is probably why it has been confusedly named "Tidal acceleration".

I looked for a more precise explanation of the mechanism but I didn't find any. Does anybody know how it works exactly?

 

The offset of the bulges produces a net forward pull or acceleration of the Moon in its orbit.  The key here is that the Moon is in orbit.  Giving something in orbit a forward push causes it to climb higher in its orbit, but in doing so, it trades off kinetic energy for gravitational potential energy. 
So let's say that you are trying to move a satellite from one orbit to a higher one.  This can be done by a two step process. 
First you give the satellite a forward boost. This changes its orbit into an elliptical one with its present orbit at perigee and the new orbit at apogee.  Wait for 1/2 orbit, and the satellite will be at apogee.  But in climbing away from the Earth, it loses speed. By the time it gets to apogee, it is moving slower than it needs to be to hold a circular orbit at that distance (even though this speed is less than the speed it had in the original orbit.).
so now you give it another forward boost to get it up to circular orbit speed.   Two forward boosts gets you to a higher orbit but moving at a lower speed.
With the Moon, it is under a small constant thrust.  It is an acceleration because this forward push always has it moving just a bit faster than it needs to for its altitude,  which causes it  to climb away.   And because of the trade off between climbing and speed, it loses orbital velocity in the process.( Though its velocity always remains a bit more than circular orbital velocity at any time.)  It's total orbital energy increases even though the kinetic energy portion decreases.

With the Interaction between Earth and Sun, it would be the tidal bulge created by the Earth on the Sun that would have an effect on the Orbit of the Earth around the Sun.  There are a number of factors that come into play here.  The Sun is a gaseous body, Which changes the drag between Sun and bulge, also the Sun doesn't rotate all in one piece.
There are also other planets orbiting the Sun creating their own bulges.  Jupiter, 5 times further away and with 333 times the mass, has 2 2/3 more tidal effect on the Sun than the Earth does.  Venus, Though slightly less massive is close enough to the Sun to have twice the tidal effect of the Earth. Since these tidal bulges will be following their respective planets, (sometimes complimenting each other and sometimes opposing) and not the Earth, you have a complicated scenario.
With the Galaxy you are dealing with a structure that is entirely composed of Orbiting objects.    There is no equivalent of the tidal friction you get with a planet.   And even if you had a central body to raise tides on, since the stars of the galaxy are even distributed around the center,  you wouldn't get a bulge  with unidirectional lobes ( like those pointing towards and away from the Moon)  All the stars would be trying to create a bulge in their direction and they would cancel each other out.  No lobes, no tidal acceleration.

[Le Repteux (OP)]

So let's say that you are trying to move a satellite from one orbit to a higher one.  This can be done by a two step process.
First you give the satellite a forward boost. This changes its orbit into an elliptical one with its present orbit at perigee and the new orbit at apogee.  Wait for 1/2 orbit, and the satellite will be at apogee.  But in climbing away from the Earth, it loses speed. By the time it gets to apogee, it is moving slower than it needs to be to hold a circular orbit at that distance (even though this speed is less than the speed it had in the original orbit.).
so now you give it another forward boost to get it up to circular orbit speed. Two forward boosts gets you to a higher orbit but moving at a lower speed.

Hi Janus,

The theory says that when the first boost is forward, the second one must be backward because the satellite is then going too fast to execute a circular orbit at this new higher altitude, and that this second boost must also happen immediately if we want the orbit to immediately get circular.

With the Moon, it is under a small constant thrust. It is an acceleration because this forward push always has it moving just a bit faster than it needs to for its altitude,  which causes it to climb away.

That's what I thought too when I saw the offset bulges at Wiki.

And because of the trade off between climbing and speed, it loses orbital velocity in the process.( Though its velocity always remains a bit more than circular orbital velocity at any time.)  It's total orbital energy increases even though the kinetic energy portion decreases.

Your trade off between kinetic energy and gravitational potential energy looks as if no backward boost was needed to slow down the speed. It is so if we want the trajectory to become elliptic, but adding constant tiny forward boosts to a satellite which just got at the perigee of an elliptic trajectory doesn't slow down its speed. To increase its altitude constantly while slowing its speed constantly, we need to accelerate it a bit and immediately decelerate it to its new orbit, and do it over and over constantly. That's what we would need to do with a satellite, and that's also what we would need to do if it was at the place of the moon.

To me, that tidal acceleration theory looks a bit circular: it means that the force exerted by the moon on the earth would finally profit to the moon. I wouldn't mind if it was clear, but I still don't get it. If I would simulate it for instance, it certainly wouldn't produce a recessing, and if a motion can't be simulated, it can't be calculated either, which is actually the case of the calculations on the wiki page: they are only about the constant forward torque produced by the offset bulges of the earth on the moon, nothing about any backward one.

[Le Repteux (OP)]

The solar tides on Earth do drag the total angular momentum of the Earth/moon system down

Hi Halc,

What you say means that the boost from the moon's offset tides on the earth's orbital speed has to be equivalent to the boost from the earth's offset tides on the orbital speed of the moon. But is it really? In principle, the bulges get forwardly offset by the forward proper rotational speed of the planets, and the proper rotational speed of the moon is far from being the same as the earth's one, so its bulges get lot less offset while also getting a lot less massive. Of course, the earth needs to get less orbital speed than the moon for them to stay inline with their common barycenter because it is nearer to it, but it is a lot more massive than the moon too, so it is a lot more difficult to move around. If the moon's offset bulge was the same as the earth's one, the two boosts might be equivalent, but with less offset, I don't think they could. Besides, I explained to Janus how a backward boost was lacking in the "tidal acceleration" explanation, and it would still be lacking even if both accelerations were equivalent.

[Janus]

So let's say that you are trying to move a satellite from one orbit to a higher one.  This can be done by a two step process.
First you give the satellite a forward boost. This changes its orbit into an elliptical one with its present orbit at perigee and the new orbit at apogee.  Wait for 1/2 orbit, and the satellite will be at apogee.  But in climbing away from the Earth, it loses speed. By the time it gets to apogee, it is moving slower than it needs to be to hold a circular orbit at that distance (even though this speed is less than the speed it had in the original orbit.).
so now you give it another forward boost to get it up to circular orbit speed. Two forward boosts gets you to a higher orbit but moving at a lower speed.

Hi Janus,

The theory says that when the first boost is forward, the second one must be backward because the satellite is then going too fast to execute a circular orbit at this new higher altitude, and that this second boost must also happen immediately if we want the orbit to immediately get circular.

No.   You always end up with less speed when you reach the higher orbit altitude then needed to hold a circular orbit there.    Orbital velocity at any point of an orbit is found by
 V = sqrt(u(2/r-1/a)
Where u is the gravitational parameter ( 3.987e14 m^3/s^2 for the Earth)
r is the present radial distance of the orbit
a is the semi-major axis

start with a circular orbit with r= 8,000,0000 m
you want to move to an circular orbit of r = 10,000,000 m
This makes a=9,000,000 m
Circular orbital velocity at original orbit = sqrt(u/r) = 7059.6 m/s
Perigee velocity of orbit with perigee at 8,000,000 m(r=8,000,000m) and apogee at 10,000,000m = 7441.4 m/s
Meaning a delta V requirement of + 381.8 m/s to leave the original orbit and enter the transfer orbit.
Apogee velocity of transfer orbit (r=10,000,000) = 5953.2 m/sec
Circular orbital velocity at 10,000,000m = 6314.3 m/sec
Meaning a delta V requirement of +361.6m/sec to achieve that circular orbit.
a total forward delta V of 742.9 m/s  but a net loss of 745.3 m/s in terms of orbital velocity.

With the Moon, it is under a small constant thrust. It is an acceleration because this forward push always has it moving just a bit faster than it needs to for its altitude,  which causes it to climb away.

That's what I thought too when I saw the offset bulges at Wiki.

And because of the trade off between climbing and speed, it loses orbital velocity in the process.( Though its velocity always remains a bit more than circular orbital velocity at any time.)  It's total orbital energy increases even though the kinetic energy portion decreases.

Your trade off between kinetic energy and gravitational potential energy looks as if no backward boost was needed to slow down the speed. It is so if we want the trajectory to become elliptic, but adding constant tiny forward boosts to a satellite which just got at the perigee of an elliptic trajectory doesn't slow down its speed.

Increasing velocity at perigee increases the apogee. Increasing speed at apogee increases perigee.  If you are at apogee and want to circularize your orbit, you want to raise the perigee up to your present altitude. This requires an increase in velocity.

To increase its altitude constantly while slowing its speed constantly, we need to accelerate it a bit and immediately decelerate it to its new orbit, and do it over and over constantly. That's what we would need to do with a satellite, and that's also what we would need to do if it was at the place of the moon.

Once again, NO. 
Total orbital energy is Kinetic energy plus Gravitational potential energy or mv^2/2-um/r  where m is the mass of the satellite.
Assuming a unit mass for m, we can use v^2/2-u/r
It can also be given as -u/2a  where a is again the semi-major axis. (for a circular orbit a=r)
Starting in a circular orbit, increasing v increases the KE while GPE remains constant, it also raises a.  Total E has increased. But as the mass starts to climb from its original altitude its total energy remains the same, but its GPE increases. This means it must lose KE and thus velocity.
Again going back to our 8,000,000m to 10,000,000m change in altitude.
We start with a total  specific(per kg) energy of  7059.6^2/2 - 3.987e14/8,000,000 = -24918524 j/kg
We boost the speed to 7441.4 m/s pumping up the energy to -22150283 j/kg ( less negative means an increase)
This total remains constant as the mass climbs to it apogee of 10,000,000 m so
-22150283 =  v^2/2- 3.987e14/10,000,000
v^2 = 2(3.987e14/10,000,000 - 22150283) = 35439434
v= 5953.1 m/s ( within 0.1 m/sec of the answer I got above  and well within the rounding error margin.)
Your conclusion that you would need to brake upon reaching apogee is completely the opposite of what the actual physics says.

To me, that tidal acceleration theory looks a bit circular: it means that the force exerted by the moon on the earth would finally profit to the moon. I wouldn't mind if it was clear, but I still don't get it. If I would simulate it for instance, it certainly wouldn't produce a recessing, and if a motion can't be simulated, it can't be calculated either, which is actually the case of the calculations on the wiki page: they are only about the constant forward torque produced by the offset bulges of the earth on the moon, nothing about any backward one.

The only backward one is the drag the bulges produce on the rotation of the Earth slowing its rotation rate.  There doesn't need to be a backward force on the Moon.  Its orbital velocity decreases due to its climbing away from the Earth even though there is a constant "forward" pull.    As counter-intuitive as this seems, this is how orbital mechanics works.
"Forward takes you out, out takes you back, back takes you in, and in takes you forward"

[Le Repteux (OP)]

OK. I just realized that you were both talking about the Hohmann transfer orbit, whereas I was taking about an immediate one. To get a Hohmann transfer, we have to wait till the satellite gets to its apogee, and then accelerate it again to get a circular orbit, whereas if we don't want to wait, then we can decelerate it immediately after having accelerated it. If I understand well, you mean that the moon suffers a tiny boost, waits till it gets at its apogee, and then uses another boost to stay on a circular orbit. That would be fine if no other boost would be happening in between and if the moon knew what it was doing, but it is not the case. The boosts happen all along the trajectory, so how does the moon differentiate the first boosts from the second ones. Moreover, if I had to simulate that motion, I might probably get the same kind of slow spiraling trajectory, but I can't figure out how the speed would ever get down.

Quote

To me, that tidal acceleration theory looks a bit circular: it means that the force exerted by the moon on the earth would finally profit to the moon. I wouldn't mind if it was clear, but I still don't get it. If I would simulate it for instance, it certainly wouldn't produce a recessing, and if a motion can't be simulated, it can't be calculated either, which is actually the case of the calculations on the wiki page: they are only about the constant forward torque produced by the offset bulges of the earth on the moon, nothing about any backward one.

The only backward one is the drag the bulges produce on the rotation of the Earth slowing its rotation rate.  There doesn't need to be a backward force on the Moon.  Its orbital velocity decreases due to its climbing away from the Earth even though there is a constant "forward" pull. As counter-intuitive as this seems, this is how orbital mechanics works.

As I just said, we can also decelerate a satellite immediately after having accelerated it, and it will also follow a circular trajectory. It takes more energy than waiting till it decelerates all by itself, but it works. This way, if we accelerate it continuously as it is the case with the tidal explanation, it will simply never slow down. That's what we do when we want a satellite to escape the earth's gravitational acceleration. To me, accelerating it as little as the tides would do (and also accelerating it less and less), would only take it more time to reach the same height: it would not slow it down. I have another explanation that looks promising, but I'll wait till I'm convinced that I can't understand the tidal one before talking about it, so thanks for digging it out further.

[Halc]

The solar tides on Earth do drag the total angular momentum of the Earth/moon system down

Hi Halc,

What you say means that the boost from the moon's offset tides on the earth's orbital speed has to be equivalent to the boost from the earth's offset tides on the orbital speed of the moon.

The comment above refers to the solar tides on Earth, not tides caused by the moon.  That friction slows down the spin of Earth but has no immediate significant effect on the moon.

The lunar tides transfer momentum from Earth to the moon.  Total momentum must be conserved, but not total energy, which gets lost to heat of tidal friction.

Besides, I explained to Janus how a backward boost was lacking in the "tidal acceleration" explanation, and it would still be lacking even if both accelerations were equivalent.

Your comment about backwards boost is mistaken.  Two forward boosts are required to move a satellite to a new higher orbit, as Janus correctly conveyed.  The moon on the other hand gets a small but continuous forward force which slowly adds to its mechanical energy, stealing away Earth’s mechanical energy.

If I understand well, you mean that the moon suffers a tiny boost, waits till it gets at its apogee, and then uses another boost to stay on a circular orbit.
 That would be fine if no other boost would be happening in between and if the moon knew what it was doing, but it is not the case. The boosts happen all along the trajectory, so how does the moon differentiate the first boosts from the second ones.

The force on the moon is continuous, and so the orbit is technically a continuous spiral, never a circle.  The satellites could get to a higher orbit the same way, but rockets are typically not designed to run at low power like that, and the people that paid for it want it there more quickly I suppose.

if we accelerate it continuously as it is the case with the tidal explanation, it will simply never slow down.

The speed of the moon is continuously slowing down due to gradually going into a higher orbit, as is pointed out in the last line of the wiki article you quote in the OP.

[Janus]

OK. I just realized that you were both talking about the Hohmann transfer orbit, whereas I was taking about an immediate one. To get a Hohmann transfer, we have to wait till the satellite gets to its apogee, and then accelerate it again to get a circular orbit, whereas if we don't want to wait, then we can decelerate it immediately after having accelerated it. If I understand well, you mean that the moon suffers a tiny boost, waits till it gets at its apogee, and then uses another boost to stay on a circular orbit. That would be fine if no other boost would be happening in between and if the moon knew what it was doing, but it is not the case. The boosts happen all along the trajectory, so how does the moon differentiate the first boosts from the second ones. Moreover, if I had to simulate that motion, I might probably get the same kind of slow spiraling trajectory, but I can't figure out how the speed would ever get down.

Quote

To me, that tidal acceleration theory looks a bit circular: it means that the force exerted by the moon on the earth would finally profit to the moon. I wouldn't mind if it was clear, but I still don't get it. If I would simulate it for instance, it certainly wouldn't produce a recessing, and if a motion can't be simulated, it can't be calculated either, which is actually the case of the calculations on the wiki page: they are only about the constant forward torque produced by the offset bulges of the earth on the moon, nothing about any backward one.

The only backward one is the drag the bulges produce on the rotation of the Earth slowing its rotation rate.  There doesn't need to be a backward force on the Moon.  Its orbital velocity decreases due to its climbing away from the Earth even though there is a constant "forward" pull. As counter-intuitive as this seems, this is how orbital mechanics works.

As I just said, we can also decelerate a satellite immediately after having accelerated it, and it will also follow a circular trajectory. It takes more energy than waiting till it decelerates all by itself, but it works. This way, if we accelerate it continuously as it is the case with the tidal explanation, it will simply never slow down. That's what we do when we want a satellite to escape the earth's gravitational acceleration. To me, accelerating it as little as the tides would do (and also accelerating it less and less), would only take it more time to reach the same height: it would not slow it down. I have another explanation that looks promising, but I'll wait till I'm convinced that I can't understand the tidal one before talking about it, so thanks for digging it out further.

If the forward force is high enough, then, under a constant force, the satellite can both gain altitude and velocity.  However, below a certain threshold, the velocity lost climbing exceeds the velocity that would have been gained through the application of the forward thrust over the same period.  This is the situation that the Moon finds itself in.  The forward force is below this threshold and it loses orbital velocity as it spirals away from the Earth even though it is under a constant forward force,

[Le Repteux (OP)]

If the forward force is high enough, then, under a constant force, the satellite can both gain altitude and velocity. However, below a certain threshold, the velocity lost climbing exceeds the velocity that would have been gained through the application of the forward thrust over the same period. This is the situation that the Moon finds itself in. The forward force is below this threshold and it loses orbital velocity as it spirals away from the Earth even though it is under a constant forward force,

Interesting possibility, so let's analyze it. The boost from the tides is constant even if it is getting weaker and weaker with time since the altitude increases constantly and since the tides themselves get lower. A small portion of the boost does the same thing, and a portion of that portion too. The smallest portions we can analyze all do the same thing: they succeed to bring the moon at a higher altitude while increasing its orbital speed. There is no break between those boosts that could let the speed get down. If there was any, the moon would effectively slow down and get away from the earth a bit, but there is none. Each small boost is immediately followed by another one, so to me, those boosts can do nothing else than to constantly increase the speed even if their force is constantly getting down. If you got a link to the threshold explanation, can I have it so that I can study it more thoroughly please?

[Le Repteux (OP)]

The speed of the moon is continuously slowing down due to gradually going into a higher orbit, as is pointed out in the last line of the wiki article you quote in the OP.

I'm looking for a mechanism that I could simulate, and I'm afraid there is no way to simulate that kind of deceleration but to add an ad hoc equation to the coding that fits the data.

[Le Repteux (OP)]

This thread on tides emerged from another one that I had decided to leave because it wasn't progressing, but unexpectedly, it did. Colin2B decided to answer an unanswered proposition that I had posted about the tidal phenomenon, and he agrees with it, so I refer you to it since it is precisely the one I was about to make here. Feel free to discuss it here though since this thread's subject is more specific than the other one.

[rmolnav]

The boost from the tides is constant even if it is getting weaker and weaker with time since the altitude increases constantly and since the tides themselves get lower. A small portion of the boost does the same thing, and a portion of that portion too. The smallest portions we can analyze all do the same thing: they succeed to bring the moon at a higher altitude while increasing its orbital speed. There is no break between those boosts that could let the speed get down. If there was any, the moon would effectively slow down and get away from the earth a bit, but there is none.

Please kindly note that what in bold is what is actually happening !!
Orbiting dynamics is not my forte, but I think that, apart from transient changes in the direction of earth´s pull on the moon (along each quarter cycle between apogee and perigee), with each cycle moon changes some kinetic energy into potential energy (as Janus says), it gets a little farther, and orbits with a little lower tangential speed ... , despite the fact that initially increases in tangential speed (caused by the gap between actual moon´s and "sublunar" bulge´s locations) do happen, but the bottom line is a tangential speed decrease.
 

[Le Repteux (OP)]

but the bottom line is a tangential speed decrease.

It is, and it supposedly depends on the bulges being dragged forward by rubbing against the earth's daily rotation, what offsets them a bit, and what in return accelerates the moon forward on its orbital trajectory. It would work if the moon wouldn't have to slow down while getting to a higher orbit, but it has, and the only mechanism we have to explain that is math: the moon supposedly slows down because it transforms its kinetic energy into potential energy. That would be true if we would let it go after having accelerated it a bit, it would then slow down all by itself until it would have reached its apogee, but the forward pull by the offset bulge never stops, so I can't see where the speed would have the time to get down.

[rmolnav]

... but the forward pull by the offset bulge never stops, so I can't see where the speed would have the time to get down.

The forward force is below this threshold and it loses orbital velocity as it spirals away from the Earth even though it is under a constant forward force,

You both are forgetting a small, but possibly important, detail.
Perhaps what I said on my post is not just to say "apart from ...":

...apart from transient changes in the direction of earth´s pull on the moon (along each quarter cycle between apogee and perigee)

I meant that sublunar bulge pull tangential component is actually not constant, due to the fact that only where apogee (and almost where perigee) the full pull vector is perpendicular to moon´s path ...
Perhaps the "effective" tangential bulge pull decreases from perigee to apogee more than what increases from apogee to perigee ...
Janus, your knowledge on this field is much deeper than ours (at least than mine ...). Any comment from you would be really appreciated. 

[Le Repteux (OP)]

I meant that sublunar bulge pull's tangential component is actually not constant, due to the fact that only where apogee (and almost where perigee) the full pull vector is perpendicular to moon´s path ...
Perhaps the "effective" tangential bulge pull decreases from perigee to apogee more than what increases from apogee to perigee ...

Even when the trajectory is not circular, the pulling component cannot be anything else that perpendicular, and the inertial one tangential, so even if the strength of the radial force changes all along the trajectory, it is still there, and it doesn't stop accelerating the tangential speed forward.

[rmolnav]

Even when the trajectory is not circular, the pulling component cannot be anything else that perpendicular, and the inertial one tangential

I can´t agree with that ...
Earth, located on one of the focus of an ellipse (moon´s trajectory), pulls the moon through a force vector with main component certainly perpendicular to moon´s trajectory, but also through other tangential one not null on most moon´s locations ...
The perpendicular component is the one which actually exerts as centripetal force, and points not towards the ellipse´s focus, but towards the center of curvature of the orbit at the considered point.
And the tangential component (by the way, I wouldn´t call it "the inertial one") tends to increase moon´s tangential speed along parts of the orbit, and decrease along the rest
SIMILAR THINGS can be said about the additional bulge pull we are discussing ... Its tangential component varies with the varying angle between the bulge´s center-moon´s CM and the elliptical trajectory ...
I won´t even guess whether at some parts of the orbit that bulge tangential pull can be "negative", but it´s quite clear it increases along parts of the orbit, and decreases along the rest.
And, as far as I can understand, that can affect the bottom line: after each complete cycle (28/29 days), moon is farther (though only a very tiny little bit), and its tangential speed is very, very little smaller !!

[Le Repteux (OP)]

Earth, located on one of the focus of an ellipse (moon´s trajectory), pulls the moon through a force vector with main component certainly perpendicular to moon´s trajectory, but also through other tangential one not null on most moon´s locations ...

You're right about the force not always being perpendicular to the tangential motion if the trajectory is an ellipse, but I still can't figure out how that tangential motion could slow down while constantly being accelerated. At least, with the idea that it is the orbital speed that raises the tides, the motion becomes an acceleration followed by a deceleration, and such a tangential action is precisely what is needed to change the height of a circular trajectory without waiting for the satellite to get to its apogee. This way, each point of the surface would be accelerated to a higher orbit during its transit, and then decelerated back to almost where its was at the beginning of the transit. Remains to determine why the "almost".

Let's try another explanation. One of the phenomenon we don't usually account for when studying orbital motion is aberration. We don't think of it because we don't study gravitation in terms of information traveling in void, thus in terms of light. We know that the motion of the earth produces aberration on the light from the stars, so it certainly produces some on the information from the moon. That information should pull the earth forward a bit since it is offset and it should also pull it stronger since it is blueshifted, but it is also sent backward a bit by the moon in order to hit the moving earth in time, which produces redshift, while the offset it produces is completely rectified by the one from aberration later on, so the blueshift subtracts from the redshift but not completely since the moon is going faster and produces more redshift than the blueshift the earth produces, thus when the earth uses that information, it is still redshifted a bit and so doesn't produce as much acceleration as it should to keep the earth at a constant distance from the moon. During that time, the inverse is happening to the moon: the information is sent backwards by the earth and the moon produces aberration on it, so it also appears to come at right angle but this time, the redshift produced by the earth is less important than the blueshift produced by the moon, and it should thus produce more acceleration than is needed to keep the moon at a constant distance from the earth. So it doesn't seem to work, but fortunately, since aberration is a relativistic effect, we can count on the increase in mass due to motion to justify the surplus of force applied on the moon, and to the less important increase in mass of the earth to justify the lack of force applied on the earth. This way, the orbital motion appears to be saved, but the recessing one is not.

Let's try another one then. What if the slow recessing of the earth/moon system was due to a constant slow shortening of the atoms' wavelengths? This way, the time a laser pulse takes to make a roundtrip might only appear to take longer, because it would simply be due to our clocks getting faster with time while c would stay constant. That phenomenon would produce redshift, which might explain some of the gravitational anomalies that we observe, from the probes getting slower to the galaxies fleeing away, but it would also explain gravitation itself, because all the bodies that are actually orbiting would have to use an information that looks redshifted to stay at the right orbital distance, so they would all need to accelerate towards one another to blueshift it a bit. The farther they would be from one another, the larger would be the shift but the weaker would be the force, because the shift would only increase with distance whereas we know that the intensity of the information diminishes with the square of the distance. That's my favorite explanation, but I admit it is a bit hard to swallow.

[rmolnav]

Quote from: rmolnav on 28/09/2018 11:43:20
Earth, located on one of the focus of an ellipse (moon´s trajectory), pulls the moon through a force vector with main component certainly perpendicular to moon´s trajectory, but also through other tangential one not null on most moon´s locations ...
You're right about the force not always being perpendicular to the tangential motion if the trajectory is an ellipse, but I still can't figure out how that tangential motion could slow down while constantly being accelerated .

What in bold is not always so, I guess ... The "misalignment" of sublunar bulge´s pull is very, very tiny (moon is some sixty times earth radius away ...), and that could be insufficient to give a "positive" tangential pull where the lack of perpendicularity with the elliptical orbit is "negative" ...
AND REGARDING:

Let's try another explanation. One of the phenomenon we don't usually account for when studying orbital motion is aberration. We don't think of it because we don't study gravitation in terms of information traveling in void, thus in terms of light. We know that the motion of the earth produces aberration on the light from the stars, so it certainly produces some on the information from the moon. That information should pull the earth forward a bit since it is offset and it should also pull it stronger since it is blueshifted, but it is also sent backward a bit by the moon in order to hit the moving earth in time, which produces redshift, while the offset it produces is completely rectified by the one from aberration later on, so the blueshift subtracts from the redshift but not completely since the moon is going faster and produces more redshift than the blueshift the earth produces, thus when the earth uses that information, it is still redshifted a bit and so doesn't produce as much acceleration as it should to keep the earth at a constant distance from the moon. During that time, the inverse is happening to the moon: the information is sent backwards by the earth and the moon produces aberration on it, so it also appears to come at right angle but this time, the redshift produced by the earth is less important than the blueshift produced by the moon, and it should thus produce more acceleration than is needed to keep the moon at a constant distance from the earth. So it doesn't seem to work, but fortunately, since aberration is a relativistic effect, we can count on the increase in mass due to motion to justify the surplus of force applied on the moon, and to the less important increase in mass of the earth to justify the lack of force applied on the earth. This way, the orbital motion appears to be saved, but the recessing one is not.

Let's try another one then. What if the slow recessing of the earth/moon system was due to a constant slow shortening of the atoms' wavelengths? This way, the time a laser pulse takes to make a roundtrip might only appear to take longer, because it would simply be due to our clocks getting faster with time while c would stay constant. That phenomenon would produce redshift, which might explain some of the gravitational anomalies that we observe, from the probes getting slower to the galaxies fleeing away, but it would also explain gravitation itself, because all the bodies that are actually orbiting would have to use an information that looks redshifted to stay at the right orbital distance, so they would all need to accelerate towards one another to blueshift it a bit. The farther they would be from one another, the larger would be the shift but the weaker would be the force, because the shift would only increase with distance whereas we know that the intensity of the information diminishes with the square of the distance. That's my favorite explanation, but I admit it is a bit hard to swallow.

YOU ARE RIGHT on what in bold ...
And all those things relative to forces on earth and/or moon caused by "optical" phenomena caused their speeds are, to me, utterly absurd !!

[Le Repteux (OP)]

And all those things relative to forces on earth and/or moon caused by "optical" phenomena caused their speeds are, to me, utterly absurd !!

Optical phenomenon are measured with instruments made out of atoms, and since those atoms move according to the direction and the intensity of the gravitational force applied on them, then we can deduce that the direction and the speed they take while being accelerated depend on any optical phenomenon they suffer during that moment. When we observe a star, we have to tilt our telescope a bit in the direction the earth is moving to account for the aberration phenomenon, so it means that the atoms of the earth are actually accelerating in this direction with regard to the star because the pulling from the star is also bent by the aberration phenomenon. It wouldn't be the case if gravitation was instantaneous, but it's not.

What in bold is not always so, I guess ... The "misalignment" of sublunar bulge´s pull is very, very tiny (moon is some sixty times earth radius away ...), and that could be insufficient to give a "positive" tangential pull where the lack of perpendicularity with the elliptical orbit is "negative" ...

When the direction of the force is not perpendicular to the trajectory of a body, the tangential component affects its tangential speed, and the perpendicular one affects its direction. In the case of an orbital trajectory, these two components produce an elliptical trajectory whenever they are not perpendicular to one another. Adding a small offset force should affect the elliptical trajectory of a body the same way it produces it: where the body is moving away from the force, it should slow down its tangential speed a bit faster than it is actually slowing and curve its direction a bit faster than it is actually curved, and where it is moving closer, it should increase its tangential speed a bit faster than it is actually increasing and curve its direction a bit faster than it is actually increasing too. It's hard to tell what the resulting trajectory would be, but it helped me to realize that I did not account for the perpendicular component in my analysis of the tidal acceleration when the orbit is circular. In this case, while the tangential component would be accelerating the tangential speed, the perpendicular one would be pulling the moon stronger since it is closer to it. This way, the bulges would only produce a lower orbit, and the increased speed would be normal, but once in a lower orbit, the bulges would increase, and so would the speed and the force, which means that the orbital distance would contract instead of expanding, which is not what we observe.

Know what? I think it is clearer to consider that the cg of the earth is not affected by the bulges, and safer to discard the tidal acceleration hypothesis. Here is wiki's unclear explanation using the conservation of energy principle, which is not a mechanism anyway:

The gravitational torque between the Moon and the tidal bulge of Earth causes the Moon to be constantly promoted to a slightly higher orbit and Earth to be decelerated in its rotation. As in any physical process within an isolated system, total energy and angular momentum are conserved. Effectively, energy and angular momentum are transferred from the rotation of Earth to the orbital motion of the Moon (however, most of the energy lost by Earth (−3.321 TW)[citation needed] is converted to heat by frictional losses in the oceans and their interaction with the solid Earth, and only about 1/30th (+0.121 TW) is transferred to the Moon). The Moon moves farther away from Earth (+38.247±0.004 mm/y), so its potential energy (in Earth's gravity well) increases. It stays in orbit, and from Kepler's 3rd law it follows that its angular velocity actually decreases, so the tidal action on the Moon actually causes an angular deceleration, i.e. a negative acceleration (−25.858±0.003"/century2) of its rotation around Earth. The actual speed of the Moon also decreases. Although its kinetic energy decreases, its potential energy increases by a larger amount.

[Halc]

As in any physical process within an isolated system, total energy and angular momentum are conserved.

Angular momentum is conserved yes, but a pure Earth/moon system is hardly a closed system.  Total energy is always lost to friction.  I'm surprised to find that wording on a wiki page.  Yes, angular momentum of Earth is transferred to the moon, and the effect is very measurable.  Discard the tidal acceleration hypothesis if you want, but then you need to explain the moon moving away a very measurable 4 cm each year.

Edit: OK, the wiki does admit that only a 30th of the energy is transferred to the moon, and the rest is lost to friction.
The initial comment sort of said otherwise, but I guess 'isolated' system doesn't mean a closed one.