[Janus]

#17 Janus says:
"... As the tangential acceleration affects the Moon, in climb away from the Earth, this involves an increase in gravitational potential energy. This increase is actually greater than the energy it gained from the tidal acceleration, and this is made up for by a loss of kinetic energy for the Moon; it decreases its orbital speed".
I had not seen your post until now ... I have to ruminate it more quietly.
It´s clear that Moon´s angular speed has to decrease, but I´m now not quite sure about tangential speed.
In any case, I can´t actually  understand your reasoning: how if Moon continuously suffers the tangential acceleration due to Earth tidal bulge tangential attraction, it climbs away from the Earth, and it gets "too high" potential energy increase, and then it has to decrease its kinetic energy and velocity, to compensate ...
I´m realizing now that gravitational energy should actually decrease with the "climb away from the Earth", because it is mass times g and times distance. Distance increases, but g decreases inversely to the square of the distance, doesn´t it?

mgh for gravitational potential is only a close approximation that works when you are dealing with small values of h over which g changes little. Even then, it tells you the change in potential caused by lifting a mass a distance of h against gravity.   Thus using this formula raising 1kg by 1 meter near the surface of the Earth, increases its gravitational potential by ~9.84 joules.   Raising the same 1 kg by 1 meter at the distance of the Moon's orbit using this same formula gives  1kg x~ 2.7e-3m/s^2 x 1 = ~2.7e-3 joules for the increase in gravitational potential energy, A smaller increase, but still an increase. 

The more accurate formula for finding  the difference in gravitational potential energy that takes into account the change in g over the height h is
  GMm(1/r-1/(r+h)) ,  where M is the mass of the Earth in this case, and r is the starting distance from the center of the Earth. 

If you try to argue that gravitational energy decreases with an increase in height due to a decrease in g, it would apply to objects near the surface of the Earth as well. (even more so, since the drop off in g over a 1 meter change in altitude near the surface of the Earth is larger than the same drop off in g  over the same change in altitude at Moon orbit, as 1 meter is a larger percentage of the Earth radius than it is off the radius of the Moon's orbit).   Since we don't conclude that raising a mass 1 meter near the surface of the Earth decreases its gravitational potential, we cannot conclude that raising a mass at Moon orbit distance decreases its gravitational potential.

[rmolnav]

#20 Janus
O.K.: potential energy is not an absolute value, but a magnitude always referred to an agreed zero level, and your examples put clear it is positive if we pass from a zero level distance to a higher distance ...
But I´m seeing this concept is more tricky than what it appears.
Gravitational potential energy is linked to a gravitational field, in our case consisting in all "g" values, Earth´s pull on the unit of mass at each point.
Those pulls are forces always radial. in the direction of Earth´s C.G. It is clear that if e.g. we throw a stone vertically upwards in a vacuum (no friction), initial kinetic energy will change into potential energy until maximum hight, due to "g" negative acceleration. And the addition of both energies would keep constant. And similarly when falling back.
But Moon in orbit around Earth has constantly zero "vertical" speed. Earth´s pull does not change any speed vertically, nor does the tangential component of the tidal bulge´s pull we are considering. If this one increases Moon´s tangential speed, we should not consider this increase has to be slowed back because Moon´s distance increases and potential energy increases too.
They are stuff linked to vectors perpendicular to each other, which can operate independently ...

[Janus]

#20 Janus
O.K.: potential energy is not an absolute value, but a magnitude always referred to an agreed zero level, and your examples put clear it is positive if we pass from a zero level distance to a higher distance ...
But I´m seeing this concept is more tricky than what it appears.
Gravitational potential energy is linked to a gravitational field, in our case consisting in all "g" values, Earth´s pull on the unit of mass at each point.
Those pulls are forces always radial. in the direction of Earth´s C.G. It is clear that if e.g. we throw a stone vertically upwards in a vacuum (no friction), initial kinetic energy will change into potential energy until maximum hight, due to "g" negative acceleration. And the addition of both energies would keep constant. And similarly when falling back.
But Moon in orbit around Earth has constantly zero "vertical" speed. Earth´s pull does not change any speed vertically, nor does the tangential component of the tidal bulge´s pull we are considering. If this one increases Moon´s tangential speed, we should not consider this increase has to be slowed back because Moon´s distance increases and potential energy increases too.
They are stuff linked to vectors perpendicular to each other, which can operate independently ...

Any acceleration in the direction of orbital velocity will result in an increase in distance from the planet.  This is just a fact of orbital mechanics.  In fact, acceleration along this direction is the most efficient means of  gaining altitude. 
You can show this in the following example.
The total energy of a orbiting craft is it is the sum of it kinetic energy and it Gravitational potential energy.   In orbital mechanics GPE is taken to be -GMm/r  (The zero point is set at infinite distance and become more negative (less) as r becomes smaller)
Thus the orbital energy is mv^2/2-GMm/r 
v for a circular orbit can be found by v= sqrt(GM/r) as along as we assume that m is small compared to M
If we substitute this for v we get E= -GMm/2r
Now it also turns out that this equation works for non-circular orbits if use 'a', the semi-major axis of the orbit instead of r (for a circular orbit a=r) a is also the mean orbital radius of the orbit.

Now let's work out an example for a craft of 100 kg with an initial orbit with radius of 7000 km (altutude of ~622 km)
This gives us a orbital velocity of  7547 m/s, and an orbital energy of -2847857143 joules.

Now let's give the craft a instant boost of 1000 m/s.  This puts it into a new elliptical orbit with its present altitude the perigee.  This also increases its orbital energy to -2043153836 joules ( add 1000 m/s to the the orbital velocity and use the first equation I gave for orbital energy. )

Using this new value, we can solve for the semi-major axis of the new orbit, which turns out to to be ~9735 km. the average radius of the orbit has increased by 2735 km.

Now what happens if we apply this sudden boost in a direction directly away from the Earth?  Now we have to use vector addition to get the new orbital velocity.  This will be sqrt(1000^2+7547^2)= 7613 m/s

using this new velocity to find the new orbital energy, we get -2797853836 joules. 

Solving as we did before for the new semi-major axis, we get 7125 km, or an increase of just 125 km compared to the 2735 km increase we got by adding the boost along the direction of orbital motion.

Also, by thrusting directly away form the Earth, the new orbit will have a perigee closer to the Earth. In fact, in this case, the new perigee would be closer to the center of the Earth than the surface of the earth is. Your craft would not be able to complete one full orbit without crashing into the Earth. So applying a boost away from the Earth could eventually lead to crashing into it.

Orbital mechanics is not always intuitive, and may not get the results you might first expect if you are not familiar with how orbits behave.

[rmolnav]

#22 Janus
I have not checked your numbers, but I am one "not familiar with how orbits behave", and have realized I did something wrong ...
[/size]You know, since one or two years ago I´ve been discussing topics such as sea tides, centripetal and centrifugal forces, reason why we always see same face of the Moon[/size], etc.
[/size]For all those topics, for the sake of simplification, I always considered circular orbits instead of elliptical ones, what is sufficiently accurate.
[/size]But if we are considering the very, very tiny misalignment of tidal bulge, which produces the tangential pull on the Moon that makes it climb away from the Earth, we must be similarly accurate in our reasoning.
[/size]And what I said about perpendicularity between g and tangential speed (tangential to the orbit) is not sufficiently accurate, apart from when just at four ellipse "corners" ... [/size]Even if the orbit were a "circular-spiral", tangential speed would not be exactly perpendicular to g ...
[/size]The transformation of the increase of kinetic energy into an increase of potential gravitational energy happens for same reason as in the case of the stone thrown upwards, though in a much subtle way ...
By the way: that tiny misalignment was relatively much bigger when Moon and Earth were "children", they were much, much closer to each other, sea tides were much stronger, Earth rotated much faster ... The pace of the climbing away we are considering has been decreasing since the beginning.[/font]
So, coming back to the question " [/font]Will we eventually lose the moon?", I think the answer is NO, at least for that climbing away.
We should also keep in mind that after sufficient time of Moon-Earth distance increasing, Sun related tidal effects would get bigger than Moon´s ... If Sun not evolved as we know it will, a "fight" between Sun and Moon would happen ... Would Earth end synchronized to the Moon (as Moon is already), or to the Sun? 
[/font]

[zx16]

If the Moon does get further away from the Earth, where will it eventually go? Will it become, in millions of years of time, a new satellite of Jupiter?

[rmolnav]

#24 zx16
As said on #23, being the Moon´s getting further away from the Earth an "asymptotical" movement (its pace tends to zero), it should not go sufficiently away to stop being our satellite.
If even so we wanted to answer your question, I think it would be a kind of lottery, because it would depend on several unknown factors: exact planetary scenario in the moment of going out of its orbit around us, which physical fact would cause that, exact velocity of the Moon in that moment ...

[Janus]

We should also keep in mind that after sufficient time of Moon-Earth distance increasing, Sun related tidal effects would get bigger than Moon´s ... If Sun not evolved as we know it will, a "fight" between Sun and Moon would happen ... Would Earth end synchronized to the Moon (as Moon is already), or to the Sun? 

Theoretically,  if we assumed no changes in the Sun mass (which there will be, as the Sun is constantly losing mass) We could predict the following scenario:

The Moon will continue to move outward from the Earth while the Earth rotation slows.  A some point, they will match up. ( in some tens of billions of years. It is hard to come up with an exact value, as there ar a lot of factors involved, such as how continental drift changes the drag effect the Earth has on the tidal bulges).     The only thing that would prevent that is if the Moon were to recede far enough away from the Earth to bring the Hill sphere into account.  One reason that this isn't likely is because the Earth is actually losing rotational energy faster than the Moon is gaining it. Most of the braking slowing the Earth rotation is due to tidal friction and this energy is turned to heat which the Earth radiates away. 
Once the two match up, the tidal acceleration from the Earth-Moon interaction will be zero. The Sun will still be applying tidal braking to the Earth.  The Earth will now be trying to rotate slower than the Moon orbits. This sets up a reverse tidal acceleration effect between the two, were the Moon will begin to circle in towards the Earth giving up angular momentum to the Earth fighting the slow down.  At first, the Sun will win and Earth will continue to slow.  But as the Moon gets closer, its influence will increase and the Earth will start to speed up again. This will continue until the Moon reaches the Roche limit and is torn apart forming a ring.  The even distribution of the ring material will result in no more tidal bulge caused by that material and the Sun will take over again, slowing the Earth.    Whether or not this results in the Earth ever tidal locking with the Sun depends on if we are considering the effects of the other planets. The Earth is far enough away from the Sun that perturbing effects from other planets will prevent it from ever coming into a purely locked situation with the Sun.

In reality, the Sun is constantly losing mass and as a result, the Earth will move away from the Sun weakening its tidal influence.  ( the loss of Solar mass and resulting expanding orbit makes it hard to predict whether or not the Sun will be engulfed by the Sun when it expands to a Red giant.  The Earth could be as far as way as Mars is now, and the Sun might not expand out that far.)