<p>You missed the the most important factor, which is, how close the sun is and as a result of the distance to the sun and relative mass, the Hill sphere and the approximate distance where <a href="https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability" rel="nofollow noreferrer">the stable orbital distance ends</a>.</p>

<p>The angular momentum to rate the moon gets pushed away from the planet is an interesting combination of formulas, and could probably be calculated based on the factors you mentioned, density, mass, size, shape and mass of the liquid oceans on the surface, rate of rotation, etc.  There's probably a not all that complex approximation that could be worked out for difference scenarios to show how far a moon would migrate before the system tidally locks and how long it would take. (some less interactive systems might take trillions or quadrillions of years or longer to reach a relatively final position, compared to the approximately 50 billion or so timetable for the Earth-Moon system to tidally lock.</p>

<p>But if we look at the Sun-Earth-Moon as an example, the most important on whether the Earth loses it's moon is how far the stable part of the Hill Sphere extends, which, if the masses of all 3 objects are what they are now, the primary factor is how close the Earth is to the Sun.   If the Earth was as close to the Sun as Venus, the Moon probably would escape, but it would take some time.   If the Earth was as close to the sun as Mercury (on average), it would have lost the Moon long ago.</p>

<p>OK, I think I have a simple inroad to your other questions, at least a good place to start.   Angular momentum.</p>

<p>If we consider a two body orbital system then the tidal effect on the orbit is fairly straight forward.  Considering systems where the bulge rotates ahead of the moon the moon is slowly pushed away from the planet and the planet's rotation gradually decreases. </p>

<p>In this two body system the moon never escapes because in a two body system, orbits remain stable over an enormous distance, the question essentially becomes, how far does the moon move away and how long does it take until the system is tidally locked where it, at least mathematically, reaches it's end state.</p>

<p>The 3rd body in the picture (the Sun) has an effect in a couple ways.   The relation between the Sun and Planet's relative mass and distance determines the approximate boundary of the stable region of the Hill Sphere, beyond which the moon likely escapes.   The Sun also tugs on the Moon, creating perturbations in it's orbit and making it non-curcular, where, in comparison, the two body system with the moon being pushed away moved towards becoming circular.   </p>

<p>For Red-dwarfs, where planets are close to the star, the planet becoming tidally locked to the red-dwarf star becomes much more likely (though some partial locking like Mercury's 3/2 orbit to rotation ratio may happen occasionally), but in most cases, with a red dwarf star and planets relatively close proximity, the planet becomes tidally locked to the star and effectively rotating slower relative to the moon, leading to a decaying orbit for moons towards their planets in red-dwarf systems.   The stable orbital region in such systems would also be much smaller.   The sun is a significant factor in the motion of moons to or from planets.</p>

<p>But getting back to the 2 body system and estimating the tidal effect, knowing that the moon will never escape from a closed two body system, the question then works out to how far the Moon can move from the planet and how long it will take until mutual tidal locking is reached.  </p>

<p>How far the Moon can travel from the planet can be estimated by the total angular momentum of the system, so you're correct that the initial rotation speed is a factor, so is the relative mass between the Planet and Moon.  A more massive moon creates a larger tidal bulge on the planet, which results n the Moon moving away from the planet faster than a less massive moon.  At the same time, the more massive moon requires more angular momentum as it moves away from the planet, so it moves less far.   A less massive moon moves further but takes much much longer to get there and the tidal lock also takes longer to happen.  </p>

<blockquote>
  <p>A more fluid planet would experience stronger tides, which would again
  slow it down faster, but definitely accelerate the satellite faster. A
  more viscous fluid than water would experience weaker tides, but I
  think they'd be farther ahead. I think more fluid is definitely more
  effective, but it seems less obvious than the first point.</p>
</blockquote>

<p>It's not correct to say a more fluid planet would experience stronger tides.  Saying that it would experience higher tides than if it was solid would be more accurate.  The tides the planet experiences depend on the mass and distance of the Moon and the size of the planet, so the Moon being equal, the tides would be the same whether the planet is liquid or solid.   Fluids are less rigid, so they are better at making tides, but there are tides on land too. On Earth they are called <a href="https://en.wikipedia.org/wiki/Earth_tide" rel="nofollow noreferrer">Earth-tides</a>.  </p>

<p>The higher tide does effectively push the Moon better and slow the planet more quickly and liquid responds to tides more efficiently, so planets with a liquid surface do move their moons more efficiently and faster than planets without a liquid surface.  The shape of the oceans matter too.   The wider Pacific has bigger tides than the narrower Atlantic. </p>

<p><a href="https://i.sstatic.net/VCTqn.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/VCTqn.jpg" alt="enter image description here"></a>   </p>

<p><a href="http://marinebio.org/oceans/currents-tides/index.aspx" rel="nofollow noreferrer">Source</a></p>

<p>I don't believe the viscosity of the liquid matters.   Liquids flow, solids don't and father ahead isn't correct either.   Planets generally rotate as a unit, though there are small discrepancies between the rate of rotation between the core and mantle, the entire planet rotates and the tidal bulge rotates with it.  Liquids are more efficient and bulging, but not either way the bulge moves ahead with the planet's rotation at equal angles ahead of the moon.</p>

<blockquote>
  <p>A larger planet would have its orbit slowed less by tidal
  acceleration, but have a stronger pull to escape. A larger satellite
  would cause stronger tides and more tidal force, but would require
  more force to accelerate, and would slow the planet down more quickly.
  I really can't tell which has the stronger effect...</p>
</blockquote>

<p>Here you have to be careful because it's a little more complicated.  The tidal force a planet experiences from it's moon has to do with the mass of the moon, the distance of the moon, but also the mass of the planet and the radius of the planet.</p>

<p>Looking at the numbers:</p>

<p><a href="https://web.njit.edu/~gary/320/assets/tide_fig1.gif" rel="nofollow noreferrer"><img src="https://web.njit.edu/~gary/320/assets/tide_fig1.gif" alt="enter image description here"></a>   </p>

<p><a href="https://web.njit.edu/~gary/320/Lecture12.html" rel="nofollow noreferrer">Source</a></p>

<p>the 2nd part of the formula is pretty tricky, $[(1 + R/d)^{-2} - (1 - R/d)^{-2}]$</p>

<p>but if you work out the math for fairly distant moons, the difference in the tidal force from the far side of the planet to the near side is about:</p>

<p>$$\Delta F = \frac{-4GMmR}{d^3}$$</p>

<p>So looking at the Earth-Moon system, ignoring the sun for now.</p>

<p>If you make Earth the same size but twice the mass, the tidal force doubles but Earth's surface gravity would double as well which would give the ocean water twice the weight with equal mass, the force on the Moon should be similar, but a heavier Earth would take longer to slow down, so the effect would be that the Moon would move further, not less far.</p>

<p>If you keep the density the same and make the Earth larger, say 8 times the mass, so twice the diameter.  The tidal force the Moon exerts on our new super-earth would be 16 times greater but the gravity on the surface of the Earth just twice as much.   What that means is that by making the planet larger, the Moon actually moves away faster.   The larger (but equal density) planet should slow down at roughly an equal rate but as the Moon moves away that rate would slow down.</p>

<p>Some of the math on tides gets confusing for me.  The inverse tidal force that moves moons away from planets drops off significantly with distance.  The tidal force drops off with the cube of the distance.  The force from the tidal bulge may drop off even faster than that because the angle decreases with distance, (I would guess it drops off by the 4th power), so, as the Moon's distance from the Earth doubles, the speed it moves away drops by (roughly), 16 times.   (((I think))).</p>

<p>So, if you make the Earth 8 times as massive, and twice the radius, the tidal force increases by 16, but if the Moon moves twice as far, the tidal force drops by 16, which is kind of neat that it cancels out.   The distance to gravity ratio, assuming the Moon doesn't change and the planet's density doesn't change, is 1 to 1, which is what I'd expect.</p>

<p>That's probably longer than it should be and it probably needs a lot of clean-up, but I need to call it a day, so I'll look at this tomorrow.  </p>

CC BY-SA 3.0
2018-01-08T15:47:10.110Z
Thanks for this great answer! On the schema, is it the sun or the moon that you picture as the small disc?
J. Chomel
CC BY-SA 3.0
2018-01-08T16:33:35.107Z
I copied the picture from a website, so it's not mine, but in the picture it's the sun (yes they got the size wrong, I know), but for the mathematical formula it doesn't matter, it's only the distances that are represented by "R", "d" etc. That are needed for the formula.   You could cross out sun and write moon and the picture still works.
userLTK
CC BY-SA 3.0
2018-01-26T10:55:36.897Z
Excellent answer of the first question, very detailed and an interesting read! I will select the other answer, since it includes the second question. I'm embarrassed that it wasn't obvious to me that the satellite can only be ejected by the influence of a 3rd body, or that the final state would be based only on the system's initial angular momentum with everything else only affecting time frame. Thanks! I did know there are land tides. So while solids/viscous liquids would take longer to rebound, the tides wouldn't be as high in the first place, which is why they wouldn't get "farther ahead"?
user1410910
CC BY-SA 3.0
2018-01-06T15:39:03.257Z
2018-01-08T11:23:03.020Z
2018-01-08T11:23:03.020Z
userLTK
<blockquote>
  <p>So that's question 1: Is it really possible for tidal acceleration to
  throw a satellite into space before the planet is tidally locked, and
  if so, does either objects' size impact it, or just fluidity and
  relative periods?</p>
</blockquote>

<p>During the locking process, the angular momentum of one is fed into the other until the periods match.  The final destination depends on the initial momentum of both objects and the moment of inertia.  As an example, if you make the moon less massive, or the earth more massive or with a faster spin, then the final configuration of the system would have a larger orbit.</p>

<p>You could certainly create a system where the final theoretical configuration has a distance to the satellite that is beyond the primary's Hill sphere.  Such a system would be expected to eventually separate.</p>

<p>Fluidity might change the rate at which the final destination is reached, but not its configuration.</p>

<blockquote>
  <p>Will all non-retrogade tidally decelerating satellites eventually
  crash into their planet, or is it possible for the planet to catch up
  given the right starting conditions?</p>
</blockquote>

<p>Each object is transferring the same amount of angular momentum.  But the response of each object depends (at least initially) on the relative magnitudes of their moments of inertia.  For a system where the satellite is (relatively) small, or where the difference in rotation rates is large, impact is inevitable.  The primary cannot speed up fast enough to keep up with the increasing speed of the descending partner.</p>

<p>But when the difference in periods is small, and the primary is small enough/far enough, then the objects would lock before impact.  </p>

CC BY-SA 3.0
2018-01-26T11:13:24.087Z
Thanks! Answered both questions (though the other answer went into more detail on the nuances of how fluidity would affect the speed of momentum exchange). Basically, the final state is entirely based on initial angular momentum of the system, so simple! Plus reading up on Hill spheres made me think about sub-satellites, which lead me to read why the Moon couldn't have a long term sub-satellite, so that's a bonus :p
user1410910
CC BY-SA 4.0
2023-07-23T19:42:07.910Z
This answer requires a minor correction. Instead of the Hill sphere, we must consider the *reduced Hill sphere*. https://astronomy.stackexchange.com/questions/36918/as-the-moon-and-the-earth-are-predicted-to-get-into-tidal-lock-how-slow-would-t/50836#50836
Michael_1812
CC BY-SA 3.0
2018-01-06T08:41:21.330Z
2018-01-06T08:41:21.330Z
BowlOfRed
<p>So, I know the Moon experiences tidal acceleration from the Earth. And, from what I've read, if not for the fact that the sun would boil away the oceans and engulf both of them first, about 50 billion years from now the Moon would be geosynchronous.</p>

<p>I was curious whether tidal acceleration could ever cause a satellite to escape, and short answer seems to be, yes, it's possible. Is that accurate? And if so, which factors add to it?</p>

<ol>
<li>A faster rotating planet also results in tides farther ahead, and has more energy to lose before being totally locked. Very closely related, a farther satellite orbits more slowly, resulting in tides farther ahead, and requires less energy to escape. This one seems obvious: the faster the planet is still rotating while the satellite is farther away, the more likely it escapes.</li>
<li>A more fluid planet would experience stronger tides, which would again slow it down faster, but definitely accelerate the satellite faster. A more viscous fluid than water would experience weaker tides, but I think they'd be farther ahead. I think more fluid is definitely more effective, but it seems less obvious than the first point.</li>
<li>A larger planet would have its orbit slowed less by tidal acceleration, but have a stronger pull to escape. A larger satellite would cause stronger tides and more tidal force, but would require more force to accelerate, and would slow the planet down more quickly. I really can't tell which has the stronger effect...</li>
</ol>

<p>So that's question 1: 
Is it really possible for tidal acceleration to throw a satellite into space before the planet is tidally locked, and if so, does either objects' size impact it, or just fluidity and relative periods?</p>

<p>Then, I wondered about tidal deceleration, and crashing into the planet. It seems obvious that any satellite orbiting retrograde would never stop decelerating, so all would eventually get torn apart and crash onto their planet. For non-retrogade satellites, an article I read implied that all of them will have the same fate... Which implies that the planet's rotation will always speed up more slowly than the satellite's falling orbit. Is that true? If not, then it seems there must be cases where the planet's rotation catches up, and they again end up tidally locked.</p>

<p>So that's question 2:
Will all non-retrogade tidally decelerating satellites eventually crash into their planet, or is it possible for the planet to catch up given the right starting conditions?</p>

<p>Edit: Haven't really found any more information on this. Is there someone familiar with relevant equations, who knows the answer?</p>

<natural-satellites><orbital-mechanics><tidal-forces><tidal-locking>
Eventual outcome of tidal acceleration and deceleration
user1410910
# Eventual outcome of tidal acceleration and deceleration

(Asked by: username_0 on 2017-12-14)

So, I know the Moon experiences tidal acceleration from the Earth. And, from what I've read, if not for the fact that the sun would boil away the oceans and engulf both of them first, about 50 billion years from now the Moon would be geosynchronous.

I was curious whether tidal acceleration could ever cause a satellite to escape, and short answer seems to be, yes, it's possible. Is that accurate? And if so, which factors add to it?

1. A faster rotating planet also results in tides farther ahead, and has more energy to lose before being totally locked. Very closely related, a farther satellite orbits more slowly, resulting in tides farther ahead, and requires less energy to escape. This one seems obvious: the faster the planet is still rotating while the satellite is farther away, the more likely it escapes.
2. A more fluid planet would experience stronger tides, which would again slow it down faster, but definitely accelerate the satellite faster. A more viscous fluid than water would experience weaker tides, but I think they'd be farther ahead. I think more fluid is definitely more effective, but it seems less obvious than the first point.
3. A larger planet would have its orbit slowed less by tidal acceleration, but have a stronger pull to escape. A larger satellite would cause stronger tides and more tidal force, but would require more force to accelerate, and would slow the planet down more quickly. I really can't tell which has the stronger effect...

So that's question 1:
Is it really possible for tidal acceleration to throw a satellite into space before the planet is tidally locked, and if so, does either objects' size impact it, or just fluidity and relative periods?

Then, I wondered about tidal deceleration, and crashing into the planet. It seems obvious that any satellite orbiting retrograde would never stop decelerating, so all would eventually get torn apart and crash onto their planet. For non-retrogade satellites, an article I read implied that all of them will have the same fate... Which implies that the planet's rotation will always speed up more slowly than the satellite's falling orbit. Is that true? If not, then it seems there must be cases where the planet's rotation catches up, and they again end up tidally locked.

So that's question 2:
Will all non-retrogade tidally decelerating satellites eventually crash into their planet, or is it possible for the planet to catch up given the right starting conditions?

Edit: Haven't really found any more information on this. Is there someone familiar with relevant equations, who knows the answer?

**Tags:** natural-satellites, orbital-mechanics, tidal-forces, tidal-locking

---

## 2 answers

### Answer by username_3 on 2018-01-06. Score: 5 | Accepted answer

> So that's question 1: Is it really possible for tidal acceleration to
> throw a satellite into space before the planet is tidally locked, and
> if so, does either objects' size impact it, or just fluidity and
> relative periods?

During the locking process, the angular momentum of one is fed into the other until the periods match. The final destination depends on the initial momentum of both objects and the moment of inertia. As an example, if you make the moon less massive, or the earth more massive or with a faster spin, then the final configuration of the system would have a larger orbit.

You could certainly create a system where the final theoretical configuration has a distance to the satellite that is beyond the primary's Hill sphere. Such a system would be expected to eventually separate.

Fluidity might change the rate at which the final destination is reached, but not its configuration.

> Will all non-retrogade tidally decelerating satellites eventually
> crash into their planet, or is it possible for the planet to catch up
> given the right starting conditions?

Each object is transferring the same amount of angular momentum. But the response of each object depends (at least initially) on the relative magnitudes of their moments of inertia. For a system where the satellite is (relatively) small, or where the difference in rotation rates is large, impact is inevitable. The primary cannot speed up fast enough to keep up with the increasing speed of the descending partner.

But when the difference in periods is small, and the primary is small enough/far enough, then the objects would lock before impact.

> **Comments:**
> **username_0**: Thanks! Answered both questions (though the other answer went into more detail on the nuances of how fluidity would affect the speed of momentum exchange). Basically, the final state is entirely based on initial angular momentum of the system, so simple! Plus reading up on Hill spheres made me think about sub-satellites, which lead me to read why the Moon couldn't have a long term sub-satellite, so that's a bonus :p  
> **username_4**: This answer requires a minor correction. Instead of the Hill sphere, we must consider the *reduced Hill sphere*. https://astronomy.stackexchange.com/questions/36918/as-the-moon-and-the-earth-are-predicted-to-get-into-tidal-lock-how-slow-would-t/50836#50836  

---

### Answer by username_1 on 2018-01-06. Score: 4

You missed the the most important factor, which is, how close the sun is and as a result of the distance to the sun and relative mass, the Hill sphere and the approximate distance where [the stable orbital distance ends](https://en.wikipedia.org/wiki/Hill_sphere#True_region_of_stability).

The angular momentum to rate the moon gets pushed away from the planet is an interesting combination of formulas, and could probably be calculated based on the factors you mentioned, density, mass, size, shape and mass of the liquid oceans on the surface, rate of rotation, etc. There's probably a not all that complex approximation that could be worked out for difference scenarios to show how far a moon would migrate before the system tidally locks and how long it would take. (some less interactive systems might take trillions or quadrillions of years or longer to reach a relatively final position, compared to the approximately 50 billion or so timetable for the Earth-Moon system to tidally lock.

But if we look at the Sun-Earth-Moon as an example, the most important on whether the Earth loses it's moon is how far the stable part of the Hill Sphere extends, which, if the masses of all 3 objects are what they are now, the primary factor is how close the Earth is to the Sun. If the Earth was as close to the Sun as Venus, the Moon probably would escape, but it would take some time. If the Earth was as close to the sun as Mercury (on average), it would have lost the Moon long ago.

OK, I think I have a simple inroad to your other questions, at least a good place to start. Angular momentum.

If we consider a two body orbital system then the tidal effect on the orbit is fairly straight forward. Considering systems where the bulge rotates ahead of the moon the moon is slowly pushed away from the planet and the planet's rotation gradually decreases.

In this two body system the moon never escapes because in a two body system, orbits remain stable over an enormous distance, the question essentially becomes, how far does the moon move away and how long does it take until the system is tidally locked where it, at least mathematically, reaches it's end state.

The 3rd body in the picture (the Sun) has an effect in a couple ways. The relation between the Sun and Planet's relative mass and distance determines the approximate boundary of the stable region of the Hill Sphere, beyond which the moon likely escapes. The Sun also tugs on the Moon, creating perturbations in it's orbit and making it non-curcular, where, in comparison, the two body system with the moon being pushed away moved towards becoming circular.

For Red-dwarfs, where planets are close to the star, the planet becoming tidally locked to the red-dwarf star becomes much more likely (though some partial locking like Mercury's 3/2 orbit to rotation ratio may happen occasionally), but in most cases, with a red dwarf star and planets relatively close proximity, the planet becomes tidally locked to the star and effectively rotating slower relative to the moon, leading to a decaying orbit for moons towards their planets in red-dwarf systems. The stable orbital region in such systems would also be much smaller. The sun is a significant factor in the motion of moons to or from planets.

But getting back to the 2 body system and estimating the tidal effect, knowing that the moon will never escape from a closed two body system, the question then works out to how far the Moon can move from the planet and how long it will take until mutual tidal locking is reached.

How far the Moon can travel from the planet can be estimated by the total angular momentum of the system, so you're correct that the initial rotation speed is a factor, so is the relative mass between the Planet and Moon. A more massive moon creates a larger tidal bulge on the planet, which results n the Moon moving away from the planet faster than a less massive moon. At the same time, the more massive moon requires more angular momentum as it moves away from the planet, so it moves less far. A less massive moon moves further but takes much much longer to get there and the tidal lock also takes longer to happen.

> A more fluid planet would experience stronger tides, which would again
> slow it down faster, but definitely accelerate the satellite faster. A
> more viscous fluid than water would experience weaker tides, but I
> think they'd be farther ahead. I think more fluid is definitely more
> effective, but it seems less obvious than the first point.

It's not correct to say a more fluid planet would experience stronger tides. Saying that it would experience higher tides than if it was solid would be more accurate. The tides the planet experiences depend on the mass and distance of the Moon and the size of the planet, so the Moon being equal, the tides would be the same whether the planet is liquid or solid. Fluids are less rigid, so they are better at making tides, but there are tides on land too. On Earth they are called [Earth-tides](https://en.wikipedia.org/wiki/Earth_tide).

The higher tide does effectively push the Moon better and slow the planet more quickly and liquid responds to tides more efficiently, so planets with a liquid surface do move their moons more efficiently and faster than planets without a liquid surface. The shape of the oceans matter too. The wider Pacific has bigger tides than the narrower Atlantic.

[![enter image description here](https://i.sstatic.net/VCTqn.jpg)](https://i.sstatic.net/VCTqn.jpg)

[Source](http://marinebio.org/oceans/currents-tides/index.aspx)

I don't believe the viscosity of the liquid matters. Liquids flow, solids don't and father ahead isn't correct either. Planets generally rotate as a unit, though there are small discrepancies between the rate of rotation between the core and mantle, the entire planet rotates and the tidal bulge rotates with it. Liquids are more efficient and bulging, but not either way the bulge moves ahead with the planet's rotation at equal angles ahead of the moon.

> A larger planet would have its orbit slowed less by tidal
> acceleration, but have a stronger pull to escape. A larger satellite
> would cause stronger tides and more tidal force, but would require
> more force to accelerate, and would slow the planet down more quickly.
> I really can't tell which has the stronger effect...

Here you have to be careful because it's a little more complicated. The tidal force a planet experiences from it's moon has to do with the mass of the moon, the distance of the moon, but also the mass of the planet and the radius of the planet.

Looking at the numbers:

[![enter image description here](https://web.njit.edu/~gary/320/assets/tide_fig1.gif)](https://web.njit.edu/~gary/320/assets/tide_fig1.gif)

[Source](https://web.njit.edu/~gary/320/Lecture12.html)

the 2nd part of the formula is pretty tricky, $[(1 + R/d)^{-2} - (1 - R/d)^{-2}]$

but if you work out the math for fairly distant moons, the difference in the tidal force from the far side of the planet to the near side is about:

$$\Delta F = \frac{-4GMmR}{d^3}$$

So looking at the Earth-Moon system, ignoring the sun for now.

If you make Earth the same size but twice the mass, the tidal force doubles but Earth's surface gravity would double as well which would give the ocean water twice the weight with equal mass, the force on the Moon should be similar, but a heavier Earth would take longer to slow down, so the effect would be that the Moon would move further, not less far.

If you keep the density the same and make the Earth larger, say 8 times the mass, so twice the diameter. The tidal force the Moon exerts on our new super-earth would be 16 times greater but the gravity on the surface of the Earth just twice as much. What that means is that by making the planet larger, the Moon actually moves away faster. The larger (but equal density) planet should slow down at roughly an equal rate but as the Moon moves away that rate would slow down.

Some of the math on tides gets confusing for me. The inverse tidal force that moves moons away from planets drops off significantly with distance. The tidal force drops off with the cube of the distance. The force from the tidal bulge may drop off even faster than that because the angle decreases with distance, (I would guess it drops off by the 4th power), so, as the Moon's distance from the Earth doubles, the speed it moves away drops by (roughly), 16 times. (((I think))).

So, if you make the Earth 8 times as massive, and twice the radius, the tidal force increases by 16, but if the Moon moves twice as far, the tidal force drops by 16, which is kind of neat that it cancels out. The distance to gravity ratio, assuming the Moon doesn't change and the planet's density doesn't change, is 1 to 1, which is what I'd expect.

That's probably longer than it should be and it probably needs a lot of clean-up, but I need to call it a day, so I'll look at this tomorrow.

> **Comments:**
> **username_2**: Thanks for this great answer! On the schema, is it the sun or the moon that you picture as the small disc?  
> **username_1**: I copied the picture from a website, so it's not mine, but in the picture it's the sun (yes they got the size wrong, I know), but for the mathematical formula it doesn't matter, it's only the distances that are represented by "R", "d" etc. That are needed for the formula.   You could cross out sun and write moon and the picture still works.  
> **username_0**: Excellent answer of the first question, very detailed and an interesting read! I will select the other answer, since it includes the second question. I'm embarrassed that it wasn't obvious to me that the satellite can only be ejected by the influence of a 3rd body, or that the final state would be based only on the system's initial angular momentum with everything else only affecting time frame. Thanks! I did know there are land tides. So while solids/viscous liquids would take longer to rebound, the tides wouldn't be as high in the first place, which is why they wouldn't get "farther ahead"?  

---

