<p>To answer this question, we have to recall the definition of the Laplace surface. Conventionally called <em>Laplace plane</em>, this surface is not actually a plane. In the vicinity of a planet, it coincides with this planet's equatorial plane, while further from the planet it coincides with the planet's orbit about the Sun. As <a href="https://en.wikipedia.org/wiki/Laplace_plane" rel="nofollow noreferrer">Wikipedia</a> says,</p>
<ul>
<li><em>The Laplace plane arises because the equatorial oblateness of the parent planet tends to cause the orbit of the satellite to precess
around the polar axis of the parent planet's equatorial plane, while
the solar perturbations tend to cause the orbit of the satellite to
precess around the polar axis of the parent planet's orbital plane
around the Sun. The two effects acting together result in an
intermediate position for the reference axis for the satellite
orbit's precession.</em></li>
</ul>
<p>Locally, the Laplace plane (or, better to say, Laplace surface) is perpendicular to the axis about which the orbital plane of the satellite is precessing.</p>
<p>Being remote from Earth, the Moon &quot;feels&quot; the pull of the Sun stronger than it &quot;feels&quot; the Earth's oblateness. So its (local) Laplace surface is close to the ecliptic. The normal to the lunar orbit precesses about the normal to ecliptic. (Actually, so does the lunar spin axis, because the Moon is in a Cassini state.)</p>
<p>Now, it we place the Moon on a slightly more inclined orbit (say, <span class="math-container">$i_0=$</span>10<span class="math-container">$^{\circ}$</span>), the tidal forces will start working to reduce <span class="math-container">$i$</span>. However, due to the presence of the Sun, they will reduce <span class="math-container">$i$</span> not to zero but to the present configuration, where the orbit will be precessing about the Laplace surface.</p>
<p>As our collocutor eshaya rightly pointed out, a larger moon raises a larger tidal bulge on the Earth. Since the tidal interactions are caused by the tidal bulge, the timescale of the tidal evolution would be shorter for a more massive moon.</p>
<p>All said about the Moon pertains to the outher satellites of the giant planets.</p>
<p>The case of the inner satellites or Phobos and Deimos is different. They &quot;feel&quot; the oblateness of their host planet stronger than the pull of the Sun. So, in their case, the tidal interaction is working to nullify the inclination.</p>
<p>Now, how about Triton? Aside from being retrograde, its orbit is strongly inclined to Neptune's equator. How can it remain stable? To solve this puzzle, we may have to challenge the common belief that the tidal forces are always working to reduce the inclination. To see that this is not always so, please have a look at eqn (165) in <a href="https://arxiv.org/abs/1904.02253" rel="nofollow noreferrer">this paper</a>, and read the subsequent paragraph. If you manage to prove, using this knowledge, that Triton's orbit is tidally stable, your result will be publishable in <em>Nature</em>, certainly in <em>Nature Astronomy</em>. For accurate analysis, you will have to use the full expression (162), and will have to develop a physically reasonable tidal model for Neptune. (In the first approximation, I would probably neglect the tides in Triton, though I am not 100% sure if this is permissible.)  Also, you will have to demonstrate that the entire set of initial conditions contains a finite
-- <em>and not negligibly small!</em> -- subset leading to the present configuration. I would estimate this work as 1/3 - 1/2 of a good PhD thesis. Could be a whole <em>bona fide</em> thesis, if contains a detailed tidal model for Neptune.</p>

CC BY-SA 4.0
2022-06-20T22:03:19.267Z
2023-12-03T14:58:00.477Z
2023-12-03T14:58:00.477Z
Michael_1812
<p>Discussions <a href="https://chat.stackexchange.com/rooms/119459/discussion-between-pm-2ring-and-uhoh">elsewhere</a> suggest that this happens and that smaller moons would tend to low inclination with respect to the planet's equator faster than larger ones.</p>
<p>For example, Mars' small moons are inclined only by 1° or so while Earth's moon's inclination varies between 23-5° and 23+5° with an 18.6 year period.</p>
<p>But this is not a good example because the origin and history of these moons is very different.</p>
<p><strong>Question:</strong> So if we had two Earths around two Suns at 1 AU each, and each had a rocky moon like our own inclined at say 10° with respect to the Earth's oblate equator at the same distances <em>except that one moon was big and one was small:</em></p>
<ol>
<li>Would they migrate to low inclination over time?</li>
<li>If so, does the time it takes depend on the Moon's size?</li>
<li>If so, which would tend to migrate to low inclination faster?</li>
</ol>

CC BY-SA 4.0
2021-02-09T23:39:17.680Z
I believe this is an unsolved problem in astronomy.  It is considered to be too much of a coincidence that Mars's moons ended up with such a low inclination, but I don't think a conclusive mechanism for evolution to that state has been proposed.
Connor Garcia
CC BY-SA 4.0
2021-02-10T00:14:38.127Z
@ConnorGarcia thanks for reinforcing what I've written about that: "But this is not a good example because the origin and history of these moons is very different." While I cites a few contrasting/extreme examples, the question can be answered based strictly on theory and does not need to address any particular object. That's not what this question is about.
uhoh
CC BY-SA 4.0
2021-03-22T17:36:39.810Z
Are the Moons and Earth in this hypothetical question "rocky" to mean they don't have symmetric density/mass distribution w.r.t. the Earth's equatorial plane and any plane passing through the axis of rotation?  If I remember correctly, the Earth has more mass in the Northern Hemisphere than the Southern Hemisphere.
Connor Garcia
CC BY-SA 4.0
2021-03-22T23:06:19.267Z
@ConnorGarcia [A terrestrial planet, telluric planet, or rocky planet is a planet that is composed primarily of silicate rocks or metals. Within the Solar System, the terrestrial planets accepted by the IAU are the inner planets closest to the Sun, i.e. Mercury, Venus, Earth, and Mars. Among astronomers who use the geophysical definition of a planet, the Moon, Io and Europa may also be considered terrestrial planets.](https://en.wikipedia.org/wiki/Terrestrial_planet) I don't mean to specify that it must not be symmetric, but ya they certainly can have some odd-numbered harmonics.
uhoh
CC BY-SA 4.0
2021-03-22T23:13:34.153Z
See [Is Earth's “pear shape” mostly J₃?](https://space.stackexchange.com/q/45348/12102) and also [1959 Peanuts cartoon about the Fischer ellipsoid (Earth is “pear-shaped”)?](https://scifi.stackexchange.com/q/234294/51174) and [Which side of the Moon faces Earth, the heavy side or the lighter side?](https://space.stackexchange.com/q/22520/12102) I still don't understand how there can be more mass on one side of the center of mass than the other, but I"m probably missing something.
uhoh
CC BY-SA 4.0
2021-03-23T18:28:59.980Z
I think the usual equator definition is a plane orthogonal to the rotation axis but equidistant from the poles, rather than passing through the center of gravity.  The more symmetric the planet, the closer the center of gravity will be to the equator.
Connor Garcia
CC BY-SA 4.0
2021-03-30T19:35:31.487Z
A larger moon raises a larger tidal bulge on the Earth.  Since the torques are caused by the tidal bulge, I would think that the timescale for evolution would be shorter for a more massive moon.
eshaya
CC BY-SA 4.0
2021-04-03T05:02:15.117Z
@uhoh this is a really great question. I wish I knew enough to answer it.  Deep respect---
Connor Garcia
CC BY-SA 4.0
2021-04-03T05:10:12.143Z
@ConnorGarcia ha! your respect is sorely misplaced. I'm simply a Stack Exchange junkie constantly looking for my next fix. I don't mean to make light of the addiction, I use SE to avoid my others. This question simply comes directly from the linked discussion; I always ask "How do you know..." in order to see if a new question is possible. That's my *modus operandi* See for example [this](https://chat.stackexchange.com/transcript/message/57517883#57517883). In that case I chose to [write an answer](https://biology.stackexchange.com/a/100100/27918) because it was faster
uhoh
CC BY-SA 4.0
2021-08-31T16:18:18.907Z
Wouldn't this depend on the higher order moments of the multipole expansion of gravitational fields?  That is, the oblateness of a planetoid affects the moments differently than a spherical planetoid, so it would alter the gravitational field model, right?
honeste_vivere
<solar-system><orbital-mechanics><tidal-forces><planetary-science>
Do moons of rocky oblate planets migrate to low inclination over time? If so, does the time it takes depend on the Moon's size?
uhoh
# Do moons of rocky oblate planets migrate to low inclination over time? If so, does the time it takes depend on the Moon's size?

(Asked by: username_0 on 2021-02-08)

Discussions [elsewhere](https://chat.stackexchange.com/rooms/119459/discussion-between-pm-2ring-and-username_0) suggest that this happens and that smaller moons would tend to low inclination with respect to the planet's equator faster than larger ones.

For example, Mars' small moons are inclined only by 1° or so while Earth's moon's inclination varies between 23-5° and 23+5° with an 18.6 year period.

But this is not a good example because the origin and history of these moons is very different.

**Question:** So if we had two Earths around two Suns at 1 AU each, and each had a rocky moon like our own inclined at say 10° with respect to the Earth's oblate equator at the same distances *except that one moon was big and one was small:*

1. Would they migrate to low inclination over time?
2. If so, does the time it takes depend on the Moon's size?
3. If so, which would tend to migrate to low inclination faster?

**Tags:** solar-system, orbital-mechanics, tidal-forces, planetary-science

> **Comments:**
> **username_1**: I believe this is an unsolved problem in astronomy.  It is considered to be too much of a coincidence that Mars's moons ended up with such a low inclination, but I don't think a conclusive mechanism for evolution to that state has been proposed.  
> **username_0**: @ConnorGarcia thanks for reinforcing what I've written about that: "But this is not a good example because the origin and history of these moons is very different." While I cites a few contrasting/extreme examples, the question can be answered based strictly on theory and does not need to address any particular object. That's not what this question is about.  
> **username_1**: Are the Moons and Earth in this hypothetical question "rocky" to mean they don't have symmetric density/mass distribution w.r.t. the Earth's equatorial plane and any plane passing through the axis of rotation?  If I remember correctly, the Earth has more mass in the Northern Hemisphere than the Southern Hemisphere.  
> **username_0**: @ConnorGarcia [A terrestrial planet, telluric planet, or rocky planet is a planet that is composed primarily of silicate rocks or metals. Within the Solar System, the terrestrial planets accepted by the IAU are the inner planets closest to the Sun, i.e. Mercury, Venus, Earth, and Mars. Among astronomers who use the geophysical definition of a planet, the Moon, Io and Europa may also be considered terrestrial planets.](https://en.wikipedia.org/wiki/Terrestrial_planet) I don't mean to specify that it must not be symmetric, but ya they certainly can have some odd-numbered harmonics.  
> **username_0**: See [Is Earth's “pear shape” mostly J₃?](https://space.stackexchange.com/q/45348/12102) and also [1959 Peanuts cartoon about the Fischer ellipsoid (Earth is “pear-shaped”)?](https://scifi.stackexchange.com/q/234294/51174) and [Which side of the Moon faces Earth, the heavy side or the lighter side?](https://space.stackexchange.com/q/22520/12102) I still don't understand how there can be more mass on one side of the center of mass than the other, but I"m probably missing something.  
> **username_1**: I think the usual equator definition is a plane orthogonal to the rotation axis but equidistant from the poles, rather than passing through the center of gravity.  The more symmetric the planet, the closer the center of gravity will be to the equator.  
> **username_2**: A larger moon raises a larger tidal bulge on the Earth.  Since the torques are caused by the tidal bulge, I would think that the timescale for evolution would be shorter for a more massive moon.  
> **username_1**: @username_0 this is a really great question. I wish I knew enough to answer it.  Deep respect---  
> **username_0**: @ConnorGarcia ha! your respect is sorely misplaced. I'm simply a Stack Exchange junkie constantly looking for my next fix. I don't mean to make light of the addiction, I use SE to avoid my others. This question simply comes directly from the linked discussion; I always ask "How do you know..." in order to see if a new question is possible. That's my *modus operandi* See for example [this](https://chat.stackexchange.com/transcript/message/57517883#57517883). In that case I chose to [write an answer](https://biology.stackexchange.com/a/100100/27918) because it was faster  
> **username_3**: Wouldn't this depend on the higher order moments of the multipole expansion of gravitational fields?  That is, the oblateness of a planetoid affects the moments differently than a spherical planetoid, so it would alter the gravitational field model, right?  

---

## 1 answers

### Answer by username_4 on 2022-06-20. Score: 6 | Accepted answer

To answer this question, we have to recall the definition of the Laplace surface. Conventionally called *Laplace plane*, this surface is not actually a plane. In the vicinity of a planet, it coincides with this planet's equatorial plane, while further from the planet it coincides with the planet's orbit about the Sun. As [Wikipedia](https://en.wikipedia.org/wiki/Laplace_plane) says,

* *The Laplace plane arises because the equatorial oblateness of the parent planet tends to cause the orbit of the satellite to precess
  around the polar axis of the parent planet's equatorial plane, while
  the solar perturbations tend to cause the orbit of the satellite to
  precess around the polar axis of the parent planet's orbital plane
  around the Sun. The two effects acting together result in an
  intermediate position for the reference axis for the satellite
  orbit's precession.*

Locally, the Laplace plane (or, better to say, Laplace surface) is perpendicular to the axis about which the orbital plane of the satellite is precessing.

Being remote from Earth, the Moon "feels" the pull of the Sun stronger than it "feels" the Earth's oblateness. So its (local) Laplace surface is close to the ecliptic. The normal to the lunar orbit precesses about the normal to ecliptic. (Actually, so does the lunar spin axis, because the Moon is in a Cassini state.)

Now, it we place the Moon on a slightly more inclined orbit (say, $i\_0=$10$^{\circ}$), the tidal forces will start working to reduce $i$. However, due to the presence of the Sun, they will reduce $i$ not to zero but to the present configuration, where the orbit will be precessing about the Laplace surface.

As our collocutor username_2 rightly pointed out, a larger moon raises a larger tidal bulge on the Earth. Since the tidal interactions are caused by the tidal bulge, the timescale of the tidal evolution would be shorter for a more massive moon.

All said about the Moon pertains to the outher satellites of the giant planets.

The case of the inner satellites or Phobos and Deimos is different. They "feel" the oblateness of their host planet stronger than the pull of the Sun. So, in their case, the tidal interaction is working to nullify the inclination.

Now, how about Triton? Aside from being retrograde, its orbit is strongly inclined to Neptune's equator. How can it remain stable? To solve this puzzle, we may have to challenge the common belief that the tidal forces are always working to reduce the inclination. To see that this is not always so, please have a look at eqn (165) in [this paper](https://arxiv.org/abs/1904.02253), and read the subsequent paragraph. If you manage to prove, using this knowledge, that Triton's orbit is tidally stable, your result will be publishable in *Nature*, certainly in *Nature Astronomy*. For accurate analysis, you will have to use the full expression (162), and will have to develop a physically reasonable tidal model for Neptune. (In the first approximation, I would probably neglect the tides in Triton, though I am not 100% sure if this is permissible.) Also, you will have to demonstrate that the entire set of initial conditions contains a finite
-- *and not negligibly small!* -- subset leading to the present configuration. I would estimate this work as 1/3 - 1/2 of a good PhD thesis. Could be a whole *bona fide* thesis, if contains a detailed tidal model for Neptune.


---

