Or was it always at about 5.14 degrees inclination or has the inclination changed over time?

See diagram.

James K's answer to this question got me thinking about this, and I don't mean to call him out, but this part of his answer doesn't seem quite right when he wrote:

Our moon is unique in being close to the plane of the ecliptic, and not in the plane of the equator, which suggests its formation was not like that of other moons in the solar system.

For starters, I imagine the physics of the initial orbit of an impact moon would be somewhat complicated, but it seems likely, given an angled impact like the one that's believed to have formed our moon (and hitting at an angle is statistically more common than a direct impact anyway),

So, in this scenario, the planet gets significant angular momentum and this angular momentum should dictate the planet's new equator and axial tilt relative to it's orbit. I would think the moon should form roughly along that same equatorial plane, but I'm just guessing. Perhaps it could form several degrees off - not sure.

The 2nd point, is, a rapidly rotating planet with a large equatorial bulge and a close moon, if the Moon formed off the equator, would it's orbit migrate over the equator where the gravitation was greatest and would that happen relatively quickly or not at all?

That's basically the question. Was the Moon always at roughly 6 degrees off the Earth's equator or has it only moved off an orbit over Earth's equator more recently, perhaps due to the gravitational effect of the Sun?

Or are there other factors. Mars' axis is thought to have changed rather significantly due to Jupiter's gravitational effects perhaps 100,000 years agao and it's Moons orbit over Mars' equator which suggests that Mars' equatorial bulge dragged the Moons with it? - or is my thinking way off on that?

My thinking is that a planet's equatorial bulge would drive moons towards a 0 degree inclination around it's equatorial bulge and our Moon is different because of proximity to the sun which also has a strong gravitational effect. The 5.14 degrees of inclination is a balance between the Earth's equator and the solar gravity.

  • $\begingroup$ Note that the linked diagram shows the angle between the Moon's orbital plane and the ecliptic, the Earth's orbital plane, not the Earth's equatorial plane. The obliquity of the ecliptic is ~ 23.5° and the Moon's orbital plane inclination to the ecliptic is ~5.4°, so its inclination to the equatorial plane can vary from 18.1° to 28.9°, and it precesses between those limits over an 18.6 year cycle. See en.wikipedia.org/wiki/Lunar_precession particularly the diagram near the top of that article, and the section on nodal precession. $\endgroup$
    – PM 2Ring
    Dec 6, 2020 at 15:32

2 Answers 2


Yes, the Moon was likely formed in an orbit over the equator. I'm going to recycle my answer to a different question about the Moon's orbit as most of it is relevant to this one as well.

The main concept here is the Laplace plane, which is the plane about which a satellite's orbit precesses. In general, the inclination of a satellite with respect to this plane gets damped to small values. Close to the planet, the main influence on the Laplace plane is the equatorial bulge of the planet, further away the Sun (and to a lesser extent, the other planets in the Solar System) becomes a more significant influence.

The transition between satellite orbits aligned with the planet's equator and satellite orbits aligned with the planet's orbit occurs around a distance called the Laplace radius $r_{\rm L}$, from Nesvorný et al. (2014), which can be thought of as the scale of a planet's regular satellite system. This distance is given as:

$$r_\mathrm{L}^5 = J_2' R_\mathrm{p}^2 a_\mathrm{p}^3 \left(1-e_\mathrm{p}^2 \right)^{3/2} \frac{M_\mathrm{p}}{M_\odot}$$

where $R_{\rm p}$ is the planetary radius, $a_{\rm p}$ is the planet's semimajor axis, $e_{\rm p}$ is the planet's orbital eccentricity, $M_{\rm p}$ is the mass of the planet and $M_\odot$ is the mass of the Sun.

$J_2'$ is the quadrupole coefficient of the planet $J_2$, adjusted for the presence of $n$ equatorial inner satellites:

$$J_2' R_\mathrm{p}^2 = J_2 R_\mathrm{p}^2 + \frac{1}{2}\sum_{i=1}^n a_i^2 \frac{m_i}{M_\mathrm{p}}$$

In the case of the Moon, there are no inner satellites so $J_2' = J_2$.

The angle between the planet's spin axis and the Laplace plane $\phi$ is given by:

$$\tan 2\phi = \frac{\sin 2\theta}{\cos 2\theta + 2r_\mathrm{L}^5/a^5}$$

Where $\theta$ is the planet's obliquity and $a$ is the semimajor axis of the satellite's orbit. For $a \ll r_{\rm L}$, the denominator becomes very large, resulting in $\phi \approx 0$. For $a \gg r_{\rm L}$, the $2r_\mathrm{L}^5/a^5$ term becomes very small, and the equation becomes $\tan 2\phi \approx \sin 2\theta / \cos 2\theta$, giving $\phi \approx \theta$, which is the behaviour described earlier.

Using the $J_2$ value of Earth of 1.08263×10-3 from the NASA Earth Fact Sheet, the Laplace radius works out to 53600 km, or 8.4 Earth radii. Note that this may not necessarily have been the exact value billions of years ago, due to differences in the Earth's shape caused by rotational spindown, deformations due to ice ages, etc.

The Moon likely formed closer to the Earth than this distance, so would have been aligned with the equator. As it migrated outwards across the transition region, the orbit would have undergone rapid changes in orientation, which would also have an effect on the Earth's obliquity, potentially allowing the Earth to transition from a high-obliquity state which might explain the Faint Young Sun paradox (Jenkins, 2000) to the less-severe obliquity we observe today.

Sarah Stewart-Mukhopadhyay created a video of a simulation of the migration of the Moon across the transition region, with the Earth initially in a high-obliquity state.

  • $\begingroup$ "In general, the inclination of a satellite with respect to this plane gets damped to small values." I read that as, the satellite eventually orbits within the Laplace plane. But then it would not precess. What am I missing? $\endgroup$ Dec 7, 2020 at 1:29
  • $\begingroup$ @KeithMcClary - yes, for the exact case, the plane of the orbit does not move (at least not according to the model's approximations). Nevertheless, the direction of perihelion can still rotate within the plane even while the plane remains fixed, which also counts as precession of the orbit. $\endgroup$
    – user24157
    Dec 7, 2020 at 8:23
  • $\begingroup$ But the "nodal precession" does not get damped? (I'm not sure whether the damping is necessarily part of the answer.) $\endgroup$ Dec 8, 2020 at 3:04
  • $\begingroup$ @KeithMcClary - in the situation where the orbit is exactly within the Laplace plane, the nodal precession has been fully damped: in fact there is no line of nodes, and the rotation that would cause nodal precession is degenerate with the rotation of the perihelion around the orbit normal. This is the same phenomenon as "gimbal lock" which is inherent to specifying orientation in terms of Euler angles. $\endgroup$
    – user24157
    Dec 8, 2020 at 6:52

The plane of the Moon's orbit experiences an 18.6 year Nodal Precession, so the angle between its orbital plane and the equatorial plane varies. Even if the planes coincided at some time, it would be momentary.

Wikipedia discusses the origin of the orbital inclination:

The mean inclination of the lunar orbit to the ecliptic plane is 5.145°. Theoretical considerations show that the present inclination relative to the ecliptic plane arose by tidal evolution from an earlier near-Earth orbit with a fairly constant inclination relative to Earth's equator.[11] It would require an inclination of this earlier orbit of about 10° to the equator to produce a present inclination of 5° to the ecliptic. It is thought that originally the inclination to the equator was near zero, but it could have been increased to 10° through the influence of planetesimals passing near the Moon while falling to the Earth.[12] If this had not happened, the Moon would now lie much closer to the ecliptic and eclipses would be much more frequent.[13]

  • $\begingroup$ It seems that the OP misinterpreted the diagram linked in the question to indicate that the lunar orbit is inclined at 5.4° to the celestial equator rather than the ecliptic. $\endgroup$
    – PM 2Ring
    Dec 6, 2020 at 15:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .