I like to think I have a good enough understanding of astrophysics but there's still one thing that I just can't explain.

Why does the Moon orbit the Earth the way it does?

Many other moons in the Solar System orbit their planet along the equator and have a nearly circular orbit. After researching, it seems to boil down to two main causes:

  • Since every planet rotates on its axis, it has an equatorial bulge, and that extra mass around its equator creates an unequal gravitational field that over millions of years aligns the satellite's orbit around it.

  • There's a phenomenon called tidal circularization, which on long timescales, acts to dampen the satellite's eccentricity and make its orbit nearly circular.

Okay, that makes sense for most moons. But our own Moon has an orbit that is notably eccentric and it's also inclined to Earth's equator by an amount that somehow changes over time from 18° to 28°. Why is the Moon's orbit so complicated?

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    $\begingroup$ I don't have enough confidence or detail to make this an answer, but I think the main things are that the Moon is big compared to the Earth, far from the Earth compared to the size of the Earth and the system as a whole is close to the Sun. $\endgroup$ Jun 12, 2020 at 8:30
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    $\begingroup$ "Literally every other moon in the Solar System orbits its planet along the equator and has a nearly circular orbit" - Iapetus, Triton, and a whole bunch of outer satellites of the gas giants would disagree... $\endgroup$
    – user24157
    Jun 12, 2020 at 8:43
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    $\begingroup$ As I've said before, Lunar theory is complicated. ;) Regarding the nodal precession, it helps to think of the Moon's orbit being slightly inclined to the ecliptic and perturbed by the Earth's equatorial bulge. I suppose I should put that in a proper answer... Ah, James beat me to it. $\endgroup$
    – PM 2Ring
    Jun 12, 2020 at 8:45
  • $\begingroup$ Also related: astronomy.stackexchange.com/q/10946/16685 $\endgroup$
    – PM 2Ring
    Jun 13, 2020 at 4:42

3 Answers 3


The moon is so big that the processes that circularize and reduce the equatorial inclination would take much longer. The moon is big because of how it formed: a huge collision in the early solar system. (Unlike, say the Galilean moons that probably formed along with Jupiter, or Triton, that looks like a captured TNO)

The other fact that makes its orbit complex is that the moon is the closest moon to the sun, so solar perturbation is more significant than for other moons. Solar perturbation is the main cause of the periodic variation in inclination and the various other forms of precession seen in the moon's orbit.

The sun also pulls the moon out of a equatorial orbit and into a roughly ecliptic orbit as the perturbations of the sun become more significant than those of the equatorial bulge. As noted elsewhere at lunar distance, the sun provides more than 50% of the gravitational field.

So the reason the moon's orbit is complicated is that it does follow the laws of physics, and with three (or more) bodies, those laws have complex effects.

  • 3
    $\begingroup$ The Moon's orbit was likely equatorial until it migrated outwards across the Laplace radius, beyond which the perturbations from the Sun (which tends to align the orbit with the Earth's orbital plane) are more significant than the Earth's equatorial bulge (which tends to align the orbit with the equator). $\endgroup$
    – user24157
    Jun 12, 2020 at 8:48
  • $\begingroup$ @PM2Ring Surely that doesn't matter because the Moon's orbit around the Sun is almost the same as the Earth's orbit around the Sun. What matters is the difference in gravity when the Moon is on the Sun side of the Earth and when it's opposite the Sun side. That difference would be much smaller than the effect of the Earth's gravity. $\endgroup$
    – CJ Dennis
    Jun 13, 2020 at 4:11
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    $\begingroup$ @PM2Ring "the Sun pulls roughly twice as strongly". Since the Sun is "pulling" the Earth as well as the Moon, this statement has less relevance than it appears. $\endgroup$
    – CJ Dennis
    Jun 13, 2020 at 5:15
  • $\begingroup$ @PM2Ring Yes, that is what I was getting at. $\endgroup$
    – CJ Dennis
    Jun 13, 2020 at 6:25
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – PM 2Ring
    Jun 13, 2020 at 7:16

In addition to what @JamesK said, I would like to point out that the statement that the moon's inclination "somehow changes over time from 18° to 28°" is rather misleading.

Even if the moon orbited the Earth in a perfect circle, in exactly the same plane as the Earth orbits the Sun (known as the ecliptic plane), you would seem to see the moon sometimes traveling north of the equator and sometimes south, from the Earth's perspective, just as the noontime Sun seems to move north and south from day to day over the course of a year, for the simple reason that the Earth's north pole does not point perfectly perpendicular to the plane in which the Earth orbits the sun. As the spinning Earth travels around the sun, its axis of spin keeps pointing in essentially the same direction (roughly towards Polaris, the "North Star", at least for the next few thousand years or so until precession gradually moves it), while the Sun's (and solar system's) axis points in a different direction, with the result that at opposite points in its orbit, the Earth's axis effectively tilts about 23.5 degrees towards the sun's axis (at the summer solstice) and similarly 23.5 degrees away from it (at the winter solstice).

The plane of the Moon's orbit around the Earth is only tilted about 5 degrees or so from the ecliptic plane. The pair of points at which the Moon's orbit crosses the ecliptic plane (the orbital nodes) slowly rotate around the earth every 18.6 years or so. (When one of the nodes passes between the Earth and the Sun and the moon passes through either of the nodes, then you get a solar or lunar eclipse.) The Moon will be at maximum angular deviation from the ecliptic plane at times when it is at a one of the points in its orbit exactly between its two nodes. At such times, the Moon's inclination with respect to the ecliptic plane will be at a maximum: either +5 or -5 degrees.

The Earth's axial tilt (and hence also the tilt of the equator) with respect to the ecliptic plane will be a fairly constant 23.5 degrees, but the direction it faces with respect to the moon will vary over time, causing the effective tilt to vary between ± 23.5 degrees.

Therefore, at times when the moon is inline with the earth axial tilt, and the Moon is midway between its two nodes, the total angular deviation between the equator and the moon will be one of the values calculable from ± 23.5 ± 5 degrees.

Another way to look at it is to consider the Earth's equatorial tilt (with respect to the ecliptic plane) in the direction of the moon to be a sine wave ranging from ± 23.5 degrees, and the moon's tilt (also with respect to the ecliptic plane) in the direction of the Earth to be a different sine wave (at a different frequency!) ranging from ± 5 degrees. The angle between the equator and the Moon is then the sum of these two waves.

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    $\begingroup$ Another Wikipedia article related to this topic is titled Lunar standstill. $\endgroup$
    – PM 2Ring
    Jun 17, 2020 at 10:03

The reason the Moon doesn't orbit the Earth's equator is to do with the Laplace plane. This is the plane around which a satellite's orbit precesses: close to the planet, the equatorial bulge is the dominant contribution to the orbital precession, so the plane matches the equatorial plane. Away from the planet, the Sun is the main contribution. The transition between these two regimes occurs around a distance termed the Laplace radius ($r_\mathrm{L}$), which is given by:

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

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

The quantity $J_2'$ is the quadrupole coefficient of the planet and $n$ inner satellites (assuming they are in the planet's equatorial plane):

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

where $J_2$ is the planet's quadrupole coefficient and $a_i$ and $m_i$ are the satellite semimajor axes and masses respectively. For the Earth-Moon system, there are no inner satellites and $J_2' = J_2$.

The angle between the planetary 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 planetary obliquity. This shows the overall behaviour: for small $a$, the denominator tends to infinity and the angle tends to zero. For large $a$, the $2r_\mathrm{L}^5/a^5$ term vanishes, giving $\phi = \theta$. Incidentally this means that the Laplace plane should probably be referred to as a Laplace surface: it isn't actually planar, even though it can be treated as such in the extreme cases.

The above formulae are from the introduction of Nesvorný et al. (2014).

In the case of the Earth, the $J_2$ value is 1.08×10-3 (NASA Earth fact sheet), putting the Laplace radius $r_\mathrm{L}$ at about 8.4 Earth radii. The Moon's orbit is at 60 Earth radii, which is well beyond the Laplace radius so the Laplace plane is very close to the ecliptic. So even if the orbit of the Moon were in the Earth's equatorial plane, precession would cause it to move out of the plane. An example of a satellite in the transition region is Iapetus (discussed in the Nesvorný et al. paper): note that in the case of Iapetus, there is a significant contribution to $J_2'$ from Titan.

The Giant Impact hypothesis for lunar formation predicts that the moon formed close to the Earth in the equatorial plane. The effects of tides caused the Moon to slowly migrate outwards, through the transition region and into its current orbit. The current ~5° inclination of the Moon relative to the Laplace plane is likely a legacy of inclination excitation during the transition. Here's a video of a simulation of the transition by Sarah Stewart-Mukhopadhyay.

  • $\begingroup$ I really like how you explained everything in detail with the equations +1 $\endgroup$
    – WarpPrime
    Dec 10, 2020 at 23:28

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