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I am interested to know whether it's possible for the moon to orbit the earth at such a rate that we would see only one New Moon and one Full Moon a year. If so, what other effects would we experience?

I at first assumed this would be a one-year period, and technically I think that would produce one New Moon, but that single New Moon would last all year (i.e., you would never see the moon). So I'm interested to know if there is an orbital period that would give us one New Moon and one Full Moon, as seen from our earth, each year.

As I hold my hands up in front of my face and experimentally rotate them around each other, I want to say a six-month period would produce this effect, but I'm not sure that's correct, so I'd appreciate a word from anyone who actually knows what they're talking about. Thanks!

(Apologies if this has been asked before; I wasn't able to find it.)

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This would naturally be achieved by a quasi-satellite. These are objects with a 1:1 orbital resonance with a different eccentricity to the planet. From the perspective of the planet, this leads to the quasi-satellite appearing to travel around the planet once a year in a retrograde direction. Starting with the quasi-satellite at its maximum distance from the Sun, this would be at full phase. Six months later, the quasi-satellite would be between Earth and the Sun, so would be at new phase. The diagrams below show an idealised version of the configuration based on Keplerian orbits in a non-rotating frame and in a frame where the Earth and the Sun are kept in a fixed position.

Diagram of the orbital motion of a planet and a quasi-satellite in the Sun's frame Diagram of the orbital motion of a planet and a quasi-satellite in the Earth's frame

Real systems will evolve dynamically due to the gravitational interaction between the planet and the quasi-satellite, leading to precession and oscillations of the alignment of the orbits.

Earth currently has several known quasi-satellites, the closest of which is (469219) Kamoʻoalewa. The known quasi-satellites in our Solar System do not have long lifetimes: they are relatively recent captures into the quasi-satellite orbit that will eventually escape. Nevertheless, there are some conditions under which quasi-satellites can be stable over long periods, and perhaps some exoplanetary systems have more favourable conditions for quasi-satellites than ours does.

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  • $\begingroup$ This is a beautiful answer! +n! It realizes almost exactly I said was impossible "copy/pasting Earth's orbit to the left". $\endgroup$ – uhoh Oct 16 '20 at 12:41
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    $\begingroup$ Is this really as stable as your animation suggests? Won't the close approaches with Earth perturb the satellite's orbit from that nice ellipse? $\endgroup$ – Barmar Oct 16 '20 at 15:16
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    $\begingroup$ @Barmar - a real system will definitely deviate from the idealised Keplerian model I used to generate the animations, the orbits will tend to precess and the alignment oscillates back-and-forth. Real quasi-satellites are also in inclined orbits, which makes things even more complicated. This is obviously less convenient for making looping animations. Nevertheless quasi-satellite systems can last quite a long time in the right circumstances despite these effects. $\endgroup$ – user24157 Oct 16 '20 at 15:24
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Very interesting question!

Yep you are right, a one year prograde orbit would appear to stand still, and a six month period prograde orbit would do the trick.

We handle synodic periods like this:

$$\frac{1}{T_{syn}} = \frac{1}{T_1} - \frac{1}{T_2}$$

where $T_1 = 0.5$ is Farmoon's half-year orbit and $T_2 = 1$ is Earth's one-year orbit around the Sun. If both were one year the period would be infinite.

See also this answer to How would one calculate the synodic period of the Earth and an elliptical orbit?

For retrograde we set $T_{syn}=-1$ and see that a parked orbit where $\frac{1}{T_1} = 0$ would also work.

Prograde apparent 1-year motion orbit solution

For the prograde orbit a distance of roughly 1.36 million km would do the job and that is just barely inside Earth's Hill sphere so would be stable short term. Any farther out and the Sun's gravity would pull it away. So close to the Hill sphere you have to take the Sun's gravity into account when calculating the optimum distance and orbit.

This answer to What is the difference between Sphere of Influence and Hill sphere? says:

Hill Sphere: given a large mass (eg Sun) and a small mass (eg Earth), can a tiny mass (eg Moon) find a stable orbit around the small mass? (If the tiny mass goes outside the Hill Sphere of the small mass, no.)

Earth's Hill sphere is about 1.5 million km.

1-year orbit fails

As mentioned above, if you had a 1 year prograde orbit, it would appear to stand still.

However, another problem is that the distance would have to be about 2.2 million km from Earth, and that's outside of the Earth's Hill sphere. In other words, the Sun's gravity would dominate and pull it out of orbit before it made it once around the Earth!

The only way to get an object to circle the Earth once a year is not to orbit it (not possible) but to park at a Lagrange point.

While the following discussion is for L1, it applies to L2 as well.

Move to 1.5 million km and you are in the vicinity of the Sun-Earth Lagrange L1 point. There are several artificial satellites there including SOHO and DSCOVR. They are actually in halo/Lissajous orbits about L1.

What's really happening there is that you are now in a Heliocentric orbit that would normally have a shorter period than one year, but the slight tug of Earth's gravity slows you down just enough that you remain roughly between Earth and Sun.

It's not really stable there. While some halo orbits are actually stable in the circular restricted three-body problem, real-world perturbations from Venus, Jupiter and the Earth's own orbit's eccentricity would destabilize it in a matter of years.

FarMoon would do as you say and appear to always be a "new Moon".

Retrograde apparent 1-year motion orbit is unphysical

$$\frac{1}{-1} = \frac{1}{T_1} - \frac{1}{1}$$

means $T_1 = \pm \infty$

To get an apparent one year retrograde NewMoon you need to maintain a constant angle from Earth in inertial (non-rotating) space. For example imagine copy/pasting Earth's orbit to the left by a hundred thousand km. and putting the Moon on that track and force it to go around once a year.

That means it would orbit a "ghost Sun" offset from the sun by the same distance.

This doesn't happen, it's not physically possible. So the only choice is the prograde six month orbit.

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    $\begingroup$ According to Domingos et al. (2006), the stable region for prograde satellites only extends out to about half of the Hill radius, so I reckon the prograde orbit barely within the Hill sphere would turn out to be unstable. $\endgroup$ – user24157 Oct 16 '20 at 12:21
  • $\begingroup$ @antispinwards Thanks for that. I'm guessing that if we put it there, it would orbit for a (human) lifetime perhaps, at least long enough to call it an orbit. But we wouldn't be likely to discover exoplanets with moons at 0.91 R_{Hill). $\endgroup$ – uhoh Oct 16 '20 at 12:40
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    $\begingroup$ However, it has been claimed that no statellite can have a stable orbit if its orbital period around the planet is longer than about one ninth of the planet's orbital period around the star. See my answer $\endgroup$ – M. A. Golding Oct 16 '20 at 19:55
  • $\begingroup$ @M.A.Golding I've just asked Help understanding why these two sub-Hill sphere moon stability limits are so different? $\endgroup$ – uhoh Oct 17 '20 at 2:47
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There are calculations that claim that no actual satellite of a planet in a stable orbit can have an orbital period around its planet that is longer than one ninth of the planet's orbital period around their star.

In "Exomoon Habitability Constrained by Illumination and Tidal Heating", Rene Heller and Roy Barnes, the habitablity of hypothetical giant exomoons orbiting giant exoplanets in other star systems is discussed.

Red dwarf stars are the most common type of stars. Exoplanets orbiting them in their circumstellar habitable zones would be so close that they would become tidally locked to their stars, wiht one side always facing the star. The climate effects of that might make such a planet uninhabitable.

But if an Earth sized moon orbited a giant planet in the habitable zone of a red dwarf star, it would be titally locked to the planet instead of to the star, as almost all natural satellites in our solar system are. The moon's day or rotational period would be the same length as its month, its orbital period around the planet.

Hellar and Barnes say in section 2 habitabilityof exomoons:

The longest possible length of a satellite’s day compatible with Hill stability has been shown to be about P∗p/9, P∗p being the planet’s orbital period about the star (Kipping 2009a).

Exomoon habitability constrained by illumination and tidal heating

The source is given as:

Kipping, D. M. 2009a, MNRAS, 392, 181

Which is probably:

Authors David M Kipping Publication date 2009/1/1 Journal Monthly Notices of the Royal Astronomical Society Volume 392 Issue 1 Pages 181-189

The relative orbital periods of the statellite and the planet are discussed in section 3.1 and in appendix B.

Transit timing effects due to an exomoon

Thus it is possible that all exomoons in stable orbits will have orital periods around their exoplanet which are equal to or less than one ninth of the orbital periods of their exoplanets around their stars.

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