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If we consider a planet with the same characteristics as the earth, (diameter, mass, rotation period) what would be the maximum possible diameter of a moon on a geosynchronous orbit? Considering the Roche limit and other considerations of which I have no clue. This moon would be made of the same materials as our moon.

Context:

The fact is that a lot of things are due to chance, for example, in our reality, seen from the earth, the apparent size of the sun and the moon are similar, and this is absolutely due to chance.

In the same idea, an alternative earth could very well have a geosynchronous moon orbiting at 36'000km, whose apparent diameter would be (or not, hence the question of the upper size limit) the same as our moon.

If the configuration is stable and life appears despite the absence of tides, a sedentary civilization* could be born, evolve, and even reach great astronomical knowledge, without even knowing that the second most luminous celestial body after the sun, is simply not visible, unless you sail far enough to find it.

In a way, the moon could, by pure chance, be discovered after the other planets of the solar system.

Here is the idea behind the question.

*A civilization that does not feel the need to leave its island or continent.

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    $\begingroup$ Can a moon be larger than the planet it is orbiting around, or would it then become the planet? $\endgroup$
    – Justin Thyme the Second
    Commented Mar 30, 2021 at 14:17
  • $\begingroup$ Could you go back and explain your idea of a geosynchronous moon, then how such a pairing would come about? After that, what might be left to query? $\endgroup$
    – Robbie Goodwin
    Commented Mar 30, 2021 at 20:54
  • $\begingroup$ @RobbieGoodwin Edited adding the idea behind $\endgroup$
    – user35577
    Commented Mar 31, 2021 at 10:46

3 Answers 3

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Simple answer within the parameters you have specified.

The Earth's rotation is not changed from its current rate. It must remain 24-hour(~ish). The moon must be geosynchronous.
The moon must be a moon not a ring.

The solution depends slightly on your definition of just what a "Moon" is.

Definition 1: The fluffy definition of "Moon" defined as an object orbiting around another, larger object. Preferably a planet.
It must be a moon. I.e. smaller than Earth.
Solution: put a second Earth at exactly 53219km distance (center-to-center).
To ensure the moon is smaller than Earth, remove one atom from it.
Sanity check: Yes, both planets are outside each other's mutual Roche limit.

Definition 2: The stricter definition of "Moon" which precludes a binary planet like above, requires that the Barycenter of the system be within the body of the larger object.
In this case, it would occur when the Barycenter is at the surface of Earth.
Earth r = 6371km
Solution2: Moon orbit R = 44484km
Moon mass = 0.168 Earth
Sanity check: Yes, the Moon is well outside the Roche limit for Earth.

This places the barycenter right at the surface (ok, about 650m above sealevel), the Moon will have an orbital period of exactly 1 day.

Appearance:
Your big geocentric moon will be 8373 km in diameter (taken as direct scale-up of our current Moon, no regard for increased density due to higher self-gravity, sorry)

From the closest point on Earth below it, it will show as a circle some 9.55 degrees wide. 19.1 Times wider than our moon, 366 times the surface area.

It will 100% guaranteed undergo a very entertaining lunar eclipse every single night, and the Earth under it will undergo a matching solar eclipse every single day, at least in the equatorial regions.

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    $\begingroup$ So, I lost interest in trying to make a better answer, especially as you and L. IJSpeert have covered all the important stuff. Have a +1 $\endgroup$
    – Starfish Prime
    Commented Mar 30, 2021 at 13:18
  • $\begingroup$ So I'm guessing the binary planets model would involve the two bodies being tidally locked to each other. Would they be habitable? I would THINK so, since each would have sun on the locked side as they rotate around each other, but IDK if there are other factors that would screw this up. $\endgroup$
    – DWKraus
    Commented Mar 30, 2021 at 13:40
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    $\begingroup$ In this very special scenario, both Earths would still have a 24-hour day. The far sides would not even notice a difference! The near sides would have a spectacular 'ceiling' view. And would experience daily Eclipses, yet be a bit warmer because of the constant very bright reflected and radiated Earthshine. $\endgroup$
    – PcMan
    Commented Mar 30, 2021 at 17:53
  • $\begingroup$ @PcMan Would the Earthshine at night really more than make up for the loss of direct solar heating due to the eclipses during the day? This also makes me curious what sorts of crazy weather patterns would develop on those planets, given how much weather and climate are tied to solar heating patterns. $\endgroup$
    – reirab
    Commented Mar 30, 2021 at 22:25
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As stated by others, your orbit does in fact depend on the mass of the moon as well, since it is not negligible with respect to Earth. As Starfish Prime said in his answer, the Roche limit for a rigid body ends up at about 9500 km assuming Earth and Moon properties. I wanted to go a little bit further and calculate the orbital separation for such a body in an orbit with Earth.

We can calculate this from Kepler's third law, and inserting the now known variables: formula

Where the sidereal day is 23h56m04s, G is Newton's gravitational constant and we use the mass of the Earth and the Moon's density as given. I plotted this with some of the involved quantities indicated, see figure below. The value at R equal to the Roche limit is about 9.55 Earth radii, or 60800 km.

[Edit] However, if you want the moon to stay a moon in the sense that the centre of mass of the orbit (barycentre) is still within the Earth (answer by PcMan), we get a lower limit. I included the measurement of the barycentre from the centre of Earth and from the intersection with 1 (R_earth) I find a maximum radius and orbital separation at that radius of:

R_moon_max = 0.65 R_Earth = 4150 km

a_moon = 6.97 R_Earth = 44400 km

enter image description here

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    $\begingroup$ Yep. A moon at geosynchronous orbit can exceed the Earth's size, before it is in any danger of passing Roche's limit... And as soon as it does so, it is no longer a moon, but Earth becomes the moon. $\endgroup$
    – PcMan
    Commented Mar 30, 2021 at 11:57
  • $\begingroup$ @PcMan Yes, I realised this while reading your answer, and I will include this into the plot as well. $\endgroup$
    – L. IJspeert
    Commented Mar 30, 2021 at 12:00
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    $\begingroup$ Nice graph, though it is a bit noisy with that big key splurged over it. $\endgroup$
    – Starfish Prime
    Commented Mar 30, 2021 at 13:20
  • $\begingroup$ Thanks, and fair point. I made an effort to not make it overlap with the important parts, but it would indeed be even nicer to just place it below the graph. $\endgroup$
    – L. IJspeert
    Commented Mar 30, 2021 at 13:22
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Orbital period doesn't depend on the mass of the orbiting object. The orbital velocity dictates the height of the orbit and its period.

The only limit is that the mass of the object has to be smaller than the one of the main body, so that one orbits the other and not vice versa.

To answer your question, the upper limit for the mass of something in a geosynchronous orbit is the mass of Earth.

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  • $\begingroup$ I agree orbital period does not depend on mass. Still this does not answer the question, since I think there are a lot of size growing forbidding factors involved before having to measure which body the barycenter is closest to. $\endgroup$
    – user35577
    Commented Mar 30, 2021 at 10:23
  • $\begingroup$ @qqjkztd: It is irrelevant "which body the barycenter is closest to". As the masses of the two bodies become more and more similar, each of them will be eventually tidally locked to the other; and at that point they will appear stationary in each other's sky. For example, Earth is stationary in the Moon sky, because the Moon is tidally locked to Earth. $\endgroup$
    – AlexP
    Commented Mar 30, 2021 at 10:52
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    $\begingroup$ Orbital period depends on the sum of the masses of the two objects. When the two objects are widely different in mass, the mass of the smaller object can be generally ignored without much loss in accuracy. $\endgroup$
    – notovny
    Commented Mar 30, 2021 at 10:56

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