3
$\begingroup$

I believe it is true (correct me if I'm wrong) that a planet with a significant moon such as ours cannot be tidally locked. The orbit of the moon would continue the rotation of the planet and must, itself, be lost to permit the planet to become fully locked to its star.

I say "significant" because any rock with a mass << than the mass of the planet would have so little influence on the rotation of the planet that it would permit locking. What the ratio of mass is below which tidal locking of the planet is possible without losing the moon is not germane to this question. Interesting question, though.

But what if I declare in my ignorance of celestial mechanics that a planet has two moons: a lower-mass moon in a near orbit and a larger-mass moon in a counter-orbit further out? Simplistically, the gravitational influence on the planet (the tendency to "tug" it in one rotational direction) is balanced out, permitting planetary tidal locking to the star without requiring the moons to be lost in space first.

Insofar as we understand celestial mechanics today, is this configuration possible? Or must moons of significant mass always be lost to space before planetary locking can be achieved?

For the purpose of this question I'm interested in a period of only 100,000 years. Thus, if the condition is unstable (unable to be achieved for significant astronomical periods), that's OK.

$\endgroup$
3
  • $\begingroup$ My naive assumption would be that those two moons could not have the same Orbital period if they are not on the same orbit. Is this a concept I need to go learn about? $\endgroup$
    – Kyyshak
    Sep 8, 2023 at 16:46
  • $\begingroup$ To a first order approximation, if the mass of the moons are very small compared to the mass of the planet, amy two moons with the same orbital semimajor axis will have the same period, even without having the same orbit. That said, having the same period without being locked into some kind of resonance is usually a recipe for orbital gravitational thunderdome. $\endgroup$
    – notovny
    Sep 8, 2023 at 18:29
  • $\begingroup$ @Kyyshak I don't know if that's an issue or not. But Mars' moon Phobos has an orbital period of 8 hours and Deimos has an orbital period of 30 hours. If a contra-orbit (orbits against the spin of the planet) can be established at all, I doubt it would be difficult to rationalize a suitable orbital period for the two moons. $\endgroup$
    – JBH
    Sep 8, 2023 at 19:58

1 Answer 1

4
$\begingroup$

So, obviously this depends a lot on the initial conditions. For simplicity, lets assume the ideal case of two moons, one half as massive as the other, on circular orbits around a central body. Like this;enter image description here

Now there are a few * to this. First of all, Universe sandbox 2 uses classical mechanics for its simulations. However, outside of extreme cases this is fine. All we really care about is that the simulation is not done on point masses, but instead takes the physical size of the body into account. As many orbits appear stable for point masses, but really are not.

Thankfully, we can actually ignore a lot of variables. Like for instance density or composition. Only the final mass and radius matters. And of those two, the mass is more important.

For now, lets just see how this specific setup works out

$M_{P1} = 1$

$M_{M1} = 0.025 M_{P1}$

$M_{M2} = 0.05 M_{P1}$

$d_{P1-M1} = 150000 km$

$d_{P1-M2} = 300000 km$

Now, if the need comes up, i can redo this using GR but i dont want to.

enter image description here

This is the state after 120 Years. As you can see the orbit of M2 has become elliptical and both bodies experience significant orbital orbital precession. The elliptical nature is important, because it means the system is losing energy as a whole.

enter image description here

And after ~250 years, M2 got kicked out. If we think about this, it makes sense. M1, while having a lower mass, has a lot more Energy in its orbit as it is faster and deeper inside of P1 gravity well. So it takes a lot more energy to change its orbit. Notice, even after M2 got kicked out, M1´s orbit is elliptical but i mean the Percenter distance is ~120000 km while the Apocenter is 200000. Sure, the orbit is not circular but its fine.

Where as M2 experienced Apogee´s around 0.006 AU (~ 1 million kilometers) at the end. Orbits with such high eccentricity are not stable as there is a huge energy imbalance. The velocity of the orbiting object constantly changes. And this energy has to come from somewhere.

But ok, lets run this again but change the fractions. So that M1 has 0.01P1 and M2 has 0.005P1. Also lets bump up the distance to 500000 km for M2. Once again, perfectly circular orbits 3...2...1 go;

So after 5300 years M2 has an orbital period of 10 years, with a minimal distance of 0.02 AU and a maxima of 0.07 AU. You have to keep in mind, the fact the orbit is so elliptical is completely unrealistic if there is literally any other object in a solar system. M2 is only still orbiting because nothing else tugs on it. Even then, now (6000 Years) the orbit has a period of 13 years. The energy M2 loses each time it passes is insane.

7300 years, and we have an orbital period of 34 years. Again, not realistic if your solar system has any other bodies.

And at long last, 8700 years into the journey M2 is adios.

Answer Section

As per your question, i think the answer is No.

While these simulations have many flaws, non of them are to your advantage. We looked at very idealized situations which would not occur anywhere. Be that perfectly circular orbits, no other bodies to interact with etc. Kind of the only thing you can look at for help is numerical precision, since the more precise your simulation is the more stable it tends to be. Especially in cartesian space and simple math. However, i dont think the results are unreasonable. And they paint a picture in which these configurations are just not stable. At least in the timeframe you want. I mean, config 2 was in essence Earths moon and half of Earths moon orbiting, and it worked for less than 10000 years. Plus, for most of that runtime M2 was so far away you would have not really seen it.

You might want to look at a different configuration. Binary planets, with a distant moon. If you have two Earth sized planets orbiting, their orbit is going to be pretty circular. A third body will still not be stable, but it will be more stable than what we have seen here. Because the gravitational center is more or less constant. Tidal forces are still different, but i would bet money on a Binary with a moon being stable for 100.000 years...

But this might not be what you aim for, since this requires a close binary with a distant moon.

$\endgroup$

You must log in to answer this question.

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