Can a natural satellite exist in a geostationary orbit?

  • While browsing through Physics SE, I noticed a question about satellites in geostationary orbit (unrelated to the one I'm asking here), and for a moment I interpreted it as referring to natural satellites (e.g. a moon). So I wondered: Could a natural satellite exist in geostationary orbit?



    Then I stopped and thought. For large gas giants, such as Jupiter, having moons too close to the planet can be fatal (for the moon). If it ventures inside the planet's Roche limit, it's toast. But there is good news: the Roche limit depends on both the masses and densities of the primary body and the satellite. So perhaps this reason is non-applicable, as a high-mass natural satellite might be able to survive. So the question changes:



    Could a sufficiently high-mass, high-density natural satellite occupy geostationary orbit over its primary body?


    I wonder if there are few inaccuracies in the question. Roach limits does not depend _both_ on density _and_ mass of the bodies in concern. Rather, it depends on density/mass of both bodies and radius of one body. See http://en.wikipedia.org/wiki/Roche_limit#Rigid-satellite_calculation

    Other thing is, for a satellite to be safe from tidal forces, it does not have to be low-mass and low-density. Rather, the satellite has to be _high-mass_ and _high-density_. Bigger (thus heavier) satellites like Pluto's Charon will tend to stay. because the Roche limit is lower for heavier/denser satellites.

    Thanks, @Krumia I can't believe I messed that up. I checked the formulas a couple times before I posted it, but I must have mixed up the primary and satellite.

    @Krumia Ah, now I know what I was thinking. A more massive satellite means more gravitational force between the two, meaning the two would be closer together, possibly negating the effects of having a smaller Roche limit. I might un-do my edit.

  • Walter

    Walter Correct answer

    8 years ago

    Of course, a natural satellite (moon) could have an orbital period equal to the spin period of its host (provided such an orbit would be accessible). However, the tidal friction that may generate such a locking is quite weak, so this would have to be a rare chance. Moreover, perturbations to the orbit from other moons or their host star may put the moon out of such an orbit.



    On the other hand, what is rather common is that a moon's orbital period equals their own (rather than their host's) spin period. This is exactly the case for the Earth Moon (you may say that Earth is on a "selenostationary" orbit) and naturally occurs form the tidal interaction of the planet with its moon.


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