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It is said that at small sizes, icy planets are more likely to be in hydrostatic equilibrium than terrestrial rocky ones. But why, as a matter of fact? Shouldn't denser bodies be more likely to collapse into equilibrium (I'm talking of equilibrium achieved by self-gravity only) and be more able to remain in that state?

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    $\begingroup$ Can you explain why you think that should be the case - what is your reasoning? $\endgroup$
    – Rory Alsop
    Jun 17 '21 at 10:34
  • $\begingroup$ Planets are in hydrostatic equilibrium unless they are getting bigger or smaller on interestingly short timescales. $\endgroup$
    – ProfRob
    Jun 17 '21 at 11:20
  • $\begingroup$ OK, I should qualify my comment. The Earth, Moon etc are not exactly as oblate as they should be for their rotation and therefore they are said not to be in hydrostatic equilibrium. But I think the phrase "collapse into equilibrium" is a bit misleading, because the departures from equilibrium are really not very big AFAIK. $\endgroup$
    – ProfRob
    Jun 17 '21 at 11:28
  • $\begingroup$ @ProfRob I'm talking of any celestial body, not just (dwarf) planets, though Vesta for example isn't considered to be in hydrostatic equilibrium despite being spheroidal and differentiated. I'm asking on why icy bodies reportedly more easily achieve spheroidal shape through self-gravity than denser ones. $\endgroup$
    – John
    Jun 17 '21 at 11:55
  • $\begingroup$ @RoryAlsop My question is asking the same vice versa. Wouldn't one expect that denser bodies are more steadfast? E.g. less dense bodies could be rubble piles. $\endgroup$
    – John
    Jun 17 '21 at 11:57
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No.

Hydrostatic equilibrium for celestial bodies occurs when the internal gravitational forces overcome the rigidity of the materials of the body.

So, a dense rigid body may not be in hydrostatic equilibrium, while a similarly-sized, less-dense, fluid body could be in hydrostatic equilibrium.

Similarly, while rock is more dense than ice, it is also more rigid. So one can't assume that the minimum radius for hydrostatic equilibrium for a rocky planet is necessarily less than the minimum radius for hydrostatic equilibrium for an icy planet. I don't know if such radii have been conclusively determined for various compositions of bodies.

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  • $\begingroup$ It might be useful to note that the potato radius of a rocky body is about 400 km while the potato radius of an icy body is about 300 km. This is because ice is considerably more ductile than is rock. $\endgroup$ Jun 17 '21 at 15:28
  • $\begingroup$ Also thanking David Hammen for the explanation concerning ice. But aren't there moons larger than that and still unspheroid? $\endgroup$
    – John
    Jun 17 '21 at 15:48
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    $\begingroup$ @DavidHammen I don't think "ductile" is the right property. It's rigidity, or more specifically "compressive-strength," which is the pressure necessary to make a material plastically deform. $\endgroup$
    – Connor Garcia
    Jun 17 '21 at 17:14
  • $\begingroup$ @john there are many moons, asteroids, dwarf planets, etc. in the solar system. But their composiitons vary in density from almost 100 % rock to almost 100 % ice. Various types of rocks and ices can also vary in density and compression strength. Because of the great variation in composition, there are not a lot of examples of objects with the exact same composition of varying sizes for astronomers to see the exact radius at which an object of that composition would become spheroidal. So the exact limits are uncertain. $\endgroup$ Jun 17 '21 at 18:21
  • $\begingroup$ @M.A.Golding All spherical moons except Io and the Moon are ice moons. Let's not count Europa either because below its ocean it is rather terrestrial too and therefore has a higher density than the other ice moons. The smallest spherical moon is Mimas at 396.4 km (236.31 mi) and the largest irregular moon is probably Proteus at 420 km (260 mi) which is a rocky moon but having a density of just 1.3 g/cc. Mimas is less dense at 1.15 g/cc. Señor Garcia was right, less density lets a celestial body be more likely spherical. $\endgroup$
    – John
    Jun 18 '21 at 4:56

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