Is this minimum mass known? or maybe, is it given in terms of density? If so, how much density is the minimum to have an spherical object due to its own gravity?
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3$\begingroup$ You would have to define "objects." Icy bodies start to become round under their own gravity at a certain mass. Rocky planets will take more. Liquids would form a sphere with miniscule mass as I assume gas proto-planets would be round as soon as they have enough gravity to be considered an "object." $\endgroup$– Robert CartainoCommented Apr 1, 2014 at 20:10
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1$\begingroup$ The minimum mass to be spherical is a tiny fraction of a gram. A drop of water is spherical. You should ask what is the maximum mass that an object could be and still be non-spherical. This depends on how quickly it is formed, because if there is not sufficient time to cool, it will melt and become round. Planets and asteroids run into this problem. $\endgroup$– eshayaCommented May 8, 2019 at 22:08
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6 Answers
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Icy objects, such as most in the Kuiper belt can reach an equilibrium if they are about 400km across, whereas the rocky asteroid Pallas, at 572km clearly has an irregular, non-spherical shape. All rocky objects larger than Pallas (and there aren't many) are spherical.
Rock tends to be stronger than ice. Rocky objects are able to withstand their own gravity for longer than icy ones. Pallas is a reasonable cut off point. The next smaller asteroids (Vesta, Hygiea etc) are round-ish, but not in hydrostatic equilibrium. On the other hand, small, icy moons such as Miranda and Mimas are in, or close to equilibrium. Mimas has a diameter of just under 400km.
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$\begingroup$ Asking for clarification may be against policy, but I feel necessary in this exchange. James Kilfiger states, "Pallas, at 572km clearly has an irregular, non spherical shape. All rocky objects larger than Pallas (and there aren't many) are spherical." However, is this diameter, 572 km, an approximate threshold for spherical bodies, or are there rocky, spherical bodies with a smaller diameter? $\endgroup$– GS1969Commented Jul 17, 2016 at 7:24
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$\begingroup$ Sorry - but while Pallas is not a perfect sphere, it is generally spherical. astronoo.com/en/pallas.html $\endgroup$– RickCommented Feb 17, 2020 at 15:34
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$\begingroup$ @Rick Unfortunately I can't upvote comments. I'd like to add that 10 Hygiea and 704 Interamnia which are smaller than Pallas also have spheroidal shapes. $\endgroup$ Commented Jan 1, 2021 at 12:35
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1$\begingroup$ Pallas is "generally spherical" but it is not in hydrostatic equilibrium. It behaves like a solid lump. By contrast the Earth acts to a high approximation as if it was liquid. The shape of the Earth is the same shape that a drop of water the same size and spin of the earth would have (barring some very small lumps and bumps like Mt Everest) This means Earth is in equilibrium. Pallas is not. It might be roughly round, but it isn't gravitationally in equilibrium. $\endgroup$– James KCommented Mar 6, 2022 at 20:52
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This question is more complicated than it seems like it should be!
There is no threshold mass or density beyond which an object becomes perfectly spherical; even supermassive stars are slightly oblong. The only exception is black holes, which are perfectly round up until you reach the quantum level. If we want a simple answer, most guesses are somewhere around $\frac{1}{10000}$ the mass of earth, or $6\cdot10^{20}$ kg, but that is very approximate and depends on the composition of the object.
answered Apr 1, 2014 at 20:10
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It collapses onto itself, becoming more spherical. This process is called Gravitational collapse, and for a cloud of dust, will happen when the dust cloud is greater than the Jeans mass.
Megan Whewell, Education Team Presenter for the National Space Centre, writes of other radiuses:
[F]or bodies made mainly of rock, the minimum size to become a self-gravitating sphere is about 600km diameter; but, for bodies mainly made of ice, the minimum size is about 400km diameter.
Obviously, some level of collapse can happen before that point, non-solid objects would need to be larger, and objects made out of stronger stuff like steel will also need to be larger, but that gives some idea of the necessary scale for solids at least.
It’s worth noting that a hollow sphere won’t collapse under its own gravity, because the net force of gravity at any point inside a sphere is zero, the greater mass of the more distant part of the shell counteracting the lesser mass of the closer part.
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Beside of mass, the rotation also affects the shape of an object. The faster the object spins, the more oblate (like a flattened sphere) it is. This happens due to the centrifugal force at its equator. Examples are Haumea (a dwarf planet) and Regulus (a main sequence star).
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Presume the body is composed of one substance, particularly a solid of some mass-density and compressive strength. Those two properties and universal gravitation should be combinable in math, resulting in a threshold radius for crushing the center, and after that the maximum sustainable range of surface radius would decrease with increasing radius. You might have to choose what percentage irregularity you would accept as "round".
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There is no clear answer, because all objects have some oblateness. If you want a rough cut-off point, we could use the mass of Mimas (about 6*10-6 earth) for it, because it is one of the smallest spherical moons.
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$\begingroup$ While the question asks about spherical, the real issue is if an object achieves hydrostatic equilibrium. So if it's rotating (but not too quickly), hydrostatic equilibrium would be an oblate spheroid and in shorthand one might say "It is spherical" because that's easier to say than "It has achieved hydrostatic equilibrium and is now either a Maclaurin spheroid or a Jacobi ellipsoid" $\endgroup$– uhohCommented Sep 11, 2021 at 9:44
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