<p>No.</p>
<p>Hydrostatic equilibrium for celestial bodies occurs when the internal gravitational forces overcome the rigidity of the materials of the body.</p>
<p>So, a dense rigid body may not be in hydrostatic equilibrium, while a similarly-sized, less-dense, fluid body could be in hydrostatic equilibrium.</p>
<p>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.</p>

CC BY-SA 4.0
2021-06-17T15:28:32.847Z
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.
David Hammen
CC BY-SA 4.0
2021-06-17T15:48:27.850Z
Also thanking David Hammen for the explanation concerning ice. But aren't there moons larger than that and still unspheroid?
John
CC BY-SA 4.0
2021-06-17T17:14:40.830Z
@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.
Connor Garcia
CC BY-SA 4.0
2021-06-17T18:21:05.207Z
@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.
M. A. Golding
CC BY-SA 4.0
2021-06-18T04:56:17.900Z
@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.
John
CC BY-SA 4.0
2021-06-18T16:01:12.303Z
@John I wrote radius, not diameter. I did have the numbers wrong: It's  about 300 km for rocky bodies  and  about 200 km for icy bodies. There are bodies that are a mix of ice and rock. I would not consider Europa to be an icy moon. It's a rocky moon with a 100 km thick icy crust. Ganymede and Callisto are tough. Once again, there's not a boundary. Icy versus rocky is a spectrum. But you are classifying Europa as an icy moon and Proteus as a rocky moon? On what possible basis? Most classify Proteus as an icy body, and as such its size places it smack dab at the potato radius.
David Hammen
CC BY-SA 4.0
2021-06-18T16:07:19.487Z
@DavidHammen Alright, Proteus may be icy as it is less dense than Europa which I clearly distinguished from usual icy bodies.
John
CC BY-SA 4.0
2021-06-20T14:50:15.247Z
@John: I suggest you google the term potato radius.
David Hammen
CC BY-SA 4.0
2021-06-17T15:20:53.597Z
2021-06-17T15:20:53.597Z
Connor Garcia
<p>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?</p>

CC BY-SA 4.0
2021-06-17T10:34:57.923Z
Can you explain why you think that should be the case - what is your reasoning?
Rory Alsop
CC BY-SA 4.0
2021-06-17T11:20:14.130Z
Planets *are* in hydrostatic equilibrium unless they are getting bigger or smaller on interestingly short timescales.
ProfRob
CC BY-SA 4.0
2021-06-17T11:28:13.423Z
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.
ProfRob
CC BY-SA 4.0
2021-06-17T11:55:12.947Z
@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.
John
CC BY-SA 4.0
2021-06-17T11:57:48.707Z
@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.
John
CC BY-SA 4.0
2021-06-17T12:13:53.640Z
I was confused by use of the word "planets".
ProfRob
CC BY-SA 4.0
2021-06-17T12:31:21.190Z
@ProfRob That's because of which bodies _I_ call planets. Vesta is such an example.
John
CC BY-SA 4.0
2021-06-17T14:55:24.187Z
@John Vesta is significantly triaxial ($b/a = 0.973$, versus $b/a = 1$ for a perfect sphere or oblate spheroid), which is inconsistent with hydrostatic equilibrium. (For comparison, Ceres' triaxiality is $b/a = 0.996$; http://planetary-mechanics.com/2018/12/23/rotational-stability-of-ceres-and-vesta/)
Peter Erwin
CC BY-SA 4.0
2021-06-17T14:56:46.333Z
@PeterErwin And what's the triaxiality of Haumea, a recognized dwarf planet that looks even weirder?
John
CC BY-SA 4.0
2021-06-17T15:22:52.400Z
@John As I understand it, Haumea (which is indeed more triaxial) is spinning so rapidly that triaxiality is a valid equilibrium shape, given its density and probable composition; see [Dunham et al. 2019](https://ui.adsabs.harvard.edu/abs/2019ApJ...877...41D/abstract). Vesta, on the other hand, isn't spinning fast enough for that to be true (the only equilibrium shape is an oblate spheroid), so its triaxiality indicates it *isn't* in hydrostatic equilibrium (e.g., [Konopliv et al. 2012](https://www.lpi.usra.edu/meetings/lpsc2012/pdf/2600.pdf)).
Peter Erwin
CC BY-SA 4.0
2021-06-17T15:51:36.750Z
@PeterErwin Haumea's sideral day is about 4 hours and Vesta's about 5 and a half. How much equilibrium is enough for the IAU to declare a body a (dwarf) planet? The Moon isn't considered to be in equilibrium either as ProfRob points out. If it were to change places with Mercury, there would be an extreme outcry if it wasn't recognized as a planet.
John
<density><hydrostatic-equilibrium><differentiation>
Aren't denser bodies more likely to collapse into hydrostatic equilibrium?
John
# Aren't denser bodies more likely to collapse into hydrostatic equilibrium?

(Asked by: username_0 on 2021-06-17)

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?

**Tags:** density, hydrostatic-equilibrium, differentiation

> **Comments:**
> **username_1**: Can you explain why you think that should be the case - what is your reasoning?  
> **username_2**: Planets *are* in hydrostatic equilibrium unless they are getting bigger or smaller on interestingly short timescales.  
> **username_2**: 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.  
> **username_0**: @username_2 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.  
> **username_0**: @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.  
> **username_2**: I was confused by use of the word "planets".  
> **username_0**: @username_2 That's because of which bodies _I_ call planets. Vesta is such an example.  
> **username_3**: @username_0 Vesta is significantly triaxial ($b/a = 0.973$, versus $b/a = 1$ for a perfect sphere or oblate spheroid), which is inconsistent with hydrostatic equilibrium. (For comparison, Ceres' triaxiality is $b/a = 0.996$; http://planetary-mechanics.com/2018/12/23/rotational-stability-of-ceres-and-vesta/)  
> **username_0**: @PeterErwin And what's the triaxiality of Haumea, a recognized dwarf planet that looks even weirder?  
> **username_3**: @username_0 As I understand it, Haumea (which is indeed more triaxial) is spinning so rapidly that triaxiality is a valid equilibrium shape, given its density and probable composition; see [Dunham et al. 2019](https://ui.adsabs.harvard.edu/abs/2019ApJ...877...41D/abstract). Vesta, on the other hand, isn't spinning fast enough for that to be true (the only equilibrium shape is an oblate spheroid), so its triaxiality indicates it *isn't* in hydrostatic equilibrium (e.g., [Konopliv et al. 2012](https://www.lpi.usra.edu/meetings/lpsc2012/pdf/2600.pdf)).  
> **username_0**: @PeterErwin Haumea's sideral day is about 4 hours and Vesta's about 5 and a half. How much equilibrium is enough for the IAU to declare a body a (dwarf) planet? The Moon isn't considered to be in equilibrium either as username_2 points out. If it were to change places with Mercury, there would be an extreme outcry if it wasn't recognized as a planet.  

---

## 1 answers

### Answer by username_4 on 2021-06-17. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_5**: 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.  
> **username_0**: Also thanking username_5 for the explanation concerning ice. But aren't there moons larger than that and still unspheroid?  
> **username_4**: @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.  
> **username_6**: @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.  
> **username_0**: @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.  
> **username_5**: @username_0 I wrote radius, not diameter. I did have the numbers wrong: It's  about 300 km for rocky bodies  and  about 200 km for icy bodies. There are bodies that are a mix of ice and rock. I would not consider Europa to be an icy moon. It's a rocky moon with a 100 km thick icy crust. Ganymede and Callisto are tough. Once again, there's not a boundary. Icy versus rocky is a spectrum. But you are classifying Europa as an icy moon and Proteus as a rocky moon? On what possible basis? Most classify Proteus as an icy body, and as such its size places it smack dab at the potato radius.  
> **username_0**: @DavidHammen Alright, Proteus may be icy as it is less dense than Europa which I clearly distinguished from usual icy bodies.  
> **username_5**: @username_0: I suggest you google the term potato radius.  

---

