I took this question from Quora because it seems interesting and this community would have some fun with it. I would assume that one would use the modulus formulas for force to counteract the gravity, but given that the question details allow for "a solid cube, although internal structure is fine (honeycomb, etc)," this might get more complicated.

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/58023 $\endgroup$ – alemi Aug 13 '14 at 9:05
  • 1
    $\begingroup$ A duplicate on Astronomy: astronomy.stackexchange.com/q/5918 $\endgroup$ – alemi Aug 13 '14 at 9:11
  • 6
    $\begingroup$ The easy observational way, no first principles required: larger than oddly shaped asteroids, smaller than round moons. $\endgroup$ – user10851 Aug 13 '14 at 9:54
  • 3
    $\begingroup$ From what material is this hypothetical cube constructed, specifically? That can make a tremendous difference. For ready example, a cube made up of room-temperature water molecules need only be microscopic before gravity and surface tension round it. A cube made up of diamond, on the other hand, could theoretically be extremely large before gravity would have any effect. $\endgroup$ – TDHofstetter Aug 13 '14 at 19:30
  • 1
    $\begingroup$ Giant hunk of metal? Carbon fiber? Diamond. Super dense neutron star material? I don't know -- I haven't taken material science classes and my understanding of shear etc. is shaky, but why is the assumption that the largest object has to be a rock/iron so implicit? $\endgroup$ – Thoth19 Aug 15 '14 at 8:14

I'll take the question to be referring to solid rock. In reality, I think small asteroids are loose jumbles of rubble with a lot of vacuum between the rocks, and larger bodies like Ceres may have been liquid when they formed.

Googling turned up [Scheuer 1981], which can be found online for free by googling. S/he estimates the maximum height of a mountain to be

$$ \sim h_1 = \frac{4Y}{\rho g}, $$

for a mountain with an ordinary shape, or, for special Eiffel-tower shapes specially cooked up for the purpose, about $5h_1$, where the 5 is $\ln(E/Y)$. Here $Y$ is the shear strength and $E$ is the Young's modulus. Let's say that our cube is close enough to a sphere so that we can estimate $g$ at its surface as for a sphere, $g\approx (4\pi/3)G\rho r$. The corners of the cube can be considered as mountains of height $h=\alpha r$, where $\alpha\sim 0.1$. The result is

$$ r=\frac{1}{\rho}\sqrt{\frac{3Y}{\pi \alpha G}} $$

Scheuer gives $Y/g_{earth}=1.5\times10^6$ kg/m2 for granite, i.e., $Y = 1.5\times 10^7$ N/m2, and $\rho\sim 2.65\times10^3$ kg/m3. Plugging in numbers gives $r\sim500$ km.

This seems to be sort of roughly in the right ballpark. Vesta has a radius of about 250 km and is shaped like a potato. Ceres is 480 km in radius and is very spherical.

If the only relevant dimensionful variables are $Y$, $G$, and $\rho$, then the expression for $r$ is constrained to have this form, except for dimensionless factors, regardless of the other physics we use in order to derive it. I think MariusMatutiae and I, after some discussion in comments, arrived at the same answer by inputting similar physics, while I think Johannes' similar result came about because the 1000 m/s he used for the thermal velocity at vaporization is a characterization of the strength of chemical bonds, which makes it sort of equivalent to the information contained in $Y$, to within an order of magnitude.

Scheuer, "How high can a mountain be?," J. Astrophys. Astr. (1981) 2, 165–169.


I am rather surprised that neither link posted above gives a simple discussion of the effect, so here it goes.

Let us consider many asteroids of cubic shape, of constant density $\rho$, and of varying side $l$. We ask when, roughly, self-gravity will be able to perturb this shape into a spherical one. A cube of side $l$ has the same volume as a sphere of radius $l_2 =(3/4\pi)^{1/3} l \approx 0.62 l$. If we draw this sphere with the same center as the cube, we see that the cube, compared to the sphere has peaks (8, corresponding to the vertices) and troughs (6, corresponding to the face centers). Thus the question becomes: when can self-gravity slide the peaks into the troughs?

We can see this in terms of force: every point close to a peak will feel a component of the gravity force along the surface; only the vertices will feel a purely radial force. This component of the gravity force along the surface is a shear, and materials tend to fracture whenever the shear exceeds some critical value, which is generally close to their Young modulus.

Now we assume we have to relocate a fraction $q$ ($\approx 0.1$) of the total mass $M$ from the peaks to the troughs; the acceleration this matter feels is a fraction $q'\approx 0.1$ of the local acceleration of gravity $GM/l^2$, so that the total stress $\sigma$ (i.e., force per unit surface) becomes

$$ \sigma \approx \frac{q q' G M^2}{l^2 4\pi l^2} $$

which is to be compared to the rock's critical stress, $\sigma_{crit}$, which we can take safely to be $\sigma_{crit} \approx E$, the asteroid's average Young modulus. Using the approximation of fixed density, $M = 4\pi \rho l^3/3$, we see that the gravitational shear exceeds the rock's critical shear for radii exceeding a critical radius $l_{crit}$, which agrees with our intuitive feeling that small rocks may have arbitrary shape, while the Earth and Mars are spherical.

Also, we find that reorganization of the asteroid's shape occurs for

$$ l > \left(\frac{9}{4\pi}\frac{\sigma_{crit}}{q q' G\rho^2}\right)^{1/2}\;. $$

Using values adequate to rocks, we find

$$ l > l_{crit} \approx 1000 km\;, $$

which jibes nicely with the fact that the asteroid Itokawa, of approximate dimension $0.5 km$, does not display a spherical shape.

  • $\begingroup$ 6 troughs, not 8. A cube has 8 vertices, but only 6 faces. $\endgroup$ – Logan R. Kearsley Aug 13 '14 at 17:52
  • $\begingroup$ @LoganR.Kearsley Oops, thanks for the gentle reminder... $\endgroup$ – MariusMatutiae Aug 13 '14 at 18:03
  • 1
    $\begingroup$ This seems considerably too small. For instance, Vesta has a diameter of 500 km, but it has a highly irregular potato shape. $\endgroup$ – Ben Crowell Aug 13 '14 at 20:02
  • $\begingroup$ Ben's post is correct. There are many asteroids much larger than the 10km limit, which are far from spherical. Did you underestimate the shear that compact materials can withstand? $\endgroup$ – CuriousOne Aug 13 '14 at 20:06
  • 1
    $\begingroup$ After my first comment, it looks like MariusMatutiae changed $l_{crit}$ from 10 km to 1000 km and inserted a G in the equation. After those changes, the answer looks similar to my own answer, so I think we're in agreement. $\endgroup$ – Ben Crowell Aug 13 '14 at 20:55

To answer this difficult question, we must first establish a mechanism that converts a solid cube into a sphere. Several answers have already been given here, in which the Young's modulus is used for evaluation. But this module characterizes elastic properties. What does an elastic cube compressed by gravity look like? I considered a 3D model of elastic body for which the elastic properties are not violated under strong compression. Figure 1 shows an elastic cube compressed by gravity. We see that even with very strong gravitational compression, the cube remains a cube. Figure 1 Can be assumed this is due to the fact that the gravity of the cube is not isotropic. This can be seen using the analytical model of gravity of the cube, published on https://arxiv.org/abs/1206.3857v1 However, when I used the isotropic gravitational potential, the result was similar - Fig. 2. It may be necessary to add viscosity to the model so that the protrusions can spread. Figure 2

This model implies an estimate for the size of the cube, which could undergo noticeable gravitational compression $$L>\frac {k}{\rho}\sqrt {\frac {E}{G}}$$ Here $k$ is a numerical coefficient,$\rho$ is a density, $E$ is a Young's modulus, and $G$ is a gravitational constant. From here we can determine that $L=10^7 m$ for $\rho=5515.3 kg/m^3$ and $E=200 GPa$ (Earth parameters) which is comparable to the diameter of the Earth.

  • $\begingroup$ I don't understand some of your explanation. 1: in Figure 1, how much empty space is in the object, what is its solid incompressible minimum? (we can assume it is too small to undergo electron degeneracy) 2: in the formula, what substance is that? I don't know anything that's 55 tons/m^3. 3: your formula seems to say that a cube made of carbon fiber (E=200+ GPa, ρ = 2000 kg/m^3) could be substantially larger than the Earth, which doesn't sound right to me. $\endgroup$ – Foo Bar Sep 28 at 23:47
  • 1
    $\begingroup$ 1. This is a hypothetical elastic body that contracts in a gravitational field. We are interested in body shape. 2. It was a typo, fixed. $\endgroup$ – Alex Trounev Sep 29 at 1:29

Some simple scaling relations suffice to determine the size beyond which gravity prevents non-spherical rocks from forming:

A molecule of mass $m$ is bound to a mass $M$ of linear size $R$ with gravitational binding energy approximately equal to $G M m / R$. If this gravitational binding energy far exceeds the molecular binding energy $E_b$, gravity will prevent any shape other than a sphere from forming.

Using $M \approx \rho R^3$ and $E_b \approx k_B T_b$ with $T_b$ the temperature at which molecules boil off, it follows that the gravitational binding exceeds the molecular binding if approximately $$R \sqrt{G \rho}> \sqrt{k_B T_b/m}$$

Here, $1/\sqrt{G\rho}$ represents a timescale of the order of the minimum orbital period for objects gravitationally bound to a mass with density $\rho$ (depending on the density typically some 1000 seconds) and $\sqrt{k_B T_b/m}$ representing the speed at which molecules manage to escape their molecular bonds and boil off (typically less than 1000 m/s).

It follows that the size $R$ beyond which gravity dictates spherical shapes is of the order of 1000 km.

  • 1
    $\begingroup$ Although this does give a number in the right ballpark, I don't think the logic is correct. The gravitational binding energy you're using is the difference between the surface and infinity, which isn't relevant to the object's internal structure. What I think you've actually estimated here is the size of an object whose gravity well is deep enough so that if further material accretes onto it by gravitational collapse (starting from rest, far away), the material will be vaporized on impact. $\endgroup$ – Ben Crowell Aug 13 '14 at 19:56
  • $\begingroup$ @BenCrowell - the key physics is that of rupturing molecular bonds due to gravitational rearrangements. Also, remember this is an order of magnitude estimate: rearranging matter over distances 10% of the objects size and over infinite distances yield - within an order of magnitude - the same gravitational energy change. $\endgroup$ – Johannes Aug 14 '14 at 14:35

I'm going to take a stab at this question myself. However, I am not at all a structural engineer, so I'm probably doing something wrong.

I researched materials for a bit, and discovered that although diamond is very strong, it's brittle and can easily crack from a sudden jolt. Carbon fiber isn't quite as strong, but it's more able to withstand impacts, so that's what I'll work with.

Carbon fiber has compressive strength and tensile strength each about 100 MPa, and shear strength over 200 MPa. I'm taking this to mean it can withstand 10^8 N of force per m^2.

Carbon fiber's density is 2000 kg/m^2. A sphere (hold this thought for a minute) of carbon fiber would produce 6*10^-5 gee (6*10^-4 m/s^2) per km of radius, and a columnar mass of 4*10^6 kg/m^2 per km of radius (2 km of diameter). Naively multiplying mass * gravity, we get 2.4*10^3 Pa per km of radius. 10^8 / 2.4*10^3 = 4*10^4 km. 40000 km?

Working with a cube instead of a sphere is harder, because the corners are under a lot of pressure while the middle of the edges and faces are being pushed outward. Dividing by 2 or even 4 for a margin of safety still leaves us with a cube 20000 km across. And we haven't even considered structural engineering yet, using hollow girders and other mass-saving techniques.

Wow, it's looking like @Alex Trounev's formula might be right. Insane aliens might be able to build a near-perfect cube much larger than the Earth.


Believe the question is too broad - we have different states of matter with different behaviours - plasma, gas, liquid, solid.

If one considers solids (rocks) one of the hardest known substances is diamond. If we were to imagine a diamond as big as the earth (with a diameter more than 1000 km as mentioned in other answers), if the structure holds for the first few minutes after creation (lets say we simply managed to drop it somewhere in space just like that) then don't see it degrading it to a sphere.

Update 1


  1. Am assuming that other forces such as centrifugal force experienced by a spinning object in space are not in scope.
  2. If liquids were to be considered then the force caused by surface-tension would be a factor as well.
  3. There could be other states of matter with more density - such as neutron stars. However not sure if we understand the inter-molecular bonds (or its equivalent for neutrons).
  4. The reason I said "if it does not degrade in the first few minutes" is that since it would be possible that the weight of the structure could still be very high that it collapses on itself. Apparently the density of earth's core is around 13 gm/cm3 where as its 3.51 gm/cm3 for diamond. So the earth's core is definitely denser than a diamond and if it turns out that the pressure is higher (especially at the core) than the inter-molecular bonding between the carbon-atoms of the diamond, then what it is that we'd have - the core may collapse but not sure what it will become after collapsing. The earth's radius is 6371km on an average.

So guessing 4117km as the maximum radius based on diamond's and earth core's density before the structure collapses on itself.

Given : 
Earth's-Radius = 6371
Earth's-Volume = 1.08×10^12
Earth's core density = 13 
Diamond's density = 3.51
Ratio of diamond's density to that of earth's core : 0.27
Max-volume of diamond-based-cube : 292.465*10^9 (ratio*earth_volume)
Length of one side of the cube : 6637
Radius of the cube if it were a sphere (same volume): 4117




This is probably a moot question anyway, since what objects are around it will greatly alter the formula. The moon, for example, has the gravity of the Earth to contend with which will cause more friction within the rock of the moon. If an object was orbiting Jupiter with all of its moons, it would have many gravitational factors which would not be present if it was in deep space, away from other bodies. These would greatly accelerate the "rounding" process making the possible size of a cube relative to its placement.


protected by Qmechanic Aug 14 '14 at 20:02

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.