Now a solid body of rock will collapse itself to a round shape when it hits about 600 km in diameter (400 km for ice). Now, the Second Death Star is estimated to be between 160 and 900 km. How big can a space ship be made of metal but still with "large open" living spaces. I assume it would still be filled with gases which would have its own gravity. Can a a spaceship be much larger than a 600 km sphere? To do so, would the infrastructure need to be primarily aluminum?

Adding from the comments.

a Dyson Sphere doesn't count, it should have internal structure.

Metal was suggested, but any material strong enough to build the self same craft is allowed.

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    $\begingroup$ I suspect the answer then will be infinite. You can imagine a graph with the x-axis being strength-to-weight of material, and the y-axis being the maximum size before it starts collapsing on itself. The graph will curve upward, and because gravity decreases on r^2 at some point it will go infinite - any materials past that point you can build infinitely large. This is especially true when you have open spaces to decrease the effective mass. I'll try and put an answer together later showing this, if I can find the equations and no one else beats me to it. $\endgroup$ Oct 7, 2015 at 21:10
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    $\begingroup$ @DanSmolinske No way. Even Hydrogen collapses on itself after a sufficient amount is present. Everything does, sturdy materials even more so. $\endgroup$ Oct 7, 2015 at 22:32
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    $\begingroup$ The article doesn't say 400 km for water. It says 400 km for ice. For water, I'm pretty sure just two drops of water would be enough to merge into just one sphere of water. $\endgroup$ Oct 8, 2015 at 4:43
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    $\begingroup$ @DanSmolinske Hydrogen is the least dense and the least massive element. If it collapses at a given point then everything that is denser and/or more massive will collapse as well. $\endgroup$ Oct 8, 2015 at 19:17
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    $\begingroup$ @AngeloFuchs It'll just skip straight to a black hole. There's no way to provide enough pressure to stop it from doing anything else. $\endgroup$
    – HDE 226868
    Oct 8, 2015 at 23:21

3 Answers 3


The defining equation of hydrostatic equilibrium - the state a celestial body must be in to maintain some semblance of a spherical shape - is $$\frac{dP}{dr}=-\frac{GM(r)\rho(r)}{r^2}$$ where $P$ is pressure, $r$ is radius, $M$ is mass, $\rho$ is density, and $G$ is the universal gravitational constant. Assuming constant density here - which is actually a problem because there are gaps - we say that $$\frac{d\rho}{dr}=0$$ and, after a quick derivation (see here for an example), we find $$P(r)=\frac{2\pi G\rho^2}{3}\left(R^2-r^2\right)$$ where $R$ is the radius of the body. At $r=0$, we have $$P(0)=\frac{2\pi GR^2\rho^2}{3}$$ Given that $$\frac{dP}{dr}<0$$ it is clear that $P$ is at a maximum at $r=0$. Re-arranging, we have $$R=\frac{1}{\rho}\sqrt{\frac{3P(0)}{2\pi G}}$$ $$$$ For $R$ to be maximized, we want the ratio $\frac{\sqrt{P(0)}}{\rho}$ to be maximized. We can say that $P(0)$ is the ultimate compression strength of a material.

Let's take a look at the strengths of various materials. The metal with the highest ratio is pre-stressed steel, at $$R=\frac{\sqrt{3,757,000,000}}{{1440}}\cdot\sqrt{\frac{3}{2\pi G}}=3,600\text{ kilometers}$$ That sounds pretty good to me.

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    $\begingroup$ Something seems wrong with the denominator. (3/2*pi*G)^(1/5) ~ 10^-5, and the density of steel is around 10^4. This should give you a denominator of around 0.1, which will increase your radius by a factor of about 10,000. $\endgroup$
    – ckersch
    Oct 8, 2015 at 0:40
  • $\begingroup$ @ckersch Oh, shoot, I calculated the radius directly from the ratio. My bad. I'll fix that. $\endgroup$
    – HDE 226868
    Oct 8, 2015 at 0:41
  • $\begingroup$ Your original sphere radius seemed quite small, so I figured there was something wring with one of the equations. My hollow sphere estimate, on the other hand, is absurdly huge. Do you see any math problems with my answer? $\endgroup$
    – ckersch
    Oct 8, 2015 at 1:27
  • $\begingroup$ @ckersch I made a comment. The analyses are completely different, so I'm trying to figure out if one of us made an incorrect assumption somewhere. Your math seems to check out, though. $\endgroup$
    – HDE 226868
    Oct 8, 2015 at 1:28
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    $\begingroup$ Another fun line of enquiry (nerd-trap): What happens if you spin it up? $\endgroup$
    – Joe Bloggs
    Dec 9, 2016 at 16:27

Going a slightly different direction from HDE226868, I'm going to design my ship to be as big of a sphere as possible. To do this, I'm going to put all of the living space on the outer surface of a big hollow steel sphere full of vacuum.

I'm going to have a lot more steel sphere, mass wise, per square meter than I will living accomodations on the outside of it, so my question essentially becomes this: how big can I make a hollow steel sphere before it is crushed by its own gravity? Now it's time for equations.


$g = GM_{sphere}/r^2$

Where $g$ is acceleration due to gravity, $G$ is the gravitational constant, $M_{sphere}$ is the mass of the sphere, and $r$ is the radius of the sphere.

Mass of sphere

$M_{sphere}=4\rho\pi r^2t$

Where $t$ is the thickness of the sphere and $\rho$ is the density of steel.

Pressure on the sphere

$p = g\rho t$

This is a conservative estimate, since only the outermost portion of the sphere actually feels the full weight of its gravity. The actual pressure involves solving a simple integral that I don't feel like doing right now .


$\sigma = pr/2t$

This is the equation for stress in a thin walled pressure vessel.

Final equation

Putting this all together, we get:

$\sigma = 4\pi G{\rho}^2r^3t^2/r^2t $

Or, simplified and solved for $r$,

$r = \frac{\sigma}{4\pi t G{\rho}^2}$

Plugging in values for the density of steel (8000$kg/m^3$), the ultimate stress of steel (3,757,000,000), and G ($6.67408 \times 10^{-11}$), we get a total maximum size of around 70,030,000km, given a thickness of 1m. Our ship's radius is inversely proportional to its thickness, so we can make it bigger if we make it thinner.

Of course, our giant sphere-ship will only be able to lurk about in deep space. Tidal Forces (Differences in the force of gravity between one side of the ship and the other) would destroy it if it came close to a large body like a planet or a star.

  • $\begingroup$ Continuum mechanics is not my thing, so I have to ask - how did you derive the equation for stress? $\endgroup$
    – HDE 226868
    Oct 8, 2015 at 1:35
  • $\begingroup$ I'm assuming it's a hollow sphere, so mass is the surface area of the sphere times density times thickness. The expression for stress is the equation for stress in a thin-walled pressure vessel. $t$ is the thickness of the shell, which I'm assuming is fairly small. $\endgroup$
    – ckersch
    Oct 8, 2015 at 1:35
  • $\begingroup$ Yeah, I re-read the answer and realized I had completely missed the mass formula. Thanks for the stress clarification. I think that's where our results differed. The other split would be the shell-vs.-solid model difference, which is actually probably more important. $\endgroup$
    – HDE 226868
    Oct 8, 2015 at 1:36
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    $\begingroup$ For anyone interested in astronomical distance comparisons, by the way, this radius is a bit under half the distance from Earth to the Sun. $\endgroup$
    – HDE 226868
    Oct 8, 2015 at 2:57
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    $\begingroup$ Are you sure the formula $p=g \rho$ is correct? The units of that are $m/s^2$ times $kg/m^3$, or force per volume. I think you might need another length factor. $\endgroup$
    – Obie 2.0
    Oct 8, 2015 at 15:50

Even though they are not solid, does the concept of a Dyson Sphere fit within your question?


Ahh, seeing the response...

It seems that most superstructure mega ships in theory have to deal not only with their own gravity of the structure, but with creating it for the inhabitants. I could see upwards of 900km depending on the solutions of internal and structural strain. The sphere of course comes to mind as the collapsed and near rest structure, one way of dealing with the stresses of gravity is to create pockets of open space that would essentially reduce the over all gravity stress as it's reduced by the open space.

I believe there has been discussion around these concepts on http://hieroglyph.asu.edu/ but I cannot search it right now.

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    $\begingroup$ No. I actually mentioned that in the comments. I'll add it to the question. $\endgroup$
    – bowlturner
    Oct 7, 2015 at 23:18
  • $\begingroup$ Ships (c.f. habitats) also have to go somewhere, so they have to not collapse under the acceleration of their own engines. A warship presumably has to accelerate fast. Otherwise, micro-gees will get you close to the speed of light surprisingly soon. $\endgroup$
    – nigel222
    Oct 8, 2015 at 16:38

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