Let's say a space-faring society wants to make a space station that has a large volume filled with air (or other gas), but no gravity. Using normal pressure tanks will require gathering an amount of material proportional to the volume, which is a ratio set by the tensile strength and desired pressure. It's likely that some other method for containing gas would be more economic for anything over a certain volume.

Could you confine a large volume of air in space using self-gravitation to hold the container together? Well obviously you can. Consider: a spherical volume of air surrounded by a solid sphere of matter through which the air cannot leak or diffuse and the pressure balances the gravity of the walls. Think of a big balloon ball in space, or alternatively, injecting the middle of the moon with air until it starts expanding. A hollow planet, if you will. There will be little gravity within the air because the walls don't contribute to the gravity inside and air has a low density.

So here is what I'd like to ask:

  • For a given pressure, what surface mass density ($kg/m^2$) would you need? What would the wall thickness be?
  • Going by the amount of material required, at what size would this approach be more economic?
  • How stable would this thing be?
  • Hypothetically, could you use the same principles to drape a airtight tarp over Mars and keep atmosphere from leaking out? It would be the surface mass density that determines the altitude at which it rests, right?

Disclosure type statements:

I can do a lot of the calcs for this myself, but I don't want to because I have doubts about certain parts and I want to avoid influencing other people with my potentially incorrect thought process. If you look at my Physics SE activity, you might notice that I'm fascinated by self-gravitation problems. I came up with this question reading Keeping air in a well.


2 Answers 2


Parameters of the surface

In order to calculate the size of the sphere we need
the equation for gravity field: $$ \nabla\vec{g} = -4\pi G \rho \qquad (1) $$ and the hydrostatic version of Newton's 2nd law: $$ \rho \vec{g} = \nabla p \qquad (2) $$ where $G$ is the gravitation constant,
$\rho$ is the density of the "cover",
$p$ is the pressure inside the "cover".

In order to solve the first equation one can use Gauss's theorem.
In spherical coordinates this will give $\vec{g} = \bigl(-g(r), 0, 0\bigr)$.

The second equation will give the distribution of pressure in the "cover". On the internal surface it is equal to the pressure of the gas. On the external surface it is zero.


The surface described by the equation (2) acts like it is made of water. My intuition says it is unstable.

If we put a small piece from the external surface to the internal one it will have less gravitational potential energy. The change of the pressure will be negligible especially for small particles. This is known as Rayleigh–Taylor instability (thanks to @mmc).

In order to get a stable surface we need some solid material. In that case additional forces appear in equation (2). Then if we increase the pressure of the gas, the cover will expand but not fly away. It will be held by the elastic forces.

In this case, though, the pressure should be so high that the balloon stability would be based on the strain not gravity. So this would be a usual balloon.

  • $\begingroup$ My thinking for stability was that is a part of the wall fell inward, there would be less inward gravity, and the pressure would push it back out. You mentioned that there is a pressure gradient through the wall, but do you also see there being a gravitational field gradient? Meaning, the inward gravity would be stronger on the outside than on the inside. $\endgroup$ Commented Feb 22, 2012 at 14:54
  • $\begingroup$ Yes there will be gravitational field gradient, it can be derived from equation (1). As for stability, I'm going to add some explanations concerning liquid cover to my answer. $\endgroup$ Commented Feb 22, 2012 at 15:53
  • 2
    $\begingroup$ If you use a liquid cover, there will be problems with the Rayleigh-Taylor instability. And I don't think a solid cover can act in a very different way due to the magnitude of the stresses involved, but this is pure speculation :D $\endgroup$
    – mmc
    Commented Feb 22, 2012 at 16:09
  • $\begingroup$ @mmc I think you hit the nail on the head. Supposing that there is no barrier between the air and the wall material means they will mix. It's energetically favorable, the only thing that would push against the transition is surface tension, which matters almost none on these scales. Nonetheless, lining the wall with basically a plastic tarp would prevent the air and water (if that's what the wall material is) from mixing. If you assume the mixing barrier exists, I think that globally the mechanics of the system are defensible. $\endgroup$ Commented Feb 22, 2012 at 16:48
  • $\begingroup$ @Zassounotsukushi I think we can represent your plastic tarp as a kind of "surface tension" and there is a certain value that stabilizes the system. It would be interesting to compare the required strength with the naive "balloon solution", but I need to read more about the topic to be able to do the calculations. $\endgroup$
    – mmc
    Commented Feb 22, 2012 at 17:24

My answer will be shamelessly Newtonian and Physics 101 in formulation. To start off the assumptions, I'm going to assume the air has no mass. To what extent is this valid? Air has about 1000x the density of other materials like rock and concrete, so we're looking at about that same volume ratio before the air mass becomes significant compared to the wall and as you'll see further into the calculations, this won't quite be the case until the object really is close to the size of the Earth.

The gravity at the surface of the balloon will be the following.

$$ g = \frac{G M }{R^2} $$

Here I have used the M variable to refer to the total mass of the wall. Now, this isn't the field that acts on the wall due to my prior arguments. Here's the part I was most unsure about: I divide this by two. Why? Well, the outside of the wall has $g$ act on it, but the inside of the wall has no gravitational field act on it at all (since I neglect the effect of the air). Average that out to get $1/2$. How does that translate into pressure? Introduce $\mu = \rho t$ where $t$ is the thickness of the wall, and you have the surface mass thickness in $kg/m^2$. This is what we need. Multiply that by the gravity and you have the same equation used on earth to find fluid head.

$$ P = \frac{1}{2} g \mu = \frac{G M}{2 R^2} \rho t$$

The point is that we already have a value of $P=14 psi$ that we wish to satisfy. For assumptions about density, $\rho$, my favorite approach is to assume it's made out of asteroid material with $\rho=1.3 g/cm^3$. Next, I'll introduce another easy equation, which is to multiply the mass thickness by the area to get mass.

$$M = 4 \pi \mu R^2 = 4 \pi t \rho R^2$$

These equations, with known density, by themselves can predict the shell thickness in what I call the "large limit". This assumed that the thickness is small relative to the total radius. So for any large space balloon made out of asteroid-density material the thickness is dictated by:

$$ t = \sqrt{ \frac{P}{2 G \pi \rho^2}} = 12.0 km$$

Since we know the thickness we may specify the radius or the mass. I thought it most appropriate to just say we have some given mass to work with. I took the mass of the asteroid 87 Sylvia, which is $1.5 \times 10^{19} kg$. Getting the rest is easy.

$$ R = \sqrt{ \frac{M}{4 \pi t \rho}} = 277.0 km$$

Yes, this is very big. However, the diameter is still about half that of Ceres. And 87 Sylvia is about the 18th largest by mass. Note that in the discussed configuration, the wall would occupy about 6.7% of the total volume.

Now I'm going to seaway into a different part of the answer where I ask "what if the balloon is fairly small?" We will start by defining $R$ to be the inner radius of the shell, which is the boundary of the air-filled region. To quickly get an answer, assume that $R\approx 0$, this forms the "small limit". You basically have a spherical asteroid and a negligible amount of air in the center. Integrate to find the fluid head, which will be set equal to 1 atmosphere.

$$ P = \int_0^t g(r) \rho dr = \int_0^t G \frac{4}{3} r \rho^2 dr = \frac{2}{3} G \pi \rho^2 t^2 $$

Now we get a definable limit for the smallest object we can make out of asteroid material that gets 1 atmosphere of pressure in its center.

$$ t = \sqrt{ \frac{3 P }{ 2 G \pi \rho^2 }} = 20.7 km $$

Obviously, this is larger than the previous large limit thickness, which is just due to geometrical factors. Now, how do we transition between the small limit and large limit values? We set up a more complex geometry, where the inner radius of the shell is $R$ and the outer radius of the shell is $R+t$. I struggled with this part of the problem a good deal, but I now have high confidence in this answer. To set it up, it's appropriate to say that the field within the rock is equal to the field you would have if the entire thing was solid (4/3 pi G rho r), minus the field you would get if the air were rock. This is using the superposition principle to subtract the rock in the middle which was "cut out".

$$ g(r) = g_{solid}(r) - g_{center}(r) = \frac{4}{3} \pi G \rho r + \frac{ G \rho \left( \frac{4}{3} \pi R^3 \right) }{ r^2 } \\ = \frac{4}{3} G \rho \pi t \frac{ \left( 3 R^2 + 3 R t + t^2 \right) }{ \left( R + t \right)^2 } $$

$$ P = \int_R^{R+t} g(r) \rho dr = \frac{2}{3} \pi \rho^2 G t^2 \frac{ 3 R+t }{ R+t} $$

This equation is easy to solve in terms of R, but not so easy to do in terms of t. It can also relatively easily be put in terms of P, M, and t.

I had graphs, but they were done when the equations were wrong, so I just wanted to get the math error corrected for now. Maybe I'll add more later.


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