At what size will self-gravitation contribute more to stability than surface tension?

The governments of Earth have embarked on an experiment to place a massive ball of water in orbit. (umm... special water that doesn't freeze)

Imagine this to be a fluid with a given density, $\rho$ ($kg/m^3$), surface tension, $\sigma$ ($J/m^2$), and formed in a sphere of radius $R$ ($m$). I think that the viscosity $\mu$ is not needed for this question, but correct me if I'm wrong.

At what size will the restorative forces from gravity (after some small perturbation) become more significant than that from surface tension? Would the type of perturbation make a difference?

Just for fun, here's a video of a ball of water stabilized by surface tension.

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@Georg doing so only gives me greater confidence that I used it correctly, so you'll need to be more specific if your intent was to show that something is amiss. – AlanSE Oct 26 '11 at 12:44
That use of afford was somewhat unusual to me. Is this homework? Another question: what is Your idea of "stability" in this context? Surface tension leads to an internal pressure of the same magnitude throughout, whereas self-gravity leads to a pressure which is zero at surface and maximum in the center of the sphere. How would You compare both? Self-gravity/surface tension of a water blob was a topic here some time ago. – Georg Oct 26 '11 at 15:35
@Georg I can't find a prior question that fits this.  I've asked various things about self-gravitation and 1/r^2 integrals before, but in no relation to surface tension.  This isn't homework.  I mean, you could give it as homework, I don't know how to solve it for one.  Without viscous friction, both surface tension and self-gravitation should lead to oscillations in response to a disturbance.  Formalizing the criteria for when it oscillates versus tears apart is something I also don't know how to do. – AlanSE Oct 26 '11 at 15:48
– Georg Oct 26 '11 at 15:57
BTW--until your blob gets very big indeed you'll need special water that doesn't boil as well: water is not stable as a liquid at very low pressures. – dmckee♦ Nov 28 '11 at 1:01

2 Answers

Let us do a quick estimation.

Let $R_{cr}$ be a critical radius of the ball so that the condition of stability for the ball is expressed as $$R<R_{cr}$$ What can $R_{cr}$ depend on? The most important properties are inertia, gravity and surface tension, which are characterized by the density $\rho$, the gravitational constant $G$ and surface tension coefficient $\sigma$. So $R_{cr}$ can be a function only of mentioned parameters: $$R_{cr}=f(\rho , \sigma , G)$$ By dimensional analysis, the dimension of $R_{cr}$ is meter. The combination $\left (\frac{\sigma}{\rho^2 G}\right)^{1/3}$ has also the dimension of meter. Therefore we can write:

$$R_{cr}=C\left(\frac{\sigma}{\rho^2 G}\right)^{1/3}$$ where $C$ is a dimensionless constant of orders of magnitude close to 1.

For water, $\sigma=0.07\frac{J}{m^2}$, $\rho=10^3\frac{kg}{m^3}$ and also $G=6.67\cdot10^{-11}\frac{Nm^2}{kg^2}$

So, a rough estimation:$$R_{cr}\approx\left(\frac{\sigma}{\rho^2 G}\right)^{1/3}=10m$$

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This might be a "realistic" size for a swimming pool at the International Space Station. :=) – Georg Oct 27 '11 at 9:53
The fact that the answer seems to be "about a swimming pool" makes this question far more entertaining that what I anticipated. – AlanSE Oct 27 '11 at 16:54

The answer by Martin is good, but I still want to continue along the thought path I had in mind. I hope I can give a different physical basis for confirmation of the number.

I'm sure there are better ways to do this, but I want to do it using only the information I have. I want to consider the transfer of some amount of mass ($m$ for now) from near the surface of the sphere to the inside of the sphere (fully integrating it). In both cases we can asses some amount of energy difference between the spherical blob of mass $M+m$ and the state of the sphere with mass $M$ with the $m$ mass hovering just above the surface. So the two states under consideration are:

• State 1: (M+m) big ball
• State 2: (M) ball next to (m) ball

I have no problem assuming $m \ll M$. Now, I want to write expression for both the gravitational binding energy as well as the surface binding energy of a ball. I'll do this for a generic sphere with a mass of $M$ and uniform density.

$$E_g(M) = - \frac{3 G M^2}{5 R(M)}$$

$$E_s(M) = - 4 \pi R(M)^2 \sigma$$

I'm leaving it in this form because we'll all agree that given the mass and the density, finding $R(M)$ isn't a problem. Now I want to write expressions for the difference in energy from state 2 to state 1. This is straight forward for the surface tension energy because the bodies are non-interacting. However, for the gravitational binding energy, there is still a binding energy between the large ball and the small ball that must be included. Keep in mind that state 1 is the lower energy state.

$$\Delta E_s = E_s(M) + E_s(m) - E_s(M+m)$$

$$\Delta E_g = E_g(M) + E_g(m) - \frac{G M m}{R(M)} - E_g(M+m)$$

Now, obviously, the idea would be to set these equal, assume that $m$ is small, and then find the $M$ as a solution to that equation. But that doesn't work! I think I have a major conceptual flaw in this approach, where the scaling of the surface area of the small $m$ blob just doesn't follow a scaling that works. I didn't know what to do, so I just removed the $E_s(m)$, abandoning all logical reasoning behind my work. But when I did this and used Martin's values, I obtained the following:

$$M=1.05 \times 10^6 kg$$

This was assuming $m=0.1 kg$. If I change that value it doesn't change $M$ very much, which is encouraging. This is a satisfying answer for me, because Martin's answer comes out to around 0.5 million kg.

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