Why do water drops form spheres in space? [duplicate]

Question

When water is poured out in space, why does it always take a spherical ball-like shape?

Answer 1

No, it's not because of gravity. You need to take quite a lot of water in order for gravitational effects to become significant.

It's because of surface tension. Sphere is a shape which minimizes the surface for a given volume. The surface-tension-related potential energy of the water is proportional to the surface, so spherical shape minimizes the potential energy.

Answer 2

Minimizing energy. If there is a small amount of water, then surface tension wants to try and minimize the surface area of it, and the minimum surface area for a given volume material is a sphere. For really large volumes of water (if you, for instance, sucked all the water out of the oceans and placed it somewhere far away in space in the standard mad-scientist way), then you also get a sphere, but for a different reason: the mass of water wants to minimize its (self-)gravitational potential energy and this is also done when it is spherical. If such a volume is in the presence of external gravitational field (for instance if it was orbiting the Earth) then it would not be completely spherical: this is one of the reasons the Moon has a slightly odd shape, for instance.

In between these two regimes -- if you had a few thousand gallons of water for instance, then although it would eventually end up spherical in the absence of other influences, this would take a very long time.

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Quantifying the effects

It is interesting to try to quantify the differences between the effects. One way to do this is to consider a spherical ball of water (or anything else, but I'll stick to water because numbers are easy to come by) and consider what force you would need to bisect the sphere and move the two halves apart. Then we can compute the force that would be needed to break the surface tension, and that needed to overcome the gravitational attraction of the two halves.

Surface tension

Let the radius of the ball be $R$, and surface tension be $T$: $T$ has units of force per length. So the total force we need to exert when splitting the sphere is simply the total force exerted by surface tension around a circumference of the sphere, and we can see immediately that this goes like $R$.

$$F_T = 2\pi R T\tag{T}$$

For water, $T = 7.3\times 10^{-2}\,\mathrm{N/m}$ approximately.

Gravity

This is more complicated. First of all we can say something about the behaviour of the force: the masses of the two hemispheres go like $R^3$, and the separation like $R$, so it's immediately obvious that the force is going to go like $R^3\times R^3 / R^2$: like $R^4$ in other words. Gravity is going to win as $R$ gets big!

But we can actually get a number, even though the hemispheres are not, well, spheres, and thus hard to treat gravitationally: if you think about the surface which cuts the the ball into two hemispheres, then what is preventing the ball collapsing inwards across this surface is pressure. So the gravitational force between the two halves of the ball, when they are touching, must be equal to the integral of the pressure over that surface (it took me ages to realise this trick!).

Let's assume the density is uniform, which it won't be for really large objects but it will be for reasonably small ones. Call the density $\rho$. Then we can work out the gravitational acceleration at radius $r$ from the centre, relying on the shell theorem and knowing the mass inside $r$ is $m(r) = 4/3 \pi r^3$.

$$g(r) = \frac{4\pi}{3}G\rho r$$

And this gives us the pressure at $r$, just by integrating $g$ from $r$ to $R$:

$$ \begin{align} p(r_0) &= \int\limits_{r_0}^R \rho g(r)\,dr\\ &= \frac{4\pi}{3} G \rho^2 \int_{r_0}^R r\,dr\\ &= \frac{2\pi}{3} G \rho^2 \left[R^2 - r_0^2\right] \end{align} $$

or

$$p(r) = \frac{2\pi}{3} G \rho^2 \left[R^2 - r^2\right]$$

And finally we can integrate this over the surface to get the total force:

$$ \begin{align} F_G &= \int\limits_0^R 2\pi r p(r) \,dr\\ &= \frac{4\pi^2}{3}G\rho^2 \int\limits_0^R R^2r - r^3 \,dr\\ F_G &= \frac{\pi^2}{3} G \rho^2 R^4\tag{G} \end{align} $$

(I hope this is right: it is dimensionally OK but I might have missed factors somewhere.)

Compared

So, then, given $\rho = 10^3\,\mathrm{kg/m^3}$, $G = 6.7\times 10^{-11}\,\mathrm{m^3/(kg s^2)}$, we can solve for the radius $R$ where $F_T$ = $F_G$, and the answer is about $12.8\,\mathrm{m}$. I was surprised how small this is (and I am worried I have made a mistake therefore).

So if this is right, it means that gravity starts beating surface tension for a ball of water which is about $13\,\mathrm{m}$ in radius, and beyond that it wins rather rapidly due to the dependence on $R^4$. What this does not tell you is anything about how long it takes something to become spherical: I think that would be a bunch harder to work out.

Answer 3

I’m sure that a chemist could give a deeper answer. Or from Wikipedia we get, the surface tension occurs because water has hydrogen bonding.

Because of its polarity, a molecule of water in the liquid or solid state can form up to four hydrogen bonds with neighboring molecules. These bonds are the cause of water's high surface tension and capillary forces.

The key are the 4 possible hydrogen bonds to other water molecules of liquid water. The water molecules are bonded to each other like a free dimensional mesh.

Heating water, the water could be sprayed out in space to small pieces of course. The hydrogen bonds are weak (compared to metallic bonds) and under the influence of heat transfer the kinetic energy of the water molecules increases and the hydrogen bonds get broken.