# Does surface tension play a role in planet's shape?

I'm thinking that when the planets and stars were forming along with gravity surface tension also could have played a role in making them spherical.
Am I correct?

• A planet, by definition, is a body that is massive enough to be rounded by its own gravity... Commented Mar 20, 2016 at 8:39
• ... and surface tension is ridiculy weak compare to that. Commented Mar 20, 2016 at 10:00
• Surface tension is at a liquid/gas (usually air) interface. In a forming celestial body it's not clear what the liquid or gas would be. Commented Mar 20, 2016 at 21:53
• @lemon With a couple of additional criteria to ensure that e.g. a star isn't also a planet, and to exclude dwarf planets. Commented Mar 21, 2016 at 10:41

Let's throw some numbers at this. The Eotvos (or Bond) number is a dimensionless ratio of the body forces to surface tension forces often used in the sciences to characterize certain flows regimes. This number is given by:

$$\mathrm{Eo}=\frac{\Delta\rho g L^2}{\sigma}$$

where $\Delta\rho$ is the density differences between two phases, $g$ is the acceleration of gravity, $L$ is some length scale and $\sigma$ is the surface tension.

Now we need some numbers and some simplifications, let's assume the earth is 100% water then $\Delta\rho\sim10^3\:\mathrm{kg/m^{3}}$, and $\sigma\sim10^{-3}\:\mathrm{N/m}$. The radius of the earth is estimated at $L\sim10^7\:\mathrm{m}$. Together with a value of $g\sim10\:\mathrm{m/s^2}$, it is easy to see that $\mathrm{Eo}\gg1$ or that body forces are MUCH more important than surface tension on the scale of planets and stars.

TLDR: Surface tension is negligible compared to gravity on the scale of planets.

• re " let's assume the earth is 100% water ": And what if we assume the Earth is 100% granite, which I see as a much more reasonable assumption. Commented Mar 20, 2016 at 15:00
• Well, unless the surface tension of granite (if such a thing can be considered) is a whopping $10^{20}$ times bigger (which i doubt), the ratio is still much larger than one. Commented Mar 20, 2016 at 16:07
• @PieterGeerkens Surface tension is at a liquid/gas interface. The melting point of granite is 650-1250 C (depends on if it's wet or dry) at which point its crystalline structure is destroyed and it stops being granite. For a high surface tension look to mercury which has about 8x the surface tension of water. Commented Mar 20, 2016 at 21:56
• @Schwern: Water is not made from silica, and neither is mercury. At least granite, or any other silica based rock, is, and thus roughly approximates the actual material of which the earth is made. OP's lack of understanding on the limits of surface tension should not hinder the rendering of an accurate explanation. I suspect OP really means shear modulus when specifying surface tension. Commented Mar 20, 2016 at 22:10
• @PieterGeerkens, it doesn't matter how well you approximate the earth or any other solar body, you will always find that surface tension is neglible to gravitational forces at these scales. My assumptions just made it easy for me to give an estimate within a couple of minutes. Refining these assumptions to more realistic values (if they can even be estimated), is a waste of time because it will not provide you with new information. If you don't agree, you are free to provide an answer with more accurate values. Commented Mar 20, 2016 at 22:37

The gravitational binding energy for a spherical object of mass $M$ and radius $R$ is given by: $$E_{grav}=\frac35 \frac{GM^2}{R}$$ The interfacial energy for a spherical droplet is simply proportional to its surface area: $$E_{surf}=4\pi \sigma R^2$$ Here $\sigma$ denotes the droplet's surface tension. Taking the ratio of the two energies and using $M = \frac{4 \pi}{3}R^3 \rho$, it follows that $$\frac{E_{grav}}{E_{surf}} = \frac{GM\rho}{5\sigma}$$

The mass $M_c$ above which gravitational binding dominates over surface tension is: $$M_c = \frac{5 \sigma}{G\rho}$$

Considering $G=6.7 \times 10^{-11}~\frac{\text{Jm}}{\text{kg}^{2}}$ and typical values $\sigma \approx 10^{-3}~\frac{\text{J}}{\text{m}^{2}}$ and $\rho \approx 5 \times 10^3~\frac{\text{kg}}{\text{m}^{3}}$, it follows that $M_c \approx 1.5 \times 10^4~\text{kg} =$ 15 tonnes. Therefore, in planet formation surface tension is entirely negligible.