# Crying in microgravity

So I looked around the net and found that you really cant "cry" as we do on earth in space floating around, because the tears would just stay near your eyes, because the pull of gravity is not strong enough.

What about in an environment like the moon or something with even less gravity? Do tears fall down (albeit very slowly?). Is the surface tension enough to essentially make the tear drop stick down right all the way down our face and chin ? or will it still be like the earth where the tear drops fall off our lower face?

Or do they still stay near the eyes? Does the "friction" or surface tension of water able to overcome microgravity ?

PS: I am not a physicist, so please do forgive my ignorance.

• Did you ask this because you're curious, or did you ask because you have a secret dream of being sad in space? (And it would depend on the local gravity. Something like on an asteroid they would stay near your eyes. I'm not sure about larger bodies like the moon though.) – CoilKid Jul 29 '15 at 16:42
• @CoilKid Tears of joy perhaps? – Rover Eye Jul 29 '15 at 16:45
• @JohnRennie How is this not a physics question? I interpret his question as "Can the surface tension of water be strong enough to overcome local gravity?" – CoilKid Jul 29 '15 at 16:55
• @CoilKid To be fair, I had edited the question to reflect more accurately what I was after after John's comment was made. Sorry about that. – Rover Eye Jul 29 '15 at 16:58
• Gravity at the surface of the moon is stronger than in a microgravity environment. The answer of whether or not gravity can overcome surface tension depends on the amount of water in the drop and the intensity of gravity. A curious fact is that what we usually call microgravity is not small gravity, it can be shown that the gravitational pull in the ISS is near 90% the gravitational pull in Earth's surface. What really happens is what we call Weightlessness, which is related to Einsteins equivalence principle. – Andre Maizel Jul 29 '15 at 17:35

Short answer: A rigid hemispherical tear in a vacuum will drop instead of hang when $$g > \frac{\gamma}{\rho r^2},$$ where $$g$$ is the constant of gravity, $$\gamma$$ the surface tension of the tear, $$\rho$$ the density of the tear, and $$r$$ the radius of the tear. I estimate that a real tear will drop when $$g > \frac{40\gamma}{\rho r^2}.$$

This is just a back of the envelope calculation. It relies on a simplification of the situation, which may or may not be appropriate, so let me explain.

We will model a tear on the eyesocket as follows. We have a container filled with tear fluid. On the side there is a circular hole in the wall of the container, with radius $$r$$. By some magic the tear fluid slightly bulges out of the hole, forming a perfect hemisphere on the wall, without all our precious tear fluid draining out of the hole.

On the one hand we have the surface tension keeping the hemisphere attached to the tear basin. On the other hand, gravity is trying to pull the hemisphere down. We will ignore the deformation of of the hemisphere due to the gravity (it is therefore a rigid tear). It either sticks to the basin or it simply slides down the wall of the container. The question then becomes, how strong must gravity be to make the hemisphere slide?

Suppose the hemisphere slides. As it slides, the fluid behind hole in the wall will be exposed. This will increase the total surface area of the fluid. The basic physics of surface tension tells us that the increase of this surface area is related to the work done by gravity on the hemisphere, $$\Delta E = \gamma \Delta A.$$ Here $$\Delta E$$ is the difference in energy (the work that is done), $$\gamma$$ the surface tension, and $$\Delta A$$ the difference in surface area. What we are going to do is transform this difference equation into a differential equation ($$\mathrm dE = \gamma\,\mathrm dA$$) in terms of the height of the hemisphere $$h$$. The hemisphere that is still attached to the basin has $$h=0$$ and the height decreases as it falls.

Let's do the energy difference first. The gravitational energy of the hemisphere is $$E(h)=mgh$$, with $$m$$ the mass of the hemisphere and $$g$$ the constant of gravity. Taking the derivative of $$E$$ w.r.t. $$h$$ gives

$$\frac{\mathrm dE}{\mathrm dh}=mg\Rightarrow \mathrm dE=mg\,\mathrm dh$$

The area difference is a bit trickier. Let's say we have two overlapping circles of equal radius $$r$$ (the flat side of the hemisphere over the circular hole). As the circle on top slides over the other circle, a crescent will appear. This is the area that we are looking for. First we find out what this area is and then how it varies as the top circle moves. Let $$h$$ be the distance between the centers of the circles (equivalently, the height of hemisphere). The area of the crescent is $$A(h)=\pi r^2 - (2r^2 \cos^{-1}\frac{h}{2r} - \frac{h}{2} \sqrt{4r^2-h^2}),$$ and you can interpret it as: The area of the crescent is the area of circle ($$\pi r^2$$) minus the area of the overlap of the cirles (the expression within the parentheses; you can look up the formula for circle overlap). We differentiate the area w.r.t. the height and obtain $$\frac{\mathrm dA}{\mathrm dh}=-\frac{h^2}{2\sqrt{4r^2-h^2}}+\frac{r}{\sqrt{1-\frac{h^2}{4r^2}}}+\frac{1}{2}\sqrt{4r^2-h^2}.$$ We evaluate the derivative at $$h=0$$, because that is where hemisphere starts out, and we find $$\mathrm dA=2r\,\mathrm dh.$$

Now we know that $$\mathrm dE = \gamma\,\mathrm dA$$, so $$mg\,\mathrm dh = 2r\gamma\,\mathrm dh \Rightarrow g = \frac{2r\gamma}{m}=\frac{3\gamma}{\pi \rho r^2}\approx \frac{\gamma}{\rho r^2},$$ where $$\rho$$ is the density of the tear. The second to last equality holds because the mass of the hemisphere equals the density $$\rho$$ times the volume $$\frac{2}{3}\pi r^3$$. The approximation is good enough because our model is just a crude approximation of reality.

So, then, the tear falls when $$g$$ is greater than the value we have just found, as was claimed. Under the assumptions of this model, you can now answer the question yourself. You can google the values for the constants in the equation. What you could try to figure out is how big the tear can get for a given value of the gravitational constant. On the moon, for example, $$g_{\text{moon}}\approx 1.62\ \ \mathrm{m\ s^{-2}}$$. In any case, you can try plugging in some values and check for yourself if the model is reasonable.

You will find that the maximal radius of the tear on earth ($$g_{\text{earth}}\approx 9.8 \ \mathrm{m\ s^{-2}}$$) with $$\gamma\approx 0.072 \ \mathrm{N\ m^{-1}}$$ and $$\rho\approx 1000 \ \mathrm{kg\ m^{-3}}$$ is about $$r\approx 70\ \mathrm{\mu m}$$, which is about fourty times lower than the value $$r\approx 3\ \mathrm{mm}$$ that I expected. Because the physical units of the equation can not change, the factor of $$40$$ has to be unitless, a constant. Therefore, I estimate that a real tear can become fourty times as large as a rigid hemispherical tear, as was claimed.

One more thing. We could have obtained the answer in one paragraph. You realized that the relevant quantities in the problem are the gravitiational constant, the radius of the tear, the mass of the tear and the surface tension of the tear, which have physical units $$\mathrm{m\ s^{-2}}$$, $$\mathrm{m}$$, $$\mathrm{kg}$$, and $$\mathrm{N\ m^{-1}=\mathrm{kg\ s^{-2}}}$$. Try to form an equation with these quantities with $$g$$ on the left hand side of the equation. You will quickly find that the only possible equation which has equal physical units on both sides is our equation for the hemisphere! The only way to change the equation is by multiplying one side with a unitless quantity, a constant. By observing tears on earth we can deduce that this constant must be about $$40$$. This is the power of dimensional analysis. We can get answers to questions, even if we do not understand the underlying principles of the problem, merely by observing the physical units of the quantities under inspection.