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 envelop 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. 707
What we are going to do is transform this difference equation into a differential equation ($dE = \gamma dA$) 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 gravitiational 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
$$ \frac{dE}{dh}=mg\Rightarrow dE=mg dh $$
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 $$ \frac{dA}{dh}=-\frac{h^2}{2\sqrt{4r^2-h^2}}+\frac{r}{\sqrt{1-\frac{h^2}{4r^2}}}+\frac{1}{2}\sqrt{4r^2-h^2}. $$ We evaluate the derivative at $h=0$, because that is where hemisphere starts out, and we find $$ dA=2rdh. $$
Now we know that $dE = \gamma dA$, so $$ mg dh = 2r\gamma dh \Rightarrow g = \frac{2r\gamma}{m}=\frac{3\gamma}{\pi \rho r^2}\approx \frac{\gamma}{\rho r^2}, $$ 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 $\frac{2}{3}\pi r^3$. 707