# Explaining the shape of a raindrop and a drop of mercury

1. I saw on wikipedia that the shape of a raindrop is explained by using Laplace pressure. But why? Since the drop is in motion, we shouldn't be able to use an hydrostatic law, am I wrong?

1. The bigger a drop of mercury is, the lesser spherical its shape is. Ideally, if we consider only the surface tension, the drop should be spherical. What energy should you take into account in order to understand the problem?

I know the surface tension tends to minimise the fluid surface.

I also know it will tend to minimise the potential energy of the drop. So we should take into account the gravitational potential energy. But how can we take the minimum of two different constraints (fluid surface and gravitational potential energy)? Is there a potential linked to the surface tension?

• Comment to the post (v1): Shape of a 1. falling drop and 2. a drop resting on a surface should be two different posts. Mar 12, 2016 at 13:41
• Related: physics.stackexchange.com/q/218504/2451 and links therein. Mar 12, 2016 at 13:41
• What are the forces acting on a drop of mercury sitting on a table? Do these forces play a role in determining the shape of the drop? Mar 12, 2016 at 19:33
• You can perhaps consider two different parts of the mercury bubble; the curved edge (where surface tension dominates), the flat middle part (where gravitational forces dominate). As with many things in engineering, there are dimensionless numbers which characterize these droplets, in this case the Bond number (which is the ratio of body forces to surface tension forces) and the Morton number. Mar 12, 2016 at 21:45
• I haven't seen those numbers, so I guess I have to find something more simplistic. I've researched and I've found that the surface tension dimension is also an energy per unit area (which our teacher never told us). I now think it can be roughly explained with this energy and with the capillary length. My first question still holds. Mar 15, 2016 at 21:01