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Why are raindrops spherical when falling through the air, but lose their spherical shape when they are on a flat surface?

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The answer is surface tension.

When a droplet falls through the air, you only have one type of boundary that you need to account for (water-air $\sigma_{WA}$). In contrast, for a droplet on a flat surface such as a table, you have two types of boundaries: water-air $\sigma_{WA}$ and water-table $\sigma_{WT}$ (there is also table-air $\sigma_{TA}$ surface tension, which you might need to include in your calculations).

Depending on the relative values of those parameters, the raindrop on a table might maintain some of its droplet shape (a cut of a sphere) or form a thin layer (complete wetting)

A simple calculation for the minimization of the surface energy gives the wetting angle $\theta$: $$ \cos(\theta) = \frac{\sigma_{TA} - \sigma_{WT}}{\sigma_{WA}},$$

Complete wetting corresponds to the situation where the value for $\cos\theta\ge1$.

A raindrop falling through the air forms a spherical shape - since there is only one boundary, and thus, the goal is simply to minimize the volume (I am neglecting the effects of gravity or drag forces, which for a small droplet is a good approximation - see https://physics.stackexchange.com/a/442766/353463)

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To add: If the surface upon which the droplet of water lands is tilted, then gravity will try to pull the droplet in the down direction. The leading contact angle of the sliding droplet will be large because the fresh surface towards which it is creeping is "unprepared" for being wetted. The trailing contact angle will be small because once wetted, the surface wants to stay in contact with water.

The net result is to yield the classic asymmetric "teardrop shape" as the droplet slides down the surface.

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