Twitter person David Cox (@dcox21) asks:

Read a random fact yesterday that said the "average rain drop falls at 17mph." Is that reasonable?

Let the physics begin. You might think: hey, wont' the speed depend on how high the water started? Well, it would if air resistance on the water drop were not important. However, I suspect that the rain will fall at terminal velocity. Terminal velocity is the case when the air resistance on the object is equal to the gravitational force on the object. When this happens, the net force is zero (the zero vector) and the object falls at a constant speed.

Here is a diagram of a water drop at terminal speed.

Since the air resistance force depends on the speed of the object (but the gravitational force does not), there is one speed for which these two forces add up to the zero vector. Near the surface of the Earth, the magnitude of the gravitational force can be modeled as:

Where g is the local gravitational field (not the acceleration due to gravity - that is a non-good name for it). And what about the air resistance? It can probably be modeled as:

Where:

ρ is the density of air (about 1.2 kg/m 3 ).

). A is the cross-sectional area of the object. If the object was a sphere, this area would be the area of a circle.

C is the drag coefficient. This depends on the shape of the object. A cone and a flat circle will have the same A, but different drag coefficients.

v is the magnitude of the velocity of the object with respect to the air.

It won't matter for this case too much, but the direction of the air resistance force is in the opposite direction to the velocity.

At terminal velocity, the magnitudes of these two forces will be equal. I can write that as:

Now, what about the mass (m)? Let me assume that it is made of water (like most rain) and is spherical (even though that is not likely - it would probably be "rain drop shaped"). If I call the density of water ρ w and the radius of the drop r, then the mass would be:

Putting this into the "weight = air resistance" expression above as well as an expression for the cross-sectional area in terms of r, I get:

The cool thing here is that the terminal speed of the water drop depends on the size (radius). Larger drops will have a larger terminal velocity. So, could you just make a water melon sized water drop? No. Why not? Because at some point, the force from the air on the drop is going to break the water drop apart. The surface tension holding the drop together just won't be enough to maintain its drop status.

Then how big can it get? I have no idea. Oh, and then there is the problem of real drop instead of spherical drops. Let me look at that first. Wikipedia lists the coefficient of drag for a smooth sphere as 0.1. A rain drop should be less than this - but how much less? Well, a rain drop would take some of the water to form some sort of tail. This would decrease the cross sectional area as well as decrease the drag coefficient. I am not sure how to calculate the volume of a non-spherical rain drop, so for now I will just use a spherical drop with a drag coefficient of 0.08. I know that is wrong, but it will give me an idea about the terminal speed.

Now, how big should it be? How about I don't decide. Instead I will plot the terminal speed for a range of rain drop sizes. Let me look at drops from 0.5 mm to 5 mm. Here is that plot.

Well, the original question asked about the speeds in units of miles per hour. Here is the same plot but with different units.

Based on my estimations, 17 mph would be on the low end - but possible. It could be likely that I grossly overestimated the size of a raindrop.

Homework: Yes, there is homework. If the rain drop has a radius of 0.5 mm, from how high would it have to drop to get pretty close to the terminal velocity?

Update ——

As usual, I rush into things without exploring things in more depth. My assumption of a raindrop shaped raindrop appears to be bogus. Who would have guessed that? Anyway, here are some very useful links from commenters (Jens and Charles) and a large thanks to them.