Question

How are gravity and air friction related to a raindrop’s terminal velocity?

Answer 1

Answer:

To answer this question, let's start by clarifying that every body or object that  freely falls in a fluid (in this case the air) has a terminal velocity.

In the particular case of a raindrop (which we assume in a spherical shape) that begins its fall from a certain altitude; it will accelerate (with gravity acceleration) until reaching the terminal velocity, just at the moment when the air friction compensates its weight.

To understand it better, when a raindrop falls, two forces act on it:

1. The force of air friction, also called "drag force" $D$:

$D={C}_{d}\frac{\rho V^{2}}{2}A$

Where:

$C_{d}$ is the drag coefficient

$\rho$ is the air density  

$V$ is the velocity

$A$ is the frontal or transversal area of the object (the raindrop)

So, this force is proportional to the transversal area of ​​the falling element (the raindrop) and to the square of the velocity.

2. Its weight due to the gravity force $W$:

$W=m.g$

Where:

$m$ is the mass of the raindrop

$g$ is the acceleration due gravity

Then, at the moment when the drag force equals the gravity force:

$D=W$

The raindrop will have its terminal velocity.

This also means, the larger the raindrop size, the higher its terminal velocity.

In other words, as the drop falls, gravity force pulls it down all the time, but along the descent a force in the opposite direction - upwards - acquires importance, due to the air friction.