Everyone loves clouds. Although they are gigantic and beyond our reach, they inspire us with their wide variety of shapes. I mean, who hasn't played the game of guessing what a cloud looks like?

Of course, cloud watching can lead to some more scientific questions. If clouds contain water, shouldn't they be heavier than the air around them? And if so, why do they float?

Answering those questions takes a few steps, so let's get to it.

Why Do Things Float?

If you take a helium-filled balloon from a party and let it go, it doesn't fall. It will likely rise up in the air—although it's possible that it’s so perfectly weighted that it neither rises nor falls, but simply hovers. We often call this floating. But how does it work?

The simplest way to understand it is to consider air floating in air. (Yes, air floats.) Imagine you have a block of air that is 1 cubic meter. Air has mass, so there is a downward-pulling gravitational force on this air. If it's near the surface of the Earth, this force has a magnitude equal to the product of the mass of air (m) and the gravitational field (g = 9.8 newtons per kilogram). If there’s no wind, and the block of air just stays in place, then the total force on it must be equal to zero newtons. There has to be an upward-pushing force that's equal to the gravitational force. We call this upward-pushing force the buoyancy force.

The buoyancy force is actually the result of the air around the cube pushing against it in all directions. Since the density of air increases as you get closer to the ground, the force from the air pushing up from the bottom of the cube is greater than the force from the air pushing down on the top. The result is a net upward-pushing force.

If I know the density of air (ρ = 1.2 kg/m^{3}), then I can calculate the magnitude of this buoyancy force. (Remember that m = ρV, where V is the volume.) I'm going to write the density as ρ_{air} so that we don't confuse it with other things later.

Using known values for volume, density of air, and the gravitational field, this gives a buoyancy force of 11.8 newtons, or 2.7 pounds.

Now, let's replace that block of air with another block that’s identical in shape and size. But this time, suppose it's 1 cubic meter of water with a density of ρ_{water} = 1,000 kg/m^{3}.

Since it has the same volume as the floating air, this block will have the exact same buoyancy force. It doesn't matter what you put in that space, if it has a volume of 1 m^{3}, it's going to have a buoyancy force of 11.8 newtons. But for this cube of water, that's not enough to let it float. The gravitational force pulling it down will be much larger—it’s 9,800 newtons. The water cube is just going to fall.

In order for buoyancy to be greater than the gravitational force, you need to fill that space with a substance with a density that’s lower than air’s. There are two common methods to get this to work in real life. One is to use a thin rubber container filled with a low-density gas. (Think of a helium balloon.) The other is to use a low-mass container to hold hot air, which is less dense than cold air and will rise above it. (Think of a hot-air balloon.)

So if you want a cloud to float, it has to have a density lower than that of air. But how can that density be lower if the cloud contains both air *and* water?

It’s because clouds don't really float.

Why Does the Size of Water Matter?

Let's say that a cloud consists of air plus a bunch of very tiny water drops. The size of the drops is important. You might be surprised to learn that even if they are both made of water and have the same shape, small drops don't behave like large drops. In order to understand the difference between them, we need to look at air resistance.

Let’s start with a quick demonstration. Stretch out your arm in front of you with your hand open. Now swing your arm back and forth so that your hand moves quickly through the air. Do you feel anything? It might be slight, but there should be an interaction between your hand and the air, a backwards-pushing force that we call air resistance or air drag. (You will definitely notice it if you stick your hand out the window of a moving car.)

We can model the air resistance on a moving object with the following equation:

Just like the buoyancy force, this force depends on the density of air (ρ_{air}). But it also depends on the cross-sectional area of the object (A), a parameter that depends on the shape (C), and on the velocity with respect to the air (v).

(A quick note on the drag coefficient, C: This means that even though a sphere and a cylinder could have the same cross-sectional area, they are different shapes and so they will have different coefficients. For our calculations, we can assume that the drops of water are spheres with a drag coefficient of 0.47.)

OK, so a small drop of water starts at rest inside of a cloud. Since it's at rest with a velocity of zero, there's no air drag force on it. There is only a downward-pulling gravitational force and an upward-pushing buoyancy force. It doesn't matter whether the drop has a diameter of 1 millimeter or 1 meter—the gravitational force is going to be *much* greater than the buoyancy force.

I will write this out as an equation. Newton's second law says that the total force in the vertical direction must be equal to the mass multiplied by the acceleration in the vertical direction. Since the mass depends on the volume, I can write this as:

But wait! Some stuff cancels, like the volume. Solving for the acceleration:

The density of water is about 1,000 times greater than the density of air—so basically the acceleration is just negative g. What does this all mean? It means that if we want to look at the net force on a stationary drop of water, we can ignore the buoyancy force. It really doesn't do too much. Also, it doesn't matter if it's a big or small drop, since the volume gets canceled, so we can continue to ignore the buoyancy force.

However, once the drop starts moving, size matters—a lot.

Suppose I have a spherical drop with a radius of r. I can calculate both the volume (needed for the mass) and the cross-sectional area (needed for the air resistance). If you look at a sphere, it looks like a circle, so we can use the area of a circle.

Now I can again write Newton's second law to get an expression for the acceleration. Notice that I'm leaving off the buoyancy force since it's super tiny.

This is a fairly difficult equation to deal with—not because of the volume and area, but because of the velocity in the air drag term. If the total acceleration is in the negative direction, this means that this acceleration is in the same direction as the drop’s velocity. So it will speed up as it moves downwards. But as it increases in speed, the air drag force increases and changes the value of the acceleration.

One way to deal with this problem is to break it into small time intervals. During each time interval, we can assume the acceleration is constant (which is approximately true) and then use that to find the new position and velocity. Then we can just do the same thing for the next small time interval. This is called a numerical calculation—and it usually means creating some computer code to do all the boring math.

OK, let's model some falling drops of water—in fact, let’s model three with different sizes. The smallest drop will have a radius of 100 micrometers (1 μm = 1 x 10^{-4} m). The medium-sized one will have a radius that’s twice that, and the biggest one has a radius four times larger.

Notice that as they fall, at first all of the drops increase in speed. However, eventually they will reach a speed at which the air resistance is equal to the gravitational force. This means that the net force is zero and the drop will stop increasing in speed. For any falling object, the value of that final speed is called the terminal velocity.

The smaller drop has a terminal velocity of around 1.5 meters per second (3.4 miles per hour) compared to the largest drop at 3 m/s (6.7 mph). That is not very fast—that's like the average walking speed for a human.

But why does the smaller drop have a lower terminal velocity? There are really two competing forces here: the air resistance and the gravitational force. At terminal velocity, these two forces are equal. So, what happens when you double the size of a drop?

Since the air resistance depends on the cross-sectional area of the drop, doubling the radius increases the area by a factor of four, such that you get four times the air drag for a particular speed. The gravitational force depends on the mass of the drop, which can be found from the volume of a sphere. If you double the radius of a drop, you increase the mass by a factor of eight! So, a double-sized drop must fall faster to increase the air resistance to the same magnitude as the gravitational force. Big drops fall faster.

It's also possible for these tiny drops of water to not fall. Remember that the air resistance force depends on the relative speed between the air and the object. Let's go back to the example of the forces acting on your hand: If you stick your hand out of a car moving at a speed of 5 meters per second, there will be an air resistance force. You can get the *exact* same air resistance force if your hand is at rest but there is a large fan blowing air at it at the same speed. All that matters is the relative speed.

So, imagine a wind blowing a tiny drop of water upward at a speed of 1.5 m/s. The upward-pushing air resistance can have the same magnitude as the downward gravitational force. The drop will have zero velocity and zero net force. It will just stay there.

So this is what is happening with the clouds: The water droplets are small enough that the upward-pushing force of the air can keep them suspended aloft. But it can’t keep them at the same altitude forever. Any droplet with a large enough radius will eventually get overwhelmed by the downward pull of gravity.

Basic physics shows that clouds don't have to float—they fall, but they fall really slowly.