The effect of air resistance

In our discussion of gravity and objects moving under the influence of gravity, in order to focus on how gravity works, we chose the highly simplified (toy) model of free fall — the motion of an object when the only force acting on it is gravity.

But biological organisms for whom gravity is important live in the air (or water), so we ought to improve our model somewhat to get an idea of how motion is affected by the combination of gravity and a resistive force. 

Motion in air

For this section, we'll focus on objects moving in the air. We've discussed that for an object moving in a fluid, the resistive forces the fluid exert on the object can be pretty well modeled as the sum of two forces: viscosity, arising from the fluid sticking to the sides of the object and the object dragging the fluid with it, and drag, arising from the object pushing the fluid out of the way in front of it.  

As we saw in our discussion of the Reynolds number, Re, it gives the ratio of the drag force to the viscosity force. For Re >> 1 drag dominates, while for Re << 1 viscosity dominates. We showed that the Reynolds number has the structure of a product of the parameters of the fluid (density/viscosity = $\rho / \mu$) times the parameters of the object (size * speed = $Rv$). In units of meters and second, $\rho / \mu$ for air is about 60,000 so unless $Rv$ in meters and second is much less than 0.000016, drag will dominate.

(Note also that when an object moves within a fluid, in addition to the resistive force it feels from the fluid, the fluid also exerts an upward buoyant force on the object arising from the fact that the fluid is also pulled down by gravity. This is very important for objects moving in water, but since the effect is 1000 times smaller in air, we'll neglect that here.)

Motion as a result of gravity plus drag

Our gravitational force, $\overrightarrow{F}^{grav}= m\overrightarrow{g}$, always points down, but our drag force, $F^{drag}= \frac{1}{2}C \rho A v^2$, can point in any direction since it points opposite to the direction of the velocity. ($A$ is the area the object that is pushing through the air, $v$ is its speed, and "C" is the drag coefficient.)

We only write the magnitude of the drag force. We didn't write is as a vector. We can do that, by noting that $v^2 = vv$ and $\overrightarrow{v}$ = magnitude of the velocity times the direction. Therefore, if we write

$$\overrightarrow{F}^{drag}= -\frac{1}{2}C \rho A v\overrightarrow{v} $$

it will have both the correct magnitude of the drag force and the correct direction. (The minus sign flips the direction from along the direction of the velocity to opposite to it.)

So adding air resistance to free-fall gives the following Newton's second law equation:

$$m \overrightarrow{a} = m\overrightarrow{g} -\frac{1}{2}C \rho A v\overrightarrow{v} $$

This represents 2 differential equations for the position and velocity of the object (or 3 depending on whether we're doing a 2D or a 3D problem). These can be solved in general using Newton's second law as a stepping rule on a spreadsheet.

This is now quite tricky. Since the "$v$" that appears in the drag force is the speed, it involves both the x and y velocities: $v = \sqrt(v_x^2 + v_y^2$), the horizontal and vertical motions are not independent. The x speed affects the y acceleration  and vice versa.

To get a better conceptual idea and to think about the physics rather than just crunching it in a spreadsheet, let's restrict our considerations to an object just moving up or down in the air.

Terminal velocity

Let's tell the story of how something falls in air. Consider a falling object, choose y as the vertical coordinate with + being up. Then our N2 equations reduce to a single one relating the up and down acceleration to the up and down velocity:

$$ma = -mg + \frac{1}{2}C \rho A v^2$$

The drag term has a plus sign because, if the velocity is down (negative), the drag force points up. And "$v$" only represents the speed. We have to put in the velocity direction by hand with an extra minus sign.

Dividing by $m$ gives

$$a = -g + \bigl( \frac{C \rho A}{2 m} \bigr) v^2$$

The mess in parentheses is a constant that depends on the properties of the air and the object.

Let's tell the story of what happens to a falling object.

Suppose the object starts from rest, so $v=0$. The only force acting on the object is gravity and it begins to accelerate with an acceleration $a = -g$, speeding up moving down. 

But as it begins to speed up, the speed increases. Now there is a drag force pointing upwards — a small one, since we've just started and $v$ is small. This reduces the acceleration so the object keeps speeding up, but not as fast. As $v$ increases, the drag force keeps getting bigger and bigger and reducing the acceleration.

At some point, $v$ is large enough that the drag almost exactly cancels the force of gravity. At this point the net force is 0 and the object has no acceleration: it maintains its speed. We call the speed at which this occurs the terminal velocity, $v_T$. We can find this by putting it into the equation for the acceleration, setting the acceleration equal to 0, and solving for $v_T$. This gives

$$a = -g + \bigl( \frac{C \rho A}{2 m} \bigr) v^2$$

$$0 = -g + \bigl( \frac{C \rho A}{2 m} \bigr) v_T^2$$

$$g = \bigl( \frac{C \rho A}{2 m} \bigr) v_T^2$$

$$v_T^2 = \frac{2mg}{C \rho A}$$

$$v_T = \sqrt{\frac{2mg}{C \rho A}}$$

From this we can see a number of implications:

• As the mass gets bigger, so does the terminal velocity.
• As the density of the fluid gets bigger, the terminal velocity gets smaller.
• As the area the object presents against the fluid gets bigger, the terminal velocity gets smaller.

Now try to tell the story if you enter the fluid with a velocity that is greater than the terminal velocity. (Imagine a meteor or space capsule hitting the atmosphere!) Go through the force arguments and describe what happens to the object's speed. Does it make sense what the equations say have to happen?

Joe Redish 2/4/19