Free Fall with Quadratic Drag: From Newton's Second Law to Dimensionless Form

We start with Newton's second law for an object falling under gravity with a quadratic drag force:

$$ m \frac{dv}{dt} = mg - kv^2 $$

Here:

$m$ is the mass
$g$ is gravitational acceleration
$v$ is the velocity
$k v^2$ is the drag force (opposing motion, $k$ constant)

We have

$$ \frac{dv}{dt} = g - \frac{k}{m} v^2 $$

Identify the natural velocity scale (terminal velocity). Terminal velocity occurs when acceleration is zero

$$ g - \frac{k}{m} v^2 = 0$$ $$\therefore \, v_t = \sqrt{\frac{mg}{k}} $$

Since $ \displaystyle \frac{k}{m} = \frac{g}{v_t^2} $, substitute into the equation:

$$ \frac{dv}{dt} = g \left(1 - \frac{v^2}{v_t^2}\right) $$

We now introduce nondimensional variables

$$ u = \frac{v}{v_t}, \quad \tau = \frac{g t}{v_t} $$

where:

$u$ is a dimensionless velocity, it measures the velocity relative to terminal velocity.
$\tau$ is a dimensionless time, it measures time relative to the characteristic timescale $ v_t/g $.

Using the chain rule

$$ \frac{dv}{dt} = \frac{dv}{du} \frac{du}{d\tau} \frac{d\tau}{dt} $$

Since $v = v_t u$, we get

$$ \frac{dv}{dt} = v_t \frac{du}{dt} $$

And since $ \displaystyle \tau = \frac{g t}{v_t} $, we have

$$ \frac{d\tau}{dt} = \frac{g}{v_t} $$

$$ \frac{dv}{dt} = v_t \frac{du}{d\tau} \frac{d\tau}{dt} = v_t \frac{du}{d\tau} \frac{g}{v_t} = g \frac{du}{d\tau} $$ $$\therefore \hspace{3mm} g \, \frac{du}{d\tau} = g \, (1-u^2)$$

we obtain the dimensionless equation

$$ \frac{du}{d\tau} = 1 - u^2 $$

The equilibrium solution $ u = 1 $ corresponds to terminal velocity.

This equation is universal for all objects under gravity with quadratic drag. The specific physical parameters $m$, $g$, and $k$ no longer appear explicitly; they are encoded entirely in the scaling definitions of $u$ and $\tau$. Different physical systems therefore differ only by their characteristic scales, not by the structure of the dimensionless dynamics.

For example, a feather and a skydiver have very different masses, drag coefficients, terminal velocities, and characteristic timescales. In ordinary physical units, their motions therefore appear very different. However, after non-dimensionalization, both systems satisfy the same equation $ \, \displaystyle \frac{du}{d\tau} = 1-u^2 $. The difference between the systems is encoded entirely in the scaling definitions of $u$ and $\tau$, while the underlying dimensionless dynamics remain the same.