Terminal velocity gravity against first-order drag force

Question

Let's say that we have a falling object with two forces acting on the object: gravity $F_g = mg$ and a first order drag force $F_d = kv$.

Terminal velocity will be reached when gravity is equal to the drag force.

It turns out that for larger $k$, 1) a lower terminal velocity will be reached and 2) the terminal velocity will be reached faster. You can see this by solving the following differential equation:

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

I am wondering if this is actually a trivial statement, i.e. can you conclude this with 100% certainty without explicitly solving the differential equation (so conceptually/with logic, at the same time with 100% guarantee)?

Here is my attempt: In order to reach terminal velocity, the drag force needs to be equal to gravity, i.e. $mg = kv$. If $k$ is large, then you only need a small $v$ in order for the two forces to be equal. If $k$ is small, you need a large $v$ in order for the two forces to be equal.

However, at the same time, if $k$ is small, that means that the acceleration of the object is larger at any given time (due to a greater difference between gravity and drag), thus the velocity will increase faster. Does this mean that the drag force will increase faster or possibly at the same rate? Well, not necessarily, because $k$ is smaller as well. So if we look at $F_d(t) = kv(t)$, then $v(t)$ will increase faster, but $k$ is smaller. So the question is, whether overall $F_d$ will grow faster or slower for smaller $k$. Can we guarantee this without any explicit calculation?

In other words, we have:

$$ \frac{dF_d(t)}{dt} = k\frac{dv(t)}{dt}.$$ For large $k$, we have that $\frac{dv(t)}{dt}$ is smaller for any $t$. However, we also have a linear scaling of $k$ and the $k$ is larger. So we have a competition between $k$ and $\frac{dv(t)}{dt}$.

Note: what I mean with reach terminal velocity faster is reaching a particular threshold faster, e.g. 99% of terminal velocity.

Answer 1

I guess we can assert that with only few calculations, the very same you already did, maybe with few more considerations.

For a given set of parameters, there's only one value of the "terminal velocity", represented by the "equilibrium velocity", i.e. determined by the condition $\frac{dv}{dt} = 0$,

$$0 = m g - k \overline{v} \qquad \rightarrow \qquad \overline{v} = \frac{m \, g}{k} \ ,$$

no matter the transient dynamics (that anyway it's a first order dynamics, i.e. an exponential trend of the velocity). That's the terminal value you're looking for.

Non-dimensional equations. A non-dimensional form of the equation reads

$$ \frac{m V}{T} \frac{d \tilde{v} }{d \tilde{t}} = m g - k V \tilde{v} \ ,$$

and choosing the reference velocity $V = \overline{v} = \frac{m g}{k}$, the equation can be recast as

$$ \frac{m V}{T m g} \frac{d \tilde{v}}{d \tilde{t}} = 1 - \tilde{v} \ .$$

Thus, you get the very same non-dimensional equation,

$$\frac{d \tilde{v}}{d \tilde{t}} = 1 - \tilde{v} \ .$$

for all values of the parameters, if the reference time is taken to be $T = \frac{V}{g} = \frac{m g}{k g} = \frac{m}{k} \ .$ The solution of the non-dimensional equation has the form $\tilde{v}(\tilde{t})$, so that the dimensional solution reads $v(t) = V \tilde{v}(\tilde{t}) = V \tilde{v} \left(\frac{t}{T} \right)$.

From this latter expression, you can easily see that the threshold

$$0.99 = \frac{v_{0.99}}{V} = \frac{v(t_{0.99})}{V} = \tilde{v}\left(\frac{t_{0.99}}{T}\right)$$,

is reached at time $t_{0.99} = \tilde{t}_{0.99} \cdot T = \tilde{t}_{0.99} \frac{m}{k}$. Being $m$ given, and $\tilde{t}_{0.99}$ not a function of the parameters, from dimensional analysis it's possible to conclude that the time to reach a threshold is inversely proportional to $k$.