Atmospheric Drag

A body falling in air with velocity v⃗ is subject to a drag force of magnitude γ̂|v⃗|^{2} and direction opposite to that of v⃗, where γ̂ is a constant that may be empirically determined. Here we shall study the one-dimensional problem of a body falling directly downward.

Let x(t) be the height of the body above ground at time t. When the body is falling, the force on it is

m \frac{d^{2}x}{dt^{2}} = F = −mg + γ̂ (\frac{dx}{dt})^{2} .

So the equation of motion can be written as

x′′ − γ(x′)^{2} + g = 0,  (1)

where primes indicate time derivatives and

γ = \frac{γ̂}{m} .

(Is (1) also correct when the body is rising?)

Since (1) is not linear, it is hard to solve exactly. Let us consider the possibility that a useful approximate solution can be found in the form of the first two terms of a Taylor expansion of x as a function of γ:

x(t) = x_{0}(t) + γx_{1}(t) + γ^{2}x_{2}(t) + · · · .  (2)

(We might expect that this will work if γ is sufficiently small. The basic assumption being made here is that the correction terms γx_{1}+⋯ are small compared to the “unperturbed” solution x_{0} .) We substitute (2) into (1) and simplify, neglecting all terms of seond order or higher:

(x′′_{0} + γx′′_{1} + ⋯) − γ(x′_{0} + ⋯)^{2} + g = 0,

so after combining terms involving the same power of γ, we get

[x′′_{0} + g] + γ[x′′_{1} − (x′_{0})^{2}] + ⋯ = 0.  (3)

Since x_{0} and x_{1} themselves are not supposed to depend on γ, the only way (3) can hold is that each bracketed term separately equals 0:

x′′_{0} + g = 0,  (4)
x′′_{1} − (x′_{0})^{2} = 0.  (5)

Equation (4) is the familiar equation for a falling body without drag. Its solution is

x_{0}(t) = h_{0} + v_{0}t − \frac{1}{2}gt^{2},  (6)

where h_{0} and v_{0} are the initial height and velocity of the body. Recall that h_{0} ≥ 0 and v_{0} ≤ 0, because of our assumption that the body is falling toward ground level.

Substituting (6) into (5), we get

x′′_{1} = (v_{0} − gt)^{2} = v_{0}^{2} − 2gv_{0}t + g^{2}t^{2}.

We can solve this equation by integrating twice:

x_{1}(t) = h_{1} + v_{1}t + \frac{1}{2}v_{0}^{2}t^{2} − \frac{1}{3}gv_{0}t^{3} + \frac{1}{12}g^{2}t^{4},  (7)

where h_{1} and v_{1} are the constants of integration.

Substituting (6) and (7) into (2), we obtain our first-order approximate solution,

x(t) ≈ (h_{0}+γh_{1})+(v_{0}+γv_{1})t− \frac{1}{2}(g−γv_{0}^{2})t^{2}− \frac{1}{3}γgv_{0}t^{3}+ \frac{1}{12}γg^{2}t^{4}. (8)

This formula becomes more transparent if we consider the special case where the body is simply dropped (initial velocity 0) from height h_{0} . That is, v_{0} , h_{1} , and v_{1} are all 0:

x(t) ≈ h_{0} − \frac{1}{2}gt^{2} + \frac{1}{12}γg^{2}t^{4}.  (9)

The last term in (9) is the modification to the motion caused by drag; in a given time the body does not fall as far as it would in the absence of air resistance.

Notice that this term increases as t^{4}, so when t is big enough, it will overwhelm the first two terms. This means that, no matter how small γ is, our initial assumption that the drag effect is small will be wrong for large times. There are ways to get around this problem, but they are too sophisticated to be treated in a freshman course. Our power-series solution should be accurate provided that both γ and t are sufficiently small.

Now let us pose the problem of how fast the body must be thrown down so that it will hit the ground at exactly the same time that it would have landed if it were dropped in the absence of drag. (This is as close as we can come in a one-dimensional problem to a “targeting algorithm”.) Let T be the time of landing in the unperturbed problem:

0 = x_{0}(T) = h_{0} − \frac{1}{2}gT^{2}.

(Thus T = \sqrt{2h_{0}/g}, but we shall not need to use this formula.) The generalization of (9) to include a negative value of v_{1} is

x(t) ≈ h_{0} + γv_{1}t − \frac{1}{2}gt^{2} + \frac{1}{12}γg^{2}t^{4}.  (10)

Thus

0 = x(T) ≈ γv_{1}T + \frac{1}{12}γg^{2}T^{4},

or

v_{1} ≈ − \frac{1}{2}g^{2}T^{3}.  (11)

The additional downward initial velocity needed is γv_{1} (in this firstorder approximation).

One final remark: We have not proved that (10) actually is a good approximation to the exact solution. We can’t use Taylor’s remainder formula, because we have no obvious way of calculating the factor

max_{c∈[0,γ]} |\frac{d^{2}x}{dγ^{2}}(c)|

in it. We can see that the approximation is bad if either γ or t is too big; if γ and t are both fairly small, it seems highly plausible that the approximation is good — worth the trouble of testing experimentally or by numerical calculations for a few values of the parameters. Practical engineering can’t wait for every theoretical question to be totally settled with mathematical certitude.