Stokes Drag

Jun 30, 2022

Transient 1D

Total acceleration comprises of constant forward thrust a_o and linear drag against velocity:

$\begin{aligned} \frac{dv}{dt} &= a_o - \mu v \\ &= A(v)\end{aligned}$

Define velocity v(t_o)=v_o and v(t_o+\Delta t)=v_o+\Delta v, we can integrate the above equation to find velocity progression over time:

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$\begin{aligned} dt &= \frac{1}{a_o - \mu v} dv\\ \rightarrow \int\limits_{t=t_o}^{t=t_o+\Delta t}dt &= \int\limits_{v=v_o}^{v=v_o+\Delta v} \frac{1}{ a_o - \mu v} dv\\ \rightarrow \Delta t &= \int\limits_{v=v_o}^{v=v_o+\Delta v} \frac{1}{A}dv \\ &= \int\limits_{v=v_o}^{v=v_o+\Delta v} \frac{1}{A}\frac{dv}{dA}dA \\ &= \int\limits_{A=A(v_o)}^{A=A(v_o+\Delta v)} \frac{1}{A}\left(-\frac{1}{\mu}\right)dA \\ &= \left[-\frac{1}{\mu}\ln A\right]_{A=A(v_o)}^{A=A(v_o+\Delta v)} \\ &= -\frac{1}{\mu}\ln\left(\frac{a_o-\mu (v_o+\Delta v)}{a_o-\mu v_o}\right) \\ \rightarrow v_o + \Delta v &= a_o\frac{1-e^{-\mu\Delta t}}{\mu} + v_o e^{-\mu\Delta t} \end{aligned}$

$\implies v(t) = a_o\frac{1-e^{-\mu(t-t_o)}}{\mu} + v_o e^{-\mu(t-t_o)}$

Now that we have an explicit expression of velocity over time, finding displacement vs time is just a matter of integrating v(t) with respect to t:

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$\begin{aligned} \Delta d = \int\limits_{t=t_o}^{t=t_o+\Delta t}v(t)dt &= \int\limits_{t=t_o}^{t=t_o+\Delta t} a_o\frac{1-e^{-\mu(t-t_o)}}{\mu} + v_o e^{-\mu(t-t_o)} dt \\ &= \left[a_o\left(\frac{t}{\mu}+\frac{e^{-\mu (t-t_o)}}{\mu^2}\right) + v_o\left(-\frac{e^{-\mu (t-t_o)}}{\mu}\right)\right]_{t=t_o}^{t=t_o+\Delta t} \\ &= a_o\frac{\Delta t-\frac{1-e^{-\mu\Delta t}}{\mu}}{\mu} + v_o\frac{1-e^{-\mu\Delta t}}{\mu} \end{aligned}$

$\implies d(t) = a_o\frac{(t-t_o)-\frac{1-e^{-\mu(t-t_o)}}{\mu}}{\mu} + v_o\frac{1-e^{-\mu(t-t_o)}}{\mu} + d_o$

If we define d(t_o)=0 and t_o=0 then the kinematic equations simplify to:

$\begin{aligned} d(t) &= a_o \frac{t-\frac{1-e^{-\mu t}}{\mu}}{\mu} + v_o \frac{1-e^{-\mu t}}{\mu} \\ v(t) &= a_o \frac{1-e^{-\mu t}}{\mu} + v_o e^{-\mu t} \\ a(t) &= (a_o-\mu v_o) e^{-\mu t} \end{aligned}$

Notice that when \mu=0 (frictionless case) the equations reduce to constant-acceleration kinematics, i.e.

$\begin{aligned} d(t)&=a_o\frac{t^2}{2}+v_o t \\ v(t)&=a_o t+v_o \\ a(t)&=a_o \end{aligned}$

We can therefore summarize the kinematic equations as:

$\begin{aligned} d(t) &= a_o f_2(t) + v_o f_1(t) \\ v(t) &= a_o f_1(t) + v_o f_0(t) \\ a(t) &= (a_o-\mu v_o) f_0(t) \end{aligned}$ $\begin{aligned} \mu\ne0:& \begin{cases} f_0(t)=e^{-\mu t}\\ f_1(t)=\frac{1-f_0(t)}{\mu}\\ f_2(t)=\frac{t-f_1(t)}{\mu} \end{cases} &\mu=0: \begin{cases} f_0(t)=1\\ f_1(t)=t\\ f_2(t)=\frac{t^2}{2} \end{cases} \end{aligned}$

Or even more succinctly using prime notation:

$\begin{aligned} q(t) &= \begin{cases} \frac{t}{\mu}-\frac{1-e^{-\mu t}}{\mu^2} & \mu\ne0 \\ \frac{t^2}{2} & \mu=0 \end{cases}\\ d(t) &= a_o q(t) + v_o q'(t) \\ v(t) &= a_o q'(t) + v_o q''(t) \\ a(t) &= a_o q''(t) + v_o q'''(t) \\ \end{aligned}$

Transient 2D

In the 1D case the ODE was

$\begin{aligned} \frac{dv}{dt} &= a_o - \mu v \\ &= A(v)\end{aligned}$

Generalizing to vector form, the differential equation becomes:

$\frac{d\vec v}{dt} = \vec a_o - \mu \vec v$

Since every term is linear, the components are independent of each other, so the result is simply the 1D transient solution applied to each component separately.

$\begin{aligned} \vec d(t) &= \vec a_o f_2(t) + \vec v_o f_1(t) \\ \vec v(t) &= \vec a_o f_1(t) + \vec v_o f_0(t) \\ \vec a(t) &= (\vec a_o-\mu \vec v_o) f_0(t) \end{aligned}$ $\begin{aligned} \mu\ne0:& \begin{cases} f_0(t)=e^{-\mu t}\\ f_1(t)=\frac{1-f_0(t)}{\mu}\\ f_2(t)=\frac{t-f_1(t)}{\mu} \end{cases} &\mu=0: \begin{cases} f_0(t)=1\\ f_1(t)=t\\ f_2(t)=\frac{t^2}{2} \end{cases} \end{aligned}$

Since at any given situation, only the directions of a_o and v_o are involved, the resulting motion is always in the plane formed by your thrust and momentum vectors.

In other words, the problem can always be formulated in terms of the direction receiving full thrust, which is simply the 1D solution, and the direction in the plane that receives no thrust:

$\begin{aligned} \begin{pmatrix} d_{\parallel} \\ d_{\perp} \end{pmatrix}&= \begin{pmatrix} a_o\\ 0 \end{pmatrix}f_2(t)+ \begin{pmatrix} v_{o\parallel}\\ v_{o\perp} \end{pmatrix}f_1(t)\\ \begin{pmatrix} v_{\parallel} \\ v_{\perp} \end{pmatrix}&= \begin{pmatrix} a_o\\ 0 \end{pmatrix}f_1(t)+ \begin{pmatrix} v_{o\parallel}\\ v_{o\perp} \end{pmatrix}f_0(t)\\ \begin{pmatrix} a_{\parallel}\\ a_{\perp} \end{pmatrix}&= \begin{pmatrix} a_o-\mu v_{o\parallel}\\ 0-\mu v_{o\perp} \end{pmatrix}f_0(t) \end{aligned}$