What model equations assumptions are made in order to claim that the behavior of solutions (trajectories) achieve some terminal behavior?

Question

From maybe Calculus I or an undergrad physics course, one learns briefly about something called "terminal velocity", which seems to be accepted as true, although I've never looked for a proof of it.

I've been reading a paper that gives some model equations that quantifies the aerodynamic forces of a rigid body falling in air. It seems that the authors claim that the solutions, i.e. the trajectories, always reach some sort of terminal behavior, e.g. if the rigid body happens to be a meteorite that is tumbling (rotating + drifting) to earth, after a sufficient amount of time has passed, then it will tumble forever, until it reaches the ground.

What model assumptions, e.g. steady state assumptions, could the authors be making to support the claim of an existence of a terminal behavior?

For simplicity, assume that there's no wind / flow velocity -- just still air in the background (which is an assumption in the paper, too).

Thanks

Answer 1

Using terminal velocity as an example, a standard assumption would be that all the constants involved are in fact constant. E.g., air has constant density despite falling through the atmosphere, the coefficient of drag doesn't depend on velocity, the object has a constant surface area and shape, and so on. Then the problem simplifies to dealing with the two equations $$F_d=C_1v^2$$ for some constant $C_1$ which wraps up all of the different constants in the standard drag formula, and $$F_g=C_2$$ for some constant $C_2$ which wraps up all the components of gravitational forces. Newton's laws then give $$\frac{\partial v}{\partial t}=C_2-C_1v^2,$$ assuming a vertical fall. Asymptotically, solutions to this differential equation approach $$v=\sqrt{\frac{C_2}{C_1}}.$$ In practice objects approach this asymptotic behavior quite quickly, and your particular application determines how much time needs to elapse before you're "close enough" to the asymptotic behavior.

Adding in additional variables like the trajectory not being vertical, with a little more complexity you can prove the existence of steady-state solutions and that all starting positions approach those.

Once you add in more details like gravitational pull changing depending on how far away you are from the earth, air density changing with altitude, coefficient of drag varying with velocity, the fact that the object is not uniform and has some sort of average tumbling behavior, and whatnot the problem becomes more complicated.

For "standard" objects at reasonable velocities, the effects of all of these things are relatively small. You can incorporate them as parameters and perform a sensitivity analysis on the solutions of the differential equation to determine if the effects are small enough, or you can even numerically solve the resulting equations. For many practical purposes (like estimating the behavior of a skydiver well enough to determine when chutes should be deployed), these variables have a small impact relative to the things we actually care about, and we can consider the skydiver as effectively having a terminal velocity. More accurate applications might not be able to make such approximations.

Answer 2

Let an object with mass $m$ and initial velocity $v_0$ fall down from height $h$, so according to second law of Newton $ma=mg$, then \begin{eqnarray*} m\dfrac{dv}{dt} &=& mg\\ dv &=& g\mathrm{dt}\\ \int_{v_0}^v dv &=&\int_0^t g\mathrm{dt}\\ v-v_0 &=& gt\\ v &=& gt+v_0 \end{eqnarray*} now we add air resistance in our assumption with factor $-k$ (minus means the resistance of air against object motion), where $k\geqslant0$ hence \begin{eqnarray*} mg-kv &=& ma\\ ma+kv &=& mg \,\,\,;\,\,\,a=v'\\ v'+\frac{k}{m}v &=& g \end{eqnarray*} we solve this first order equation with integrating factor $\displaystyle I=e^{\frac{k}{m}t}$ then $$e^{\frac{k}{m}t}v=\int e^{\frac{k}{m}t}g\,\mathrm{dt}+C=\frac{mg}{k}e^{\frac{k}{m}t}+C$$ this gives the velocity equation $\displaystyle v=\frac{mg}{k}+Ce^{-\frac{k}{m}t}$, and for falling object $v_0=0$ gives $\displaystyle C=-\frac{mg}{k}$ and finally $$\displaystyle v=\frac{mg}{k}(1-e^{-\frac{k}{m}t})$$ is the velocity of object which falls on earth. After while (depend on fluid viscosity), the object reaches to terminal velocity $\displaystyle v_T=\frac{mg}{k}$, because $$|v|=\Big|\frac{mg}{k}\big(1-e^{-\frac{k}{m}t}\big)\Big|\leqslant\frac{mg}{k}=v_T$$

this matter occur for every falling object like a meteorite or skydive and in fact equations show that $$\displaystyle \lim_{t\to\infty}v=\lim_{t\to\infty}\frac{mg}{k}(1-e^{-\frac{k}{m}t})=v_T$$ for objects with low density the terminal velocity is height. Also the displacement of object during fall will be \begin{eqnarray*} y &=&\int v\mathrm{dt} + C\\ &=&\int v_T(1-e^{-\frac{k}{m}t})\mathrm{dt} + C\\ &=& v_T(t+\frac{m}{k}e^{-\frac{k}{m}t}) + C \,\,\,;\,\,\,y(0)=h\\ &=& v_Tt-\frac{m^2}{k^2}g(1-e^{-\frac{k}{m}t})+h \end{eqnarray*}