So I was answering a question on the WorldBuilding SE site and one aspect of my answer made me wonder.

Imagine you are on a massive space station of unlimited size. If you were to fire an arrow from a bow, how far would that arrow go before air resistance slowed it past being deadly? How far would it go before it just stopped and hovered in air?

I know this is dependent on the speed and shape of the arrow, but generalization is fine.


  • It looks like the average arrow weights 20 grams.
  • It looks like the average arrow travels 329 feet per second.
  • Air resistance?
  • $\begingroup$ unlimited size? $\endgroup$
    – Qmechanic
    Jul 1 '16 at 23:10
  • $\begingroup$ @Qmechanic - Well, if not "unlimited", then at least as long as the longest distance the arrow could travel. I guess the question could be reworded to ask "If I wanted to shoot an arrow, blah blah, blah ... what is the minimum length I must build my space station?". $\endgroup$ Jul 2 '16 at 3:35
  • $\begingroup$ @Qmechanic Unlimited, as in "there is no limit to its possible size", not as in "it's infinite, ha ha ha!". $\endgroup$
    – wizzwizz4
    Jul 2 '16 at 9:33
  • $\begingroup$ If the station is of unlimited size, it would immediately collapse into a singularity so the answer is "the arrow would travel an infinitely small distance but would be instantly deadly for all of eternity". $\endgroup$
    – Richard
    Jul 2 '16 at 10:13
  • $\begingroup$ What effort have you made to find an answer eg by searching the internet? $\endgroup$ Jul 2 '16 at 12:30

How far would it go before it just stopped and hovered in air?

I'll offer a second opinion. At some point the arrow will effectively come to a stop, and even after an infinite amount of time, the arrow will only have traveled a finite distance.

The assumption that drag is proportional to the square of velocity applies only at high Reynolds numbers, where turbulence dominates over viscosity. (Aside: It doesn't even quite apply then. There are a number of assumptions that go into this derivation. The coefficient of drag is not constant. It instead depends on velocity.)

Turbulence is essentially non-existent at low Reynolds numbers, which means that all the different terms that lead to a (roughly) velocity squared drag term vanish. Viscous drag, aka Stokes' drag, is the dominant source of drag in low Reynolds number regimes. Here, drag is (roughly) proportional to velocity rather than the square of velocity: $$ \frac{d^2x(t)}{dt^2} = -k\frac{dx}{dt}$$

This means that if an object subject to viscous drag has a velocity relative to the fluid, the object will come to a stop (at $t=\infty$) after having traveled a distance of $\frac v k$.

What the exact distance is where the object has come to a stop is an engineering rather than physics problem. The same goes for the first question, the distance at which the arrow is no longer fatal.


The drag force on an arrow is proportional to the square of the velocity for high velocity and proportional to the velocity itself for low velocity. Assuming the initial velocity $v_0$ is high we can neglect the drag proportional to $v$ and the equation of motion, $\vec F=m\vec a$, is simply $$\frac{dv}{dt}=-\frac{b_2}{m}v^2,$$ for an arrow of mass $m$ in the $Ox$ axis. We integrate this equation $$\int_{v_0}^v\frac{dv}{v^2}=-\frac{b_2}{m}\int_0^t dt,$$ to obtain $$v(t)=\frac{mv_0}{m+v_0b_2t}.$$

Notice that the velocity goes to zero as time goes to infinity, i.e. the arrow actually never stops.

Writing $v=dx/dt$ we can integrate the above equation, $$\int_0^x dx=\int_0^t \frac{mv_0}{m+v_0b_2t}dt,$$ which gives $$x=\frac{m}{b_2}\ln\frac{m+v_0 b_2t}{m}.$$ If you want to know the distance $x$ as a function of the speed $v$ you just eliminate $t$ from the two equations. Then knowing what is the speed necessary to kill someone (which I think is an off-topic) you plug the numbers and get the distance.

Edit: When the arrow has achieved a low velocity, the drag force is proportional to $v$. The equation of motion reads $$\frac{dv}{dt}=-\frac{b_1}{m}v,$$ whose solution is $$v=v_0e^{-b_1t/m}.$$ Integrating once again we get $$x=\frac{mv_0}{b_1}\left( 1-e^{-b_1t/m}\right).$$ The total distance covered in the limit of low velocities ($v\ll b_1/b_2$) is therefore $mv_0/b_1$.

  • $\begingroup$ The arrow will not travel an infinite distance. This answer ignores Stokes' drag. $\endgroup$ Jul 1 '16 at 22:57
  • $\begingroup$ It's probably worth mentioning that under these assumptions, velocity as a function of distance has the form $v(x)=v_0\exp(-bx/m)$. (You can see this most easily by using $\frac{dv}{dx}=\frac{dv}{dt}\frac{dt}{dx}=\frac{1}{v}\frac{dv}{dt}=-bv/m$.) This makes it clear that after a certain distance (of order $m/b$), the arrow will have effectively stopped. $\endgroup$
    – jjc385
    Jul 1 '16 at 23:03
  • $\begingroup$ @DavidHammen Thank you for pointing this out. I should not neglect the drag proportional to $v$ when the arrow has low velocity. $\endgroup$
    – Diracology
    Jul 1 '16 at 23:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.