## Falling Bullets and the Density of Air

Matt discusses a Mythbusters episode where they drop a bullet and shoot one horizontally from a gun, and see which falls first. I want to take this up where he left it off

They find that to within their experimental error, the two bullets hit the ground at the same time. It doesn’t matter how fast you move in the x-direction; gravity still gives you the same acceleration in the y-direction.

It turns out that with standard assumptions about air resistance, this is not true. What I hope to do here is, by assuming that bullets falling and shot horizontally fall at very nearly the same rate, put a bound on the density of air.

First we need a model for air resistance. If the bullet is a non-rotating sphere and air is non-viscous, then the bullet slows down because air smashes into it. We can imagine the bullet leaving a cylindrical trail behind it as it flies. Let’s say that as the bullet flies through its path, all the air it encounters gets brought from rest up to some set fraction $x$ of the bullet’s speed, and that $x$ does not depend on the bullet’s speed $v$.

Then in one incremental unit of time $\Delta t$, the bullet intersects a mass of air that is the density of air times the volume of its path over that time, and the volume of the path is given by the velocity of the bullet times is cross-sectional area. This is $\rho \pi r^2 v dt$ grams of air, where $\rho$ is the density of air, $v$ is the speed of the bullet, and $r$ is its radius.

The momentum imparted to the bullet is the same as that imparted to the air. The momentum imparted to the air is the mass of air encountered times the speed it’s being sped up to. This is $(\rho \pi r^2 v dt) x*v = dp$.

The force on the bullet is the time derivative of its momentum, so $dp/dt = F = x \rho \pi r^2 v^2$. For the time being, set $k = x \rho \pi r^2$ so the force on the bullet is just $kv^2$.

Now let’s make some simplifying assumptions. We imagine that air resistance plays a minor role, because otherwise the Mythbusters would not have found so close to a tie as they did. This means the bullets have very nearly the same y-height at any time, and that it’s the same as the normal, physics 1a parabola falling under gravity. Let the bullet fired out of the gun have the same horizontal velocity throughout its flight. Also assume the bullet goes much faster in the horizontal than in the vertical direction. Using these assumptions, we’ll try to find the relative acceleration of the two bullets, rather than finding both of their paths and then subtracting.

For the bullet falling straight down, all motion is in the y-direction, and the upward force is $kv_y^2$. For the bullet fired out of the gun, the total force is $k(v_x^2 + v_y^2)$, but only part of this is in the y-direction, so the force in the y-direction is $k v_y\sqrt{v_x^2 + v_y^2}$. That can be simplified to $k v_y*v_x$ because velocity in the y-direction is small.

Because air resistance plays a minor role, assume the bullets fall for almost the same amount of time, $t_f = \sqrt{\frac{2g}{h}}$. Over that time, there is a difference in the force on the bullets given by $F_{shot} - F_{drop} = k(v_xv_y - v_y^2) = kv_y(v_x - v_y) = (kv_x)v_y$. Notice that this is equivalent to assuming the dropped bullet doesn’t get slowed down at all. The shot bullet experiences so much more air resistance that we are essentially assuming only air resistance on the shot bullet matters.

The bullets are in free fall, so $v_y = gt$ and the total distance between the two bullets when they fall comes from integrating their relative acceleration once for their relative velocity, and again for their relative displacement. When the mass of the ball is $m$, we have

$v_{rel}(t) = \int_0^t accel_{rel}(t')dt' = \int_0^t kv_x v_y(t')/m dt' = \frac{k v_x}{m} \int_0^t gt' dt' = \frac{1}{2m}kv_x g t^2$

$y_{rel}(t) = \int_0^t v_{rel}(t') = \frac{1}{6m} k v_x g t^3$.

Plugging in $t_f$, the falling time, for $t$ gives the distance between the bullets as they land. Dividing by their speed at that time $gt_f$, gives the difference in their falling times. This comes to

$\Delta t = \frac{x \rho \pi r^2 v_x h}{3mg}$

To check, the dimensions work out to time. The difference increases with the density of air, with the radius of the ball (for fixed mass), with the horizontal firing speed, with the height dropped, and with the $x$ factor of air resistance. It decreases with increasing gravity and mass of ball (for fixed radius). All that seems plausible enough.

We can solve this for $\rho$, the density of air. Since $\pi = 3$ I can cancel those.

$\rho = \frac{\Delta t m g}{x r^2 v_x h}$.

Let’s make some guesses. They fired the bullet from about $1m$ off the floor. It traveled some $300m$ down the hall in the $1/3 s$ it had to fall, so $v_x = 1000m/s$. $g = 10 m/s^2$, $m = 100g$ and $r = .01m$. $x = .5$.

That gives

$\rho = \Delta t \frac{20 kg}{s m^3}$.

The Mythbusters think $\Delta t < .1s$ (they said 39 milliseconds), which implies

$\rho < \frac{2 kg}{ m^3}$.

And it turns out the density of air is indeed less than this – about 1 kg/m^3.

I should admit, though, that the first time I crunched through these numbers, I got that the density of air should be less than $0.1 kg/m^3$. That was plugging in $5g$ for the bullet weight, thinking of a $1cm^3$ bullet about as dense as a typical rock. But the bullet I hypothetically used had radius 1cm, not diameter, and also bullets are made of metals that are pretty dense. I googled "bullet weight" and learned that google automatically interprets that to mean the weight of rugby player Tom James, nicknamed "The Bullet". (His weight was 15 st 8 lbs).

Incidentally, I think my calculation worked by accident. The aerodynamics of a bullet are probably much more complicated than I've assumed here, since the bullet spins and probably entrains air and makes eddy currents and does all kinds of ungodly things. $v_x$ isn't constant, but according to my model it should drop by less than $50m/s$ in the time it takes the bullet to fall. I guessed kind of wildly on things like the height of the drop and the factor by which air is sped up when the ball passes through. Also, the experiment has error in terms of simultaneous dropping and how level the gun is. But things seem to have worked out to order of magnitude.