How far downwind of a covid-19 emitter has a 1000-fold reduction in PPM?

As I said in my comment, viruses aren't my area (concerning units). But the answer is actually unitless. The exact answer you seek is dependent on a couple of different variables:

• The speed of the exhalation (cough or breathing)+speed of wind
• The atmospheric stability

Of course, the viral load of the exhalation is also important, but since you're asking for a ratio, I won't list that.

The Gaussian Plume model can be used to describe pollutants. The Gaussian Plume model is: $$frac{C}{Q}=frac{1}{2pi U sigma_y sigma_z}expleft(-frac{1}{2}left[frac{y}{sigma_y}right]^2-frac{1}{2}left[frac{z-H}{sigma_z}right]^2right)$$,

where $C$ is the concentration, $Q$ is the emission rate, $U$ is the wind speed (or worst case-scenario is cough speed), $x$ is the downwind distance, $z$ is the distance from the ground, and $sigma_x, sigma_z$ are the horizontal and vertical diffusion constants.

$sigma_y, sigma_z$ are functions of stability and wind speed. The actual formulas are... iffy. Theoretically they can be described as $$sigma_{y,z}=sqrt{frac{2K_{y,z}x}{U}}$$, where $K$ is the eddy viscosity. There are empirical formulas based on stability classifications too.

Lets assume that the receiver is directly downwind and at the same height (z=H and y=0). Therefore the equation collapses to: $$frac{C}{Q}=frac{1}{2pi U sigma_y sigma_z}$$. Since the distance downwind is expressed only by $sigma_{y,z}$, then to answer your question, we need to isolate $sigma_{y,z}$. $$sigma_y sigma_z=frac{1000}{2pi U}$$

Let's just say, for the sake of argument, consider two different scenarios:

1. Empirical $sigma_{y,z}$ (refer to this for values of constants)

$$(ax^b)times (cx^d+f)=acx^{b+d}+afx^b=frac{1000}{2pi U}$$

And this is complicated to solve. So I'll leave solving for $x$ as an exercise to the reader.

2. Constant $K_{y,z}$

$$sqrt{frac{2K_{z}x}{U}}sqrt{frac{2K_{y}x}{U}}=frac{1000}{2pi U}$$ $$frac{4K_{z}K_{y}x^2}{U^2}=left(frac{1000}{2pi U}right)^2=frac{10^6}{4pi^2U^2}$$ $$x=sqrt{frac{10^6 K_{z}K_{y}}{16pi^2}}$$ $$x=frac{10^3}{4pi}sqrt{K_{z}K_{y}}$$

If we assume that $K_z=K_y=K$, then the equation simplifies further. $$x=frac{K10^3}{4pi}$$ There are many ways on how to calculate $K$. One simple formula for $K$ can be used if mixing length theory is invoked

$$K=frac{ku_*z}{phi(frac{z}{L})}$$, where $k$ is the von-Karman constant, $u_*$ is the friction velocity, $phi_M(frac{z}{L})$ is the Businger-Dyer function for momentum, and $L$ is the Monin-Obukhov length. If we assume neutral atmospheric stability,$phi_M(frac{z}{L})=1$ and the resultant equation can be written as:

$$x=frac{ku_*z10^3}{4pi}$$

More assumptions! Let's assume z=1.5 m and $u_*=1.11$ m s$^{-1}$ (corresponds to a 5 kt wind speed at 10 m with a roughness length of 0.1 m under neutral stability). Plugging those numbers in gets

$x=35$ m

Of course, there were a LOT of assumptions to get to this point. But this is what I have to offer.