# How does drag change at very low gas density?

The usual expression for the drag force $$F_d$$ of an object with velocity $$v$$, cross-sectional area $$A$$, and drag coefficient $$c_d$$ is $$F_D=\frac{1}{2}\rho v^2 A c_d$$ where $$\rho$$ is the density of the gas the object is moving through.

Is this still valid for very low densities, i.e. as $$\rho \rightarrow 0$$?

In particular, I was wondering if anything changes when the mean free path of the gas molecules becomes larger than the dimensions of the object. In case it matters, I'll note that I am interested in the situation for residual argon gas inside a vacuum chamber at temperatures $$\sim 300$$ K, not earth satellites in high orbits.

There is no reason why it would not be valid for low densities. The drag force becomes proportionally smaller. This is because there is less and less gas molecules per unit volume decreasing the number of collisions, and of course in the limit $$\rho\rightarrow 0$$ there is no gas and so no drag.
I would think there is no general $$^1$$ relationship between MFP (mean free path) and drag, but one could argue that as the density increases, the molecules are closer and so the MFP decreases. In other words, a lower value of MFP implies a higher density and therefore drag force.
$$^1$$ The mean free path is $$l=\frac{1}{n\sigma}$$ where $$n$$ is the particle number per unit volume and $$\sigma$$ is the effective cross sectional area. So given a particular setup, you may be able to formulate an expression for the drag force.