# Linear or Quadratic Air Resistance

I'm trying to model a rocket's position with air resistance as a factor. The rocket is fairly low-powered (Estes A8-3, 9.7N of thrust for 0.7 seconds), and I estimate it won't be going faster than 50 metres per second.

Should I be using linear or quadratic air resistance for this?

I found online that linear air resistance is used for relatively low velocities. How low is 'low velocity'?

• Hi, welcome to physics SE! This question could potentially be considered an engineering question, which would make it off-topic, even though I don't think it actually is engineering. Can you rephrase it a bit to focus more on the physics of the circumstances under which air resistance is linear?
– user191954
Commented Jun 3, 2018 at 8:52

The quantity to check is the Reynolds number $$Re = \frac{\rho v L}{\mu}$$ where $\rho$ is the density of air, $v$ is the relative velocity, $L$ the typical dimension of the rocket and $\mu$ the (dynamic) viscosity.
For small Reynolds numbers, i.e. $Re\ll 1$, viscous forces dominate and you have laminar flow. In this regime, the drag force is proportional to the velocity. For large Reynolds numbers, drag is dominated by turbulences and is proportional to the velocity squared.
Plugging in some numbers: $\rho \approx 1~\textrm{kg}~\textrm{m}^{-3}$, $\mu\approx 2~ 10^{-5}~\textrm{kg}~\textrm{m}^{-1}\textrm{s}^{-1}$, $v\approx 20~\textrm{m}~\textrm{s}^{-1}$, $L\approx 1~\textrm{m}$, you get a Reynolds number on the order of a million, which is much larger than 1. This suggests (as expected) that you are in the turbulent regime and should use the force proportional to the velocity squared.