# Explanation that air drag is proportional to speed or square speed?

A falling object with no initial velocity with mass $m$ is influenced by a gravitational force $g$ and the drag (air resistance) which is proportional to the object's speed. By Newton´s laws this can be written as:

1. $mg-kv=ma$ (for low speeds)
2. $mg-kv^2=ma$ (for high speeds).

I assume that $k$ is a positive constant that depends on the geometry of the object and the viscosity. But how can one explain that the air resistance is proportional to the velocity? And to the velocity squared in the second equation?

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There is a decent treatment of how drag is modeled here en.wikipedia.org/wiki/Drag_equation – joshphysics Apr 3 '13 at 16:56
We've certainly had plenty of other questions that have touched on this. This one came to my mind for the turbulent case physics.stackexchange.com/questions/14052/… it shouldn't be too difficult to get a comprehensive answer for both cases. – Alan Rominger Apr 4 '13 at 18:05

One's naive expectation would be that as the object moves through the medium, it collides with molecules at a rate proportional to $v$. The volume swept out in time $t$ is $A v t$, where $A$ is the cross-sectional area, so the mass with which it collides is $\rho A v t$. The impulse in each collision is proportional to $v$, and therefore the drag force should be proportional to $\rho A v^2$, with a constant of proportionality $C_D$ (the drag coefficient) of order unity.

In reality, this is only true for a certain range of Reynolds numbers, and even in the range of Reynolds numbers for which it's true, the independent-collision picture above is not what really happens. At low Reynolds numbers you get laminar flow and $C_D\propto 1/v$, while at higher Reynolds numbers there's turbulence, and you get $C_D$ roughly constant.

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If $Re \ll 1$, the inertial terms are very small in relation to the stresses and body forces in the Navier-Stokes equations. If this is the case, the flow is called Stokes Flow. Just a stab in the dark: perhaps the reason the drag is proportional to $v$ in this flow regime is due to the linearity of the equations without inertial terms? – OSE Apr 4 '13 at 14:05
Intriguing...(+1) Two questions though: 1) If $k$ is equal in both cases, how do you explain the difference in units? Because in the $v^2$ case, you're adding force to force, but in the $v^1$ case, you're adding force to massflow, which doesn't make much sense in my mind...And if $k$ should be different somehow, then...how? 2) Can you point us to a good resource that quantifies the dependence of $C_D$ on $Re$ more precisely than the wiki article, for a few (simple) objects? – Rody Oldenhuis May 31 '13 at 7:46
@RodyOldenhuis: (1) The $k$ defined by the OP in equation 1 is not the same as the $k$ defined in equation 2. (2) There's a graph in the Feynman Lectures. – Ben Crowell May 31 '13 at 14:17

To put it in simple terms, at slow speed the drag is just due to the viscosity of the fluid.

At high speed, the momentum you're imparting to each parcel of air is proportional to the speed, and the number of parcels of air per second you're doing it to is also proportional to speed.

Since force is momentum/second, that's why it's proportional to speed-squared.

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