# 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?

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.

• 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. – user4552 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.

Newtons laws of motion provide a simple explanation for why drag is proportional to velocity squared of a moving object. Use the example of a falling skydiver that is pulled downwards by gravity; but this explanation also applies to cars, trains, airplanes, etc .... that are being pushed through the air.

The skydiver will fall through a mass of static air ('m') which they push out of their way and accelerate ('a'). This creates a downward force (Force = ma) by the skydiver on the air, according to Newtons 2nd law of motion. The 'equal and opposite' upward force is called drag (Newtons 3rd law of motion). Drag equals the downward force exerted by the falling skydiver to push the air out of their way.

OK. So, if (prior to terminal velocity), the skydiver was to double their downward velocity. Then: (i) The mass of air fallen through each second will double (m x2). (ii) The skydiver will hit each air molecule with twice the momentum as before and thus double the acceleration of each air molecule hit (a x2). The combined effect of theses two is to quadruple the downward force (Force x4 = 2m x 2a). Consequently, the 'equal & opposite' drag on the sky diver will quadruple (Drag x4). Simple. If the skydiver's velocity doubles then drag will quadruple. This explanation isn't in any textbook.

In this process, energy and momentum is transferred from the skydiver to the air. So no net loss or gain of mass, momentum or energy in this process.

The skydiver hits terminal velocity when gravity (the force acting on the skydiver) can no longer accelerate the skydiver to a higher velocity. But don't let gravity or terminal velocity confuse you, they are not critical to explaining the relationship between drag and the skydiver's velocity.

This explanation can be applied to any object falling through a fluid; Such as a stone falling through water or air.

• This is basically wrong because you state it as if it's a rigorous proof of something that is always true. In fact it's at best a decent approximation over some range of Reynolds numbers. This explanation isn't in any textbook. Not true. Many textbooks give this explanation. – user4552 Oct 14 '19 at 20:19

I think the simplest way to express it is too think that if something goes faster, then more fluid molecules will interact with the surface, thus increasing the drag. This is definitely by far the simplest explanation. If you want a visual explanation, check out that veritasium vid on him swimming in a pool filled with shade balls.