# Why does air resistance increase with speed?

1. Jul 5, 2010

### ap_cycles

HI,

My question as per above.

A quick check with Physicsclassroom.com reveals that a (perhaps too simplistic) reason is that more and more air molecules are hitting the body as the body falls through the air. But i am confused. I thought when the body first falls through the air fluid, the number of air molecules hitting the body is proportional to the surface area of contact, which as we all know, is fixed (body shapes dont change right!).

I understand that air resistance is a form of drag force, and this drag force is dependant on velocity of the falling body. What i want to know, however, is how we can explain this phenomena in terms of particle theory.

Fellow forummers help me please? A high school physics teacher here.

2. Jul 5, 2010

### arildno

Now, let us have a body with a front surface with area A moving with velocity V.

In a time interval dt, then, the region of space the surface has traversed has the volume A*V*dt.
Multiplying this with the molecular density yields the amount of molecules displaced during that time interval.

If you reduce the velocity, fewer molecules will be displaced in the same time interval, and thus, give rise to less air resistance.

3. Jul 5, 2010

### Pengwuino

If you have still air, as you could say you have at the beginning of the fall, the air molecules still have velocities associated with them when they hit you. Their average velocity is just 0, but they still have velocities individually! So sure, it makes sense that the only thing that matters if how much surface area you have, but that's only at rest. Now consider, to make the point clearer in my opinion, not a person falling through the atmosphere at 200mph, but a person standing still being pummeled by 200mph wind. Now, of course more surface area would still mean more wind bouncing off of your, but 200mph plus or minus each individual molecules velocity when the air is at rest as a whole is a far greater force then at rest.

4. Jul 5, 2010

A moving object has to push the air out of the way and transfer some kinetic energy to that air.You can feel this when you stand by the side of the road and a speeding car goes past.If the object moves faster it moves a greater mass of air per second and gives that air a greater increase of speed.Perhaps you could ask your students to make fans and then feel the effects of fanning themselves at different speeds.

5. Jul 5, 2010

### Phrak

Yes, well, drag is, to first approximation, in proportion to the square of the velocity.

6. Jul 5, 2010

### Ranger Mike

I thought when the body first falls through the air fluid, the number of air molecules hitting the body is proportional to the surface area of contact, which as we all know, is fixed (body shapes dont change right!).
don't forget that at the height the object was dropped, the density of air molecules is a lot less than that at ground level. air is " thin" at jump altitude. It is a lot ' thicker ' at ground level. Air also exerts about 14 psi pressure at at ground level. the object will fall at an increasing rate until it reaches terminal velocity.

7. Jul 5, 2010

### ap_cycles

Thank you very much for all of your replies!

Now i understand.... :!!)

8. Jul 5, 2010

### arildno

That depends on what sort of drag you are talking about. It really is a very complicated phenomenon.

If you take Stoke's law of resistance for a sphere moving through a viscous fluid, that drag is linear in the velocity. (The drag due to viscous forces is a different mechanism than molecular displacement). Stokes' law is extremely accurate, and has many practical applications .

If you work with an inviscid fluid, where the fluid is locally stationary in respect to a non-accelerating body, then you'll end up in..D'Alembert's paradox.

It is cerainly true that for relatively high speeds, a square velocity law is generally best, but for low speeds, a linear model for air resistance is often accurate enough.

Thus, I chose not to delve into intricacies of modelling, but to point to one omni-present feature that points in the direction of the relation increased speed-increased air resistance.

I didn't tell the whole story, not the least because many features of that story is not known by me.

Last edited: Jul 5, 2010
9. Jul 5, 2010

### sophiecentaur

What is happening at the front of the object is only a small part of the story. With laminar flow, the energy you impart to the molecules at the front, by shifting them out of the way, (force times distance, effectively) should be returned at the back end as they come together again (particularly true if you take an object tapered at both ends, to make it simple and with a fluid of low/no viscosity).
But, once you start gong fast, the energy used in pushing the air aside at the front, is lost in the form of turbulence and it's not 'available' to push back together at the back. This is detected as a drop in pressure at the back (drag).
It's often worth while using an intermediate step in reasoning about these things and not making yourself explain everything in terms of particles, the whole way through.

10. Jul 5, 2010

### Staff: Mentor

Why can't it be proportional to both? Or proportional to one and proportional to the square of the other...?

11. Jul 5, 2010

### arildno

In high school, russ, "proportional" means there exists a proportionality constant, rather than that the ratio between two proportional quantities should be independent of those quantities..

12. Jul 5, 2010

### Staff: Mentor

13. Jul 5, 2010

### arildno

The factor of proportionality between air resistance&area might (and does) vary with some third variable (but not with either of those two).
If OP thought the proportionality factor was a constant, and couldn't be a function of, say, velocity, then his question makes some sort of sense.

Just speculating..

14. Jul 5, 2010

### sophiecentaur

You are talking as if the molecules of air are all stationary. In fact, the mean speed of the air molecules is much higher than that of a falling object. Molecules will be hitting the back as well as the front of your object. Being so simplistic about the situation doesn't help at all with understanding what's going on.
You have to consider turbulence or you would have to say that all objects of the same cross sectional area would have the same drag factor. A number of Formula One designers might disagree with you there.

15. Jul 5, 2010

### Staff: Mentor

If you hold area constant and vary the speed, then you have drag as proportional to velocity squared. If you hold velocity constant and vary the area, then you have drag proportional to the area.