# Why do heavier objects fall faster in air?

We all know that in an idealised world all objects accelerate at the same rate when dropped regardless of their mass. We also know that in reality (or more accurately, in air) a lead feather falls much faster than a duck's feather with exactly the same dimensions/structure etc. A loose explanation is that the increased mass of the lead feather somehow defeats the air resistance more effectively than the duck's feather.

Is there a more formal mathematical explanation for why one falls faster than the other?

• My guess is that it has to do with the tendency of light objects to reduce their cross-section when falling, while heavy objects of the same shape often consist of rigid material, e.g. metal, therefore their cross-section always remains constant. – auxsvr Jun 6 '14 at 11:09
• Also, lighter objects can be carried away by wind currents and turbulence more easily. – auxsvr Jun 6 '14 at 11:12
• Short answer: Gravitational pull is proportional to mass, whereas air drag is a function of area. Therefore, stuff with low area for the mass (like a lead ball) will have relatively less air drag for the same gravitational pull. A feather has a large area for its mass. – Olin Lathrop Jun 6 '14 at 12:11
• I'd say denser spherical objects fall faster with air drag. Try dropping a 100g ball with R=100mm and a 10g ball with R = 1mm. – Francisco Presencia Jun 7 '14 at 11:13
• And wait, here's a critical point: it's almost a certainty that you could design an object A which is lighter than object B, but in fact object A would reach the ground much faster than object B. – Fattie Jun 8 '14 at 9:52

## 5 Answers

We also know that in reality a lead feather falls much faster than a duck's feather with exactly the same dimensions/structure etc

No, not in reality, in air. In a vacuum, say, on the surface of the moon (as demonstrated here), they fall at the same rate.

Is there a more formal mathematical explanation for why one falls faster than the other?

If the two objects have the same shape, the drag force on the each object, as a function of speed $v$, is the same.

The total force accelerating the object downwards is the difference between the force of gravity and the drag force:

$$F_{net} = mg - f_d(v)$$

The acceleration of each object is thus

$$a = \frac{F_{net}}{m} = g - \frac{f_d(v)}{m}$$

Note that in the absence of drag, the acceleration is $g$. With drag, however, the acceleration, at a given speed, is reduced by

$$\frac{f_d(v)}{m}$$

For the much more massive lead feather, this term is much smaller than for the duck's feather.

• Force is a vector quantity, if F comes out to be negative in sign. Then, duck's feather falls faster than lead feather! – Immortal Player Jun 6 '14 at 11:44
• @Godparticle, the context is clear is it not? The two objects are dropped with zero initial speed. Thus, the drag force on each is initially zero and the terminal speed is approached from below. The acceleration asymptotically approaches zero as does the net force. F will not come out negative. – Alfred Centauri Jun 6 '14 at 11:57
• atmospheric pressure is acting dowwards, upwards and from every side. The only applicable atmospheric metric here is density (more fluid density, more drag). Try to repeat this experiment under water (lot more dense than air) – jean Jun 6 '14 at 12:24
• What you are missing here is pressure act from every side, ever. When we speak about atmospheric pressure we are, usually, speaking about a delta, a difference between pressures applied in two sides. In general the difference between internal and external sides. – jean Jun 6 '14 at 12:41
• @Godparticle what's the difference? Alfred just selected the axes such that the force projection is positive. Rotation doesn't change equations of motion, due to Galilean invariance. – Ruslan Jun 6 '14 at 15:34

A good approximation of the drag force for an object falling through the atmosphere is $-cv^2$, with $c$ a constant independent of mass. Thus, $$m \dot{v} = mg - cv^2$$ is the equation of motion with initial condition $v(0)=0$. We write $$t = m \int_0 ^{v(t)} \frac{d v}{mg -cv^2}$$ and the final result is $$v (t) = \sqrt{\frac{mg}{c}} \tanh \left( t \sqrt{ \frac{gc}{m}} \right),$$ which is a function increasing as $m$ increases for $t$ constant, therefore heavier objects fall faster than lighter ones in presence of drag due to air. The terminal speed is $$\lim_{t\to \infty} v(t) = \sqrt{\frac{mg}{c}}.$$ For a person in free fall with drag, the terminal speed is about 50 m/s.

The previous analysis depends on the fact that the cross-section of the falling object remains constant, which is often far from true and alters the result significantly, since, for example, a feather curves when falling while a feather of the same shape made of metal will not curve and will be heavier, making the difference in falling speed more pronounced. Indeed, a plot with varying $c$, which is $\propto A^{-1}$ with $A$ the cross-section, indicates that the effect of the cross-section on the speed is much more important than that of the different masses. Also, we assume that wind currents and turbulence are negligible, another assumption that may change the result in real conditions significantly.

## Edit:

This analysis, as anticipated, may fail spectacularly if one takes into consideration that an asymmetric object in general rotates in a chaotic fashion if it exceeds certain threshold angle, which can be said to depend on the density, cf. this article

• Shouldn't the limits be: $$t_2 - t_1 = m \int_{v_1}^{v_2} \frac{d v}{mg -cv^2}$$ and not $\int_0^t$ – John Alexiou Jun 6 '14 at 13:21
• @auxsvr: I am not sure that was a correction. Velocity is the dependent variable here. The free variable is time, so it indeed should be $0$ to $t$. And the numerator should be $dt$. Because in the "d" notation, the first equation is $m\frac{dv}{dt} = mg - cv^2$ and you "multiply" by $dt$ to integrate. – Jan Hudec Jun 6 '14 at 13:47
• @ja72: I believe that's worse. The left hand was correctly $v(t)$, so the integral must have $t$ as free variable. $m\int_0^t\frac{dt}{mg - cv^2}$. – Jan Hudec Jun 6 '14 at 13:50
• @JanHudec Whether velocity is the dependent or independent variable depends on the variable of integration. I've verified that this is the correct result. – auxsvr Jun 6 '14 at 13:53
• Note - that approximation has almost no relationship, in fact if I'm not mistaken no relationship, whatsoever to the "insanely complicated tumbling physics" of a feather falling from above. (Ask any computer graphics simulation engineer!) I appreciate you "mentioned this" in the last paragraph but it's the germane point here. – Fattie Jun 8 '14 at 9:50

Short answer: air drag!

Gravity is acting in both feathers the most massive receives a stronger pull to down. Air drag is counter acting that movement and is proportional to the velocity (in a very complex way, references here and here)

That stronger pull helps to overcome the increasing drag opositing force. That's why the lead feather ill accelerate faster and reach a bigger terminal velocity.

The same principle is applied to race cars. Two cars, same shape, the one with the most potent engine can accelerate more and reach a greater max velocity.

Another Example: Skydivers usually dresses something to increase air drag and stands in a position to help the drag to lower the terminal velocity and increase the fall time. A stading up jumper ill fall a lot faster.

Edit

After some discussion on the buoyance effect I searched a while about a bird's feather density, a value not easy to get. I found this reference (it'a a .pdf document) about chiken's feather and contains a lot of considerations about density. After the lecture we can use a value of 0,89g/cm3 and that almost as dense as water. So any buoyance effect is negligible. If one still want's to discuss negligible forces we can pick also the gravity variation on altitude or the effect off relativistic physics over the acceleration of a body.

• @Trengot: This answer is correct. The terminal velocity only comes about due to drag, since buoyancy does not increase with speed. You are right that buoyancy has to be included in the question, but for a feather drag is more significant than buoyancy. – Jan Hudec Jun 6 '14 at 13:04
• Buoyancy is usually negligible and I guess it's negligible for even a birds feather. Can you find some reference about that value (bird feather density) – jean Jun 6 '14 at 13:05

# Linear momentum!

I think the easiest way to gasp the concept is to think at atomic level about momentum of the objects and the air's atoms that cause friction.

Linear momentum is equal to $$m\times v$$ so heavier object have higher momentum. Imagine an atom that comes from the opposite direction of the falling object and collide with the object.

For making the things easier we suppose an inelastic one-dimensional collision. Basically when the atoms collide with the object the object lose part of its momentum, mass is constant so this cause the object to slow down.

Imagine two objects with different mass but same velocity, the object with greater mass (e.g. a metal sphere )will lose only a little bit of its linear momentum hence it will keep going with a velocity near to the initial one, in the other hand the object with smaller mass (e.g. a ballon) and hance momentum will lose a great portion of its linear momentum and so it will slow down considerably!

Whether the lead feather falls faster or the ducks feather falls faster, depends on the direction of external force per unit area acting on them, mass per unit area of each of them.

Probability of lead feather falling faster is greater than the ducks feather, because of greater probability of less downward external force on both.

Credits: $_1$ Modern's ABC of Physics-2012 Edition-Page No.695.

• Why the downvote? This is too vague to be useful, but it isn't wrong. – auxsvr Jun 6 '14 at 12:18
• Not at all. Since both feathers got the same shape the air drag acting in both are the same when both are at the same velocity. The force on the lead is bigger only because it got more mass (becuase it got the same volume but it's more dense) – jean Jun 6 '14 at 12:21
• It has nothing to do with probability – Holloway Jun 6 '14 at 12:26
• @JanHudec Probability makes sense if the object is rotating, therefore the cross-section, hence the drag, change significantly. – auxsvr Jun 6 '14 at 13:25
• @JanHudec I'm fairly confident that the result does change, but this will require further complicated and lengthy analysis. One way to see this is that an asymmetric object will try to reach equilibrium w.r.t. rotation because this configuration minimizes its energy, but this will be unstable in general and the lighter object will be accelerated more due to drag, therefore will reach chaotic rotation before the heavy object, for if the angle exceeds certain threshold, chaos ensues, cf.this article. – auxsvr Jun 6 '14 at 14:22