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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?

Simd
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    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 06 '14 at 11:09
  • Also, lighter objects can be carried away by wind currents and turbulence more easily. – auxsvr Jun 06 '14 at 11:12
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    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 06 '14 at 12:11
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    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 07 '14 at 11:13
  • The answer couldn't be simpler. The air currents and air resistance are a force ("F"). As you know F=ma. It's just that simple. – Fattie Jun 08 '14 at 09:41
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    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 08 '14 at 09:52

5 Answers5

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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.

Alfred Centauri
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    Force is a vector quantity, if F comes out to be negative in sign. Then, duck's feather falls faster than lead feather! – Sensebe Jun 06 '14 at 11:44
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    @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 06 '14 at 11:57
  • What would you like to say about atmospheric pressure? Force is acting downwards, isn't it? then, I think the lead feather must accelerate towards the earth to a less extent than the duck feather (considering mass per unit area doesn't affect much). Thus, duck feather should fall first than lead feather. – Sensebe Jun 06 '14 at 12:05
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    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 06 '14 at 12:24
  • @jean: Have you seen Torricelli's experiment? – Sensebe Jun 06 '14 at 12:35
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    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 06 '14 at 12:41
  • @jean: Dear friend, you didn't answer my previous question. I understand you have a point. Its important to proceed the discussion in a systematic manner. – Sensebe Jun 06 '14 at 12:49
  • @Godparticle: Force is indeed vector quantity. A drag force, that is discussed here, is always opposite direction from the velocity of the object relative to the fluid. So in still air, the drag of falling object is always up. If the air was moving down faster than the object, then the drag force would be directed down and the feather would indeed accelerate faster. – Jan Hudec Jun 06 '14 at 12:55
  • @Godparticle: Atmospheric pressure only comes into question by creating buoyancy due to difference of pressure on the top and bottom surface of the object. Since pressure increases with depth, buoyancy always acts upwards. But even for feather buoyancy is still considerably smaller than drag. – Jan Hudec Jun 06 '14 at 12:59
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    @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 06 '14 at 15:34
  • @AlfredCentauri I know it may be obvious to many, but you should mention that the drag force doesn't depend on the mass; that assumption is key to your conclusion. – jub0bs Jun 07 '14 at 15:13
  • While this is sort of correct I'm very surprised at how, let's say, "inelegant" the answers are here to such a simple question. The air, the shape of object, and extremely subtle engineering concepts relating to turbulence and laminar flow ......... CAUSE FORCES. It's just that simple. Then, F=ma and you're done. Amazing nobody has pointed this out. – Fattie Jun 08 '14 at 09:43
  • The nature of the forces on a feather tumbling, are staggeringly complicated, and the subject of the most cutting-edge research. Indeed, it would not be an exaggeration to say "the nature of the forces on a tumbling feather are tremendously beyond current physical and engineering understanding." However, they're just "F" forces and "F=ma". – Fattie Jun 08 '14 at 09:46
  • At root, the question is as trivial as asking - say, for example - "Why does a light car with engine X, get around more quickly than a heavy car with engine X?" The answer is just "oh, observe that this is simply an F=ma issue..." – Fattie Jun 08 '14 at 09:47
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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

auxsvr
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  • 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 06 '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 06 '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 06 '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 06 '14 at 13:53
  • @auxsvr: The result is correct. But it is expressions for $v(t)$ and $lim_{t\to\infty} v(t)$. So the integral should also be expression for $v(t)$, not $t$. – Jan Hudec Jun 06 '14 at 14:00
  • @JanHudec when integrating over speed as an independent variable then the limits have to be in speeds. You are answering the question, how long does it take to move from speed 1 to speed 2? Why is there confusion on this issue? $\int_0^v \ldots {\rm d}u$ is the proper form of the integral. – John Alexiou Jun 06 '14 at 14:03
  • @ja72: But the results are answering the question what is the speed after time $t$. – Jan Hudec Jun 06 '14 at 14:18
  • So after integration the resulting function needs inversion. – John Alexiou Jun 06 '14 at 14:47
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    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 08 '14 at 09:50
  • Finally a decent answer! Thank you. – 52d6c6af Jun 19 '23 at 09:59
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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.

jean
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    @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 06 '14 at 13:04
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    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 06 '14 at 13:05
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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. enter image description here

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!

Community
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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.

Downward force should not be neglected considering atmospheric pressure, air pressure can hold up a $10m$ high column of water (Torricelli's law). Quite a tall glass! Click here to see the video.$_1$

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

Sensebe
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    Why the downvote? This is too vague to be useful, but it isn't wrong. – auxsvr Jun 06 '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 06 '14 at 12:21
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    It has nothing to do with probability – Holloway Jun 06 '14 at 12:26
  • I am joining the downvote. This does not answer the question. – Jan Hudec Jun 06 '14 at 12:51
  • @JanHudec: I think I have answered the question. If you can explain where you need explanation, I can try to explain. – Sensebe Jun 06 '14 at 13:00
  • You have not mentioned either drag nor buoyancy, both of which can be easily quantified. And the talk about probability just does not make sense. Probability is not involved; the example in the question is feather and a lead (forged into the same shape) falling in still air from zero velocity compared to the ground and the outcome is perfectly deterministic there. – Jan Hudec Jun 06 '14 at 13:08
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    @JanHudec Probability makes sense if the object is rotating, therefore the cross-section, hence the drag, change significantly. – auxsvr Jun 06 '14 at 13:25
  • @auxsvr: Rotating the object won't change the fact that drag always acts in opposite direction from velocity and buoyancy always acts in opposite direction of gravity. So the acceleration will differ, but not the qualitative result. – Jan Hudec Jun 06 '14 at 13:30
  • Whether probability is one, zero or in between them, it makes sense everywhere. – Sensebe Jun 06 '14 at 14:17
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    @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 06 '14 at 14:22
  • -1 because, as it is, this is too vague to be useful – Jim Jun 06 '14 at 14:26