Discussion started in the Pascal’s Wager thread in GD. There is disagreement.
I’m not sure what the original discussion is because I was told I misunderstood, so we’ll just go with the question in the title.
F=ma
ma=mg-drag force
The drag force is not dependent on mass, so mass/weight cannot cancel out and the acceleration of a falling object is dependent on weight.
In the other thread I had said that in the majority of instances weight played a part, somebody else pointed out that the majority of instances were in places that didn’t involve an atmosphere. They’re right about that, I’m wrong about that (I was thinking about the majority of instances in human experience).
So, in the Earth’s atmosphere, does weight matter for falling objects?
If atmospheric drag is a consideration, then weight does matter for the reasons that you state.
If atmospheric drag is a consideration, two spheres with the same volume but different masses will fall at different rates, such as a 1-L solid steel sphere compared to a 1-L hollow steel sphere. These two spheres will also have different terminal velocities.
You are correct. Air resistance depends on the shape of an object, but not its mass. As a result, any calculation for terminal velocity or acceleration will include a mass dependence.
No they won’t. A 1-ft diameter lead ball will fall much faster than a 0.1-inch diameter lead ball, even though they have the same shape and density. Because as you increase the size of the ball, the weight increases as cube of diameter, while air resistance is proportional to frontal area and thus only increases as square of diameter.
It really is about all three factors: weight, shape and size.
You are correct that I was wrong. I should have specified that the shape you choose also matters. That is, if I take a twenty-pound barbell weight and drop it from a a thousand feet; then take two more identical weights, weld them one on top of the other, and drop them from the same height; the forty-pound weight will fall at the same rate as the twenty-pounder, because the frontal area remains the same.
At any rate, it’s not the weight that determines the speed of falling. It’s easy to manipulate an object so that its weight is unchanged but its rate of fall decreases or increases. As I mentioned in the other thread, try dropping the same piece of paper from a height – flat the first time, folded in various ways, and then crumped.
No, it’s about weight. The equation for motion is
ma = mg - drag
where, up to second order, we may write
drag = bv + cv²
with b and c defined as two constants called the linear and quadratic drag coefficients, respectively.
For a sphere,
b = B × D
c = γ × D²
D is the diameter of the sphere.
B and γ depend on the nature of the medium and the object and have a units (force * second / meters squared). For example, for a spherical projectile at STP, they have approximate values
B=1.6 × 10^-4 Ns/m²
γ=0.25 Ns/m²
So we now have
ma = mg - BDv - γD²v²
so the acceleration of the object is
a = (mg - BDv - γD²v²)/m
There is no term that looks like mass/volume here, so density isn’t relevant. There is a mass*gravity term, though, so acceleration is weight-dependent.
No, it depends on both. If your argument were valid I could also say that its easy to manipulate and object so that its shape is unchanged but the rate of fall decreases or increases. Just scale it up to a thousand times the size, but the same shape.
spenczar, I am going to clip most of your post, not to dismiss it, but because it doesn’t address the point I was trying to make. Please don’t take it as me DISMISSING your point, because I am not; your post was excellent and as I look it over quickly, entirely correct.
That’s sort of my point.
When I brought this up (as an analogy) in the original thread, I was trying to get to the idea that weight, by itself, is not the only or even the prime determinant of how quickly an object falls. But it seems to me that shape and density are far more important. Bear in mind also that in the post above yours I was speaking of incrasing the weight of a falling body without changing the rate at which it falls, while changing the shape of it and definitely changing the rate at which it falls.
Obviously a solid sphere of density X and volume Y is going to fall less quickly (in atmosphere) than a solid sphere of density X and volume 2Y, because the latter sphere has more surface area to face air resistance. But that’s only one way of changing the weight of the body in question.
Is it your position that if we change the weight as I descibed earlier, so that the heavier body has the same frontal area as the lighter one, that the heavier body will fall more quickly? If so, please explain further, because I’m not seeing it.
Yes, because the upward force from the air will be the same for both, but the downward force from gravity will be greater for the heavier object. Get massive enough, and you can almost ignore the atmosphere entirely. Compare, for instance, dropping a single sheet of paper with dropping a ream of paper.
Could be true, but it depends strongly on the medium. Look again at the equation for motion, a = (mg - BDv - γD²v²)/m
In a thin atmosphere, you are entirely correct, since in that case B and γ are very small, and the acceleration can be approximated by a≈ mg/m = g.
In a thick medium, though, B and γ are large, so this approximation is no longer valid. Notice, though, that we divide B and γ by m, the mass - so the larger the mass, the less important that atmosphere becomes. This is independent of the shape of the object. The atmospheric parameters are directly divided by the total mass of the object.
Consider dropping a balloon filled with air and a bowling ball. Assume for the sake of argument that both are exactly the same size and shape, so the only difference between the two is their masses. The bowling ball will always fall faster, which actually (for me at least) is intuitively right.
So those are the mathematical and thought-experimenty explanations. The final one I can think of is the physical one: Air resistance is caused by many things physically, but one of the most significant is that the object moving through the air has to push away all the air molecules in its way. Those transfers of momentum cause the falling object to slow down. It loses energy with every collision. In fact, the total amount of momentum lost per second will depend only on the shape of the falling object and its speed, as you might expect.
However, the total velocity that it loses with each collision depends on its mass. An 18-wheeler that hits a mailbox won’t slow down much at all; a six year old on a tricycle will fall to a dead stop. This is because their masses are different - they lose the same amount of momentum, but the 18-wheeler has more total momentum, so it doesn’t lose nearly as much velocity.
The same thing happens with a falling object - a heavier one will lose less velocity with each collision, so it will end up going faster.
First, I’d point out that the equations I wrote above apply only to spherical drag - there are some linear terms that have to do with a thin layer of air that falling objects carry with them, and those would change in an unpredictable way in your example.
But that’s obviously not the point - the point is, “would doubling the weight but changing nothing else have an effect?” and the answer to that is yes, it would.
I realize that you were only talking about spherical drag. That’s why I snipped it above.
:: shrugging ::
My point was never that the weight is irrelevant. It was that it is one factor among several, and not the most important. A hundred-point lead javelin, dropped point first from a thousand feet up, is going to fall a lot faster than a hundred pound lead sphere, no?
Well, in determining which factor is most important, you have to have some way of comparing them. That is, if I make the same amount of change to the shape or to the mass, which one has more effect on the falling speed? Well, how do you measure an amount of change to shape? What counts as the same amount of change?
And my point is that “most important” is not well defined. The difference you are talking about in shape is enormous. That lead sphere is about a foot in diameter, the javelin is maybe half an inch. A 1kg ball of lead is going to fall a lot faster than a 40g sphere made out of balsa wood, no?
