It’s a fundamental tenet of gravity that weight has nothing to do with how fast objects fall, right? Heavy objects do NOT fall faster than light objects. Ignoring air resistance, a hammer and a feather would both fall at the same speed.

In that case – imagine two identical balloons. You inflate one balloon with regular air (not helium) and you fill the other something heavy (maybe water) so that they’re exactly the same shape and size. Now the balloons are the same shape, size, and texture, so it’s not clear why air resistance would affect one more than the other. But if you drop both balloons, the heavy one will fall faster everytime. Why?

At first I thought it was random air currents affecting the lighter balloon more due to inerertia. But that makes no sense cause random air currents are just as likely to push the balloon down as up.

So why does the air balloon fall slower than the water balloon?

The air resistance on each will be exactly the same, as will the resulting upward force. Remember F= MA? Force equals mass times acceleration. A little rearranging gets you A = F/M which shows you clearly that the acceleration due to the upwards force will be greater on the less massive object.

The amount of air resistance is the same, but as a proportion to the balloons’ mass, it is much, much greater on the one filled with air, so it has a bigger effect.

When two objects fall through a medium (usually air or water on Earth) several forces need to be in equilibrium.

Down - Gravity acting on the object
UP - boyancy / resistance - The object has to push the medium out of the way. If the medium is denser than the object, gravity acting on the medium will displace the object instead of the other way around. This is why balloons and ships float.

The difference between these forces is accelerations (F=MA)
So your air ballon essentially pushes the air out of the way more slowly than the water balloon because it is less massive and exerts less force. If it were filled with helium, it would not be massive enough to push the air below it out of the way and instead the air would displace the balloon, pushing it upwards.

Two objects in a vacuum don’t fall at the same rate because the forces on them are equal: They’re not. They fall at the same rate because all of the forces on them are proportional to their mass. The more massive object has a greater gravitational force on it, but it also has more mass to get moving, which cancels out. By contrast, air resistance is not proportional to mass, so it doesn’t cancel out.

>it’s not clear why air resistance would affect one more than the other

I haven’t heard this said quite plainly enough: the issue isn’t whether air resistance would affect one more than the other. It’s whether the air resistance affects either of them enough to matter. Your heavier balloon will be almost completely dominated by weight, whereas your lighter one will have such a small weight that that same air resistance is dominant.

The ideal situation where objects fall at the same rate regardless of their mass and their shape requires that the only factors important enough to matter are the mass (which resists change in velocity) and the weight (which creates change in velocity).

An interesting point, here, by the way - it’s sometimes said that Isaac Newton did not get the second law right, because he said force is mass times acceleration, and in Einsteinian relativity this isn’t correct any more. But Newton was very careful to say that force would equal the time derivative of momentum, and to speculate whether this was equivalent to F=ma without asserting it was. He spoke at some length about the importance of mass, about whether it was always proportional to weight, and about whether it was invariant for an object that was moving and accelerating.

The buoyancy of an air filled balloon is negligible. Buoyancy is defined as the difference in force applied upwards and downwards to an object. The difference in pressure over one foot of altitude is not significant compared to the forces of gravity and air resistance.

The way this was explained to me was that Newton didn’t write F = m * dv/dt. Instead, he wrote F = d (m * v) / dt. I probably got the nomenclature wrong, but it makes more sense to me that way.

If you want to be hyperpicky you could say that the force of gravity is proportional to the products of the two masses falling towards a common center of gravity. So the Moon would fall towards the Earth faster than a rock would fall towards the Earth. But for all terrestrial objects the difference is too small to measure.

>Newton didn’t write F = m * dv/dt. Instead, he wrote F = d (m * v) / dt.

I don’t remember exactly how he wrote his equations and expressions, but as you and I are both saying, he spoke of the time derivative of momentum, which using today’s customs can be written d(m*v)/dt. I think he probably didn’t write it that way, and am guessing Liebniz would have been the first to write it that way, or somebody since Liebniz.

“Observed from an inertial reference frame, the net force on a particle is proportional to the time rate of change of its linear momentum: F = d (mv) / dt.”

“Since the mass of the system is constant, this differential equation can be rewritten in its simpler and more familiar form: F = ma.”

“However, Newton’s laws (combined with Universal Gravitation and Classical Electrodynamics) are inappropriate for use in certain circumstances, most notably at very small scales, very high speeds (in special relativity, the Lorentz factor must be included in the expression for momentum along with rest mass and velocity) or very strong gravitational fields.”

The wikipedia treatment seems a bit unsporting, because it seems they quote Newton accurately, and then make their own claim that mass is invariant, and then point out the incorrectness of their new version of Newton’s treatment. However, I’m not doing whatever it takes to establish that using Newton’s original treatment properly fixes this; Einsteinian relativity was complicated enough I never felt like I knew what I was doing with it. So, they may actually be justified in calling Newton wrong - anybody with more insight on this than I can muster???

That was my initial instinct, but now I’m not so sure.

Here’s a thought experiment:

Suppose you have a balloon of mass M (kiligrams) and volume V (cubic meters) when inflated. The density of air is p (kg per cubic meter). In that case, what is the force due to gravity on a balloon which is inflated?

It would be roughly (M + Vp) * 10 kg m/ s^2

The buoyancy force would be Vp * 10 kg m /s ^2. Subtracting, it would seem taht the total force due to gravity should be M * 10kgm/s^2, which is basically the same as if the balloon were uninflated. (I am ignoring the volume occupied by the balloon.)

So if buoyancy is making the difference, one might expect an inflated balloon to fall at roughly the same rate as an uninflated ballon, which it clearly won’t.

Ok, here’s another thought experiment:

Imagine you have 2 fully inflated balloon of the same weight and volume, except that one of them is one of those long skinny hot-dog shaped balloons, and the other is spherical.

If you point the long balloon straight up and down, will it fall faster than the spherical balloon? My guess is that it will. My recollection is that those long skinny balloons actually fall pretty fast at first until they start to lean over.

If that’s the case, then it’s almost certainly air resistance at work.

The amount of force gravity exerts on an object does depend on it’s mass. That’s what weight is. It just so happens that for a bowling ball and a BB gravity exerts as much force on each to accelerate at the same rate. Essentially gravity will exert enough force on a mass to accelerate everything at the same rate. Something more massive requires more force to accelerate it at the same rate compared to something smaller. Gravity exerts differing forces depending on mass.

Now when you think about wind resistance you have an opposing force. The amount of force imposed on two equal-sized balloons IS the same because they occupy the same shape and space, etc. So the reason why the air-filled balloon falls slower is becuase the ratio of gravitational pull to wind-resistance is much higher than it is for the water-filled balloon. The water filled balloon has more mass, so gets more pull from gravity.

In mechanics the way you figure these things out is you sum up the forces in all directions to get a net overall force. The net overall force considering gravity and wind-resistance is lower with an air-filled balloon.

Now imagining that you could somehow put an air-filled balloon in a vacuum the air resistance is zero so gravity is not counteracted so they fall at the same speed.

Incorrect. The difference in pressure over the height of the balloon is only negligble if the density of air is negligible compared to the density of the balloon, which it clearly isn’t: The air inside the balloon contributes more to its mass than the skin of the balloon does.

For simplicity, let’s consider a “balloon” which is a cube 1 foot on a side. The mass of the balloon is equal to 1 cubic foot times its density, and its weight is its mass times g: W = rho[sub]b[/sub]g(1 foot)[sup]3[/sup]. The difference in pressure between the top and bottom of the balloon is the difference in height times the density of air times g: DeltaP = rho[sub]a[/sub]g1 foot. The force exerted by a pressure is the pressure times the area, and the area of the top or bottom of the balloon is 1 square foot. So F[sub]B[/sub] = rho[sub]a[/sub]g(1 foot)[sup]3[/sup]. In other words, the only difference between the buyonacy force and the weight force is a different average density for the balloon and for air, and a balloon is mostly air, so the densities are close to the same. In other words, the buoyant force is close to being as large as the weight force, and certainly relevant.

Now, as it happens, air resistance is also relevant, which would account for at least some of the difference between balloon-animal balloons and spherical balloons.

Do you agree that if one ignores air resistance, a deflated balloon should fall just just as fast as a balloon full of room temperature air?

In other words, the buyancy of a balloon full of room temperature air is essentially balanced by the extra weight of the air itself, by my calculation above.

So it seems to me that the buyancy force really is negligible here, but not for the reasons expressed by Santo.

Well, it depends on what you call “acceleration.” In Newtonian physics, you usually call it the second derivative of position with respect to time. In relativity, you’d call that “coordinate acceleration”, and it would not work with F=ma. But everyone knows that when you accelerate your car, you feel as if you’re being pushed back into your seat. The acceleration you feel is “proper acceleration,” and it still works with F=ma in relativity. You can proper accelerate forever in relativity, but you will never reach the speed of light because your coordinate acceleration will approach zero.

No, I don’t, but it’s a subtle point. The inflated balloon has more mass of air than the deflated one, but the same mass of latex. Approximating the volume of the latex as 0, then in the two cases:

Inflated balloon:
m[sub]air inside[/sub] = rhoV
m[sub]total[/sub] = rhoV + m[sub]latex[/sub]
W = (rhoV + m[sub]latex[/sub])g
B = rhoVg
F[sub]net[/sub] = B - W = -m[sub]latex[/sub]*g

Deflated balloon:
m[sub]total[/sub] = m[sub]latex[/sub]
W = m[sub]latex[/sub]*g
B = 0
F[sub]net[/sub] = B - W = -m[sub]latex[/sub]*g

So, they have the same force acting on them. But the acceleration is a = F[sub]net[/sub]/m[sub]total[/sub], which will be less for the inflated balloon, since it has more total mass. So even without air resistance, the deflated balloon should hit the ground first.