Is acceleration still equal in the presence of air resistance?

[MrMar]

I remember being told that heavier things fall at the same rate as lighter things when everything else is equal. But this guy does a demonstration of a Xenon balloon falling faster than a Neon balloon. Why? Also an overloaded plane may not be able to fly, but it has the same aero dynamics, it is just heavier. What am I missing? Obviously these examples are more complicated than a simple weight difference but I do not know what the mechanism is and I suspect there are different reasons for each example.

 

[etotheipi]

It's Dr Wothers! Legend.

Anyway it's true that heavier things 'fall faster' than lighter things in air. You can try your hand at a few differential equations to see properly, but qualitatively it should be evident that if drag is proportional to some power $v^n$ of speed, then the lighter one will have its weight balanced by drag at a lower speed - i.e. it has a lower terminal speed.

 

[Arjan82]

Hi MrMar,



Different objects of different mass fall at the same rate, but only in a vacuum. In air there is also air resistance. This becomes more and more significant the lighter an object is. The balloons get to terminal velocity really quickly (almost instantly). At terminal velocity the gravity force is exactly equal to the air resistance of the balloon. The air resistance scales with $v^2$ meaning higher velocity means higher drag. But the gravitational force is higher for a heavier object, this means a higher terminal velocity.

An overloaded plane may not be able to get off the ground, because the engines may not be able to overcome the (lift induced) drag of the wings at the required speed. The drag of the wings is directly related to the lift it produces, more lift, more drag.

 

[kuruman]

Have you ever wondered why you can throw a pebble farther away than an air-filled balloon of the same weight?

 

[etotheipi]

Have you ever wondered why you can throw a pebble farther away than an air-filled balloon of the same weight?

Haha, I remember like back in year 1 or something our teacher challenged each group to see who could build the aircraft that went the furthest, given a bunch of random office supplies. And because it was windy out, in a rare moment of pure genius we just smooshed everything into this awful-looking sellotape ball and won by a landslide. Perhaps not quite in the spirit of the event...

 

[Dale]

I remember being told that heavier things fall at the same rate as lighter things when everything else is equal.

It actually takes a little more than that. Everything else must be equal AND there must be no other forces besides gravity. The most common force other than gravity for falling objects is air resistance, so you have to get rid of that. Here is a video that really over-did this idea:

 

[MrMar]

Right that is key. So why do 2 identical balloons fall at different speeds? The air resistance is identical.

 

[MrMar]

It's Dr Wothers! Legend.

Anyway it's true that heavier things 'fall faster' than lighter things in air. You can try your hand at a few differential equations to see properly, but qualitatively it should be evident that if drag is proportional to some power $v^n$ of speed, then the lighter one will have its weight balanced by drag at a lower speed - i.e. it has a lower terminal speed.

Ok, thanks. That helps.

 

[MrMar]

Hi MrMar,



Different objects of different mass fall at the same rate, but only in a vacuum. In air there is also air resistance. This becomes more and more significant the lighter an object is. The balloons get to terminal velocity really quickly (almost instantly). At terminal velocity the gravity force is exactly equal to the air resistance of the balloon. The air resistance scales with $v^2$ meaning higher velocity means higher drag. But the gravitational force is higher for a heavier object, this means a higher terminal velocity.

An overloaded plane may not be able to get off the ground, because the engines may not be able to overcome the (lift induced) drag of the wings at the required speed. The drag of the wings is directly related to the lift it produces, more lift, more drag.

Thanks for the welcome. So I presume the balloons will fall at the same rate on the moon.

 

[MrMar]

Have you ever wondered why you can throw a pebble farther away than an air-filled balloon of the same weight?

No. But if I were to guess, I would say there's more momentum in the pebble than the balloon.

 

[etotheipi]

Right that is key. So why do 2 identical balloons fall at different speeds? The air resistance is identical.

The point is that, at their terminal speeds (which, as @Arjan82 points out, they both reach almost instantly), the forces of air resistance acting on each are not identical. Rather, they're equal in magnitude to the weights of the respective balloons on which they act.

 

[Dale]

Right that is key. So why do 2 identical balloons fall at different speeds? The air resistance is identical.

No, the requirement is not merely for the air resistance to be identical. It is necessary for the air resistance to be absent. There must be no other forces.

 

[hmmm27]

So why do 2 identical balloons fall at different speeds? The air resistance is identical.

Right : at any velocity, both balloons meet exactly the same aerodrag force.

But, the force of the heavy balloon pushes downwards more than that of the lightt balloon. Heavy wins the race.

 

[A.T.]

So why do 2 identical balloons fall at different speeds? The air resistance is identical.

Things fall the same way if the net force (sum of all forces) is proportional to mass:
- Gravity alone is proportional to mass.
- The sum of gravity and air resistance (which is independent of mass) is not proportional to mass.

 

[rcgldr]

Summary::an overloaded plane may not be able to fly, but it has the same aero dynamics, it is just heavier.

The limiting factor is either the engine can't produce enough power, or some structural limit has been exceeded. In the case of gliders, ballast is used to increase speed, but the glide ratio remains about the same, until speeds and/or load factors significantly compress the air (but the glider will still glide).

The fastest gliders are model gliders designed to set dynamic soaring speed records. These are heavily ballasted gliders, and much of the weight is distributed along the wing, and the wing is a one piece wing to avoid structural issues at the 120+ g's the model pulls in the turns at 500+ mph. The record as of this video was 548 mph ~= 882 kph, a bit over mach .7 . Model designed, built, and flown by Spencer Lisenby. It's hard to spot the glider unless you expand the video to full screen.

 

[rcgldr]

Summary:: Heavier things sometimes fall faster than lighter things.

"Sometimes". Consider two objects in space where the only force acting on the objects is the gravitational force between the two objects. Let the mass of each object be noted as M1 and M2. Assume an inertial frame with its origin at the center of mass of the two objects. M1 accelerates towards the common center of mass according to the gravitational field produced by M2, and M2 accelerates towards the common center of mass according to the gravitational field produced by M1.

 

[kuruman]

No. But if I were to guess, I would say there's more momentum in the pebble than the balloon.

Not a very good guess. By assumption, the coin and the pebble have the same weight (mass) and for that reason, it is safe to also assume that your hand moves at the same speed when it releases either object.

 

[hutchphd]

Here is a video that really over-did this idea:

Seems to me Dave Scott's experiment really took it to a new level!

 

[George Jones]

If I remember correctly from decades ago, when my high school physics class went on a field trip to the Toronto Science Centre, we saw a demonstration that seemed to show that light objects fall faster than heavy objects.

 

[Halc]

OK, since we're getting picking about details, if you do the computation to enough significant digits, a dense object will fall faster as will a more massive object.
@rcgldr hinted at this.

Scenario 1 (difference in mass). Assume m2 masses a significant percentage of Earth, like the moon say.
We have a controlled Earth (not spinning, a perfect sphere, no air) all alone in space (so the sun isn't accelerating it for instance) and I have a stationary moon-size rigid balloon at the same altitude as the orbit of the moon. We let go of it and time how long it takes to hit. Now we let go of something the mass of the moon (and stronger to avoid Roche effects) from the same altitude and it will hit the ground in significantly less time. It accelerates more or less at the same rate at first, but gains speed quicker because the Earth is accelerating up towards the massive moon but not the balloon moon. Each travels a different distance because the Earth is not sitting still.

Scenario 2 (density). We put two spheres with the same mass but different density on a plate a km up. We let go and the dense one hits first because its center of mass is lower and that increases the pull of gravity on it due to it being on average at a lower radius. Only in a uniform gravitational field would these two hit at the same time.

 

[berkeman]

Thread closed temporarily for Moderation...

 

[berkeman]

Thread reopened after a side discussion was split off into this thread:

https://www.physicsforums.com/threa...-two-masses-due-to-Newtonian-gravity.1002166/

 

[Elquery]

I came here thinking I knew something, then reading the thread actually confused me. That's good.

The OP's statement in post #7 is really interesting: "So why do 2 identical balloons fall at different speeds? The air resistance is identical."

It posits that the gravitational force must be greater on the heavier object for it to overcome greater resistive (air) force. I haven't really seen a good answer here to explain why there is a larger force.

Which equation should we be looking at to see that the force is larger?

F=G(Mm/r^2) (?)

In that equation, M and m are the two masses, but since the Earth's math is so much larger, a slight increase in m would yield only a miniscule increase in force. Is this the increase in force we are looking for? It seems too small. A different equation?

 

[Ibix]

Which equation should we be looking at to see that the force is larger?

F=G(Mm/r^2) (?)

Yes. Although it's more readable with LaTeX:$$F=\frac{GMm}{r^2}$$Even putting inline LaTeX delimiters around what you wrote would be an improvement: $F=G(Mm/r^2)$ (quote my post to see how it's done or check out the LaTeX guide linked below the reply box).

In that equation, M and m are the two masses, but since the Earth's math is so much larger, a slight increase in m would yield only a miniscule increase in force.

No - the force is proportional to the mass. So if you have a balloon that weighs one gram and another that weighs two grams, the force on the second one is twice the force on the first.

The drag force would behave the same for each balloon if the shape, size, and material of the balloons was identical. So the heavier balloon would have to travel faster for the drag force to balance the larger gravitational force.

 

[sophiecentaur]

F=G(Mm/r^2) (?)

In that equation, M and m are the two masses, but since the Earth's math is so much larger, a slight increase in m would yield only a miniscule increase in force.

The answer to the equation is Proportional to BOTH masses multiplied together. Increase either one by 10% and the result increases by 10%. Try looking at formulae and figure out what those symbols actually involve. Algebra may not always agree with intuition but it's usually (= always) right.

 

[Dale]

In that equation, M and m are the two masses, but since the Earth's math is so much larger, a slight increase in m would yield only a miniscule increase in force. Is this the increase in force we are looking for? It seems too small. A different equation?

That is the correct equation. If you double $m$ then you double $F$. The fact that $M>>m$ has no bearing on that.

Generally, when you have $M>>m$ that means that you can neglect $m$ in terms like $M+m$ or $M-m$, but not in terms like $Mm$ or $M/m$ or $m/M$

 

[hmmm27]

Air resistance pushes up against both balloons equally, at any given velocity.

However, due to mass consideration, the heavier balloon pushes down more against the drag than the light one : $F=ma$ [edit : as pointed out $F=mg$ is more relevant to this]

 

[Frabjous]

I think of it this way. Air resistance is a function of the density of the air, velocity of the ball, cross sectional area of the ball and a drag coefficient. Notice the mass of the ball does not appear, therefore for identical shapes and velocities the F_drag is also identical. The deceleration goes as F_drag/Mass. Therefore the more massive ball experiences less deceleration due to drag.

If you are interested in drag there is a great popularization: Shape and Flow: The Fluid Dynamics of Drag by Shapiro with movies at
http://web.mit.edu/hml/ncfmf.html

 

[Arjan82]

Air resistance pushes up against both balloons equally, at any given velocity.

However, due to mass consideration, the heavier balloon pushes down more against the drag than the light one : $F=ma$

The air resistance is usually given as $D = C_D * 0.5 * \rho * v^2$. This means that $D$, the air resistance, is a function of $v$, the air velocity. Therefore the air resistance for two objects with the same geometry is only equal at equal velocity, and different at different velocities. The question of the OP is why the velocities are different.

Furthermore, $F = mg$ is the more relevant equation, with $g$ the acceleration due to the Earth's gravitational field.

I think of it this way. Air resistance is a function of the density of the air, velocity of the ball, cross sectional area of the ball and a drag coefficient. Notice the mass of the ball does not appear, therefore for identical shapes and velocities the F_drag is also identical. The deceleration goes as F_drag/Mass. Therefore the more massive ball experiences less deceleration due to drag.

We are talking about terminal velocity, where the acceleration (or deceleration) is zero. This is a steady state situation.

 

[Frabjous]

The air resistance is usually given as $D = C_D * 0.5 * \rho * v^2$. This means that $D$, the air resistance, is a function of $v$, the air velocity. Therefore the air resistance for two objects with the same geometry is only equal at equal velocity, and different at different velocities. The question of the OP is why the velocities are different.

Furthermore, $F = mg$ is the more relevant equation, with $g$ the acceleration due to the Earth's gravitational field.We are talking about terminal velocity, where the acceleration (or deceleration) is zero. This is a steady state situation.

The velocities are different because up to the terminal velocity of the lighter object, the deceleration on the more massive body is less than or equal to that of the lighter body.Isn’t is just g=F_drag/M? I assume you could correct M for buoyancy for low density objects.

 

[Elquery]

Yes M and m are factors, of course. Makes perfect sense and a bit of a duh moment there.

I was also heading where hmmm27 went in thinking about $$F={m}{a}$$ (or g instead of a if you want to be technical, sure ) to describe why without another force (air resistance) acceleration would be equal*. Doubling mass doubles force, yet that yields no change in acceleration: $$a=\frac{F}{m}$$

And of course this is essentially what A.T. said in post #14:

Things fall the same way if the net force (sum of all forces) is proportional to mass:
- Gravity alone is proportional to mass.
- The sum of gravity and air resistance (which is independent of mass) is not proportional to mass.

*perhaps there's a question still remaining (for me, and perhaps others): is acceleration still equal in the presence of air-resistance up to the point of terminal velocity? Or does air resistance, being an opposing force to gravity, reduce acceleration slightly right 'from the get-go'?

It seems to me that we would then be discussing force subtraction, so relative magnitudes do matter. I.e. the heavier object has, for example, a downward force of 10, the lighter object 5, and the upward (resistive force due to air) is 2 at a given velocity (who cares about units?). For the heavier object, the net downward force is 10-2=8 (80% of the gravitational force). For the smaller object: 5-2=3 (60% of the gravitational force). Thus the smaller object's net force is further reduced proportional to mass at a given velocity.
Wrong, right, confused?