Do you think he could have possibly found a way to make a feather and a hammer touch ground simultaneously, as they should, according to gravity?

Can you make up a device that avoids air resistance? some ingenious modification to obtain free fall. Can you think of any effective way?

update: I already specified in a comment that conditions must be the normal conditions on Earth: no vacuum chamber, no influence on the two objects.

What do you think: if we hung an open box of the same shape and weight around the two bodies (feather and hammer (or anything)) not touching the feather, and simultaneously let the four bodies fall, what would happen? Could it work up to a certain height?

enter image description here

This is an Atwood machine without pulleys and counterweights. The feather does not touch the box, it is a shield that avoids greater air resistance on the feather. Of course the box must have the same shape and weight of the other body, weight or hammer or other.

I imagined that this could work for heights of just a few metres, do you think it is plausible? Can anyone describe what kind of turbulence would affect the feather?

note: the three bodies are suspended and are released at the same time, the feather does not touch the box, so that all 3 bodies may free-fall regularly without any interference. It was stated above that: 'we must obtain, guarantee free fall', any solution should respect this condition.

  • $\begingroup$ @bobie I am not sure what do you mean by "two massive bodies on pulleys (with appropriate counterweights to take care of friction)" if you add a pulley how would that be free fall, and even if you do, why would they take care of friction? $\endgroup$ – Wolphram jonny Dec 2 '14 at 7:52
  • $\begingroup$ @bobie: Is your ultimate goal to actually make the feather and hammer touch the ground at the same time or just to show that if they are allowed to free-fall, they will be falling at exactly the same acceleration - "side by side"? The latter seems quite easy to do, while the former is difficult, as the hammer and feather will - at the end of the travel - have to move down to the floor of the containers against the air that is inside, which involves drag, or be propped up by the floor throughout the fall, but then you cannot show it was a free-fall. $\endgroup$ – bright magus Dec 5 '14 at 10:58
  • $\begingroup$ @brightmagus, my goal is to ascertain if Galileo could in his age prove that g is the same for2 bodies as different as a feather and a hammer, without cheating: touching, forcing, pushing etc. ,interfering in any way or using modern contraptions such as a vacuum chamber or other $\endgroup$ – bobie Dec 5 '14 at 11:05
  • $\begingroup$ @bobie: So I take it you do not accept a motorized "contraption"? Because the weights in Atwood machine are prone to air drag themselves, so I do not believe they are able to provide a truly uniform acceleration, which we need for our experiment. $\endgroup$ – bright magus Dec 5 '14 at 11:16
  • 1
    $\begingroup$ @JerrySchirmer: Because it's more fun? And more difficult to cheat. :) $\endgroup$ – bright magus Dec 5 '14 at 13:47

Assuming that you want to replicate David Scott's Apollo 15 version of Galileo's Tower of Pisa thought experiment, I think that something like your experiment could indeed work.

As Martin says, the boxes would need to be very low drag shapes: presumably you could make them out of a hollow very heavy metal - natural or depleted uranium would probably make the experiment a practicable cost - and shape them long and thin like a shuttle loom.

I would add a few ideas to Martin's if you are going to get serious. Even though you specify no vacuum chamber, you would need to evacuate the shuttles to rigorously experimentally demonstrate Galileo's freefall law: otherwise you leave yourself open to the doubt that the feather could be being pushed by the air in the chamber, itself compressed by the shuttle as it falls - in practice, of course, this won't happen by dint of Galileo's freefall law: the air inside freefalls too and there is thus no such compression as would happen to the air inside an accelerating train carriage. But you presumably are trying to demonstrate the law as decisively as you can. Also there will be very slight movement of the air as it begins to freefall: before falling, its pressure is ever so slightly higher at the bottom of the shuttle than at the top; upon freefall, this pressure gradient will disappear and beget very tiny draughts inside the vessel.

Without a vacuum chamber, there will be some air resistance on the shuttles, so you need to take steps to minimise this and quantify it. Presumably you could do your experiment at very high altitude, in the Andes or the Himalayas. Also, I would place a laser ranging sensor at the top of the feather bearing shuttle and train it on a relector on the feather. This will tell you if the feather is tending to rise relative to the shuttle as it falls.

As an aside, even though you don't want vacuum chambers, look up Fallturm Bremen. The German Wikipedia page cites a maximum evacuation vacuum of $P_B=1{\rm Pa}$, so you might still need to use your method in something like this tower to replicate the Apollo 15 experiment. The moon truly has zero atmosphere - something like $10^{-9}{\rm Pa}$. Let's see how Fallturm Bremen fares.

If a feather's terminal velocity in natural air with pressure $P_A=100{\rm kPa}$ is of the order of $v_A=1{\rm m\,s^{-1}}$ (I'm guessing here, but its not far off), then its terminal velocity in Fallturm Bremen will be something like $v_B$ where (from comparing the ram pressures on the feather in the two different atmospheres at terminal velocity):

$$1=\frac{\rho_A\,v_A^2}{\rho_B\,v_B^2} = \frac{P_A\,v_A^2}{P_B\,v_B^2}$$

so that I get $v_B = \sqrt{\frac{P_A}{P_B}}\,v_A = 316{\rm m\,s^{-1}}$. Fallturm Bremen allows a 123 metre fall, after which a resistance free object is falling at $50{\rm m\,s^{-1}}$. Thus the ram drag on the feather is one thirty sixth of its weight when it hits the bottom. This is small, but the feather would certainly hit the ground measurably after a hammer hit the bottom (although the difference would be remarkably small): you would need to use your apparatus to get rid of the difference.

  • $\begingroup$ How is falling inside an evacuated shuttle-shaped body different than falling in an evacuated tube... When it finally lands it will not "hit the ground", it will hit the inside of the container! $\endgroup$ – Floris Nov 16 '14 at 13:49
  • $\begingroup$ Is the shuttle closed? Otherwise part of the air from the opening will be sucked up by the air at lesser preasure on the outside , which is moving relative to the shuttle reference frame. Plus turbulence could also help to move thr feather out of the shuttle. $\endgroup$ – Wolphram jonny Dec 2 '14 at 23:54
  • $\begingroup$ @Wolphramjonny Yes, sorry, did I not make that clear? The shuttle must indeed be an airtight vessel. $\endgroup$ – WetSavannaAnimal Dec 2 '14 at 23:56
  • $\begingroup$ Then the op will not like it (I am not sure why), but see his comments on my answer $\endgroup$ – Wolphram jonny Dec 3 '14 at 0:00
  • $\begingroup$ @Wolphramjonny I think the OP is trying to get a grip on the equivalence principle. Indeed you are thinking like me, and I think it is impossible to replicate David Scott's Apollo demonstration here without addressing air drag in some way. The only way I can think of aside from the Fallturm is what you or I are proposing. $\endgroup$ – WetSavannaAnimal Dec 3 '14 at 3:40

Using a big vacuum chamber air resistance is removed. There is a video here.

  • $\begingroup$ I regret I cannot accept your answer as you posted it long after I had specified ' no vacuum chamber'. But you are getting the votes alright! :) $\endgroup$ – bobie Nov 13 '14 at 15:24

For a heavy enough hammer, and not too high distances, the friction from the air will not be too large, because it is a function of speed, so the smaller the speed and the higre the mass the smaller the effect of the air friction. But even if there is some friction this should not be a problem (as I will explain at the end).

If in your experiment the cage is closed, so no air turbulence on the feather from above, everything inside the cage will be on free fall, even the air. The feather will "keep floating" inside the cage wich will fall at the same speed than the hammer (the feather will not be pushed down by the top of the box, in case somebody ask).

In the scenarion where there is some air friction that becomes noticelable, what will happen is that the box will no longer be strictly on free fall, but the feather initially will (I disregard air friction inside the box because it will be minimal). Thus the feather will rest on the floor of the box, actually being stopped by the box, and all, the box, the feather and the hammer wil reach the floor at the same time.

Also in my answer I neglected the buoyancy of air, which sould be minimal on this example, although helium baloons would disagree)

Update: If the box is open, turbulence can contribute, but could be minimized if you use a long enough box. The major problem will be that the lower preasure on top of the box will suck part of the air and most likely take the feather with it. But without exact calculations there no way to tell. The longer the box, and the clser to the bottom the feather, the less strong the effects. –
I would use a low weight (foam?) pipe a few meters long on top of the metal box, I would bet $1000 that the feather stays almost at rest relative to the box

  • $\begingroup$ Turbulence can contribute, but could be minimized if you use a long enough box. The major problem will be that the lower preasure on top of the box will suck part of the air and most likely take the feather with it. BUt without exact calculations there no way to tell. The longer the box, and the clser to the bottom the feather, the less strong the effects. $\endgroup$ – Wolphram jonny Dec 2 '14 at 12:50
  • $\begingroup$ @bobie I would use a low weight (foam?) pipe a few meters long on top of the metal box, I would bet $1000 that the feather stays almost at rest relative to the box. But I dont know if that stisfies your "normal" conditions. What do you mean by "normal conditions", exactly? $\endgroup$ – Wolphram jonny Dec 2 '14 at 13:08
  • $\begingroup$ @bobie I'd love too, but I'm not an expert in fluid dynamics, I can easily make some gross mistake. Plus it is no easy thing, I would take many hours to do a simulation (I am not even sure if an analytical solution is possible, but if it is, it is beyonf my capacity). I suggest you to put that as a specific question. Some expert in fluid dynamics that was not attracted to this question might know better. $\endgroup$ – Wolphram jonny Dec 2 '14 at 13:32

This reminds me of the Apollo 15 mission where a similar experiment was preformed on the moon.

Outside a vacuum, you would have to encase the feather in a reasonably aerodynamic container in order to compensate for air resistance.


It is not practically possible that the feather does not touch the box. I mean, even if you remove all the air inside the box, gravity would pull the feather to the base of the box BEFORE starting the experiment. If you clamp the feather to the center of the box, your experiment would get biased.

I suggest working on the shape of the competing weight (the hammer). Make it face as much EFFECTIVE drag as the feather. You can have a 500 grams mass of iron shaped into an extremely thin foil which, while retaining the mass, would have a far larger surface area and hence the affect of drag on it would be comparative to that faced by the feather.

Plus, instead of a feather (or other extremely low mass/surface area objects), try and use something which at least falls straightly downwards, instead of going in circles and doing a lot of acrobatics in the air due to drag. For such low density objects being pushed from a height does not mean a "free fall" for them. It means "a dance ride to ground level" for them.

Edit to add (because I cannot place a comment):

You want to present equal drag conditions to both the objects (hammer and feather). If vacuum is not a possibility, then the only possibility left is to work on the shape of the hammer to match its mass/surface area value equal to that of the feather. My limited intellect cannot think any other method.

p.s. even if the feather and the iron mass of very large surface area have the same mass/surface area value, still it is highly unlikely that both will touch the ground at the same time. That is, because as stated, for extremely less dense (wrt surface area, not volume) objects, a fall to the ground is not free fall, but a Zulu harvest dance and zigzagging around randomly in the air until they fall down. I mean, if you let two pieces of paper of same mass fall freely from a height more than 10 feet, they would hardly, if ever, touch the ground at the same time. The turbulence during the fall is far too great to calculate or predict. Use something at least as dense as a piece of wood.


As a starting point for my answer I take this comment by bobie under his question:

... my goal is to ascertain if Galileo could in his age prove that g is the same for2 bodies as different as a feather and a hammer, without cheating ...

The problem in our experiment is air. If it wasn't for the air, the experiment would show unequivocally whether a hammer and a feather fall at exactly the same acceleration caused by Earth's gravity.

OK, why air is the problem? It is a problem as there is air surrounding our objects, and this air resists their movement through it. This resistance is a force that is proportional both to the size and to the mass of the body. Now, why does this air "hang in" there? Because it cannot free fall itself, as there is more air under it, and this air below exerts pressure up not allowing air molecules surrounding the feather and the hammer fall down freely themselves.

How can we eliminate the influence of air? We need to make the air surrounding our feather and hammer fall down as freely as they do. Then it will no longer resist the movement of our objects.

Therefore all we need to do is to put the feather and the hammer in two transparent containers, and make both containers move down side by side at exactly $g$. This will remove the pressure of Earth's atmosphere working upon the air which surrounds our objects. If the floor of the containers accelerates down at $g$, the air inside the containers is allowed to fall freely.

There is another important element of the experiment. We need to assure that the feather and the hammer inside the transparent containers do not touch the walls or the floor throughout the experiment. For this purpose we should hang them on strings from the ceiling or prop them up somehow so that before we let the containers move they will be above the floor - in the middle of the container's height, say. At the moment the experiment begins, i.e. when the containers begin to accelerate down, the feather and the hammer must be immediately (with all necessary precision) released, so they can move freely inside.

All we need to do now is to allow the containers accelerate down at $g$ long enough that we can compare (is laser technique allowed?) whether either the feather or the hammer are moving with respect to the container.

If both our objects retain the same distance to the floor of their containers throughout the experiment, it means they are each free-falling at exactly the same acceleration as the containers, which is $g$.

Now, someone might object that the feather - being lighter of the two - is being "propped up" by the air that is inside the container, while for the hammer - the heavier of the two - this same air inside does not constitute the same resistance, and therefore we cannot be sure if both are really falling. The answer is: (1) we expect the hammer to fall faster of the two, therefore if it doesn't (as presumably shown through the experiment), we can assume air is not a factor here, and (2) we are not allowed to remove air from our experiment by the very conditions set by the OP, and so it will always be a suspicion that it alters the results, no matter what we do*.

Note: How do we make the containers move at uniform $g$? 1) We can use some form of the Atwood machine, one that will assure that the acceleration is really $g$. As weights used in the machine are also prone to air drag themselves, we should make them extremely heavy as compared to the total weight of the objects and the containers - heavy enough to make air drag become negligible at least throughout the experiment (i.e. for the speeds achieved by the containers). 2) To be really certain the containers are moving at exactly $g$, we should use some electronically controlled motorized machine that would assure the required smooth movement of the containers.

*I have my doubts that it is at all possible to eliminate the microscopic air movement inside the containers. Even if we make them absolutely air-tight, there is still a chance that the initial sudden movement of the containers down might cause some turbulences. Still, if we are not allowed to evacuate the containers, any proposed experiment will have to account for that.

  1. Make the hammer the same weight as the feather (I know it would be small, but it would be a hammer)
  2. Get a bigger feather.
  3. Scrunch up the feather into a ball (less resistance).
  4. Stick the feather under the hammer.

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