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Who lifted that balloon up into the stratosphere?
March 13, 2014 7:09 AM   Subscribe

If I tie a 10N weight to a balloon and fill the balloon with helium so that the weight is lifted 100m into the air, then 1kJ of work has been done. Who did that work, and when? Who lost energy?
posted by Galaxor Nebulon to Science & Nature (34 answers total) 4 users marked this as a favorite
 
It's a bit early for me, so this may have a hole in it, as it were, but this was my immediate thought:

Energy isn't lost - the system is just in equilibrium and the energy held as gravitational potential energy while that equilibrium is maintained. The work was done by whatever compressed the helium gas that you filled the balloon with.
posted by Brockles at 7:24 AM on March 13


Cool question!

The energy was lost by the 10N of air that displaced down. You could almost picture it as a 10N sphere of air with a rope looped over a pulley and connected to a 10N lead weight.
posted by ftm at 7:26 AM on March 13 [4 favorites]


Doesn't it happen when you fill the balloon with helium, and displace the air? (So really the same thing ftm said, but at the moment of filling).
posted by defcom1 at 7:37 AM on March 13


Once the balloon is 100 m in the air, if you cut the cord, the gravitational energy will come tumbling down.
posted by Monday, stony Monday at 7:42 AM on March 13


I'd say the work was done when the helium was compressed into the original tank.
posted by robverb at 7:57 AM on March 13 [2 favorites]


Whether the helium was compressed or not should not matter: it will lift regardless because it is lighter than air. Ftm's answer seems correct to me.
posted by Akke at 8:07 AM on March 13


The Helium would have to be compressed or it would not displace the air around the balloon (or stretch the balloon) to fill it.
posted by Brockles at 8:12 AM on March 13


Ftm's answer seems correct to me.

At the point the balloon is already filled, yes. But that wasn't how I read the question and also changes the parameters (because this is a question essentially about the system and where it starts and stops).

Because if the balloon is already filled, the balloon rises because of the removal of whatever force was holding it down a few seconds ago. So you'd have to include that mechanism/person in the equation to work out where the energy 'went'.
posted by Brockles at 8:14 AM on March 13 [2 favorites]


When you're standing there ready to release the balloon, there is potential energy in the system i.e. the balloon wants to go up, or more accurately the air above the balloon would have less gravitational potential energy if it were below the balloon. Gravity is "doing work" or if you think about the whole system (air+balloon), there's no energy going in or out, it's just converting from potential to kinetic.

If you go back farther in time, at some point you added potential energy to the balloon, by filling it full of helium; more accurately, someone harvested that helium so as to create a supply of helium at ground level, or even longer ago, there was a decay of nuclear isotopes in the earth's core that resulted in the formation of pockets of helium below ground, which already has that gravitational potential energy in it.
posted by aimedwander at 8:26 AM on March 13 [5 favorites]


You could have a stretched rubber balloon filled with compressed helium, or you could have a mylar balloon filled with atmospheric pressure helium, it doesn't really matter.

The net force on the balloon as it is "held down" is zero. Releasing the balloon means a non-zero force is applied, but that force is just the buoyant force of the surrounding fluid (air) on the balloon. There is no need to consider the release mechanism.

The buoyant force itself is the result of pressure gradients in the air, so the air is providing the force and doing the work. Or what ftm said.
posted by grog at 8:27 AM on March 13 [2 favorites]


If we start from the point at which the balloon is already filled, I think the work to lift it is done by gravity / the air. Gravity is the force pulling the air molecules down, creating more pressure below the balloon than above. The molecules below the balloon are pushing harder than those above it, so the balloon rises.

As noted above, the air molecules have gravitational potential energy that is transferred to the balloon as they fall and it rises. It might be fair to say that you created that potential energy when you inflated the balloon and displaced an equivalent volume of air molecules upward. That might require leaving the helium out of the system until you introduce it to inflate the balloon, though.

A lot of it does come down to what you include in your system, I think. In a pure earth/balloon/air system, the work is done by the transfer of gravitational potential energy from the air to the balloon. But then where did that energy come from? Well, something displaced the air at some point. But what contributed the energy to displace the air? ... and so on.
posted by whatnotever at 8:39 AM on March 13


Who did that work, and when?

The work was done billions of years ago by a supermassive star which no-longer exists.

The work is getting the helium down to sea level underneath our atmosphere, lifting up the atmosphere up away from earth. If you hold a ball at the bottom of a swimming pool and release it, it floats (water displaces it, the water losing energy), but that state of higher energy - pushing the water up by holding the ball underneath it - comes from the work involved in getting the ball down to the bottom of the pool.So, where did the energy come from to lift up our atmosphere? How did we create a bubble underneath it?

From wikipedia, how do we get helium down to ground level when it floats above our atmosphere:

Most terrestrial helium present today is created by the natural radioactive decay of heavy radioactive elements (thorium and uranium, although there are other examples), as the alpha particles emitted by such decays consist of helium-4 nuclei. This radiogenic helium is trapped with natural gas in concentrations up to 7% by volume, from which it is extracted commercially by a low-temperature separation process called fractional distillation.

This means the work was done in a supernova. The fusion process inside a supermassive star can produce elements as heavy as iron, the heavier elements are created by neutron capture reactions during a supernova. The debris are strewn across the galaxy, including heavy elements. In a cloud of debris and background hydrogen, another star starts to form, and planets around it. One of the planets becomes Earth, and part of its material is some of those heavy elements, which are already decaying back into lighter elements. Energy forcing mass into atoms by a supernova billions of years ago - so much energy and mass that the atoms became unstable - is the energy forcefully injecting helium underneath the atmosphere of earth, lifting the atmosphere up.
posted by anonymisc at 10:47 AM on March 13 [2 favorites]


anoymisc: The source of the helium is irrelevant to this problem.

Here's a thought experiment that will hopefully convince you: Consider a rigid balloon the same final size and weight as the helium inflated one & evacuate the air out of it. What will happen? The answer is that it will float up to the same height that the helium one would.

This ought to tell you that it's displacing the air that matters. In the problem as stated, we're using pressurised helium to do this, but there are other ways to achieve the same end. Hence the work done is the displacement of denser air, either by a vacuum (in my thought experiment) or by helium in the helium balloon.
posted by pharm at 11:07 AM on March 13 [1 favorite]


pharm: Your example confirms that the source of the helium is critical. Each balloon has been evacuated. In one case, the evacuation was energy intensive (vacuum pumps), in the other case, the evacuation was "free" energy - helium gathered elsewhere. The source of the helium is the source of the work.
posted by anonymisc at 11:10 AM on March 13


No anonymisc, the helium does work on the air because someone collected it & compressed it. They could have collected hydrogen instead, compressed that and the end result would have been the same. Or alternatively they could have heated the air in the balloon (hot air being less dense than cold air) and displaced air that way. None of these actions taps the work done by supermassive stars.
posted by pharm at 11:16 AM on March 13


If you stuck an alpha emitter in the balloon and left an open hole in the bottom, helium coming out would accrue and displace the air. If you did it that way, you're using the supernova's energy to displace the air, right?

You might have to make the balloon out of several feet of lead in order to keep the alpha particles from escaping, though.
posted by Galaxor Nebulon at 11:23 AM on March 13


In that case you would indeed be tapping the stored energy of a long past supernova to displace the air. Alpha particles are stopped by a few millimetres of air, so you wouldn't even need the lead lining either. However, you'd need a pretty hefty radioactive source to accumulate enough helium in anything like a reasonable time period!
posted by pharm at 11:39 AM on March 13


None of these actions taps the work done by supermassive stars.

They all use various different sources of energy to create the bubble. Hydrogen bubbles can be released from fossil fuels or created directly with hydrolysis. Hot air can be created with anything from wood burning (solar-powered photosynthesis) to nuclear fission (supermassive stars). With Helium, the bubble is created by the energy of radioactive decay. Whatever power source you want to use, there will be a way to float a balloon with it :)

If you stuck an alpha emitter in the balloon and left an open hole in the bottom, helium coming out would accrue and displace the air. If you did it that way, you're using the supernova's energy to displace the air, right?

That sounds like the geological process that results in deposits of the gas (and natural gas) - the gasses float up through the ground except in geology that traps them under a non-permeable layer, where they accumulate. If a drill opens a hole in that rock, gas rushes out (under even more pressure than usual - it was displacing earth, not just air), and if you had a balloon over the top of the hole, the balloon would fill, no energy input required from you. The helium and natural gas would float, the propane and heavier stuff would float in earth but not in air. It would all be mixed together for a while.
posted by anonymisc at 11:44 AM on March 13


With Helium, the bubble is created by the energy of radioactive decay

If you're getting your helium out of a tank, then no, it's not. Consider a large vat of helium at atmospheric pressure. If you take a pipe and connect one end to the vat & the other to the balloon, will the helium expand the balloon & displace air? Clearly not. Therefore the helium itself isn't the source of the work.

The balloon will only expand if the gas in it is initially under higher pressure than the air and can therefore do work by expanding the balloon. The source of potential energy for that work is the pressure differential between the gas in the tank and the ambient air. What that gas is actually made of is irrelevant, although obviously it needs to be lighter than air at s.t.p. for the balloon to experience net lift after being filled with the stuff.
posted by pharm at 12:53 PM on March 13


[...]then 1kJ of work has been done.

To clarify the numbers, the total work done on the balloon isn't 1kJ, it is 0J.

The instant the balloon is let go, its KEinit=0J. If the balloon has stopping floating upwards at 100m, then KEfinal=0J. Thus deltaKE=0J-0J = 0J, so then by the work-energy theorem, Wnet=deltaKE = 0J. Note: if 100m isn't the top of the balloon's ascent where it stops, and the velocity at that point is something other than 0m/s, then Wnet=.5*m*(vfinal)^2.

The two forces acting on the balloon are buoyant force (upwards, exerted by the fluid/air surrounding the balloon) and gravity (downwards, exerted by the earth).
Work=Fdcos(theta)

d is 100m upwards.

So the work done by gravity is: W=(10N)(100m)cos(180) = -1000J

Wnet = Wgrav +Wbuoy = delta KE
Thus the work done by the buoyant force is 1000J if the final speed is 0m/s. (If it's still moving at the 100m point, then Wbuoy=.5*m*vfinal^2 + 1000J.)

So gravity performs -1kJ of work on the balloon. The fluid surrounding the balloon exerts a buoyant force which performs 1kJ of work on the balloon.



That may help--that way your numbers are in line with the ideas above about work being done to transfer energy from one type/location to another: the potential energy due to the location of the helium stuffed into the balloon vs the air (PE of buoyance) being transferred to KE, and ultimately to gravitational potential energy.


Assumptions/oversimplifications above:
-there are no friction/nonconservation forces (ie no air resistance)
-ignored that Fgrav changes as the balloon rises; just used Fgrav=10N as that was what was given. Else should to integrate to calculate Wgrav --instead of W=Fd, use W=int(Fdx), and use Fgrav = G*Mearth*Mballoon/r^2. Note that Fbuoy also changes as the balloon rises due to changing air density.
posted by neda at 1:05 PM on March 13


An analogous question: an old-fashioned grist mill, driven by a water wheel, grinds wheat into flour. Where does the energy come from? From the fall of the water cleverly arranged to weigh down just one side of the wheel; this energy comes (very indirectly) from the solar energy driving the water cycle, or possibly from some ancient cataclysm that put a high-pressure spring near the top of a mountain.

The proximate answer for the balloon is that the displaced air which falling under the balloon offsets the energy required to lift the load. Deeper than that requires details you don't know.
posted by fantabulous timewaster at 1:45 PM on March 13 [2 favorites]


pharm, this focus on pressure is a red herring and not the source of the work. If the balloon is made of mylar rather than rubber (rubber is not specified), helium in a vat at atmospheric pressure will absolutely float into the balloon, lift it, and carry it away.

Consider: No-matter how much work you to do pressurizing argon and filling the balloon, all that work ain't ever going to make the balloon float.

Now consider that no-matter how little work you do pressuring helium, up to and including zero work (ie letting it seep out of a crack in the ground directly into the mylar balloon resting on the ground), helium will still lift the balloon and carry the weight up into the atmosphere.

The work is a property of the helium. More correctly, the work is a property of whatever got the helium underneath the atmosphere. The work involved in pressure tanks and rubber balloons are just humans losing energy to do storage and transport. Storage and transport are unnecessary and irrelevant to the work of letting helium lift a weight - it will do that right out of the ground, at atmospheric pressure (or right out of the alpha-emitter), no compression or tanks or rubber needed.

Compressors and tanks are a transporting cost, ie an efficiency loss, you are mixing up the energy lost to those unnecessary irrelevant processes with the energy used to do the lifting.
posted by anonymisc at 4:32 PM on March 13


Bubbles of helium diffusing up through the earth and reaching the surface, underneath the atmosphere, are much like bubbles of air diffusing up through lakebed and reaching the lake floor, forming air bubbles on the lake floor underneath the water. The bubbles accrete and get larger until they detach from the lake floor and rise up through the water.

Trapping some of those bubbles in a sunken upside-down cup enables work to be done - enables the sunken cup to be lifted.

In the absence of a free source of bubbles to lift things, a cup could also be raised by heating the water inside it (requiring energy), or by electrolyzing water to create bubbles (requiring energy), etc. But using the naturally-forming bubbles requires no external energy input to manufacture the lift, because the energy was already imparted by whatever is putting air underneath the lake. The bubbles will fill a discarded upside-down cup, displace the water in the cup, and lift the cup. No human interaction at all. All of the cup-lifting work comes from the process that put air under the lake.
posted by anonymisc at 5:15 PM on March 13


All the answers referring to sources of helium or work done compressing helium are not answers to the question that was asked, which is about where the work required to lift a 10N weight 100m came from.

That is made perfectly clear within the question itself, which nominates the amount of energy inquired about as 1kJ. So the question is not about kinetic energy, or the potential energy of compressed helium, or the energy required to put the helium where it was found. This is a straight question about work done to oppose a force over a distance, for which the formula is simply work done = force applied × displacement.

To the question that was actually asked, ftm's answer is correct.

If you surround the experimental setup with an imaginary boundary, and perform the experiment in such a way that negligible energy crosses that boundary in the course of the experiment, it must be the case that the total energy within the boundary before the weight is lifted is the same as that afterward.

At both the beginning and end of the experiment, the balloon+weight subsystem has zero kinetic energy. However, the balloon+weight subsystem has 1kJ more gravitational potential energy at the end than it did at the beginning; therefore, something else within the boundary has 1kJ less energy than it did at the beginning. Since the only things within the boundary are the balloon+weight subsystem and some air, it must be the case that the "something else" is indeed that air.

If the experiment is conducted in such a way that the air within the system boundary is motionless both at the beginning and the end, and has the same temperature at the beginning and the end, then the energy the air transferred to the balloon+weight subsystem must have been gravitational potential energy, basically because there isn't anything else it could have been.
posted by flabdablet at 6:18 PM on March 13 [1 favorite]


Flabdablet, your boundary simplification ignores that the starting state within the boundary is a system not in equilibrium, while the end state is when it has reached equilibrium. The natural state of your bounded system is the weight up high, work was required to force it to the required starting point of the weight being low, out of equilibrium.

Sure, you can start the observations and the story after that work has put the system out of equilibrium, that's entirely legitimate, but I think it is an overall better understanding of the work and the system to understand why the in-equilibrium exists, because that in-equilibrium is what drives the change in question.
posted by anonymisc at 6:56 PM on March 13


A 10 N weight is suspended 100 m in the air by a string. You cut the string and the weight falls to the ground. Same thing.

In the balloon's case, the atmosphere is what's being held up by the balloon. When you let the balloon go, the atmosphere falls back to where it wants to be, below the lighter balloon.
posted by ctmf at 7:09 PM on March 13 [1 favorite]


I'd just like to add that anonymisc is perfectly correct that the helium embodies potential energy simply by virtue of being at ground level & that this source of energy is what is being tapped to loft the balloon.

However it's the fact that the balloon has displaced a larger mass of air upwards that makes the system work as described - it is this air falling down as the balloon rises that forms the other side of the energy balance sheet.

Really there's two steps involved in this question:

1) Displace a mass of air upwards that is greater than the mass of balloon + weight.
2) Allow balloon to rise & air to fall until equilibrium is reached (air is less dense higher up, so at some point the balloon displaces exactly it's own weight of air and stops.)

When the questioner asks who has done work & lost energy, most of us are assuming that this question concerns step (2). anonymisc is arguing about step (1).

I think that's a fair summary?
posted by pharm at 4:33 AM on March 14


Flabdablet, your boundary simplification ignores that the starting state within the boundary is a system not in equilibrium, while the end state is when it has reached equilibrium.

Whether any of the assemblies inside some arbitrary boundary are in equilibrium at any instant doesn't matter even slightly, provided only that the boundary is far enough away from the motion inside it that negligible energy crosses it.

The question is about where the 1kJ of work required to lift a 10N weight by 100m came from. At the instant the weight has risen by 100m, it has 1kJ more gravitational potential energy than it did at its starting height. It may also have some additional energy in the form of kinetic energy, but the question isn't about that. And the helium in the balloon may have gained some additional thermal energy from the surrounding air as it expanded and cooled out of its compression tank, but the question isn't about that either.

The question is specifically about where the 1kJ of work required to moving a 10N weight 100m in the opposite direction to the gravitational force upon it came from, and the answer to that question is that it came from a loss of gravitational potential energy in the air surrounding the balloon+weight system.

The only reason this isn't instantly obvious is that both the mass and the motion of the sinking air are widespread and diffuse, and therefore easily overlooked.
posted by flabdablet at 7:01 AM on March 14


Whether any of the assemblies inside some arbitrary boundary are in equilibrium at any instant doesn't matter even slightly, provided only that the boundary is far enough away from the motion inside it that negligible energy crosses it.

Let me put it another way - you have described a boundary with a giant hole in it through which 1kj energy passes because you are thinking spatially, not also temporally - in your thought experiment, you have only switched on your boundary after the 1kJ has been put into the system (lifting the weight of the air), then you proceed to observe the movement that the 1kj of lifted air creates within the system as it sinks.

If you instead use a boundary through which energy is actually not crossing, neither in space nor earlier in time, then the weight does not move at all because the weight is already floating and everything is already in equilibrium.
posted by anonymisc at 10:20 AM on March 14


I think that's a fair summary?

Agreed. I interpreted the question as "Why is helium a lifting gas? Where did the energy for this strange lifting power come from?". I assumed that step 2 (why do bubbles rise to the surface?) is something the asker probably understands, but isn't applying it because something they can't put their finger on seems missing between that and helium balloons and they keep questioning that (correct) understanding as wrong or incomplete because step 1 isn't in the picture.
posted by anonymisc at 10:40 AM on March 14


in your thought experiment, you have only switched on your boundary after the 1kJ has been put into the system (lifting the weight of the air), then you proceed to observe the movement that the 1kj of lifted air creates within the system as it sinks.

Here is the question again:

If I tie a 10N weight to a balloon and fill the balloon with helium so that the weight is lifted 100m into the air, then 1kJ of work has been done. Who did that work, and when? Who lost energy?

The answer that ftm originally gave, and that I have been attempting to support, is that the air that falls past the balloon+weight subsytem as it rises is what's losing the gravitational potential energy that the balloon+weight subsystem gains. This is the correct answer to the question as posed.

It is also true that the act of filling the balloon with (presumably previously compressed) helium displaces enough air under atmospheric pressure to give the balloon 10+εN of buoyancy, and also adds mechanical potential energy to the balloon itself by stretching its skin. The energy required to perform that work is partly supplied by mechanical potential energy lost by compressed helium expanding out of its tank, and partly supplied by thermal energy from the surrounding air. But that is not the 1kJ of work inquired about in the question: it's some other quantity of energy that you'd need to know the local atmospheric density and pressure, and the elasticity of the balloon, in order to calculate.

So I maintain that it is appropriate to draw the imaginary boundary around the operation being inquired about i.e. the lifting of the weight by 100m, and that the only place it's reasonable to say that the work required to do the lift comes from is gravitational potential energy lost by the air that sinks past the rising balloon.
posted by flabdablet at 11:15 AM on March 14


No-one has been disputing ftm's answer. The air fell down. That is the proximate cause.
What lifted the air up (such that it was able to fall down) is the other half of the balance sheet, and is why helium seems like free energy (unlike hot air). If you think it best to not to consider that half, that's legitimate, as I said. But my suspicion is that this missing part of the bigger picture (free-energy) was part of what was throwing off the asker's instincts, leading to the question.
(compression, stretching rubber, expansion, pressure, mechanical potential energy, stoarge tanks, etc, I've already said that all those things are irrelevant or red herrings; helium will seep out of the ground at atmospheric pressure and lift an open-ended mylar cup.)
posted by anonymisc at 11:45 AM on March 14


As far as I'm concerned, it'd be fun to trace the energy all the way back to the beginning of the universe. The more I know about the history of this energy, the more fun I have.
posted by Galaxor Nebulon at 12:25 PM on March 14


Not really practical, because the further from the interactions you're interested in that you need to put your analysis boundary in order to achieve the condition of negligible energy transfer across it, the more interactions you need to consider inside that boundary; and the mathematics required for accurate modelling of all those interactions rapidly turns impossibly unwieldy. It's like trying to follow the money trail in some insanely complicated laundering scheme, only much much harder.

Even systems as conceptually simple as three solid bodies moving freely under no influence but that of their mutual gravitational field can turn out to be so sensitive to tiny inaccuracies in measurement as to become impossible to do long-term modelling for.

Try to trace the provenance of a kilojoule of energy much further back than the immediate interactions that give rise to it in the context of an atmosphere whose processes work at petajoule to yottajoule scale, and you very quickly reach the point where your kilojoule is completely swamped by tiny measurement errors in the energy surrounding it; it simply becomes indefensible to point to any particular joule and say "yeah, that was a contributor to my 1kJ in a way that this other joule over here was not".
posted by flabdablet at 5:54 PM on March 14 [1 favorite]


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