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Instead of creating vacuum in chambers on earth, why can't vacuum be brought from outer space in chambers? Outer space pressure ranges from $10^{-6}$torr to $10^{-17}$torr very very low. Is it possible at all?

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    $\begingroup$ I don't know if possible, but even if it is, it would be incredibly expensive $\endgroup$ – Hydro Guy Aug 9 '14 at 15:53
  • $\begingroup$ Many of the answers so far are getting hung up about the cost. learner didn't ask what the cost is, and there are plenty of interesting things to be learned by considering this as pure thought experiment. $\endgroup$ – Jess Riedel Aug 9 '14 at 16:19
  • $\begingroup$ I would expect on a non-cost level it remains impractical. You have to maintain the vacuum for the experiment, and for accurate results I would expect you to need good information about what is still left, which can be controlled in a lab but not in space. $\endgroup$ – zibadawa timmy Aug 9 '14 at 16:22
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    $\begingroup$ Actually, you're exporting air, not importing vacuum... $\endgroup$ – DJohnM Aug 9 '14 at 19:01
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    $\begingroup$ If you did bring back a vacuum, what could you do with it? You can't put anything into it without letting air in or creating a separate equal vacuum as a "vacuum-lock". Vacuum chambers on Earth let you put something inside it then make the vacuum. $\endgroup$ – Jim Aug 9 '14 at 19:42
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A lot of things are possible if you want to throw a lot of money at it.

Is it practical? No.

First, I would estimate that a rocket would be thousands of dollars. Low Earth orbit would get you the high range ($10^{-6}$ torr), high Earth orbit would possible get $10^{-9}$ torr. The next difficulty is to have the container returned to Earth (it needs to be shielded from the heat of re-entry) and retrieved from where ever it landed (several thousand more dollars).

Of course, the container needs a good seal since air will be trying to find its way in.

A good vacuum system (about \$2000) can reach $20\cdot 10^{-9}$ torr for a small chamber in about 30 minutes. I have one for the mass spec at work.

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  • $\begingroup$ Thousands of dollars? Think millions, minimum. $\endgroup$ – David Hammen Aug 9 '14 at 16:15
  • $\begingroup$ @DavidHammen For a manned launch, I agree with millions. For an unmanned launch, it shouldn't be more than $100,000 since there are many commercial satellites which wouldn't be up if it was that expensive. $\endgroup$ – LDC3 Aug 9 '14 at 16:19
  • $\begingroup$ @DavidHammen It's about $10000 per kg to get something into orbit. Theoretically, this could be done for under 1 million. And launching it as a secondary payload on a communications satellite would decrease the cost $\endgroup$ – Jim Aug 9 '14 at 16:20
  • $\begingroup$ For manned launches, the correct number is several hundred million dollars. For unmanned, it's in the millions. $\endgroup$ – David Hammen Aug 9 '14 at 16:20
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    $\begingroup$ @JessRiedel I don't have the same research resources and a lot of the final budgets for these return missions are not disclosed to the public by the companies that pay for it, but I think I remember a couple; I'll see what I can find $\endgroup$ – Jim Aug 9 '14 at 18:41
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To make a long story short - Yes, it is possible, but you would have no practical reason to do it.

To make a long story long, read on.

While space is not a perfect vacuum, like you stated, it is close, and many applications of earth-generated vacuums are for simulating the conditions of outer space.

In theory, one could put a pressure vessel in a orbital - or even suborbital - trajectory, have it open to the "vacuum" of space, re-seal it, and then recover the vessel planetside. However, this would be impractical for a number of reasons.

1) Planning, launching, recovering, etc. for a space mission takes months. This is much longer than generating a vacuum with a pump takes.

2) Vacuum Chambers, due to the amount of strength needed to handle the large pressure differential, are very heavy. Launching a rocket is not cheap, and the expenses increase with the increases in weight in the payload.

As far as specific costs go, I'm no expert, so I can only give very broad estimations.

Even if the cost of launching a rocket to obtain a vacuum or purchasing/installing a vacuum chamber, pumps, etc. are about the same (as I imagine they are), the vacuum chamber would be more practical because it can be used almost on-demand (no waiting for a rocket launch) could be reused multiple times (wheras a new rocket would have to be purchased and launched), and would not need to be transported from recovery site to the "site of use", like a launched pressure vessel.

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Probably not. One problem is outgassing:

http://en.wikipedia.org/wiki/Outgassing#Outgassing_in_a_vacuum

Even if you built an airtight container, took it out into space, opened it to let the gas out, sealed it, and brought it back to Earth, you would still have the vacuum destroyed by outgassing. Small atoms trapped in the material of the container itself would constantly be diffusing to the inner surface and ejected into the interior of the container.

To some extent this can be reduced by allowing the materials to outgass for a long time in a vacuum so that the impurities in the container slowly leave. I don't know how long it takes for this to reach a steady state.

(There may be other more serious problems that an expert could tell you. My knowledge about vacuum techniques is very modest.)

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  • $\begingroup$ Let it stay in space for a couple days, that'll solve most of your outgassing problems $\endgroup$ – Jim Aug 9 '14 at 19:39
  • $\begingroup$ And after sealing it, one can cool it down to liquid helium temperatures. That will reduce the vapor pressure of the metal the container is made of. The metal vapor is less of a problem because the atoms will stick to the walls after they collide with it. $\endgroup$ – Count Iblis Aug 9 '14 at 20:11
  • $\begingroup$ @CountIblis sure. Cooling stuff to 0.4K is cheap. Much cheaper than to pump out air of a container. Not to mention that the vacuum of space is literally the worst heat conductor ever. $\endgroup$ – John Dvorak Aug 10 '14 at 9:36
  • $\begingroup$ Isn't outgassing the same problem both on Earth and just above it? The container will evaporate at the same rate no matter what's around it. The only possible difference would be due to heating on reentry, but we have heat shields for that. $\endgroup$ – John Dvorak Aug 10 '14 at 9:58
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    $\begingroup$ There are two different outgassing issues, one is that you have volatiles hiding in cracks in the material, another is the evaporation of the matarial the container is made off. You can deal with the former, more serious problem easier in space. If you get rid of all the foreign atoms and moleculs, you only have the vapor of the material itself, and then the logic behind cold welding applies: en.wikipedia.org/wiki/Cold_welding So, in this case you just cool it down and the atoms will stick to the metal. $\endgroup$ – Count Iblis Aug 10 '14 at 14:46
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Never mind cost in dollars, it costs more in energy. Assume you had perfectly efficient pumps. To create a vacuum you have to push the air out of your chamber, if we assume we want $1 \text{ L}$ of vacuum, this takes $$ W = PV = (1 \text{ atm}) (1 \text{ L}) = 10^9 \text{ erg} $$ of work.

But if you want to get your vacuum from space, even assuming you had a perfectly efficient space elevator, you first have to lift your vessel into space, seal it, and bring it back. On the way down you'll have to do work against the buoyant force of your vessel. This will require work $$ W \sim \rho V g H_0 = (1 \text{ kg/m}^3) ( 1 \text{ L} ) ( 9.8 \text{ m/s}^2 ) ( 9 \text{ km}) \sim 10^9 \text{ erg} $$ Where $H_0$ is the scale height of the atmosphere, that is, the exponential decay with height. We can estimate the scale height by a simple statistical mechanical argument. If the atmosphere were in equilibrium, all of its energy states should be populated according to the boltzmann factor $e^{E/kT}$, so the atmosphere's density should be, roughly, $$ \rho(h) = \rho_0 e^{-m g h}{kT} = \rho_0 e^{-h/H_0} $$ where $mgh$ is the gravitational energy of a nitrogen molecule. This gives $$ H_0 = \frac{ kT}{mg } \sim 9 \text{ km} $$

The order of magnitude estimate comes out a wash (they are inherently the same calculation done two different ways), but they are a wash assuming we had a space elevator. We don't have space elevators, we have rockets, which loose lots of energy to drag. Rockets cost a lot more energy. So "space vacuum" should always come out behind "earth vacuum".

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  • $\begingroup$ You don't need to lift the air with you. You only need to lift the container open, and then bring it back closed. Also, perfectly efficient pumps are an unrealistic approximation. $\endgroup$ – John Dvorak Aug 10 '14 at 9:31
  • $\begingroup$ @JanDvorak good call. My bad. Fixed. $\endgroup$ – alemi Aug 10 '14 at 9:49
  • $\begingroup$ congratulations, you have just proven Archimedes' law for gases, and the rest is just a claim that space transport is much less efficient than air compressors :-) $\endgroup$ – John Dvorak Aug 10 '14 at 9:54
  • $\begingroup$ @JanDvorak I know, but I was hoping to salvage what I had before you burst my bubble (pun intended). And I was telling myself it would illustrative to see it worked both ways. Should I just scrap it instead? $\endgroup$ – alemi Aug 10 '14 at 9:58
  • $\begingroup$ You can keep the calculations around. Not sure if they are helpful, but the vox populi will tell you that :-) $\endgroup$ – John Dvorak Aug 10 '14 at 10:01
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It's physically possible, but the chemical facility maintained a large LN2 tank with an appliance like oil diffusion pump to maintain the vacuum volume. so economically it is a bad idea.

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