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Buoyancy of a vacuum-filled "space tower"?

  1. Nov 12, 2014 #1
    Hello everyone,
    I am thinking of a thought experiment about a so-called "space tower" but don't have enough information to estimate the numbers. I understand the material strength needed for such high structures is huge, because of the weight of the structure. But what if the "tower" would be vacuumed and would have an open end extending to a near-vacuum height "above" atmosphere, while the bottom would be tightly closed (access via airlock)? Actually that would be more like a very long tube or gun barrel maybe, that could contain an electromagnetically propelled elevator (like a vertical maglev).
    1. How tall should the tower be so that the pressure inside could be easily kept constant (vacuum) using pumps? I am not aiming for perfect vacuum but if (I suppose) the decrease in atmospheric pressure with height is not linear, a compromise could be found so that the tower is tall enough so that it could be easily kept vacuumed using pumps but also not too tall so that its weight would be kept as low as possible...
    2. Because the tower is "full of vacuum" (pardon my expression) and would be displacing a huge mass of air (even if the upper atmosphere is much less dense) would the buoyancy force help reducing its weight and consequently would reduce the needed material strength? Does it matter that the top end is open?
    3. If instead of having a foundation, the tower would be supported by huge pillars, would that make any difference regarding buoyancy?
    Thanks for any input!
     
  2. jcsd
  3. Nov 12, 2014 #2

    Nugatory

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    Staff: Mentor

    The weight is not the problem. The enormous material strength is needed to resist the tidal forces within such a hypothetical structure.
     
  4. Nov 12, 2014 #3
    But leaving buoyancy aside (I've done some calculations below and the results are not good) and the stability of the tower, would this "vacuum" system work? I mean an open end in space (40 km tall, 0.4 kPa atmospheric pressure), a tightly closed end at ground level, and the pressure is equalized with that from above (at 0.4 kPa) and should stay constant "forever", right? (with a little help from some pumps)...
    Like a hugely tall glass (or, if you wish, a laboratory graduated cylinder) submersed in water but with its top outside, below the water level. It would have inside the same pressure as the one just above water level...

    You mean the tidal effect... caused by the moon, right? But it's not like the tower is really floating... And also is not THAT close to the moon (360.000 km distance iirc?). I was thinking more like 40 km tall. In which way would the tower influenced? Might the tidal force rip it appart?

    The atmospheric pressure is about 0.4 kPa at 40 km altitude and 248 K, which I think is considered "medium vacuum". If the tower (or tube) would have this height, let's say it's vacuum enough for my thought experiment. The pressure is 250 times lower than sea level so I assume 250 lower drag for my maglev elevator.
    I read somewhere that the 1km tower would in Saudi Arabia woudl weight about 1 million tons (with a section area of 319,000 m^2). If using a smaller diameter (like 10,000 m^2 section) for the "space tower" let's assume we can half this weight per km. For 40km we'd have a total weight of 20 million tons (2x10^10 kg).
    The displaced volume would be 40,000 m x 10,000 m^2 = 400,000,000 m^3
    If I could integrate the specific weight of the air bewteen 0 and 40 km altitude...
    Looking at this chart on wikipedia, I would guesstimate since the variation is not linear, a mean value would be closer to the sea level value (actually I could count the little squares). Let's say a density of 0.4 kg/m^3.
    So the weight of the displaced air column would be 160,000,000 kg. For the sake of simplicity, I'm rounding this to 2x10^8 kg.
    That's a bummer...
    So we could only reduce the weight to an apparent weight smaller by only 1%....
    If we double the cross section area and keep the same weight (hey we use some high tech materials there)...
    Then we get double displaced volume, double buoyancy... 2% reduction in weight.
    Looking again at the density chart and counting the little squares... the 5 km line would be a better aproximation (about 9 square above and 9 below) so a more correct average density would be ... (still with me?)... 0.8 kg/m^3 ...
    Thats double the density initially assumed, so double the displaced weight... double the buoyancy.
    4% reduction in weight....
    Thats all I can do with the current data.
    I'm not sure what material would such a tube need to be able to keep the vacuum and sustain its weight...
     
  5. Nov 12, 2014 #4

    russ_watters

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    The easy way to do this problem starts with sea level pressure: 101,000 newtons per square meter. That's also the buoyancy of your tower (per square meter of cross section).
     
  6. Nov 12, 2014 #5

    Nugatory

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    No, this has absolutely nothing to do with moon, and tidal effects are a far more general phenomenon. Any time that you have an object large enough that the gravitational forces on one end are different from the gravitational forces on the other end, the two ends will be pulled together or apart, and only the structural strength of the object stops this from happening.

    Consider the top of the tower for a moment. It remains in a fixed position relative to the base of the tower, right? That means that it is moving with whatever speed and centripetal acceleration keeps it in that position relative to the base, which is moving with the rotation of the earth. Compare that speed and acceleration with the speed and acceleration of a satellite orbiting at the height of the tower; that's the free-fall trajectory along which the the top of the tower would move if it were not constrained by the forces in the structure that are holding the top of the tower to the base. The required force is enormous.
     
  7. Nov 12, 2014 #6

    mfb

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    It does matter, air would flow in and within a short timescale you have the same pressure inside and outside (with the same pressure gradients as outside). You could simply close the top.

    Instead of vacuum, you can use helium. ~87% of the effect, 0% of stability issues with vacuum. It is certainly possible to construct a lightweight structure (you save a lot of material if it does not have to support its own weight) where buocancy is sufficient.
    What do you do with an extremely lightweight tower 40km tall?

    Gravity would be a bit weaker at the top of the tower. So what? That is helping construction here (but it is an effect of 1%).
     
  8. Nov 12, 2014 #7

    NTW

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    There would be no tangential, shearing forces acting on the tower; only compressive forces on the structure due to gravity, and tensile stresses due to centrifugal force, both forces canceling out at the geostationary altitude. Special materials would be needed for its construction, and it's true that those ultra-resistant materials do not exist at the present time, but may be discovered in the future...
     
  9. Nov 13, 2014 #8
    Thank you all for your replies.

    At 40km altitude, it's almost vacuum. Yes, the air will flow in but the pressure would be 0.4 kPa.
    I was not sure if the buoyancy would apply though, if the system is open at top. But then if you submerge a tall glass in water with its open end above, the buoyancy indeed works and we have a similar situation.
    But as estimated above, the apparent weight would be only 4% lower (unless very low-density high-strength materials would be used).

    Steel has an average density of 7800 kg / m3 (depending on composition) while carbon nano-tubes have been synthesized as low as 1600 kg / m3.
    That's almost 5 times lower. If this kind of material could be used, the weight of the structure would be 5 times lower so the apparent weight would be reduced by 20% for the same displaced air mass. Now we are approaching some more visible results.

    Fill it with an electromagnetically-propelled elevator :D

    Regards,
    Paul
     
    Last edited: Nov 13, 2014
  10. Nov 13, 2014 #9

    NTW

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    Unless you put a lid on your tube, it will slowly fill up with the cascade of air entering at the top. And, once equilibrium is restored, the pressure distribution of the air inside the tube will be the same as outside.
     
  11. Nov 13, 2014 #10
    Damn! I had a feeling something is not right! Of course, I can't have the same 0.4 kPa throughout the whole tube length because the air would compress due to gravity in the lower part just like in atmosphere.
    Then I put a lid and access is done by airlocks.

    Someone above mentioned geostationary... but that is over 35.000 km while my tower is only 40 km. Not even low earth orbit.
    I was thinking 40km that would be a good place to launch space shuttles using some other thrusting method.
    But since Earth radius is 6371 km, that's not even 1% reduction in distance to Earth's center.
    Since gravity is inverse proportional to the square of distance, the gravity reduction is insignifiant.

    The only use of a 40km tall vacuumed tube would be to have a starting place for a space shuttle without needing to overcome so much air drag.
    If the air density is 252 times lower (101 kPa / 0.4 kPa) the air drag should be also 252 times lower... Not sure if the costs of the tower worth the difference.
     
  12. Nov 15, 2014 #11

    mfb

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    There would be wind.
    Carbon nanotubes (if we can manufacture them in a proper way) would be good for tensile strength, but they don't resist compression enough.

    The tower would basically have to be a 40km high balloon.
    Yes air drag is an annoying thing of spacecraft launches, but getting the required horizontal speed needs much more fuel. And I don't see how an extremely fragile tower would support a massive rocket including its launch.
     
  13. Nov 15, 2014 #12

    NTW

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    Of course, wind forces should be allowed for, but I was just making reference to centrifugal and gravitational forces.
     
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