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Lift of a helium spherical balloon

  1. May 6, 2013 #1
    So I was pondering this question: On a conceptual level, how does a perfectly spherical helium balloon rise?

    I understand that the density of helium gas is lower than that of our atmospheric composition of gases, but that is not giving the full perspective for me. On a molecular level, I feel like it WOULDN'T rise (despite my experiences in person). My logic behind this is:

    -A perfect spherical balloon has no net gas collisions with its container, as they are happening in all directions with equal force.
    -There is a pressure disparity between the helium and oxygen, but the atmosphere would be pushing equally over the entire surface of the balloon. No net pressure would imply that lift is not possible
    -Gravity is the only outside force that interacts with the spherical balloon (with no net pressure/molecular collisions). Thus, there is only a net downward force.

    I'm positive that the gas density disparity between helium and our atmosphere relates to this, but how does that work on a molecular scale?
  2. jcsd
  3. May 6, 2013 #2
    Clearly this is not true. The higher you go the lower the pressure is. You experience this when you climb a mountain. There is more pressure on the bottom than the top. You experience this when you dive into a deep pool. This is the origin of buoyancy.


    edit - maybe this picture is helpful too,
  4. May 6, 2013 #3
    Ah, thank you so much. That link really helped me.

    But theoretically, in a bizarre world where there is no subtle pressure disparity between the top and bottom of a balloon, it would not rise?
  5. May 6, 2013 #4


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    Balloons rise not because of pressure but due to the difference in the weight of the helium inside the balloon envelope and the weight of air displaced by the balloon.

    Ultimately, balloons rise due to Archimedes Principle.
  6. May 6, 2013 #5
    With no pressure disparity what direction would you define as "rise"?

    In a space craft either orbiting or far from a gravitational field a balloon would not rise, as you deduced. Gravity creates the pressure gradient which gives us buoyancy. No pressure gradient, no buoyancy.
  7. May 7, 2013 #6


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    Hey cool, it's an example of loss of accuracy when subtracting two numbers that are very nearly the same. The pressure at the top of the balloon and the bottom are very nearly the same. The balloon can do that subtraction very accurately indeed, getting the net difference in up and down forces.

    Computer programmers are familiar with this subtraction thing. Floating point numbers in computers have finite accuracy. So if you have two numbers that are different by only a very small amount, and you subtract them, you lose accuracy. If you wind up doing that a few times you can get results that are pretty much of no accuracy at all.
  8. May 8, 2013 #7


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    Buoyancy is caused by a pressure gradient. If there is no pressure gradient in the displaced fluid, there will be no buoyancy.
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