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Suppose I inflate a balloon on the moon and then let it go. What would

  1. Dec 13, 2009 #1
    Suppose I inflate a balloon on the moon and then let it go. What would happen?

    Would it:

    1) slowly release air and fall to the ground of the moon


    2) release air faster and fly away faster than on earth

    My answer:
    I reasoned that the key to this question is the pressure difference, for the balloon will fly away because of an overpressure, correct? Now because the overpressure is larger than on the moon and because of an absence of air answer 2 is correct? Am I right?
  2. jcsd
  3. Dec 13, 2009 #2


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    Science Advisor

    Re: balloon

    Air will rush out of the open nozzle of the balloon, and the balloon will fly off in the other direction. How fast it goes will depend on the velocity of air coming out. You'll need a little bit less air to inflate the balloon because there's no pressure from outside. I suspect the balloon might be a bit faster on the Moon because there's no air resistance, but I think the exit velocity of the air should be pretty similar to Earth. The pressure difference for a given balloon size should be about the same.

    Once all the air is out, the deflated balloon will follow a parabolic path the same as a rock back to the ground.

    Cheers -- sylas
  4. Dec 13, 2009 #3
    Re: balloon

    Think about Newton's third law
  5. Dec 13, 2009 #4
    Re: balloon

    I believe it will fly away much faster on the moon than on the earth. Air resistance on earth for something as weak and big as a balloon is a huge deal.
  6. Dec 14, 2009 #5
    Re: balloon

    So I'm wrong that the main reason for going faster it the pressure difference? I thought that the reason the balloon releases air faster was due to the pressure difference which is larger on the moon than on earth.
  7. Dec 14, 2009 #6
    Re: balloon

    If you were with Neil Armstrong in 1969 and if you released the baloon after your hatch opened, the baloon would 'blow up' to an enormous size, then take off in a direction that's opposite the baloon's orifice.
    It may also happen that the baloon will blow up to such a large radius that it may burst before it even has a chance to take flight. To find out for sure, determine the pressure the baloon's rubber can withstand, per square inch. Multiply this by the square inch surface area of the baloon. The rubber will have some kind of constant ('rebound constant') that tells you how quickly this rubber can 'snap back' to its original shape after undergoing shear strain, allowing you to determine the rate at which the air will leave the orifice of the baloon for a known orifice diameter. If the rubber can't exhale all of its air while staying below its rebound constant, it will pop. If it can, then it will, as previously stated, become very large, float away while releasing its air, then drop to the ground as a flat baloon when all air is released.
  8. Dec 14, 2009 #7
    Re: balloon

    And if you read my question again you'll find out that I'm inflating he balloon whilst on the moon. So no hatch is involved. But the pressure on the moon itself is much lower than on earth. Apparently I'm dont understand the mechanism involved since I'm keep getting back to the pressure-story.
  9. Dec 14, 2009 #8


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    Re: balloon

    If you inflate the balloon to the same size on the moon as on earth, the pressure difference between inside and outside is exactly the same.
  10. Dec 15, 2009 #9
    Re: balloon

    The pressure differential required to inflate the balloon to the same size on the Moon as on Earth is the same.

    However, there's less mass in the balloon due to 0 ambient pressure. However, there's also no backpressure at the nozzle, just as there is no air friction. Therefore, I've a hunch the balloon will accelerate faster, due to the lower mass and higher exit velocity while retaining the same intermal/external pressure ration, and encountering no air friction, the latter two issues which, each and of themselves would result in greater acceleration, all factors being equal, but, when combined, eclipse the first factor.

    Well, it's just a hunch, but it's quite a good run-on sentence, don't you think? LoL...
  11. Dec 15, 2009 #10


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    Re: balloon

    How about this expriement? You blow up the balloon, then tie off the nozzle. Then you drop the balloon and a hammer from the same height at the same time. On the moon, they both land at the same time, on the earth, the large amount of drag on the balloon causes it to reach a slow terminal velocity in a matter of inches from the release point, and the hammer lands first.

    Back to the OP, the balloon released with an untied nozzle on the moon is going to take off like a small rocket because there's no aerodynamic drag.
  12. Dec 16, 2009 #11
    Re: balloon

    And what makes you think Neil Armstrong wasn't on the moon when he opened the hatch, stepped on the surface of... Ummm... that huge thing in the nighttime sky that gives us the tides, and said that it was "One small step for man..."?

    Anyway. A baloon that's inflated on the moon and in its vacuum will have the same pressure as one on the earth, provided they are both inflated to the same volume.
    Snce it takes far fewer air molecules to inflate the baloon on the moon (provided it's being blown-up in vacuo) to the same diameter as that baloon that's on the earth, the baloon on the moon will have far less 'fuel' (air) for it to scoot along.
    The moon's lower gravity does not have a significant impact on the length the baloon will travel.
    The absence of an atmosphere on the moon can also be mitigated by using a narrow baloon, meaning we can lower the impact of atmospheric drag for the baloon that's released on earth.

    Therefore, the baloon on the earth will travel considerably farther than the one on the moon.
  13. Dec 17, 2009 #12
    Re: balloon

    Well...I do think neil was there.

    Ok, so this agrees what other people say in this thread, the pressure difference is the same.

    So the mass of the balloon is smaller than on earth given that both balloons have the same volume?

    So gravity doesn't has much influence because the mass difference is small but the air resistance is the key factor, right?
  14. Dec 17, 2009 #13
    Re: balloon

    No, the mass of the baloon remains unchanged. The mass of the air in the baloon + the mass of the baloon itself is less on the moon than it is on earth, for baloons blown up to equal volumes.
    Correct. It's all dependent on applied materials. Compare a baloon made with Rubber X which is a 'space-age' rubber that's both feather-light and robust, with Rubber Y which is just your ordinary vulcanized rubber--the stuff of bicycle inner tubes. Both can be used in our earth/moon experiment, but for the former (Rubber X), gravity isn't a factor in propultion distance, whereas gravity is a factor in propultion distance with Rubber Y.
    Air resistance is a significant factor because at sea-level, there's so much compressed air that we can do things like fly helicopters and other aircraft.
  15. Dec 29, 2009 #14
    Re: balloon

    Ok, the answers are in it seems that Neo anderson explanation was the only and correct answer!
  16. Dec 29, 2009 #15


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    Homework Helper

    Re: balloon

    This ignores the fact that the atmosphere greatly reduces the effective exhaust velocity of the air expelled out the nozzle of the balloon (compared to the moon case), which greatly reduces the specific impulse (impulse per unit of mass expelled). Also the effect on terminal velocity of the greater mass is related to the log of the mass ratios, while directly related to the effective exhaust velocity.

    Even if the narrow balloon was housed in a non-deforming, very low drag, container to reduce the drag, it's my guess that the reduction of specific impulse at sea level is larger than the reduction of mass of the air contained by the balloon on the moon, and so the balloon on the moon achieves a higher terminal velocity when it's air runs out. In formula form:

    VE = effective exhaust velocity
    M0 = initial mass of balloon and air
    M1 = final mass of ballon after air expelled
    terminal velocity = VE ln(M0/M1)

    VEmoon ln(M0moon/M1) > VEearth ln(M0earth/M1)


    Last edited: Dec 30, 2009
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