Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

My son has asked me a question, and I'm afraid my high school maths are not up to the task:

Is it possible to calculate the height that a given balloon (filled with a known quantity of helium) will attain when released, assuming it won't burst before?

My line of reasoning is as follows (and may already be flawed, so please point out my errors):

  • The balloon will ascend as long as the mass of the air that it displaces is heavier than the mass of the balloon. It will stop when both these masses are equal, so that's what we need to find out.
  • We know the mass of the balloon and assume that it will remain constant throughout (no leaks).
  • We know the air pressure on the ground. For simplification, we can assume that air pressure drops linearly (?) when ascending (no inversion weather etc.), so we can calculate the air pressure at any given height.
  • The effects of gravity on the balloon are reduced the farther up it goes.
  • The balloon expands when ascending because of the reduced air pressure, thereby displacing more air (which adds to its buoyancy), but this air is lighter (which takes away buoyancy). <-- This part is giving me the most trouble.

Are there any other factors that need to be taken into account? Shouldn't it be possible to calculate a height where equilibrium is reached and the height of the balloon remains stable?

Is it as complicated as it looks to me?

share|improve this question
add comment

1 Answer 1

I guess you have mentioned all significant factors. Just a few comments.

Gravity reduction with altitude is not significant for altitudes accessible to balloons.

Pressure is a rather complex function of altitude (http://en.wikipedia.org/wiki/Barometric_formula).

You should take into account the elasticity of the balloon envelope to calculate the balloon volume as a function of altitude.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.