# How does a balloon stay in the air?

Inspired by a parody seen on twitter, and that someone replied that an elementary school teacher would know, I was wondering what the non-magical explanation was.

I can remember a couple of ideas I was taught at school:

Aristotelian: Hot air rises, so the air/less dense gas in the balloon pushes it up. ( Obviously false from a Newtonian perspective - the only forces on the gas in the balloon are gravity and the balloon itself, so if the gas is rising then the balloon must be pushing the gas up. )

Homeopathic: The balloon rises by displacement - the force on it is the difference between its weight and the weight of the air it has displaced. ( How would the air around the balloon remember that there used to be something denser there - does air remember whatever was once placed in it? If you started with a balloon in a vacuum chamber and filled the chamber with air, then would it not float? Something which is no longer present in a system can't affect the behaviour of that system. )

While I have an idea, I'd like to know if there is a common explanation using Newtonian physics for the static case (don't need to go into fluid dynamics unless you want to) rather than the magical ones above, which while empirically true do not explain how a balloon floats in terms of forces and gravity.

If the balloon were a vertical cylinder, taking zero relative pressure as the top of the balloon, the increase in pressure at the bottom on the inside due to the less dense gas is less than the increase in pressure at the bottom due to the heavier gas outside the balloon, so there is a net force on the base of the balloon. Other shaped balloons approximate the same taking only the vertical component of the pressure.

There is one type of forces you didn't take into account : pressure forces.

As the pressure diminishes when the altitude rises, it tends to push more on the bottom of the balloon than on the top. That is Archimedes's principle.

The usual formula is

$$P(z) = P(0)e^{-\frac{z}{h_0}}$$ (One reaches it through fluid statics)

Where $h_0 = \frac{k_b T_0}{mg}$

If the balloon is between $z_0$ and $z_0 +h$, with $h<<h_0$ ($h_0 = 8 \textrm{km}$)

$$P(z) \simeq P(z_0)(1-\frac{z}{z_0})$$

And you can reach through ugly calculation that the resulting force pushes up the balloon, with a strengh that equals the weight of the air it has displaced! To get an idea, you can suppose the balloon is a cube.

As the hot air is less dense, the weigh of the displaced air is higher than the weigh of the air inside the balloon : it floats!

Now if the balloon were in a vacuum chamber it wouldn't float, as there would be no pressure on it!

Edit : the last sentence was wrong : see comments (In fact, if the balloon were not rigid, it would expand and fill the whole vaccuum chamber).

• Sorry, the 'it' was ambiguous - if you filled the chamber with air. In a vacuum, a balloon would expand until the internal pressure and the tensile forces in the skin either balanced the pressure, or exceeded its elastic limit. I think solving that problem was 1st year university engineering maths, but it might have been the year before at school. – Pete Kirkham Jul 23 '17 at 17:37
• You're right, I didn't realise that, thanks! But it wouldn't float, would it? – Cabirto Jul 23 '17 at 18:18