Because the density of the balloon when filled with helium is lower than the atmosphere surrounding it by a good bit. The same could be said for a balloon filled with carbon dioxide underwater (you know if you could blow a balloon up under water and release it. It would float to the top rather quickly and then float on top of the water. Or release an ice cube from your freezer underwater, it will float to the top since it contains air pockets, its density is less than the surrounding water and it will float up.

So yes, gravity on such a small mass when the density is much lower would not overcome the tendency to float to the lower density area.

Mikey T.

I understand what you are telling me. I've seen balloons float away and ice cubes rise to the surface. But could you kindly work out the formula for me and show me the results of the earth's gravity pulling on a helium filled balloon that is 10" diameter. I haven't done much math for a long time. If you don't want to that's fine.

Thanks,

Yendor

Since I am at work I will not be able to do the math for you right now but the mass of the balloon vs the buoyancy in the air is what you need. It depends on those factors. Just saying a 10" balloon filled with helium really isn't enough per say. If you want a close approximation of the forces at work you would need to know the mass of the empty balloon and the mass of the helium inside once filled to an appropriate level. This would give you enough, with the mass of the Earth numbers to calculate the force of gravity on the balloon. You would then need the buoyancy calculations for that volume of helium for air at sea level. It's a bit of calculations and some research to do, so good luck and if I get a chance I will do the same and we can compare numbers.

This is easy. You can actually calculate the buoyant force from scratch, but I'll skip that and say that Fb=d*V*g, where d is the density of the fluid the object is immersed in, V is the volume displaced by said object and g is the gravitational constant. In this case, the only forces at work are gravity and buoyancy, so for an object to remain stationary (neither sink nor rise), the two must be equal

mg=d*V*g -> m=V*d. Any object with a mass greater than the mass of air for a same volume will sink, otherwise it will float. Let's take the example of the balloon (I will work in the metric system now, without changing the numbers; the results won't change). He has a density of 0.164 Kg/m

^{3}, and let's consider the balloon a sphere of radius 10cm.

Volume of the sphere=4/3 Pi*r

^{3}=4188 cm

^{3}. To get the mass of the He, just multiply it by the density (carefulwith the units) and you will get 0.686g. Add that to the mass of the balloon itself (let's say 2g), and the total mass is 2.686g

Now for the mass of air: at sea level and normal temperature, air has a density of 1.2 kg/m

^{3}. Multiply that by the volume and you get 5.02g

Since 5.02>2.686, the buoyant force is greater than the gravitational force, and the balloon will rise. The density of air is not constant with altitude, however, so at some point the two will equalize (if the rubber resists) and the balloon will be stationary.