As I understand, in a balloon/air balloon, a gas burner is used to heat air or using some lighter atom like helium. Since helium or hot air is lighter and less dense than the cool air around the balloon, the heated air or helium causes the whole balloon to rise. And as I understand, vacuum is a space where there are no atoms inside. So no atom inside means it's lighter than to have atom an inside? Isn't it? So why we can't make something fly by giving it a vacuum container?
[Physics] Balloon, lighter than air and vacuum
airbuoyancyvacuum
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If the volume between the two containers is a true vacuum (by definition no molecules present), taking away the air inside the inner container would have no effect.
The net pressure on the outer container would be atmospheric pressure, since the internal pressure is zero.
If the volume between the two containers is only a partial vacuum, and the inner container was ideally rigid there would still be no effect from removing the air from the innner container. But is the vacuum is only partial and the inner container elastic, removing the air from the inner container would cause the outer container to implode.
If you have a "slack" balloon (one with no elasticity, like is used for some extreme altitude work like the one Felix Baumgartner used for the highest free-fall)
then the pressure inside is the same as the pressure outside, and the balloon will not find an equilibrium position due to pressure (the volume of air displaced will change with altitude, and the weight of air displaced will be unchanged). As Peter Green pointed out, if such balloons reach a "fully inflated" state at a certain altitude, then they will stop rising (reach equilibrium) - assuming they don't burst because of the pressure differential that will build up. For weather balloons this allows for sufficiently fine altitude control; but it is nowhere near the 1-5m range you are asking about.
On the other hand, if you had a perfectly rigid container, then it's conceivable that you will see a height at which the "balloon" would stabilize. It's worth noting how small the pressure gradient is. The approximate equation (ignoring certain large-scale effects) is
$$P=P_0 e^{-mgh/kT}$$
The derivative:
$$\frac{dP}{dh}=-\frac{mg}{kT}P$$
From this it follows that the pressure changes by 1% for a height change of approximately 88 m; if you want to stabilize over a range of 1 - 5 m, you need to have a rigid balloon with a mass that is correct to within 0.05%. But of course the ideal gas law still applies, so a 1C change in temperature will change the density by about 0.3%. This means the temperature has to be known (and stable) to about 0.1C before you can even think about it...
But wait - there's more. The pressure inside an ordinary (toy) balloon drops as the balloon increases in size: this means that a "real" balloon with elastic walls that is initially too light will not only rise - it will keep rising, as the buoyancy actually goes up as the balloon gets higher since the elastic skin becomes less capable of sustaining a pressure difference as the balloon gets bigger. Typically the result is that the balloon will eventually burst - hence the need for a "slack" balloon for high altitude work.
Now if you just have a thin string hanging from the bottom of your balloon, and it drags on the ground, then the additional weight of string that the balloon carries will cause it to find a stable height - even without it being tied.
Final note: indoors, there is typically a bit of air current which significantly affects the motion of a "free" balloon. And "cheap" helium balloons have significant leakage, meaning that a balloon that has the correct buoyancy one moment will be too heavy the next. I once did an experiment where I inflated a balloon and hung a box of matches underneath; I then waited until the balloon-with-matchbox landed, would take out one match, and watch it float off again. It would take about 10 minutes for it to "land", and I repeated the cycle. That gave a neat way of estimating the leakage rate for that particular type of balloon, in "match weights per hour".
Best Answer
I assume you are asking why we are not drawing air out of a balloon like container so as to create the lower density that helium or hot air gives us.
The answer is that it is hard to maintain a vacuum with a thin enough, so as to be almost weightless, rigid contaning surface. A balloon with gas inside equalizing the atmospheric pressure with the gas pressure is easy to make air tight and stable. A vacuum in a thin enough walled container would be liable to collapse, due to the atmospheric pressure, and would be hard to make air tight.