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A helium balloon contains mostly helium which is an inert gas that is less dense than the air around it, suppose I drained all the air molecules from the balloon and note that this special balloon can still maintain the shape as if it is fully inflated.

My question is can this balloon containing nothing rise?

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    $\begingroup$ Think about its weight compared to the surrounding gas it's displacing: if, filled with helium (which has positive, if small, weight), it rises, then what do you think will happen when it's 'filled' with a comparative vacuum (which weighs even less)? $\endgroup$ – Kieran Hunt Apr 29 '15 at 9:25
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If you have an object immersed in air, then you can calculate the forces on it using Archimedes' principle.

There are two forces to consider. Firstly you have the weight of the object, which is simply:

$$ F_g = mg $$

where $m$ is the mass of the object and $g$ is the acceleration due to gravity. This force acts downwards. Secondly you have the bouyant force, which we calculate using Archimedes' principle:

$$ F_b = V\rho g $$

where $V$ is the volume of your object (i.e. the volume of air displaced), $\rho$ is the density of the air and again $g$ is the acceleration due to gravity. The bouyant force acts upwards. The net force acting on the object is calculated by taking the difference between the two forces (we take the difference because the forces act in opposite directions):

$$\begin{align} F_{net} &= F_b - F_g \\& = V\rho g - mg \tag{1} \end{align}$$

I've used the convention that an upwards force is positive, so if $F_{net} \gt 0$ the object will rise and if $F_{net} \lt 0$ the object will fall.

Now consider your experiment. You start with a balloon filled with helium and you know this rises i.e. $F_{net} \gt 0$. Now you take the helium out of the balloon but keep it the same shape. What are you changing? The volume $V$ stays the same, and obviously the density of the air and the acceleration due to gravity are unchanged. So the only thing you are changing is the mass of your balloon $m$, which is reduced. So if you take equation (1) and reduce $m$ you should be able to tell what happens to the net force.

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  • $\begingroup$ I see as long as Fnet is a positive number it will rise, earlier I was thinking about the volume parameter whether to exclude the vacuum space or not anyway thanks. $\endgroup$ – user6760 Apr 29 '15 at 10:26

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