This may be silly question, but why does a helium ballon rise? I know it rises because helium is less dense than air. But what about the material of the ballon. It is made up of rubber/latex which is quite denser than air. An empty ballon with no air in it falls, so why does a helium filled balloon rise?

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    $\begingroup$ The average density of the balloon filled with helium is smaller than the density of air. When the balloon is filled, it normally has a very large volume of helium compared to the volume of the rubber. $\endgroup$
    – WarreG
    May 30, 2019 at 9:59
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    $\begingroup$ Have you thought why ships float on water even being made of heavy metals $\endgroup$
    – Tojra
    May 30, 2019 at 15:11
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    $\begingroup$ @DavidWhite - Seems like just another way to say the same thing. $\endgroup$
    – Glen Yates
    May 30, 2019 at 18:26
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    $\begingroup$ Does an empty balloon really have no air in it? Are you suggesting that in an empty balloon, there is the same vacuum as we'd find in outer space? Or do you think there is some air in there? $\endgroup$ May 30, 2019 at 19:37
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    $\begingroup$ @EricLippert - Yes, an empty balloon has no air. This is not the same as having a vacuum, as an empty balloon also has no volume. $\endgroup$
    – Glen Yates
    May 30, 2019 at 20:16

10 Answers 10


The buoyant force* depends on the volume of the object (or at least the volume of the object submerged in the fluid) and the density of the fluid that object is in, not necessarily/directly on the density of the object. Indeed, you will usually see the buoyant force written as $$F_B=\rho_{\text{fluid}}V_{\text{sub}}g=w_{\text{disp}}$$ which just shows that the buoyant force is equal to the weight of the displaced fluid.

We usually talk about more dense objects sinking and less dense objects floating because for homogeneous objects of mass $m$ we can write the volume as $V=m/\rho$, so that when we compare the buoyant force to the object's weight (for example, wanting the object to float) we get $$m_{\text{obj}}g<F_B=\frac{\rho_{\text{fluid}}m_{\text{obj}}g}{\rho_{\text{obj}}}$$ i.e. $$\rho_{\text{obj}}<\rho_{\text{fluid}}$$ This is what we are familiar with, but keep in mind that this emerges from the buoyant force's dependency on the object's volume (not density) after we assumed that we had a homogeneous object.

If our object is not homogeneous (like the balloon), then you have to be more careful. You do not just "plug in" the density of the rubber, since it is not purely the volume of the rubber material that is displacing the surrounding air. You have to differentiate between the entire balloon and the rubber material. So, the buoyant force would be given by $$F_B=\rho_{\text{fluid}}V_{\text{balloon}}g$$ whereas the weight is given by $$w_{\text{balloon}}=(m_{\text{rubber}}+m_{\text{He}})g=(\rho_{\text{rubber}}V_{\text{rubber}}+\rho_{\text{He}}V_{\text{He}})g$$ So, if we want floating, we want $$w_{\text{balloon}}<F_B$$ $$(\rho_{\text{rubber}}V_{\text{rubber}}+\rho_{\text{He}}V_{\text{He}})g<\rho_{\text{fluid}}V_{\text{balloon}}g$$ i.e. $$\frac{\rho_{\text{rubber}}V_{\text{rubber}}+\rho_{\text{He}}V_{\text{He}}}{V_{\text{balloon}}}<\rho_{\text{fluid}}$$ We end up with something a little more complicated, but if we treat the balloon as a single object then we get a similar result to the homogeneous case. Just define the density of the balloon as


and so we end up with


It should be noted that it's not just the fact that helium is in the balloon that causes it to rise then. You still need the volume of the balloon to be large enough to displace enough of the surrounding air. However, helium is used because it's density is so low that as we add more helium to make the balloon (buoyant force) larger, we are not making the balloon weigh too much more such that the buoyant force can eventually overcome the balloon's weight.

To qualitatively summarize this, the density of the object only matters when we look at the object's weight. The volume of the object (more specifically, the volume the object takes up in the fluid) is what matters for the buoyant force. The relation of these two forces is what determines if something sinks or floats. If your object isn't homogeneous then you should look at the overall density of the object which is the total mass of the object divided by the volume the object takes up in the fluid.

* If you want to know about where the buoyant force comes from, then Accumulation's answer is a great explanation. I did not address it here, because your question is not asking about where the buoyant force comes from. It seems like you are just interested in how comparisons of densities can determine whether something floats or sinks, so my answer focuses on this.

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    $\begingroup$ Good answer, I'll just add that you can make a balloon out of material denser than latex, and still have it float so long as its volume is big enough. On the show Mythbusters, they were able to make a very large balloon out of lead foil that floated in air. $\endgroup$ May 30, 2019 at 17:56
  • $\begingroup$ @NuclearWang Right, just replace "rubber" with "material" $\endgroup$ May 30, 2019 at 18:20
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    $\begingroup$ -1 for needing ten equations & inequalities to explain "Why does a helium balloon rise?" Nothing about the helium being contained by the balloon, nor even a mention of the word pressure or its gradient. Instead of explaining, you just start with an equation "you will usually see" and manipulate terms. $\endgroup$
    – uhoh
    May 31, 2019 at 5:53
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    $\begingroup$ @uhoh I try to answer questions based on the entire body rather than just the title of the question. The OP is obviously confused with the idea you usually hear about less dense rising and more dense sinking, so I addressed this concern. Had the OP asked about pressure gradients I would have addressed that (And I think the OP knows that a blown up balloon has its helium contained). I did not write the answer for you, I wrote it for the OP, which is evident by the fact that they accepted this answer. $\endgroup$ May 31, 2019 at 11:30
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    $\begingroup$ @uhoh I don't think it's useful to address concerns the OP did not bring up just to talk about what you want to talk about. I think it's better to address the misconception behind the actual question. You are more than welcome to write up your own answer if you don't like mine, but I find it odd that you seem to suggest an answer that is already accepted isn't good enough and even going as far as to comment this on other answers. $\endgroup$ May 31, 2019 at 11:44

The high-level explanation is "buoyancy". If you want to know the actual mechanism, it's that the pressure in a fluid increases with depth: the air pressure at the top of a balloon is slightly lower than the air pressure at the bottom of the balloon. If for each point on the surface of the balloon, you take the air pressure as a vector perpendicular to the surface of the balloon, and integrate all of those vectors over the whole surface area of the balloon, you'll find that there's a net upwards force. And with some multi-dimensional calculus, it can be proven that this force always works out to be the volume of the balloon times the density of the air, i.e. the weight of the displaced fluid. It's this force integrated over the surface of the object that gives rise to what we know as "buoyancy".

If the buoyant force is greater than the weight of the object, then the object has a net upwards force. And "the buoyant force is greater than the weight of the object" is equivalent to "the object is less dense than the surrounding fluid".

An empty ballon with no air in it falls

If you have a balloon of the same volume as the helium balloon, but with no air in it, then it will float. What you probably mean is "if you have a deflated balloon, it falls". Putting helium in a balloon per se doesn't make the balloon float; rather, the helium provides a force to push the sides of the balloon away from each other, which causes it to inflate. When the balloon inflates, its volume increases. And it is this increased volume that causes the buoyancy. More volume -> more air displaced -> more buoyancy. If you could get a balloon to stay inflated without anything inside it, it would rise even faster than a balloon of the same volume with helium in it.

For a helium balloon to rise, the total weight of all the air displaced must be greater than the weight of the balloon plus the weight of the helium inside it. When the balloon is filled with helium, it's not the density of the rubber/latex by itself that matters, what matters is the density of the balloon+helium. In other words, the weight of the rubber/latex plus the weight of the helium inside, divided by the volume that the balloon is taking up. The more volume you can get the balloon to take up, the lower the total density. That's how hot air balloons work: when you heat up air, its volume increases, so its density decreases.

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    $\begingroup$ This is a much better explanation for "Why does a helium balloon rise?" than the currently accepted answer! $\endgroup$
    – uhoh
    May 31, 2019 at 5:54
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    $\begingroup$ And of course, the reason inflated empty balloons don't exist is that atmospheric pressure is actually pretty huge - it takes a lot of support to brace a volume of vacuum against the pressure of our atmosphere (just look at videos of rail tankers imploding when their contents suddenly lose pressure). And that support means more mass, which in turn means higher density and you're back to square one. It doesn't help that while we have many lightweight materials that are very strong in tension, we don't have much that's lightweight and strong in compression (especially with regards to buckling). $\endgroup$
    – Luaan
    May 31, 2019 at 6:50
  • $\begingroup$ @Luaan: videos plural? The Mythbusters had to work hard to create that result intentionally, and even then had to put a dent in the tanker before pumping a nearly full vacuum. After disabling safety valves etc. (S14E02 Tanker Crush mythresults.com/tanker-crush and google the title to find video clips from the show.) They declared the myth "busted". If you've seen multiple videos, presumably they were from very different rail tankers, because the spherical symmetry of real ones makes them very strong under uniform compression of atmospheric pressure. $\endgroup$ Jun 1, 2019 at 1:41
  • $\begingroup$ @PeterCordes Right, I think that's probably the only video out there. The myth was busted, but that's not the same as what we're talking about. Rail tankers are built to withstand atmospheric pressure - but they also don't rise in the air (even with the best vacuum we can do), because they're too heavy for their volume. The tricky part is finding something strong enough to resist atmospheric pressure while at the same time being lightweight enough to create enough buoyancy to outstrip its own weight (not to mention allow carrying payloads etc.). $\endgroup$
    – Luaan
    Jun 1, 2019 at 5:44
  • $\begingroup$ @Luaan: I was only nitpicking that one part of your comment. Other than that yes it's a helpful explanation of why using He or H2 pressure to oppose atmospheric pressure is useful for creating lift, even though they're not massless. $\endgroup$ Jun 1, 2019 at 6:37

The balloon rises because of buoyancy. The force (weight) of the helium plus the latex/rubber downward is less than the buoyant force of the volume of air displaced by the balloon acting upward.

This means the weight of the balloon material was not great enough to cause the balloon to sink in the air.

Hope this helps.


My favourite way of explaining the intuition is by comparing it to a balance scale.

A balance scale has a state which only one of the two sides can achieve (the lower position).

The heavier of the two sides falls, because the system reaches a lower energy state by having the heaviest thing as low possible, even at the cost of a lighter thing being higher up.

Fluids work the same way. The balloon, and the atmosphere around it, are contending for the same position in space, and they can't both occupy it. Since systems tend towards lowest energy states when possible, that space is preferentially taken by the denser thing (the atmosphere). As a result, the balloon has choice by to be "pushed" up, supported by the force of the gas underneath it.


This may be silly question, but why does a helium balloon rise?

In addition to the answers already given, I'd like to give another quite simple explanation of the effect:

Let's think of two objects which are "connected" in a way that if one object rises a certain distance, the other object will fall the same distance.

An example would be a beam balance (as it is drawn in position "1)" of the attached image): The gravity will pull down both objects. However, the lighter object will rise and the heavier object will fall. In the image, the iron weight will fall and the feather will rise.

If some object is moving in the air, you have to keep in mind that not only the object is moving: The air also moves!

If some object moves from a position A to a position B, air with the same volume as the object moves from position B to position A (as it is drawn in position "2)" of the image).

So we have the same situation as with the beam balance, the iron weight and the feather:

Two "objects" are "connected" in a way that if one object (the balloon) rises, the other "object" (the air) falls down.

The gravity will have influence on both "objects", the lighter one will rise, the heavier one will fall.

Objects in the air

An empty balloon with no air in it falls, so why does a helium filled balloon rise?

The empty balloon is heavier than the air that has the same volume as the empty balloon. This means that the empty balloon will fall down and the air will rise.

The helium-filled balloon is a bit heavier (!) than the empty one.

However, the air with the same volume as the filled balloon is much heavier (because the volume is much larger). It is heavier than the filled balloon.

Therefore the air will fall down and the filled balloon will rise.

  • $\begingroup$ This is a better version of my answer. I like it! $\endgroup$
    – Alexander
    Jun 1, 2019 at 2:48

The Archimedes principle: The buoyant force on an object submerged in a fluid is equal to the weight of the fluid that is displaced by that object. This also applies when an object is submerged in a gas.

The balloon and its contents have weight due to gravity. The upward buoyant force opposes the downward weight force. A balloon inflated with helium has less total mass than the air it displaces; its weight (and average density) also are lesser. Therefore the buoyant force exceeds the weight force, the net force on the balloon is upwards, and the balloon rises.

It's a simple question, so I think it's best to give a simple answer.


The air pressure on the bottom of the balloon is slightly higher than the air pressure on the top of the balloon (because of the altitude difference). This difference gives rise to the bouyant force. This force competes with gravity as the two main forces acting on the balloon (at least in this thought experiment). Whichever is strongest dictates which way the balloon will float.

If the balloon had been filled with air, and the rubber itself had been (virtually) non-existent, then the balloon would just float in mid-air, as that's what air does in air. This tells us that the magnitude of the bouyant force on the balloon corresponds exactly to the weight of such an imaginary "air-only" balloon of the same shape and size. Since what we will be comparing are the weights of an "air-only" balloon and a helium-filled real ballon of the same shape and size, average density is the most important physical quantity for determining whether the balloon floats up or down.

Helium is less dense than air, but rubber is more dense. So a balloon with little helium will still have an average density higher than air, and thus the force of graviy will be stronger than the bouyancy, and the balloon will fall. However, as you put more helium into the baloon, the average density goes down, and at some point it becomes less dense than the surrounding air. This makes it float upwards.


A too heavy balloon would never rise.

If you fill helium slowly into a light balloon, there are three different phases. At first the balloon will not rise because the weight of the balloon itself plus the filled helium is higher than the weight of the displaced air.

At a certain point both weights are equal and the balloon will neither rise nor sink.

When more helium is filled, the balloon will rise, even with a small payload.

Good balloons are made from a very thin material to keep the weight of the hull small compared to the weight of the displaced air when filled.

If you want to lift a certain weight larger than the lift force of the balloon what is better, to take more balloons of the same size or to take a larger balloon?

If you take 8 balloons, you need 8 times more helium to get 8 times the lift force.

If you take one large balloon with the double diameter you also need 8 times the helium. But the weight of the balloons hull is only 4 times the weight of the small balloon. So you get a little more lift force from the one large balloon compared to the 8 small balloons using the same amount of expensive helium.


Lots of good answers here, but none address the main issue with your questions. The balloon doesn't rise due to any direct mechanism. Instead, the denser air around the balloon pushes itself down, with the net effect pushing the balloon up.

This is similar to the misstatement that 'heat rises'. Again, it doesn't. Instead, cold falls, pushing the heat up.


im not a scientist im a mere normal dude, so if i may explain it in simple terms?

and my answer would be because helium is lighter than air... much lighter about 10x lighter

Helium gas 0.00018 Air 0.00128

that is the molecule density ....

take for an example a balloon filled with air, and submerged under water... the pressure from the density of the water will pop the ballon right to the surface... or in other terms will make the balloon float.

same effect applies to air and helium filled balloons.

and the rubber that the balloon is made of (which if expanded or filled with any gas is a about 10 microns thick that is about the thickness of a human hair... the weight on that would be very small so it won't affect much the helium gas inside and if it does all you have to do is add more helium gas) will amount just a bit under the atmospheric pressure so it will still rise up.

i know it's a oversimplification but that's how i understand it.

another cool similar effect is the temperature difference, for example hot air vs cold air a effect put to use in hot air balloons... the hot air has a much smaller density than the cold/normal air, so it pushes the ballon up taking you on a trip across the land xD and if you want to go down all you have to do is wait for the air in the balloon to cool down, and down you go.

Edited: someone also mentioned surface tension and that would be as for example all around the ballon the atmospheric pressure exerted a force right? but the force of atmospheric pressure is weaker at the top of balloon and stronger at the bottom, because the atmosphere get weaker as you go towards space imagine it being a faded out color towards space.

that structure allows for a irregular pressure around the ballon pushing the ballon upwards on top of the force of the helium inside.

the same thing happens in water as i said, the deeper you go in water the stronger the pressure will push your balloon to top.

this entire effect and behavior of the rising balloon is a bundle of different forces working together, but sure ... we can say helium is lighter than air lol... wouldn't be incorrect would just be missing a lot of information.

hope it helped in any way .. :) im a simpleton i like things to be explained simple ...