The JP Aerospace's Tandem airship achieved a record-breaking 28,982 meters for the highest airship flight.

After reaching this height, can the helium balloon be used as a propellant, the same way if you release a balloon without tying the end, it will fly through the room?

A guy used a vacuum cannon to shoot a ping-pong ball at supersonic speed by releasing air into a vacuum pipe. I believe similar conditions will meet our balloon at an altitude of 28,982 meters, so the question is will it be enough to get to the moon?

Assume all the perfect conditions for this idea.

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    $\begingroup$ @Bergi My best guess is that Ilya wants to say "using the helium gas inside the balloon as a propellant". Propel the "balloon rocket" with the helium. $\endgroup$ May 16 at 13:16
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    $\begingroup$ IIRC the gravity at ISS altitude (~500km) is still 85% of what we experience at sea level. There is no way we can create an elastic material that can generate enough force from shrinking to achieve escape velocity $\endgroup$ May 16 at 13:50
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    $\begingroup$ I just asked a question over at Space Exploration SE about the idea to use the gas in a high-altitude balloon as a propellant, effectively turning the balloon into a rocket. Now don't get your hopes high ;-). $\endgroup$ May 16 at 14:28
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    $\begingroup$ A reminder from XKCD that orbit is not far away; orbit is "fast away". $\endgroup$ May 16 at 15:47
  • $\begingroup$ Hi Ilya Gazman. Did you try to do a back-of-an-envelope-calculation? $\endgroup$
    – Qmechanic
    May 17 at 13:10

9 Answers 9


With the helium tightly sealed inside? No. Once the density of the atmosphere is as thin as the density of the helium gas, the buoyant net upward force is gone and the balloon stops going further up.

With helium gas being released? Once you're up there, the density inside and outside your balloon is the same. With helium being lighter than air, that means the pressure inside is higher, so yes, the helium will come out. But that kinetic energy is nowhere near the amount needed to accelerate the balloon to the required speed to get to the Moon, which is some 8km/s or thereabout. Actually, your "children's party helium balloon rocket engine" would use one of the worst choices for a propellant due to helium being both light (some satellites use xenon for being massive) and chemically inert (most rockets use chemical propellant). What matters in that case is momentum, i.e. both high mass and high speed, whereas the helium in your balloon has neither.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – SuperCiocia
    May 17 at 22:09

I think some people are misunderstanding your question: you're asking if a helium balloon can be used as propellant, the same way if you release a balloon it will fly through the room.

The answer is no, not even close. If you manage to get to the upper parts of the atmosphere you stil have a long way to go. One of the most important quantities in space travel is delta-v, the change in velocity that a certain propellant will give you. Starting out, the balloon is stationary. You would first have to start orbiting the earth and after that you would have to accelerate to get to the moons orbit. Quoting wikipedia to get a sense of the delta-v required it says that to start orbitting you would need about 7.8 km/s and to get to the surface of the moon you would need an additional delta-v of about 5.9 km/s.

Can deflating a balloon give you such speeds? No, there is simply not enough energy stored in the gas.

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    $\begingroup$ tnx, I edited the question. Can you please calculate the "not even close" balloon speed? $\endgroup$ May 16 at 13:41
  • $\begingroup$ @IlyaGazman: Part of the show-stopper here is that energy density for compressed gas (in terms of joules per kg) goes up by compressing it more, to higher pressure. (Since you can get it to do work by expanding). But then it's less buoyant. Even if you had a huge lifting balloon to raise a compact high-pressure balloon, the Isp (specific impulse) of your rocket, how much delta-V you get per mass, depends on the exhaust speed, and that's nowhere near as high as a chemical rocket. $\endgroup$ May 16 at 15:25
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    $\begingroup$ Here's a way to estimate the ∆v: Assuming the helium exits the nozzle at the $v_{rms}$ of a gas at 0°C, the "exhaust velocity" is about 1.3 km/s. Applying the Tsiolkovsky rocket equation with these numbers, we find that the helium would have to outweigh the balloon by a factor of $e^{7.8/1.3} \approx 400$ in order to get to orbital velocity, and a factor of $e^{(7.8+5.9)/1.3} \approx 38000$ (!) to escape the Earth altogether. And that doesn't take air resistance into account, which even in the upper atmosphere would be substantial. $\endgroup$ May 16 at 15:36
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    $\begingroup$ @MichaelSeifert: en.wikipedia.org/wiki/Cold_gas_thruster lists a specific impulse (Isp) of 165 for He in practice, theoretical max 179. (At 25C, 1 atmosphere). So that's an effective exhaust-velocity of 1618 m/s. (Or 175 5m/s theoretical max). So the full:empty mass ratio "only" has to be 124 to LEO. Still unlikely for a rubber balloon and a light gas like helium, even with best-case assumptions of a well-designed nozzle and using every last drop of He at constant pressure. $\endgroup$ May 16 at 17:05
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    $\begingroup$ This is incorrect. You do not need to start orbiting, at least not in LEO. That's how Apollo did it, but there's no need to pause in LEO before continuing on. You can simply keep thrusting upward. However, you do have to get into lunar orbit, but because it's a lot higher, it is of course a lot slower than LEO. $\endgroup$ May 17 at 0:02

I'll start from the premise that a helium-filled party balloon has risen through the atmosphere to the point of buoyant equilibrium without otherwise changing significantly, and we then let the gas escape in order to propel the empty balloon toward the moon.

To impact the moon from a stationary (w.r.t. the ground) starting point in the upper atmosphere, we need virtually all of the 11 km/s of Earth's escape velocity. That is, we need 11 km/s of $\Delta v$. We'll approach this by trying to calculate the $\Delta v$ of the balloon when the throat (assumed previously held closed) is opened.

To calculate $\Delta v$ we will use the Tsiolkovsky rocket equation:

$$\Delta v = v_e \ \mathrm{ln} \frac{m_0}{m_f}$$

where $v_e$ is the effective exhaust velocity, $m_0$ is the initial mass of the balloon plus gas, and $m_f$ is the final mass of the empty balloon.

We'll start by calculating $v_e$. If we assume the throat of the balloon forms a nozzle with perfect efficiency, then $v_e$ when the gas first starts escaping will be similar to the speed of a molecule inside the balloon. The root mean square speed of a molecule of an ideal gas is (see Physics of Music: Speed of Sound in Air and Wikipedia: Speed of Sound):

$$v_\mathrm{rms} = \sqrt{\frac{3 k_B T}{m}}$$

where $k_B$ is the Boltzmann constant, which is about $1.4 \times 10^{-23} \frac{\mathrm{J}}{\mathrm{K}}$, $T$ is the temperature in Kelvin, which we'll just assume is $300 \ \mathrm{K}$ (room temperature), and $m$ is the mass of a single molecule of gas, which for helium is 4 atomic mass units or about $6.8 \times 10^{-27} \ \mathrm{kg}$. Plugging in those numbers yields:

$$v_e \approx v_\mathrm{rms} \approx 1400 \frac{\mathrm{m}}{\mathrm{s}}$$

(To two significant figures, that's a specific impulse of 140 s. Peter Cordes observes that Wikipedia, citing Nguyen et al., gives a theoretical 179 s for helium at 298 K, but I don't know how that was derived. Perhaps my $v_e \approx v_\mathrm{rms}$ assumption needs refinement, but I'll proceed with it anyway.)

Next, according to Aerodynamics of a Party Balloon, the mass of the empty balloon alone is about:

$$m_f \approx 1.3 \ \mathrm{g}$$

Finally, $m_0$ is $m_f$ plus the mass of the helium. If the balloon was filled to a volume of 5 liters at one atmosphere of pressure and $300 \ \mathrm{K}$, it will contain about $0.8 \ \mathrm{g}$ (based on $4 \ \frac{\mathrm{g}}{\mathrm{mol}}$ and the ideal gas law). Adding $m_f$, we get:

$$m_0 \approx 2.1 \ \mathrm{g}$$

Plugging in all the numbers, we estimate (with very generous assumptions):

$$\Delta v \approx 670 \ \frac{\mathrm{m}}{\mathrm{s}}$$

This is well short of the needed $11 \ \frac{\mathrm{km}}{\mathrm{s}}$, so we can safely conclude that a party balloon cannot reach the moon by releasing its trapped helium.

Of course the practical $\Delta v$ of a helium balloon would be much less than this, for a variety of reasons, including:

  • The opening is not a good nozzle.
  • The gas cools as it escapes.
  • There is no attitude stabilization so it will spin more than accelerate linearly.
  • The balloon will have undergone changes in temperature and pressure during the rise through the atmosphere to a point of buoyant equilibrium (where we assume the throat is opened). It might burst due to pressure, or freeze and shatter, etc.
  • The point of buoyant equilibrium is still well inside appreciable atmosphere, so some $\Delta v$ will be lost due to drag, assuming it accelerates in the right direction.
  • $\begingroup$ Your estimated v_e is pretty close to what en.wikipedia.org/wiki/Cold_gas_thruster#Propellants lists as the best practically measured I_sp for He: 165 s giving 1618 m/s (Isp * 9.8 m/s). When you say "the gas cools as it escapes", might be good to clarify that you're only talking about the exhaust: the air in a deflating balloon does tend to stay near constant pressure and thus not cool, and also not lose thrust until near the very end when the rubber goes limp. (Thrust at the end, when it's lightest, is most important for Δv.) $\endgroup$ May 17 at 12:39
  • $\begingroup$ @PeterCordes Thanks for the Isp observation; edited answer. As for the gas cooling inside the balloon, I still think it should throughout because the rubber tension decreases. I'm not persuaded by the argument you linked because there must be a pressure differential to cause the flow, and that differential must come from the tension, which must decrease as it deflates. Plus, you can hear the effect of lower pressure difference as the aeroelastic flutter frequency at the throat decreases. $\endgroup$ May 17 at 15:42
  • $\begingroup$ Right, the rubber itself cools. But to get the required Δv to even make LEO, a best-case balloon needs about 124x the mass of gas to rubber. (That's likely totally impractical with a gas as light as helium.) You're right that for a typical balloon, the rubber cooling will probably cool the gas inside, especially if it's a slow enough jet that there's time for heat transfer. So good point, that is a problem. $\endgroup$ May 17 at 15:45
  • $\begingroup$ +1 for mentioning escape velocity - I'm no expert but the first thing I thought when I saw this question was that escape velocity was the biggest problem here. $\endgroup$ May 17 at 21:46
  • $\begingroup$ At 1 liter of helium the balloon won't even rise. $\endgroup$ May 18 at 7:21


This is essentially analogous to asking if an inflatable boat can fly.

A helium balloon will ascend until it reaches the atmospheric "surface", and then it will stop and float. In much the same way an inflatable boat released from the bottom of the sea will ascend to the surface and float there.

While technically the earth's geocorona extends well past the moon, a helium balloon is always going to be denser than the tenuous haze of hydrogen out there, and so won't float in it.

If you could create a bubble of True Vacuum of sufficient size, with nothing at all (not even trace hydrogen) then you could theoretically create an object that was bouyant enough to float the moon.
But if you can do that, you can do anything, because that's just plain magic.

  • $\begingroup$ How would a bubble of True Vacuum retain its bouyancy once it was entirely outside of Earth's atmosphere? $\endgroup$ May 16 at 23:40
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    $\begingroup$ @JeremyFriesner "technically the earth's geocorona extends well past the moon" $\endgroup$
    – philipxy
    May 17 at 1:17
  • $\begingroup$ Perhaps another question... how much magic true vacuum would be needed to float a person to the moon? Bigger or smaller than the earth? (just the person in their space suit with their light-weight handwavium oxygen regenerator and magic vacuumiser field device) $\endgroup$
    – Stephen
    May 17 at 2:19
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    $\begingroup$ @Stephen The article says the geocorona has 0.2 atoms of hydrogen per cubic centimetre at the moon's distance. The size of the bubble would need to displace the mass of the person in hydrogen atoms. Say your person with equipment weighs 70 kg, that means you need a bubble that displaces 70 kg of hydrogen, which is 4.2x10^28 atoms, at a density of 0.2 atoms per cm^2 that is 2.1x10^29 cm^3, which is 2.1x10^14 km^3, which is a bubble with diameter 74,000 km. The earth's diameter is 12,742 km. The bubble is about 6 times the diameter of the earth, and about 24 times the volume of the earth. $\endgroup$ May 17 at 8:03
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    $\begingroup$ @SimonBiber Dayum. That's a big bubble.. Like a beachball in a teacup. Of course, the distance from the earth to the moon is a lot bigger than that again. So it might actually work pretty well. And a magic vacuum-bubble would float in the deeper atmosphere quite well too. Maybe just scale it up appropriately to your needs rather than start with blasting the entire atmosphere off the surface of the earth from get-go. $\endgroup$ May 17 at 9:06

After reaching this height, can the helium balloon be used as a propellant, the same way if you release a balloon, it will fly through the room?

Sure. There's pressure built up in the balloon envelope, and if released in a controlled fashion, it will produce thrust in the opposite direction.

Will it be enough to get to the moon?

Not even close.

What you have to realize is that the vacuum cannon in your video had a lot of weight in the pipe and related equipment, but the pressure was only used to accelerate a ping-pong ball (2.7 grams). The airship in question weighed over 13,000 times as much, and so would receive over 13,000 times less acceleration from the same impulse. So, assuming the ping-pong ball reached 2.7 km/s*, the airship wouldn't even have reached 0.3 m/s under the same conditions.

But lets assume that we don't have to get the whole vehicle to the Moon. Assume that there's some kind of pressure cannon on the airship that uses the pressure of the balloon to launch a ping-pong ball at 2.7 km/s. Would that be enough to get to the Moon?

No. You need a minimum of about 7.5 km/s just to achieve orbit, or gravity will pull you back down into the atmosphere. The Saturn V (the launcher for the Apollo Program) achieved 2.7 km/s on its first stage alone. Assuming it was shot straight up, your hypersonic ping-pong ball will spend about 4.6 minutes attaining a maximum altitude of roughly 400 km (including the approx. 30 km height of the launch platform), which is less than 0.1% of the way to the moon.

*The guy in the video guesstimated that the ball would exit the cannon at between "three to eight times the speed of sound". I don't know where he got those numbers, but assuming he's correct, Mach 8 at sea level is about 2720 m/s.

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    $\begingroup$ There were many good answers here, but I really liked your idea of making a pressure cannon that uses the pressure of the balloon to launch a ping-pong. $\endgroup$ May 16 at 22:46

Helium is used in balloons because it is lighter than air, which is the reason why such a balloon rises. That’s why it only works up to a certain altitude. Once you’re high up enough, air becomes so thin that the balloon would expand into it and burst.

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    $\begingroup$ The gas in high altitude balloons does expand when the balloon ascends, but it doesn't burst the envelope because they make the envelope big enough to accommodate the expansion. When they're launched, they look like this: nasa.gov/sites/default/files/img_2246_3.jpeg Lots of extra room for the gas to expand into. The limiting factor is that the less-dense air at high altitude provides less buoyancy. You can make the balloon bigger to compensate for that, but the buoyancy falls off fast, and the cost of making the balloon bigger gets higher fast. $\endgroup$ May 15 at 21:47
  • $\begingroup$ Better picture: forbes.com/sites/elizabethhowell1/2019/07/05/… The picture in my previous comment probably is of a balloon that still is being filled. The one in the Forbes article actually has been launched. $\endgroup$ May 15 at 22:09
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    $\begingroup$ The problem isn't that the balloon bursts (cheap balloons do burst but we can solve that if we want by making stronger balloons - it just costs more). The problem is when there isn't any air left there's no buoyancy force, because helium isn't lighter than nothing. $\endgroup$
    – user253751
    May 16 at 9:23

A fundamental challenge you will run into is that bouyancy and rocket propulsion want to opposing things. For bouyant lift, you want low densities. This means relatively low pressures. However, for efficiency, rockets want very high exhaust velocities. This requires high pressures. If your balloon could store the helium in a pressure cylinder, it would become too dense to float.

So you would need something to pressurize it. Where does your energy come from? Helium is inert, so it can't be your source of energy, merely your propellant.

You would need something like an ion thruster to even make headway. They use electricity, such as from solar cells, for evergy. We typically use heavy gases for that because it is easier to make them efficient. Helium is at a major disadvantage there.

You also have to consider that the atmospheric drag is non negligible at floating levels. Ion thrusters have very low thrust, so they would have trouble overcoming drag.

At some point, yours balloon starts to look like a minor lift device strapped to an ultra high tech state of the art thruster. I don't think that's the spirit you are looking for.


TL;DR It is almost certainly impossible, and even if achieved it would be utterly useless.

If we assume that the balloon is made out of an indestructible plastic substance, like with a real balloon, and it holds enough helium to get it to the moon, it will eventually cease to float, before it gets to the moon.

Even if we loosely interpret the question and think about using helium as a propellant, it is ineffective and would be one of the worst rocket fuels out there. So no, it would not be a good method of propelling a rocket like thing.

To break through the earth's lower atmosphere, it would need to go much faster than what a conventional balloon.


How I would do it.
Have a decent nuclear heater onboard, somewhat like a dirigible. After max hydrostatic altitude is reached, use gas for thrust. If you could heat it up to a few thousand degrees, it could make a capable propellant.

How Randall Munroe might do it.
Randal is very technical and very creative, a great combination (https://what-if.xkcd.com/157/)

You could have two cannisters, one providing the lift (helium balloon) and another at cartoon-physics pressures to just push you to the moon. If it is compressed to being solid helium3, then it is compressed to more than 3000 kPa, and maybe you could do something with the jet that comes when it melts/boils/jets out of the thrust cannister. Current technology does not yet provide for a cannister like this.


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