Can a balloon start from Earth and fly to the Moon, using Helium for lift to the top of the atmosphere and then as propellant? 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.
 A: 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.
A: 
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.
A: 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.
A: 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.
A: 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.
A: 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.

A: No
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.
A: 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.
A: 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.
