# Would a helium-filled football/soccer ball actually fly farther than a normal ball?

DICLAIMER: I did this answer before watching the video. I am a silly American, and I took "football" to mean "American football". So my numbers are not exactly correct, but I think they are still within the same ballpark (American baseball?), so I believe my conclusions are still valid.

Before trying to draw conclusions from the data, let's think about if we would expect there to be much of a difference in how the ball travels depending on the gas it is filled with.

So there are multiple things that matter that has been mentioned either in your question or in the answers that have already been posted: The vertical force acting on the ball (essentially the effective "gravity" it feels), the effects of air resistance (the force will roughly be the same between balls, but a lighter ball will be affected more than a heavier ball), and the initial velocity of the ball (you would expect for the same kicking force that the lighter ball would start off moving faster than a heavier ball).

For each of these things, it will be useful to understand how the mass/weight of the football material compares to the mass/weight of the gas inside of it. According to this site a deflated football weighs roughly one pound which is about $$4.5\,\mathrm N$$.

The football is filled to a pressure of $$12\,\mathrm{psi}\approx8\times10^4\,\mathrm{Pa}$$. Adding in atmospheric pressure we get have a gas that is at a pressure of roughly $$2\times10^5\,\mathrm{Pa}$$. From this site we can estimate the volume of the football to be about $$280\,\mathrm{in^3}\approx4.5\times10^{-3}\,\mathrm{m^3}$$. Finally, using the ideal gas law assuming a ball at room temperature, the football will contain about $$0.36$$ moles of gas.

So let's look at the three gases in question. The molar mass of dry air is about $$30\,\mathrm{g/mol}$$, so this gives the mass of air to be roughly $$10\,\mathrm g$$ (confirmed by the first link above) and the weight of the air to be roughly $$.1\,\mathrm N$$. For helium we have $$4\,\mathrm{g/mol}$$ which gives a mass of roughly $$1.5\,\mathrm g$$ weight of roughly $$.015\,\mathrm N$$. And finally for sulfur hexafluoride we have about $$150\,\mathrm{g/mol}$$ which gives a mass of about $$50\,\mathrm g$$ and a weight of about $$.5\,\mathrm N$$.

Now for the analysis. The net vertical force (neglecting vertical air resistance) on the ball is given by $$F_y=F_b-w_B-w_g$$ where $$F_b$$ is the buoyant force, $$w_B$$ is the weight of the ball, and $$w_g$$ is the weight of the gas. The vertical acceleration is then $$a_y=\frac{F_b-w_B-w_g}{m_B+m_g}$$

The buoyant force is the same for all balls, it is just $$\rho_\text{atmosphere}V_Bg\approx.05\,\mathrm N$$. The mass and weight weight of the ball we already found as about $$.45\,\mathrm kg$$ and $$4.5\,\mathrm N$$. The mass and weight of each ball we also found above. The vertical accelerations for all of the balls is then $$a_\text{air}\approx-9.89\,\mathrm{m/s^2}$$ $$a_\text{He}\approx-9.89\,\mathrm{m/s^2}$$ $$a_\text{S}\approx-9.9\,\mathrm{m/s^2}$$

So it looks like the vertical accelerations are not very much different (at least not to the accuracy within my rounding).

What about the effects of air resistance and the kicking force on our balls? Well, this is determined by the total mass of the balls $$m_B+m_g$$, as the accelerations they feel from these forces is given by $$a=F/m_\text{total}$$. These total masses are easily determined

$$m_\text{air}\approx0.46\,\mathrm{kg}$$ $$m_\text{He}\approx0.45\,\mathrm{kg}$$ $$m_\text{S}\approx0.5\,\mathrm{kg}$$

Once again, I cannot see these making a huge difference, and this could be expected. The mass of just the ball materials is around $$0.45\,\mathrm {kg}$$ anyway, so I don't see changes of at most $$.05\,\mathrm {kg}$$ having a significant influence. Certainly we would not expect a noticeable difference between the air filled ball and the helium filled ball.

Ultimately, it looks like the material of the ball is massive enough so that the gas it is filled with will not have a huge impact on its travel. As @BobD discusses in his answer, what is likely happening here is that the "experiment" was not done with much precision in terms of similar conditions for the kicks, determining the distance of travel, etc. It is most likely just a ploy to get more views.

What would make their results better (but still not the best) is if they did a bunch of tests with the same type of ball and determined a distribution of distances for each ball. Then they should have compared the three distributions and see if the differences between the distributions were statistically significant. This would have been much better than just looking at which ball went the farthest, as this is not solely determined by the mass of the ball. I have confidence in my estimations, but as we all know, a well-executed experiment would be better and have the final say here.

Now obviously, this experiment wasn't done with much rigor (three participants at five total kicks each), but the results were still somewhat surprising: it was recorded that the helium ball went the farthest in 8 of 15 kicks, but in second place was the sulfur hexafluoride with 6 out of 15 (normal air went the farthest in only 1 of 15).

Having watched the video I'm not surprised at the results. I would find it hard to reach any scientifically based logical conclusions regarding the difference in performance between any of the balls based on the demonstration. In particular, the difference in performance between air and helium, compared to difference between helium and sulfur hexafluoride makes no sense to me.

There appears to be no measurement, or control, of the impact force associated with each kick. Nor any control over the angle of each kick relative to the ground. Some kicks resulted in launching the ball higher in the air. Others resulted in the ball flying low and bouncing along the ground. One ball hit a bush. Another hit a rock. Overall this is a terrible experiment and any attempt at a physics explanation to make sense of the results seems, at least to me, pointless.

The above being said, assuming each of the three soccer balls is identical in terms of the outer surface and the outside diameter, for a given velocity each ball should experience the same air drag force. After launch the only forces each soccer ball experiences are gravity and air drag.

The horizontal acceleration, for the same horizontal velocity, of each ball will be

$$a=-\frac{F_{r}}{M}$$

Where $$F_r$$ is the air drag force (same for each ball for the same horizontal velocity), and $$M$$ is the mass of the soccer ball (covering + enclosed gas). So for the same initial kick off velocity and same air drag force, the ball with the greatest mass should travel further. But by how much?

A regulation soccer ball has a mass of about 0.425 kg and a circumference of 70 cm. That equates an overall volume of about 0.0058 m$$^3$$. Neglecting the thickness of the wall, we can assume the volume is approximately equal to the volume of the enclosed gas. At 12 psig, assuming ideal gas behavior of each gas, and density proportional to pressure, the mass of each gas contained in the ball at 12 psig is approximately 0.0017 kg (helium), 0.013 kg (air) and 0.064 kg (sulfur hexafluoride).

Thus, the total mass of each soccer ball is about:

Helium filled: 0.427 kg; Air filled: 0.438 kg; sulfur hexafluoride: 0.489 kg.

If the initial horizontal velocity after being kicked is the same for each ball, given the small difference in masses of the balls, combined with all the other variables, it is unlikely to me there would be much of a difference between the performance of each. Perhaps the sulfur hexafluoride might travel somewhat further, all else being equal.

Hope this helps.

J Thomas gave a pretty good overview of why you might expect different results with the experiment.

I just want to point out the problem in your analysis. $$P = \rho R T$$

You seem to be looking at what wikipedia calls the molar form: $$P = \rho \frac R M T$$ where $$M$$ is the molar mass of the compound. In that case, you often define $$R_{\text{specific}} = \frac R M$$. That means the $$R$$ in your equation should actually be $$R_{\text{specific}}$$, which will change how $$\rho$$ and $$V$$ relate to each other depending on the gas (since $$R_{\text{specific}}$$ depends on the gas).