# How do you explain the “weight” of helium?

If weight is measured as M * g (mass x gravity), then how does the "weight" of a massive helium filled balloon negative? Let's say you have a 63,000 m^3 helium balloon. The mass of the helium is around 11,000 kg! 11,000 kg * 9.81 is a substantial amount of "weight", but the measured weight is negative due to the density differential with the surrounding body of fluid (air). What accounts for this mathematically? Is the "g" term negative? Is the m*g weight equation not valid in this case?

• It is like a bubble of air in water. The bubble goes up very visibly, but water goes down. – Pieter Nov 1 '18 at 0:37
• Now calculate the weight of the air the helium balloons is displacing. – Triatticus Nov 1 '18 at 0:53

Most objects have densities that are very much larger than that of air. For calculating the motion of these objects in air, the buoyancy force can be neglected without introducing substantial errors. If you define “weight” as what a scale reads, though, it is technically not exactly $$w=mg$$ unless the object is in a vacuum, but rather $$w=mg-\rho g V$$, where$$V$$ is the volume of the object and $$\rho$$ is the density of the surrounding fluid. For a helium ballon, this second term (the buoyancy force) is not only not negligible, it is larger than the gravity force. So $$w$$ is negative.
• The mass of the helium plus the mass of the balloon is $m$. The total gravitational force is $mg$. Neglecting buoyancy, that would be the weight. The buoyancy force is actually the result of higher air pressure near the bottom of the balloon than near the top, but is most easily calculated by using Archimedes' principle, which says that it's equal to the weight of the displaced air. Does this make it clearer? – Ben51 Nov 1 '18 at 22:40