Question

If you release a ball while inside a freely falling elevator, the ball remains in front of you rather than falling to the floor because the ball, the elevator, and you all experience the same downward gravitational acceleration. What happens if you repeat this experiment with a helium-filled balloon?

Short answer

Therefore, the helium balloon will not float to the ceiling on it’s down. It will be suspended in a stationary position until another force act on it, such as person pushes the balloon.

Step-by-step solution

Archimedes’ principle

When an object is partially or fully submerged in a fluid, the fluid exerts on the object an upward force called the buoyant force. According to Archimedes’s principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object:

$B={\rho }_{Fluid}g{V}_{disp}$

Where$V_{disp}$ the volume of fluid is displaced and${\rho }_{Fluid}$ is the density of the fluid.

Find what happens if you repeat this experiment with a helium-filled balloon

• Heavilyanddenseandlightdensegaseshavedifferentnaturetoitsselflikelessdensegasesrisesinnormalenvironment. Thereisreasonheliumballoonrisesinnormalenvironmentisbecauseheliumislessdensethannormalair, sogravityaffectsthetwogases. Thisiscalledbuoyancy.Conversely, ifyouplacedheliumballooninsidearoomfilledwithhelium,theheliumballoonisnomorebuoyantthantheheliumintheroom.
• In case of free-falling elevator, if the elevator reaches terminal velocity and a perfect zero gravity environment is achieved, then no object inside the elevator will have weight, because there will be no effect of gravity. Therefore, the helium balloon will not float to the ceiling on it’s down. It will be suspended in a stationary position until another force act on it, such as person pushes the balloon.