[Physics] Dethestifying the physics of weightlessness in parabolic flights

Question

Parabolic flights are often described roughly as a plane accelerating upward for a certain time and then free falling (thus stopping its thrust) and during the arc-path it takes before falling straight downward, the passengers feel weightless.
This is both used to mimic being in a spaceship in order to prepare astronauts and more importantly, physicists​ run experiments in these conditions so as to gather data under near-0-gravity conditions.

Often to explain the idea behind the perceived weightlessness feeling, the analogy with a downward accelerating elevator is made, i.e. if the elevator is accelerating downward at a rate equal to $g,$ the passenger feels weightless. But intuitively, I had always imagined this the other way around, namely, that if there's an upward acceleration equal to $g,$ the accelerated object will have a net acceleration $\vec{a}=\vec{0}$ and this would be a weightless situation because we don't feel accelerated towards anything. But admitted​ly just from the equations, e.g., $m=F/a$ this is not clear, as my described example would even imply an infinite mass…

It would be brilliant if someone could explain what is going on without relying on metaphors or too mis-leading analogies, and instead just arguing with basic Newtonian equations: what is achieved in a parabolic flight that we call "weightlessness"?

Answer 1

I think the confusion arises because you think that to feel weightless you need $\vec{a}=0$. This is not the case.

I would instead define weightlessness of an object as follows: In the objects rest frame, there are no forces acting on the object (as opposed to the sum of the forces being zero, leading to zero acceleration by Newton's law).

Let me illustrate the difference between the two by taking your elevator example.

1. You fall with the elevator. In your frame and assuming that the situation is ideal, there are no forces acting on you.
2. You stand in an elevator that is not falling. Then in your frame there are two forces acting on you. Gravity is pulling you downwards and the floor is pushing you up. Now you are gonna tell me 'but is this not the same since by Newton's law the forces add to zero and there should be no difference'. No! The reason is that you are a finite size object. The floor is pushing only on your feet, while gravity pulls (to a good approximation) uniformly on every in your body. This creates a strain gradient across your body, i.e. your feet feel your whole body above pushing down on them while your hear only feels your (hopefully still flourishing) hair on it. This is what you perceive as "weight", your feet push into your body. You won't have that in a freely falling elevator or a parabolic flight.
3. The example in the question, which is you standing in an elevator that is accelerating upwards with $g$. Well, same story as in 2, except you have twice the pressure on your feet now.

This is not only the reason why we feel weightless in a parabolic flight, it is in some cases also the reason why people want to do physics experiments in weightless conditions. Most physical systems are finite size objects and if you want to do a precision experiment where the difference in force of the different parts actually matters, you would have to either invent a way to stop them from falling without resting them on the floor, or you go into free fall (disclaimer: I don't actually know if anyone does precision measurements in parabolic flights. I doubt it, cause planes are shaky). Of course there are lots of other effects, most of which have the same origin of strain/force gradients though (e.g. the difference in the shape of flames, see this article).

EDIT: Make sure you check out MaximUmansky's comment below!