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"?

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    $\begingroup$ When you are in free fall near the surface of the earth, your acceleration is not zero: it is $g$. The weightlessness comes from the floor of the elevator not exerting any force on you (because you don't exert any force on it, since both you and the elevator are falling with the same acceleration). $\endgroup$ – NickD Jun 3 '17 at 14:58
  • $\begingroup$ People of the flat-earth persuasion are confused about this as well. What they don't get is that gravity itself is not acceleration (unless you appeal to general relativity, which really gets them steamed :) Rather it is a force acting on matter, kind of like long stringy elastic bands pulling literally everything on earth toward the center of the earth. The more massive something is, the more bands it has. When we talk about the "acceleration of gravity" we mean how fast those bands accelerate things toward the center of the earth when nothing is holding them up. $\endgroup$ – Mike Dunlavey Jun 14 '17 at 15:01
  • $\begingroup$ In fact it does not have to be parabolic; it is no different from an astronaut in a spacecraft with engines turned off moving around the Earth on elliptic, parabolic, or hyperbolic trajectory - still the astronaut will be in weightlessness. Furthermore, if there are more than a single gravitating body affecting the spacecraft then its free-motion trajectory can be very complicated - and still the astronaut inside will be in weightlessness. $\endgroup$ – Maxim Umansky Jun 16 '17 at 0:15

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!

  • $\begingroup$ Brilliant explanations, thank you. If I may ask, why free falls are given as examples of weightlessness? because there are forces still acting on us, namely gravity and even air friction. $\endgroup$ – user929304 Jun 15 '17 at 12:04
  • $\begingroup$ @user929304 thank you! Indeed gravity still acts on us, but since it applies to every bit of your body it disappears in your rest frame. (because there are no strain gradients for such a force, at least if we approximate the gravitational field as constant). Friction is there too, you're right! But we can neglect it as an approximation for a parabolic flight. Hope that helps, let me know if you have more questions! $\endgroup$ – Wolpertinger Jun 15 '17 at 14:28
  • $\begingroup$ Thanks, makes a lot of sense. So can we generalize by saying that in most cases where there s no normal force on us counteracting the gravity, we only "feel" weightless as every part of our body is subject to the same acceleration [*] and we're not pushing against anything. With quotations because it doesn't really mean that we will be easy to be pushed around as we still have our non-zero rest mass, because the momentum change needed to move a body even in vacuum is proportional to the mass of that body. Is this a sound recap? $\endgroup$ – user929304 Jun 15 '17 at 18:37
  • $\begingroup$ [*]: should we add a small caveat here as follows: we are still being accelerated, though it's constant it is still changing our momentum in time $\partial p/\partial t \neq 0$, but then why do we not feel a fictitious force acting on us in opposite direction of the gravitational force? similar to how when a car accelerates forward we are pushed backward. $\endgroup$ – user929304 Jun 15 '17 at 18:42
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    $\begingroup$ It is interesting to note that a high-tech tight-fitting ‘skinsuit’ - Gravity Loading Countermeasure Skinsuit (GLCS) - is being developed for astronauts for creating strain in the body to feel it more natural for humans in weightlessness and counter-balance negative effects of it on human health. Check out esa.int/Our_Activities/Human_Spaceflight/Astronauts/… $\endgroup$ – Maxim Umansky Jun 16 '17 at 0:29

In a parabolic flight or free fall, both the floor on which the object is placed and the object itself move with the same acceleration that is g=9.8m/s. In other words, the floor is dropping at the same rate as the object, so the object experiences free fall. Say if you are standing in a falling lift. Both you and the lift have the same acceleration g. You decide to jump. You jump and now you are supposed to fall down right, but the floor of the lift(in fact the lift itself) is falling at the same rate as you, so in your frame of reference you do not experience any net acceleration. Now say you are standing in a lift which is going upwards with acceleration g. You jump again. You fall down due to the acceleration due to gravity. Also, the floor is advancing towards you with acceleration g. So the overall result is that the both effects add up and you fall down with acceleration 2g in your frame of reference.


In terms of Newton's laws and our experience as humans think of this:

When you are standing on a floor gravity pulls you down and die to this you are pressed against the floor which gives an upwards normal reaction force. This Normal reaction force gives you the feeling of being in pressed contact with the floor.

Now consider projectile motion. The path of the projectile is a parabola and notice that the acceleration of a particle in a parabola is always g downwards so if an airplane follows this trajectory then the floor of the airplane also falls away at g downwards so thus a person is no longer pressed against the floor then( as floor seems to fall away) and so the normal reaction force is now 0. That is weightlessness.


As Nick points out, weight is the feeling you get when the floor pushes your feet upwards. What you feel is the electromagnetic repulsion in your feet that the earth exerts on you. This force accelerates you at a rate of $-g$ which sums with the gravitational pull to leave you at rest. In free falling only the gravitational pull acts. You are indeed accelerated BUT you do not feel it. You can not know it. Because there is nothing pushing your feet. And this is extremely deep and, in fact, the starting point of General Relativity.

This is called the equivalence principle and is special of gravity (in contrast with other forces). The point being that the charge under gravity (gravitational mass) is exactly the same as the inertial mass (Unlike electric charge which obviously does not coincide with the inertial mass). This in turn implies that you can't distinguish acceleration from the gravitational pull. Einstein used this insight to argue that the gravitational pull does not exist. Instead, we move in a curved space-time. Then, the gravitational force does not exist. Then indeed when you are free falling $\mathit{there \, are \, no \, forces \, exerted \, on \, you}$.

To sum up: You cannot tell that you are free falling, because you cannot tell acceleration from gravitational pull.

And about Newtons equations, you are just using wrong signs. The point is how the floor accelerates with respect to you. If the elevator has the same acceleration as you, the acceleration you $\mathit{feel}$ is the subtraction. That is, $g - ElevatorAcceleration$.



The way to understand is the concept of non-inertial reference frame. Classical mechanics (and fruthermore iniertial law) works in inertial reference frame, that that doesn't undergo acceleration (also called Galilean reference frame). If you want to use $$\vec{F}=m\vec{a} $$, the $a$ is in an inertial frame of reference. So for the falling plane, it accelerates downward at $g$. In a outside inertial frame of reference, your acceleration in this frame $\vec{a_{if}}$ is equal to your acceleration $\vec{a_p}$ in the plane plus the acceleration of your plane (non inertial reference frame), $\vec{a_{pf}} = \vec{g}$. So the Newton law in the inertial frame writes: $$ \vec{F} = m\vec{a_{if}} = m(\vec{a_p} + \vec{a_{pf}}) $$ $$\vec{F} - m\vec{a_{pf}} = m\vec{a_p} $$ the $- m\vec{a_{pf}}$ term is called a fictious force. It gives rise to centrifugale force (opposed to the centripet acceleration) and the coriolis force is your non inertial frame is now in a rotation around one axis. Now if your only force $\vec{F}$ is $\vec{g}$, you have $$ m\vec{a_p}=0$$ That is weightless. I insist, it is not a feeling, it is an actual null force state.

  • $\begingroup$ and in case of your other exemple (the lift going upward), you will experience a 2g force. You can obtain this result making the same calculus. This is actually why war plane pilot experience few g during take of : the acceleration forward create a ficteous force backward. $\endgroup$ – denis Jun 14 '17 at 15:53

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