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I doubt I can hide from the effects of gravity when I'm close to Earth. We usually don't question this biker is fully experimenting gravity:

enter image description here
Source.

But with the well-known "reduced gravity" ballistic flight path provided by the Zero-G A300, it's common to talk about micro-gravity, or reduced gravity, or 0-G.

enter image description here
Source.

It seems to me that the gravity in the A300 is not reduced at all, it's still the same than when flying straight and level. Exactly like the gravity hasn't changed for the stunt biker.

I also often read that the weight appears null, due to the free fall. But again I don't understand... According to Merriam-Webster:

Weight: The force with which a body is attracted toward the earth or a celestial body by gravitation and which is equal to the product of the mass and the local gravitational acceleration.

And that's what I learned: $w = mg$, and because mass and gravity acceleration haven't changed, then it seems weight hasn't changed either.

If gravity and weight haven't changed, then what is reduced actually in so-called reduced gravity? What does really happen to the aircraft and its occupants?

The only thing that I can see as new in free fall is that the different parts of a body (or the of the aircraft) are not anymore ordered from top to bottom by their weight. For example air will not be on top of fuel in the aircraft tanks, and blood will not be moved towards our feet without an action.

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The aircraft gives you a protected area in which you can experience free fall without any drag or other significant forces having a large effect on your body. This is essentially identical to what an astronaut experiences on the International Space Station, only with the drawback that the semi-minor axis of your orbit is so small that it has two intersection points with the surface of the Earth. Obviously, the pilot pulls up before that becomes a problem.

If you are inside the aircraft, your intuitive frame of reference is what you see around you - the interior of the aircraft. Relative to the aircraft, the acceleration is very close to zero because it falls with you and uses the engines to cancel out the drag it experiences. Relative to the surface, your acceleration is around 9.81 $\frac{m}{s^2}$, as expected during free fall. This is not proper acceleration, so it is impossible to sense or measure (unless you measure your change of speed relative to the Earth).

The term "zero g" is nevertheless accurate if it is understood as describing the g-forces. The definition of g-force from Wikipedia is:

The g-force acceleration experienced by an object is due to the vector sum of all non-gravitational and non-electromagnetic forces acting on an object's freedom to move.

So g-force, on the surface of the Earth, is actually the combination of the acceleration caused by gravity and the ground preventing you from freely moving as this acceleration would dictate. If the surface of the Eearth, or any other object in your way, is temporarily (mostly) removed, the g-force acting on your body is approximately zero. This is commonly described as a micro-$g$ environment or microgravity to highlight the fact that

  • There are sill some small forces acting upon an object because all micro-$g$ environments have imperfections.
  • Gravity is still accelerating an object in $\mu g$. It's just not perceived because every part of the object is accelerating at the same rate resulting in no internal forces (i.e. there is no proper acceleration).
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  • $\begingroup$ Thanks for the very good explanation to the layman I'm. I see that the more you learn about the details, the more the technical vocabulary shows the initial confusion around the principles, so you have to understand the limited context in which inaccurate terms are used to make sense of them (g-force is not a force, g scalar is the intensity of the acceleration due to the gravity, but do not measure gravity itself, 1 g is when only gravity acts on the object, and it directed upwards, weight is related to g-forces only, not to gravity, etc). Anyway I understand now why 0g makes sense! $\endgroup$ – mins Aug 7 '15 at 9:22
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W=mg for the motorcycle rider and the person in the airplane, whether they are in free fall or not, because, as you stated, the mass of the person is unchanged and the earth still exerts gravity on that person, during their free fall experience. When you get back to the ground, or the airplane gets back to level flight, the surface that you are standing on exerts a normal force equal to your weight, and this force keeps you from going into free fall. It is also the normal force that allows you to feel something pushing against your feet as you stand on the surface, so it is the normal force that allows you to actually feel the magnitude of the force that gravity is exerting on you. This means that the thing that is reduced in "reduced gravity" is the normal force that is acting on you.

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  • $\begingroup$ Hi David. It's a very clear explanation, thanks. Would you say the force preventing free fall is a fictitious force, due to the choice of the Earth frame of reference? If this is the case, any link that would allow me to dig further? $\endgroup$ – mins Aug 6 '15 at 17:19
  • $\begingroup$ The normal force is not fictitious. While I am standing on the floor, there is a force stopping me from falling through the floor. This force is pushing perpendicular to the floor (hence the name "normal" force) and straight up, directly balancing the force of gravity, such that there is no NET vertical force on me. If you have a physics book available, read more about the normal force or look here:khanacademy.org/science/physics/forces-newtons-laws/… $\endgroup$ – David White Aug 6 '15 at 17:27
  • $\begingroup$ The video helped. In the course of this search about 0-G, I read a a couple of articles about weight and its confusing uses. I feel this extends to (micro)gravity and weightless(ness). $\endgroup$ – mins Aug 6 '15 at 19:16
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You can't feel the action of a uniform gravitational field. The Earth's gravitational field across the extend of your body is very close to uniform. You can't feel Earth's gravity. You can only feel non-gravitational forces.

Einstein's equivalence principle says that it is impossible to perform a local experiment that can distinguish free fall in the vicinity of a gravitating body and free fall far removed from any gravitating body. Another way to put this: gravitation is a fictitious force (in the context of general relativity).

Most introductory physics texts use a definition of weight similar to the one you found in Webster's dictionary. A few use an alternate definition of weight, which is that weight is what an ideal spring scale measures. Most advanced physics texts that cover relativity use this latter definition of weight. This is the context in which the term "weightless" for those Zero-G A300 parabolic arcs is exactly correct.


The equivalence principle highlights an issue with the Newtonian concept of an inertial frame of reference. I'll look at four different scenarios:

  • Sitation A: You're in a windowless room in a rocket in deep space (i.e., far removed from any gravitating bodies) with all thrusters quiescent,

  • Sitation B: You're in a windowless room in a rocket in deep space (i.e., far removed from any gravitating bodies) with thrusters firing to yield 1g acceleration,

  • Sitation C: You're in a windowless room in a rocket in orbiting the Earth, with all thrusters once again quiescent, and

  • Sitation D: A windowless room in a rocket sitting still on the surface of the Earth.

To pass the time away, you pick up some colored sticks and connect them together, like this:

Then you ask yourself, could that that form the basis of an inertial frame of reference? Newtonian mechanics and general relativity disagree.

Situation  Newtonian  Relativistic
    A         Yes         Yes 
    B          No          No  
    C          No         Yes 
    D         Yes          No  

There is nothing you can feel, no local experiment you can perform, that will let you distinguish between situations A and C, or between situations B and D. Yet Newtonian mechanics says those indistinguishable situations are somehow very different from one another. General relativity says otherwise. Those indistinguishable situations say something profound about the nature of the universe.

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  • $\begingroup$ I'm just curious; does your answer pertain to my answer regarding "can't feel a uniform gravitational field"? $\endgroup$ – jjack Aug 6 '15 at 18:12
  • $\begingroup$ Otoliths are small stones (lithos = stone) named saccule and utricle in the internal ear which can sense gravity and other linear accelerations. I guess otoliths are ineffective in free fall, and gravity is not sensed in these conditions. $\endgroup$ – mins Aug 6 '15 at 19:02
  • $\begingroup$ @mins - Otoliths do not sense gravity. Nothing can. Your otoliths would yield the exact same output if you were in a rocket far removed from any gravitating body that is thrusting at 1 g as they do when you are standing still on the surface of the Earth. Otoliths are accelerometers. Accelerometers sense acceleration relative to an instantaneously co-moving free-falling frame. $\endgroup$ – David Hammen Aug 6 '15 at 19:12
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    $\begingroup$ I stand corrected. Yes they sense acceleration in general, including acceleration of gravity. $\endgroup$ – mins Aug 6 '15 at 19:18
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    $\begingroup$ @mins - They do not sense acceleration due to gravity. Nothing can sense that. They sense acceleration due to every force but gravity. Some MEMS accelerometers use a principle very similar to that in your otoliths. An accelerometer at rest on the top of a table will register an acceleration of 9.81 m/s, directed up. The accelerometer is sensing the normal force exerted on it by the table, but it is not sensing the gravitational force exerted on it by the Earth. $\endgroup$ – David Hammen Aug 6 '15 at 19:58
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Your body has zero acceleration relative to the aircraft. This is due to the aircraft flying a 1g-curve (parabola) with acceleration in the direction of the earth's gravitational acceleration - the normal force balancing the weight goes away. You feel essentially weightless during the curve.

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One way of looking at this is that in free fall you are still subject to gravity but there is no reaction force balancing it as you are accelerating under gravity at the same rate as your immediate environment.

In the situations described in the question this is temporary but if you are in orbit around the earth (or any other large body) the effect is sustained. By definition orbit is permanent free-fall in that the tangential velocity to the body providing the gravity is in equilibrium with the acceleration due to gravity.

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