Artifificial zero gravity environment

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Discussion Overview

The discussion revolves around the mechanics of creating an artificial zero gravity environment in aircraft that perform parabolic flights. Participants explore the physics behind the experience of weightlessness during these flights, including the roles of gravitational forces, terminal velocity, and the definitions of equilibrium in different contexts.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants explain that the artificial zero gravity effect is achieved through a plane's high-altitude dive, where everything falls at the same rate due to gravity.
  • Others argue that the plane can exceed terminal velocity due to its engines, which allows it to create the zero gravity effect for a limited time.
  • One participant presents equations to describe the forces acting on the airplane and the passengers, suggesting that passengers would eventually fall faster than the airplane if it were to dive without engine power.
  • There is a discussion about the nature of equilibrium, with some participants asserting that reaching terminal velocity constitutes static equilibrium, while others challenge this by suggesting it is a form of dynamic equilibrium.
  • Participants debate the definitions of static and dynamic equilibrium, with references to mechanical equilibrium and the implications of motion in these contexts.

Areas of Agreement / Disagreement

Participants express differing views on the definitions of equilibrium and the mechanics of the zero gravity experience. There is no consensus on whether terminal velocity represents static or dynamic equilibrium, and the discussion remains unresolved regarding these definitions.

Contextual Notes

Participants highlight limitations in the definitions of equilibrium and the nuances of how forces interact in dynamic systems. The discussion reflects varying interpretations of physical concepts without reaching a definitive conclusion.

Who May Find This Useful

This discussion may be of interest to those studying physics, particularly in the areas of mechanics and dynamics, as well as individuals curious about the physics of flight and weightlessness.

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When people go in those planes that go real high and then nose dive down toward Earth in order to create an artificial zero gravity environment for the passengers, how exactly does that work?

This is of course due to the gravitational constant of acceleration, and everything falls toward Earth while accelerating at the same rate. But the plane eventually would reach terminal velocity when its drag force is equavalent to g, and therefore putting it in dynamic equilibrium. When that happens shouldn't the passengers continue to accelerate since there is no drag force on them?

Sure, the system they are in is moving at a constant velocity once it reaches terminal velocity, but the passengers are still in Earth's gravitational field, and their bodies would have no idea that the system they are in is moving at a constant velocity until they hit a wall or touched the plane in some way (I am assuming they are just floating inside the plane while the artificial zero gravity envrionment is achieved).
 
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The jet has engines so it can "fall" faster than terminal velocity.

Of course, you are right, ride doesn't last very long. Very soon the jet reaches its maximum safe velocity and must discontinue the dive.
 
For the airplane:
[tex]mg - kv = ma[/tex] -(a)

For the person inside the airplane:

[tex]mg=ma[/tex] -(b)

From (a):

[tex]a = g - \frac {kV}{m}[/tex]

From (b):

[tex]a=g[/tex]

So put into a dive with the engines off, the people would eventually do exactly as you say, fall faster than the airplane and hit the cockpit. This is because case (a) always has a smaller acceleration. (due to minus sign).

Unless, I have made an error.

When people go in those planes that go real high and then nose dive down toward Earth in order to create an artificial zero gravity environment for the passengers, how exactly does that work?

The airplane does not exactly 'dive,' it does 40 sinusoids in one direction, turns around, and repeats over and over again. So at one point you feel very heavy, the next very light. So terminal velocity does not come into play here, it's not representational of how the airplane actually flys; because, as I have said, there is no striaght 'dive' and the engines are not turned off.

But the plane eventually would reach terminal velocity when its drag force is equavalent to g, and therefore putting it in dynamic equilibrium.

One correction, when that happens it is not dynamic equilibrium, it is static. (constant terminal velocity.)
 
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cyrusabdollahi said:
One correction, when that happens it is not dynamic equilibrium, it is static. (constant terminal velocity.)

Thats not static equilibrium. Static equilibrium would be like a ball in the bottom of a bowl. I think this example is static BECAUSE there is no velocity. I think dymamic equilibrium implies a constant velocity.0

You obviously know more about this than I do, so please let me know if I am incorrect.
And I am just wondering, but if you don't think terminal velocity is dynamic equilibrium, then what the heck do you think it is?
 
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A ball at the bottom of the bowl is stable equilibrium.
 
Then what is dynamic equilibrium?
 
I have heard that used in chemistry, not dynamics. Something about forward and backwards reaction rates.

Static equilibrium is when the sum of forces are zero, i.e. constant velocity, i.e. constant terminal velocity.

-dynamic http://en.wikipedia.org/wiki/Dynamic_equilibrium
 
cyrusabdollahi said:
Static equilibrium is when the sum of forces are zero...
I'd call that mechanical equilibrium. (It's only static if it's not moving.)
 
Doc Al said:
I'd call that mechanical equilibrium. (It's only static if it's not moving.)

Thats what I thought.
 
  • #10
Doc Al said:
I'd call that mechanical equilibrium. (It's only static if it's not moving.)

Yes, you are correct as always Doc :smile:. Mechanical is the best term, but not dynamic equilibrium.

When we take a course in 'statics', we work with objects that constant motion or are at rest. So the term is used rather loosely I suppose. It should be called statics and mechanical equilibrium or something of that nature...

wiki said:
Statics is the branch of physics concerned with the analysis of loads (force, moment, torque) on a physical systems in static equilibrium, that is, in a state where the relative positions of subsystems do not vary over time, or where components and structures are at rest under the action of external forces of equilibrium. When in static equilibrium, the system is either at rest, or moving at constant velocity through its center of mass.

Seems like wiki is not clear on this issue either...:smile:

So, either we can both be right, wrong, or agree to disagree. :-p

But all in all, I think you are right in calling it mechanical equilibrium. :wink:
 
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  • #11
It just seems funny to me to call something moving--and possibly spinning--as static. :rolleyes: (But I've never seen the term "dynamic equilibrium" applied to mechanics, except for statistical mechanics.) And you are correct, the term "statics" is used with wild abandon. But I think everyone agrees that when the sum of the forces (and torques) equal zero we have mechanical equilibrium (linear and rotational).

The problem with wiki is that you can find a quote to back anything
wiki entry on "mechanical equilibrium" said:
The special case of mechanical equilibrium of a stationary object is static equilibrium. A paperweight on a desk would be in static equilibrium. A child sliding down a slide at constant speed would be in mechanical equilibrium, but not in static equilibrium.
:wink:
 

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