Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Artifificial zero gravity environment

  1. Sep 16, 2006 #1
    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 earths 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).
    Last edited: Sep 16, 2006
  2. jcsd
  3. Sep 16, 2006 #2
    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.
  4. Sep 16, 2006 #3
    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.

    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.

    One correction, when that happens it is not dynamic equilibrium, it is static. (constant terminal velocity.)
    Last edited: Sep 16, 2006
  5. Sep 16, 2006 #4
    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?
    Last edited: Sep 16, 2006
  6. Sep 16, 2006 #5
    A ball at the bottom of the bowl is stable equilibrium.
  7. Sep 16, 2006 #6
    Then what is dynamic equilibrium?
  8. Sep 16, 2006 #7
    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
  9. Sep 16, 2006 #8

    Doc Al

    User Avatar

    Staff: Mentor

    I'd call that mechanical equilibrium. (It's only static if it's not moving.)
  10. Sep 16, 2006 #9
    Thats what I thought.
  11. Sep 16, 2006 #10
    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.....

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

    So, either we can both be right, wrong, or agree to disagree. :tongue2:

    But all in all, I think you are right in calling it mechanical equilibrium. :wink:
    Last edited: Sep 16, 2006
  12. Sep 17, 2006 #11

    Doc Al

    User Avatar

    Staff: Mentor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook