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Weightlessness in orbit

  1. Jul 14, 2011 #1
    To explain weightlessness to my students, I use the analogy of a falling elevator (lift) as folows - if we stand on weighing scales in an elevator and the elevator cable snaps, the scales will show zero since they are falling to the ground at the same acceleration as us, therefore registering no reaction force.

    However, the sensation of being in a lift hurtling to the ground (as anyone who has been on those free-fall rides at amusement parks knows) is surely not that of floating in the air as we see astronauts doing. The difference seems to be that a falling lift has only a vertical velocity while an orbiting shuttle has a horizontal (tangential) velocity also.

    In other words, if my falling lift were to have a horizontal initial velocity component also (ie, were thrown as a projectile parallel to the ground), and that component were imagined to be made larger and larger until it covered a horizontal distance equal to the circumference of the earth in the time it took to 'fall' to the ground (which it never would), then we would experience in the lift the same 'floating' sensation that the astronauts undergo, because the lift would be in orbit (at the small height above ground from which it had when thrown).

    Is that a correct way of putting it? Isn't the orbitting shuttle really only a falling lift thrown with a large enough horizontal velocity?

    Ranjit
     
  2. jcsd
  3. Jul 14, 2011 #2
    I don't think so. Constant horizontal motion feels exactly the same as no motion. Simply free falling can do the job. The reason why it doesn't feel so in amusement parks is because you are tied up in the seat and there is actually resistance to the falling.
     
  4. Jul 14, 2011 #3

    Drakkith

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    Staff: Mentor

    Yes, the space shuttle and a falling elevator are exactly alike other than horizontal movement.

    See here for more: http://en.wikipedia.org/wiki/Free_fall
     
  5. Jul 27, 2011 #4
    If the free-falling amusement ride was fully enclosed to prevent atmospheric interaction with its on-board thrill seekers and the free-falling amusement ride was actually forced to accelerate downward precisely at gravity’s rate of 9.8 m/s^2 (thereby forcing the free-falling ride to overcome the ever-increasing atmospheric air resistance normally encountered), then its on-board thrill seekers would experience the very same weightless effect that is experienced by astronauts while in a sustained earth orbit. No horizontal component whatsoever is required to experience the “weightless” effect.
     
  6. Jul 28, 2011 #5
    Hmmm. If just dropping to the ground vertically like a stone, would one be able to do large horizontal movements like the shuttle astronauts are seen doing, swimming through their space? Curious to know if skydivers can do these during the free-fall part of their drop.

    Also, the relevance of the horizontal movement surely comes into play to make the "drop" last indefinitely, doesn't it? A vertical drop from the shuttle's orbiting height (around 350 miles?) would take about 45 minutes using s = 1/2 g t^2 as an approximation. Shuttle orbits last for days.

    Ranjit
     
  7. Jul 29, 2011 #6

    Nabeshin

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    Science Advisor

    Sure, if the skydivers were not experiencing air resistance and had something to push off of, no reason they couldn't emulate the motions you see people in orbit do.
    Right, but the physical sensation of free fall is no different.
     
  8. Jul 29, 2011 #7

    Ryan_m_b

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    Object's thrown fall at the same rate as object's dropped. Ignoring air resistance if I dropped a ball from a height of 1m at the same time as launching a ball horizontally from 1m they would both hit the ground at the same time, even though the thrown ball would land further away. The difference comes in when we change the surface I'm standing on from a hypothetical infinite plane to a sphere. Now the horizontally thrown ball takes longer to fall because the curvature of the sphere increases the distance between the thrown ball and the surface, if the curvature of the sphere matches the rate of fall then we have orbit[1]. This is http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/newt/newtmtn.html" [Broken].

    [1]I've attached an illustration to help.
     

    Attached Files:

    Last edited by a moderator: May 5, 2017
  9. Jul 29, 2011 #8
    NASA has the educator edition "Exploring Space through Algebra" that addresses Weightless Wonder – Reduced Gravity Flight: http://www.nasa.gov/pdf/264005main_Algebra_Edu_C9.pdf [Broken]

    The OP might find it useful. Very informative.:smile:
     
    Last edited by a moderator: May 5, 2017
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