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Gravity in the upper atmosphere

  1. Jan 18, 2013 #1
    1. The problem statement, all variables and given/known data

    See attachment

    2. Relevant equations

    Centripetal acceleration Mv^2/r

    3. The attempt at a solution

    I guess the answer to be D

    But I am not quite sure about the same rate of acceleration.
    If we hypothetically remove the space station and visualize the astronaut orbiting around the Earth alone (in his spacesuit), again according to newton's law of gravitation he must feel weightlessness (wouldn't he ? ). This would happen when his entire weight is providing the Centripetal Force required for acceleration at a particular velocity .

    So does this explanation not render choice D as incorrect ? So the next correct choice looks "A" but again it is flawed since Earth's gravitational field is theoretically infinite .

    Attached Files:

  2. jcsd
  3. Jan 18, 2013 #2


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    You would also feel weightless in an elevator in free fall. Why is that?
  4. Jan 19, 2013 #3
    If the elevator is accelerating downwards at 9.8 ms^-2 , we would essentially be floating in the elevator. This is because the entire weight is used to provide the acceleration .
  5. Jan 19, 2013 #4


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    Sort of. How does that make you feel about answer D?
  6. Jan 19, 2013 #5
    I am not sure, but the one thing I am completely certain of is that if the space station was accelerating at v^2/r (v = velocity of space station and m = mass of space station)

    and mv^2/r = weight of the astronaut = GM/r^2 * mass of the astronaut

    then she would be weightless !

    I really cant find a connection between the elevator and this situation except that the weight provides the Centripetal force . (entirely)
  7. Jan 19, 2013 #6


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    They are both the same. The acceleration in orbit is v^2/r. Both the astronaut and the station are accelerating at the same rate since they both moving at speed v. Weightless. If you are falling in an elevator then the acceleration of the elevator is g, and you are accelerating at the same rate. So weightless.
  8. Jan 19, 2013 #7
    Actually, I am still confused about the way you approached the problem.

    Why aren't you taking the Normal reaction force from the surface of the space station into consideration ?
    Last edited: Jan 19, 2013
  9. Jan 19, 2013 #8
    There is reaction to anything that moves relative the station and bumps into its walls. Notably, the molecules of air.

    Anything that is stationary is in free fall just like the station itself, and there are no forces, save for microscopic mutual gravity.
  10. Jan 19, 2013 #9
    What the question is getting at is that both she and the space station are exactly as you analyzed it. The question is just saying that, when she is inside the space station, she perceives this "weightlessness" more than if she were out in free in space. The same goes for the elevator example.
  11. Jan 19, 2013 #10
    I would like to turn my attention towards the use of the word "perceive" .

    It shows that she would be theoretically weightless in both situations (space station and free space) but it would be easier for her to experience (notice) weightlessness in the space station .
  12. Jan 19, 2013 #11


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    My answer is E. I am accelerating at the same rate as the room I'm in, but I don't experience weightlessness.
  13. Jan 19, 2013 #12


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    Ok, good call. Is answer E the only net force acting on the system is gravity? That would define 'free fall'. I could sign on with that answer.
  14. Jan 19, 2013 #13
    Why do u think that would be so ?

    Is it because you are given the information about the magnitude of the centripetal acceleration ? And thus you would be expecting some Normal reaction force from the floor of the space station ?
  15. Jan 19, 2013 #14


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    Nothing so complicated. I sit in my room on the earth's surface and I drop a pen. It falls. Not weightless. Yet neither the room nor I are accelerating. haruspex is pointing out that answer D is defective in a way I overlooked.
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