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*If* the ISS is freely falling, doesn't that mean it's following

  1. Jun 17, 2013 #1
    ...curvatures in spacetime being created by earth's gravity?

    My understanding is that its speed offsets the gravity being created by earth, that it's freely falling towards earth but never hits it because its speed somehow causes it to keep missing. If that's the case, wouldn't that suggest that the ISS is under the influence of earth's gravity (but that it just constantly maintains its orbit because of speed)?

    Thanks.
     
  2. jcsd
  3. Jun 17, 2013 #2

    mfb

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    As seen by general relativity: Right.
    I don' think "offset" is the right word here.
    That is better.
    Of course it is!
     
  4. Jun 17, 2013 #3

    PeterDonis

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    I would say the curvature in spacetime *is* Earth's "gravity"; they aren't two separate things, one of which is created by the other. However, that's really a side issue here.

    The more important point is that it's curvature in *spacetime*, not space. In spacetime, the path of the ISS is not an ellipse; it's a helix, because time is one of the dimensions. The Earth's path in spacetime is just a line that runs upward (assuming we think of the time direction as "upward") through the center of the helix.

    Since the Earth is the primary determinant of spacetime curvature in its vicinity, yes, the ISS's trajectory, given its speed at some particular instant, is primarily due to the Earth.
     
  5. Jun 18, 2013 #4
    Peter, I have a question related to this that may seem trivial but would help me understand this some more.

    If the ISS were closer to the earth's surface, would it have to speed up or slow down to maintain its smooth orbit around the earth?
     
  6. Jun 18, 2013 #5

    PeterDonis

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    The orbital velocity gets larger as the radius gets smaller, so if the ISS were closer to the Earth, it would need to speed up to maintain orbit.
     
  7. Jun 18, 2013 #6

    WannabeNewton

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    You can answer that using Newtonian gravity. ##\frac{GM_{e}}{r^{2}} = \frac{v^{2}}{r}\Rightarrow v^{2} = \frac{GM_{e}}{r}## so smaller ##r## means larger ##v##.

    EDIT: Peter won this one :frown:
     
  8. Jun 21, 2013 #7
    This reminds me of Olympic ski jumpers and how they can jump so far. :smile:
     
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