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Space station - tension of the structure

  1. Oct 16, 2009 #1
    There is a space station of the length L and uniformly distributed mass M on the low Earth orbit. One side of it always faces the Sun. Which tension will the structure experience?


    My attempt of the solution: bottom of the station is L/2 meters below the stable orbit at that velocity. Therefore, the mass of the bottom of the station has some weight. In fact, any small mass m at the bottom of the station will weight:

    (L/2) / R = m / fg

    L - length of space station
    m - test mass
    R - orbit radius
    f - tension or weight
    g – acceleration of free fall at the height of the orbit

    m is one quarter of total space station weight. Proof: the bottom half of the station has mass M/2. The weight of an object at the height of L/4 will be half of those at the bottom of the station. Therefore, weight M/2, uniformly distributed over bottom half of the station, is equivalent to M/4 at the bottom of the station (in regards to tension).

    After solving the equation, I have got:

    (L/2) / R = M / 4fg
    L / 2R = M / 4fg
    L / R = M / 2fg
    f = M R / 2 g L

    Right or wrong?
    Last edited: Oct 16, 2009
  2. jcsd
  3. Oct 16, 2009 #2


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    In reality I would have thought the differential thermal expansion between the hot and cold side would be more than the gravitational field difference.
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