1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?

    long_space_station.png

    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

    where:
    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

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    In reality I would have thought the differential thermal expansion between the hot and cold side would be more than the gravitational field difference.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Space station - tension of the structure
  1. Rotating Space Station (Replies: 18)

  2. Smoke in a space station (Replies: 22)

Loading...