Space station - tension of the structure

  • Thread starter Privalov
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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:

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,774
13
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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