Space station - tension of the structure

[Privalov]

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

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?

 

[mgb_phys]

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

 

[astrokreng]

I can say that your solution appears to be correct. The tension experienced by the structure of the space station will be directly proportional to the mass of the station, the radius of its orbit, and the length of the station. This is because the tension is caused by the gravitational force between the station and the Earth, and the greater the mass and length of the station, the greater the force will be. Similarly, the tension will be inversely proportional to the acceleration of free fall at the height of the orbit, as a higher acceleration would result in a smaller gravitational force. Therefore, the equation f = MR / 2gL accurately represents the tension experienced by the space station.