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?