- #1

- 24

- 0

There is a space station of the length

My attempt of the solution: bottom of the station is

where:

After solving the equation, I have got:

Right or wrong?

**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 LL / 2R = M / 4fg

L / R = M / 2fg

f = M R / 2 g L

Right or wrong?

Last edited: