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This question already has an answer here:

If i have a space elevator with a space station at the point of being in geostationary orbit what would happen to all the stuff inside as the station should be at the "0g" point?

Important info

  • station is 100 floors centered on 0g.
  • all floors are built with the earth under foot.
  • normally has artificial gravity but is not working now.
  • has been left alone for 50 years.

Assumptions im making based on googled info.

Geo stationary orbit is about 35,786 km above the earths equator and below Geo sat gravity would pull everything down eventually. above Geo sat the centrifugal force would pull everything up away from earth.

Based on that info I'm not sure given the time frame what to expect the interior of the station to look like. would everything have already been pulled in there respective directions or would the forces involved be low enough to leave objects in free fall for that long of period.

EDIT. issue of a possible duplicate has been raised. to clarify i am asking specifically about the gravity effects of a geo sat orbit near the 0g point

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marked as duplicate by Separatrix, Secespitus, L.Dutch, sphennings, Ash Nov 20 '17 at 12:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I don't have the scientific knowledge to back up a full answer, but I would think that the orbital momentum would have a greater effect on free floating objects than either the Earthward microgravity or the outward centrifugal force. I think that most things would be collected, in each room, along the walls which are farthest from the orbit-path-facing "front" end of the ship. $\endgroup$ – Henry Taylor Nov 18 '17 at 9:35
  • $\begingroup$ its very very similar but i feel its not a duplicate as im asking specifically about what happens because of gravity effects in geo sat orbit near the 0g point. that question is asking about just any space station in some orbit and issues related to decompression. $\endgroup$ – Arthur Durbin Nov 20 '17 at 11:48
  • $\begingroup$ @ArthurDurbin, I can't distinguish between the answers that would be correct to these two questions, what is correct for one is correct for the other, hence considering them duplicate. The decompression is mentioned to remove issues related to air movement and resistance rather than as a specific factor. Specific orbit is also irrelevant to the fundamental calculations involved. $\endgroup$ – Separatrix Nov 20 '17 at 12:22
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As You correctly guessed the "orbit plane" is actually a surface (no thickness), so the individual atoms of the space station would, if separate, be on different orbits and drift away over (short) time; of course they are kept together by direct interaction so structural strength of your space station won't have any problem in keeping it together.

This doesn't hold true for any free-floating object which, unless prevented, will try to follow his own orbit. This means objects "above" the "mean orbital surface" will drift to the ceiling, while those "below" (in Earth direction) will drift toward the floor.

Let us try to compute the actual forces.

On orbital plane (assumed being the Station Center of gravity) the two forces, gravity ($F_g = \frac{G M m}{r^2}$) and centrifugal ($F_c = m v^2/r$) are exactly the same "by definition" . This enables us to easily find relationship between speed and radius of a circular orbit.

What we need to do is a bit different, we need to find what is the difference of force if we change "a bit" the radius leaving the speed the same.

So we have:

$\Delta F = F_g^\prime - F_c^\prime = \frac{GMm}{(r+\delta)^2} - mv^2/(r+\delta) = ma$

$a = \frac{GM}{(r+\delta)^2} - v^2/(r+\delta)$

If we substitute the relevant data:

$G = 6.67 \times 10^{-11}$

$M = 5.972 \times 10^{24} kg$

$r = 35786 Km = 35786000 m$

$v = 3.07 Km/s = 3070 m/s$

and assume for $\delta$ 50 floors about $3m$ each: $\delta = 50 \times 3m = 150m$ we get:

$a = \frac{6.67 \times 10^{-11} \times 5.972 \times 10^{24}}{(35786000+150)^2} - 3070^2/(35786000+150) \approx 8 \times 10^{-7} m/s^2 $

Note: I had to "cheat" to get the result; the values I have for all constants are way too imprecise to correctly compute the result (we are speaking about $4 ppm = 4 \times 10{-6}$). What I did is to "massage" radius till the acceleration ($\delta = 0$) was null, then I added the displacement. Someone is welcome to cross-check my computations as I might have goofed somewhere.

From basic Physics we have:

$s = 1/2 at^2$ ... $t = \sqrt{2 s / a} = \sqrt{\frac{2 \times150}{8 \times 10^{-7}}} \approx 8470s \approx 2 h 20m$

As you see the "free floating" objects will drift to floor/ceiling in a matter of hours, days at most for objects nearer to orbital plane.

Note that the forces involved are quite tiny, so even the smallest friction will prevent it, so, in particular, you can expect surface tension to keep liquids plastered to whatever surface they touch and not to drift around.

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The whole station would be in a microgravity environment. But there would be some very small effects. Objects at the exact geostationary would remain in freefall. Objects in the “upper floors” above geostationary orbit would have excess velocity compared to their position just above geostationary orbit so they would tend to move very slowly up to the ceiling. Objects on the lower floors would have insufficient velocity for their position just below geostationary orbit so would tend to sink down to the floor.

This effect should be measurable but would be extremely small and likely to be overwhelmed by many other factors such as gravitational anomalies (the earth is not a perfect sphere and the force of gravitation changes a little depending on where you are on the surface). Thermal motion of the air would probably also stir the objects around causing a much bigger disturbance than the tiny gravitational effect.

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