Yesterday I had 5 answers to this question. Today it's only 4. The missing answer had quite a few upvotes, but also involved a lot of math I was hoping to learn from but didn't have the chance to study.

Is it possible to read the missing answer? I assume the author deleted it himself.

• It was deleted by the author after getting a couple comments suggesting errors. I don't know whether the author intends to rework it and undelete or just move on. – Monica Cellio Jul 13 '17 at 18:25
• I see, thank you @MonicaCellio. It would be interesting to know the gist of the "errors", but unlikely I could verify accuracy. Respect author's wishes, but the question got both yes and no answers so I'm still not sure a correct answer. I'll wait and maybe he'll update…. In a few months maybe I can see it myself. – wetcircuit Jul 13 '17 at 19:02

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It might be unclassy of me to do this, but here are the deleted math parts. Please read my note at the bottom.

The specific energy (i.e., $\text{J}/\text{kg}$) of your habitat, which was constructed "in orbit", comes from the Vis-viva equation: $$\epsilon = \frac{1}{2} v^2 - \frac{G M}{r}$$ This is the amount of energy your habitat ring has initially, per $\text{kg}$. Now, when we decrease the radius, the energy has to remain constant (that's like, a law, bro): \begin{align} \epsilon_{new} &= \epsilon_{old}\\ \left(\frac{1}{2}v_{new}^2 - \frac{G M}{r_{new}}\right) &= \left(\frac{1}{2}v_{old}^2 - \frac{G M}{r_{old}}\right)\\ \frac{1}{2}v_{new}^2 &= \frac{1}{2}v_{old}^2 - \frac{G M}{r_{old}} + \frac{G M}{r_{new}}\\ v_{new} &= \sqrt{ v_{old}^2 - \frac{2 G M}{r_{old}} + \frac{2 G M}{r_{new}} } \end{align} Subbing in the velocity of a circular orbit: \begin{align} v_{new} &= \sqrt{ \frac{G M}{r_{old}} - \frac{2 G M}{r_{old}} + \frac{2 G M}{r_{new}} }\\ v_{new} &= \sqrt{ \frac{2 G M}{r_{new}} - \frac{G M}{r_{old}} }\\ \end{align} This is the new velocity with which your ring rotates.

Now, if your ring had been instead constructed in the lower orbit, it would have had velocity (same formula):$$v_{lower} = \sqrt{ \frac{G M}{r_{new}} }$$ Let's compare those two velocities: \begin{align} \left( v_{new} \right) &\stackrel{?}{>} \left( v_{lower} \right)\\ \left( \sqrt{ \frac{2 G M}{r_{new}} - \frac{G M}{r_{old}} } \right) &\stackrel{?}{>} \left( \sqrt{ \frac{G M}{r_{new}} } \right)\\ \sqrt{G M} \sqrt{ \frac{2}{r_{new}} - \frac{1}{r_{old}} } &\stackrel{?}{>} \sqrt{G M}\sqrt{ \frac{1}{r_{new}} }\\ \frac{2 r_{old} - r_{new}}{r_{old} r_{new}} &\stackrel{?}{>} \frac{r_{old}}{r_{old} r_{new}}\\ 2 r_{old} - r_{new} &\stackrel{?}{>} r_{old}\\ r_{old} &> r_{new} \end{align} Therefore, if $r_{old} > r_{new}$ (which it is, since we're decreasing the radius), then the ring will spin at a velocity ($v_{new}$) that is faster than the velocity ($v_{lower}$) it would have had, had the ring been constructed in the lower orbit in the first place.

This means that the ring is moving faster than necessary to stay in orbit, and objects within it must therefore be pulled inward by the habitat ring itself. They experience this as an outward acceleration.

Now on to the reason for deletion: this post is wrong. As someone pointed out in comments, the change in orbital radius involves work being done. Something has to apply the work to this station to move it from the outer orbit to the inner orbit and therefore the total kinetic + potential energy of the station is not conserved.

The correct approach is conservation of angular momentum from Kyle's answer.

I hope I'm not abusing my power to see the deleted answer here; the owner who deleted it was right and had the right to do so, but since you seemed interested in the math itself, and I like people who want to learn about math, I felt compelled to share.