3
$\begingroup$

How wide can you make an orbital ring?

This question is simple. The orbital ring is a concept in which a ring encircles the earth (or other celestial body) and revolves around it so that the centrifugal force cancels out the gravitational pull. In a nutshell. A geostationary orbit is achieved at an altitude close to 35,786 kilometres (22,236 miles) and directly above the equator. That's a circumference of about 224,850 kilometres (I don't trust my math). The thickness of the cable itself is heavily dependent on the materials being used, but I'd opt for what we use in satellites today.

I like to think of orbital rings as a train of satellites that have been linked together to form a coherent ring. I mean how else would you assemble the thing? So orbital ring modules makes the most sense to me. That means once the rings construction has been finished you could keep adding modules to the ring.

My design has hexagonal solar cells, so that they can be fitted together efficiently and in two dimensions. Essentially, after centuries what used to be a ring has turned into a branching structure. This is very similar to the geometry of snowflakes and the way they grow. Think of it as an orbital swarm coming together to form an orbital ring. Personally I like this structure because it gives a cool flat surface to walk on. That's pretty much about it.

When is too much too much?

Theoretically the orbital swarm could grow forever eventually covering the entire earth. That is if gravity doesn't come into play. My assumption is that the wider you make it the less centrifugal force you get. There should be a limit. As judges of sci-fi what is your take on this? What is the limit?

$\endgroup$

2 Answers 2

3
$\begingroup$

It's all in the details

First, no genuinely rigid orbital ring will be stable, without help. See this question - Niven had to add attitude jets. (But he took them away again...)

Second, it is true that the orbital ring must resist a force at either edge. However, this force could be managed by turning the satellites into statites, which is to say, giving them each their own method of maintaining position. Notably, mirrors have been made up to 99.9999% reflective (see some of the Breakthrough Starshot discussions...) so you could bounce a beam of light between two corner reflectors at equally displaced edges of the ring many, many times to provide a reasonably large amount of light pressure from a weak beam of light that will not harm folks who get in the way. (Because it takes many, many such beams working together, no temporary obstruction should interfere with the ring)

Additionally, your shell can be constructed like a wicker Christmas tree ball, with rings running in many directions and only intentionally empty spaces. Each ring would press against some other which it moves beneath, perhaps using light pressure, and then orbital mechanics would bring it back outward again at another location. A system of rings of this type, given suitable elevators to moderate the acceleration, could transport people throughout the entire mini-Dyson sphere quite conveniently.

$\endgroup$
1
$\begingroup$

The orbital velocity is given by the formula $v=\sqrt{GM/R}$.

If the ring section taken along the plane containing the ring diameter and the axis of rotation is a straight line, any part not on the median line of the section will be subjected to a force, given by the difference between the orbital velocity it should have due to the slightly different R it has and the orbital velocity imposed by the rigidity of the ring.

In a first approximation formula, calling h the distance from the median and $\theta$ the angle under which h is seen from the central attractor $\Delta v = \sqrt{GM/R} - \sqrt{GM/R'}$, where $R'=R/cos\theta=\sqrt{R^2+h^2}$. The acceleration produced by this differential velocity can be calculated as $\Delta v /\omega$, where $\omega$ is the rotation frequency of the orbital ring.

When this force exceeds the resistance of the structure of the ring, you have reached your limit.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .