# What was the orbital radius of the Elysium toroid?

What was the orbital radius of the Elysium toroid?

When the first group of 3 ships sent by Spider were traveling towards Elysium, the movie cut to Defense Secretary Delacourt at a party at her home. She left the party and went to the command center. Once she had arrived, there was an announcement that said the distance was 15,000 Kilometers, and at the same moment, it flashed to a read out of

ETA 12:43.999 seconds

Later when Kruger was traveling to Eylisium, he spoke to her briefly before departure, and she said:

See you in 19 minutes.

Spider left shortly thereafter in one of his own ships, and arrived shortly after Kruger (suggesting the two craft were about equivalent in acceleration/speed).

So it should, I suspect, be possible to work out the total distance based upon that.

## Side notes

1. I had originally thought that as they were cutting from the first 3 ships still accelerating, to when someone moments later said they were approaching Elysium that the movie had made a goof, and the ships should be decelerating at the time. Given they were still 15 minutes away suggests not, and the 'approaching' did not mean 'close' but simply 'on a trajectory towards'.
2. The ships apparently had some sort of inertial dampener to counteract the effects of acceleration, as again while the 3 ships were still accelerating, the movie cut to a view inside one of the ships and showed objects floating around the cabin. (Oh, those Earthbounds, you can dress them up in a false bar-code but you can't take them anywhere without them littering the dang ship..)
• Dont they mention, at some point in the movie, that there is 12 minutes or whatnot to contact when some shuttles from earth are arriving? Im thinking your estimate of 5 minutes is waaay off. – Jakob Jan 10 '14 at 7:36
• @Jakob Thanks for the tip! I'll watch it again tonight and see if I can narrow/pin that down, since it is the most nebulous aspect of the entire equation. – Andrew Thompson Jan 10 '14 at 7:40

Andrew and I have been discussing this on the Physics SE. This is possibly not an appropriate answer hereabouts since it's science not science fiction, but in case anyone is interested here is the calculation of the orbital radius.

We don't know:

• the total distance, call this D

• the acceleration, call this a

We know:

• the total time for the journey, T

• at a time t the distance remaining is 15,000km

To find D:

The first half of the journey starts from a standstill, so the equation for the distance travelled is:

``````s = 1/2 a t^2   (1)
``````

and rearranging gives:

``````a = 2s/t^2      (1a)
``````

At the halfway stage the time is T/2 (which we know) and the distance is D/2 (which we don't know) so:

``````a = 2 D/2 / (T/2)^2 = 4D/T^2
``````

Now go back to equation (1) from above and substitute this value for a:

``````s = 1/2 4D/T^2 t^2 = 2Dt^2/T^2    (2)
``````

Now we know that with 12m40s left to go the distance left to go is 15km. The elapsed time t is 19 min - 12m40s = 6m20s or 380 seconds. The distance travelled at this time is s = D - 15,000km. So put these values into equation 2 above to get:

``````D - 15,000km = 2Dt^2/T^2
``````

and rearranging this gives:

``````D = 15,000km / (1 - 2t^2/T^2)
``````

We know t = 380s and T = 1140s so we get D = 19,285,714.29 m or about 19,300 km. We can then calculate a from eqn (1a) and it works out as a shade over 6g.

I was thinking that you could work this out by estimating the angle of view the ring takes in the sky and comparing it to it's diameter, but when I went to find the information I needed I immediately came across this:

http://blog.wolframalpha.com/2013/08/16/elysium/

He works out the distance above the earth's surface as only 4000 or so miles, which places it well below a geosynchronous orbit.

• Interesting, especially the information on the size of the toroid, thanks for the link. But one of my (eventual) points was going to be confirming that it was depicted much larger, from the ground, than it could actually appear. Note that one of the few indications of distance in the movie was an announcement while Delacourt was in the defense center that the approaching vehicles were 15,0000 Kilometers (approx. 9,320.6 miles) away. Logically Spider would have launched them when it was closest, minimizing trip time and chance of ..interception. I want to calculate distance from that. – Andrew Thompson Jan 20 '14 at 22:36
• Ughh.. by 15,0000 Kilometers I meant 15,000 Kilometers. :P – Andrew Thompson Jan 20 '14 at 22:47
• I made a comment on the discrepancy between the figures on the blog article.. – Andrew Thompson Jan 31 '14 at 5:31

I decided to take the calculations through to the logical conclusion.

## Given

Distance from Los Angeles = 19,285.71429 Km
Latitude of L.A. = 34o
Equatorial radius of Earth = 6,384 Km
Earth Mass = 5.97219*1024 Kg