Orbital Bridges 26th December 2000 Back to Bridges Let's imagine a small spherical planet with radius R and density D. The force per kilogram mass, g, at the surface, will be 4piRDG/3. Let's build a bridge around the planet, about a metre above the surface. We won't provide any supports. Because the bridge is parallel to the surface, we might think it is a beam, but because it experiences a compressive force and is curved, we might think it is an arch.
If the bridge is stationary, what is the compressive stress in it? Stationary compared with what? The usual answer is "the distant stars", so that's what we will assume. If the bridge can resist this stress, and remain unbuckled, will it be stable, or will it move until it touches the planet, as shown in the second picture? Which configuration has the lower energy? What seems certain is that if the bridge
rotates with the correct orbital speed, V = (gR) Someone getting on the bridge would slow it down, but they could speed it up later by jumping off backwards. Nevertheless the bridge would lose energy by friction incurred during the getting on an off. This energy would have to be replaced by an external source. To avoid loss of weightless passengers, the bridge should rotate at faster than the orbital speed, putting it into tension. Passengers would then be held on to the underside of it. It wouldn't be either an arch or a beam. Perhaps it would be a suspension bridge. By putting the bridge into a polar orbit, people could get from any point to any other point on the planet, by walking along the bridge and getting off at the right place. Obviously, they could also get from any point to any other point by walking. Over what fraction of the planet's area would they save walking distance by using the bridge. Given a planetary density of 5000 kg /
m
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