5 ms·
You are sidestepping the fact that in your OP you demonstrated and argued that the silk could not be used for space elevators because it could not support its w
by Maursault 4y ago
You are sidestepping the fact that in your OP you demonstrated and argued that the silk could not be used for space elevators because it could not support its weight beyond 14.6km, but I argued it would not need to since that weight will be countered by centrifugal force, so whether the silk can be used for a space elevator has nothing whatsoever to do with supporting its own weight, rather, it must support the centrifugal force.
If the silk only need be kept "only" taut, its weight will be perfectly balanced against centrifugal force, so in effect the silk will be weightless. But my suspicion is the tether would necessarily need to be kept taut with extra force, not "only" taught, but with force to put stress on the silk tether pointing outwards and upwards.
I have no idea how one would calculate the required centrifugal force, but perhaps you'll do us all the favor and determine whether or not the silk would be strong enough without being distracted by the weight of all the silk, which is irrelevant due to it being cancelled by centrifugal force.
- ncmncm 4y agoThe correct figure, as noted elsewhere, is 146 km. But literally the only thing holding up almost all of the span from geosynchronous orbit down to the ground is the pure strength of the cable. Centrifugal force would act usefully mainly on parts of the cable that extend out past geosynchronous orbit, to support the whole structure through tension in the cable. The cable inside that orbit would absolutely not be weightless. Its weight per unit mass is of course lower close to geosynchronous orbit, but most of the cable is very far from it.
- Maursault 4y agoEffectively weightless. The weight of any part of the tether must be entirely counteracted by centrifugal force, otherwise the weight of the tether would pull everything to the ground.
- ncmncm 4y ago"Counteracted" by force through tension in the cable. Which relies on the cable holding up its own weight. So, in no way "effectively weightless", unless you consider a regular bridge deck weightless because cables are holding it up.
- Maursault 4y ago> "Counteracted" by force through tension in the cable. Which relies on the cable holding up its own weight. So, in no way "effectively weightless", unless you consider a regular bridge deck weightless because cables are holding it up. No. That is ridiculous and wrong. The cable is not holding up its own weight, the centrifugal force is. The tension comes from the anchor point and the centrifugal force only. Weight is entirely counteracted and is no longer a consideration.
- ncmncm 4y agoI leave it to someone else to explain things to you. I can only pray that no one's safety depends on your understanding of physics.
- afiori 4y agoLet's ignore gravity for a moment. Think of a rotating chain in space, the chain is weightless, but each ring needs to apply enough centripetal force to counter the centripetal force. Lets focus on the barycenter of the chain. The centripetal force is limited by the strength of the chain; the centrifugal force is a function of rotation speed, length of the chain, and linear density of the chain. this means that for any combination of chain type and rotation speed if the chain gets too long it will break. With a geostationary space elevator you need to build a part of it over geostationary orbit and a part of it below geostationary orbit. the part above will pull the part below and make it "float". The problem is that the part below needs to be ~100 km long and has still has a weight. Think of the section of the tether at geostationary orbit as a giant weightless chain link. The issue is not having that chain link remain suspended, the problem is to have it not break while trying to balance the huge forces.