# Space Elevator Questions and What Ifs

1. Nov 2, 2010

### spancho

Suppose the classic elevator is created, affixed to the Earth somewhere on the equator. The elevator would be perfectly balanced so that there would be no net force i.e. the centrifugal force of the string and end station would equal the force of gravity. What would happen if the cord was severed at the base? It wouldn't be pulled down, cus its spinning fast enough, and it would fly off because it weighs to much. Thoughts?

2. Nov 2, 2010

Not sure

3. Nov 2, 2010

### Filip Larsen

Of the top of my head: In an ideal world it seems that a space elevator most likely would be designed to "float" freely above the surface by shifting weights up and down in the far end as it is loaded in the lower end, perhaps with a slight pull away from surface to allow smaller loads to climb up without needing to shift too much mass in the far end. This also seems like a good safety measure in case of a breach, as the elevator then would rise to a higher altitude where it would have several hours to shift mass around as opposed to having to respond quickly when descending in order to avoid depositing its lower end on the ground.

An actual design detailed to work in the practical world may of course have constraints that make the above observations irrelevant.

4. Nov 2, 2010

### skeptic2

Off hand it seems to me that for a satellite to support the weight of the tether and be geostationary, would have to be a lot farther than 22,300 miles away.

5. Nov 2, 2010

### spancho

@Skeptic2 -> Im pretty sure that the length wouldnt be an inhibiting factor, as you could just increase the mass and have the same effect.

@filip -> It would make sense to put some net force on the tether to increase stability, but it should also be possible to balance it perfectly in an ideal world. This is the situation Im interested in.

6. Nov 2, 2010

### Learnphysics

Well you say that the centrifugal force of the string will equal gravity to keep the thing in orbit.

Things in roughly circular motion have got tangential and normal acceleraiton. So if you decompose the acceleration vector for this thing... you get one pointing towards the center of the earth and one in the direction it's going.

If we cut the string, i think we lose a part of the normal acceleration vector. (Gravity is still pulling it in, so we don't lose all of it). In this case the object should sorta.. spiral out i think.

7. Nov 2, 2010

### spancho

I thought it would be thrown free at first too, but then I realized that since the station at the end is in geostationary orbit, the real force holding it in place is gravity, not the tether. If the tether is then cut, it wouldnt stop the force holding the station in place.

8. Nov 2, 2010

### skeptic2

If the satellite is in geostationary orbit with the tether then it won't be without the tether and vice versa. The thing holding the satellite in orbit is a balance between the earth's gravity on the satellite and the satellite's forward velocity. [You can think of it as the centrifugal force of the circular motion of the satellite in the outward direction being balanced with the earth's gravity in the inward direction.] If we add to the force of gravity on the satellite by adding a tether, then we need to add velocity to the satellite to balance the increased gravity. But if we add velocity the satellite it is no longer geostationary, so we have to put the satellite in a higher orbit.

I the tether breaks, the satellite will go sailing off into a much higher orbit and maybe even escape.

9. Nov 3, 2010

### Filip Larsen

You wouldn't have just a satellite in geostationary orbit and a tether.

The idea of a space elevator is to have the center of mass in geostationary orbit (since the orbit of a structure is "determined" by the location of the center of mass), so if part of the elevator extend down to the surface there must also be mass above geostationary height to keep center of mass in geostationary orbit.

If the elevator is free floating (that is, it does not exchange forces with the surface) then whenever a load is put on the lower end the mass in the upper end must shift up a bit to keep center of mass in geostationary orbit. If the tether is not free floating but attached and "preloaded" at the surface, for instance with a load corresponding to the maximum load you want to raise, then you don't necessarily have to shift mass around in the far end when you load it at the lower end, but in this case if the tether is severed at the surface the elevator will then (everything else being equal) periodically climb and descent for 12 hours each.

10. Nov 3, 2010

### netscr1be

I'm wondering about two other factors.

Isn't the load on the ribbon is going to vary depending on whether the climber(s) are loaded, how heavy that load is, where it is on the tether, and, the direction in which it is moving.

I realize the climber load is probably small compared to the overall mass of the tether but it seems to me that the climber has the potential to vary the relative masses of the other components possibly into/through critical phases (??).

In the case of a break inside Earth's atmosphere. I'm thinking the weather represents a potentially critical external force. Imagine the tether breaking and having a significant external force (typhoon) blowing the lower part apart. Also, attach enough raindrops and/or ice to the tether and there is yet another variable mass.

Adding to the quandary is another major effect of a break - changing the electrical properties of the tether. There are multiple electrical sources operating on the tether like the ionosphere, Van Allen belt energy, static from the climbers and the occasional lightning strike. Doesn't taking the tether out of direct electrical contact with the Earth have to have an effect?

--PB--

Space Elevator Journal http://spacelf8r.blogspot.com/

11. Nov 3, 2010

### Learnphysics

Hmm, so the teather is essentially tension-less? Just a facility for people to get up there?

If that is the case, then cutting the teather should really have no effect on the orbit of the thing.

12. Nov 3, 2010

### JDługosz

No, it's under great tension. It must support its own enormous weight, even without additional anchoring tensions.