- 14 Jun '13 16:24So I was wondering what kind of acceleration you would feel if you were on a space elevator at geo synchronous orbit, call it 22,000 miles up or 35,200,000 meters high. That plus the radius of the Earth, call it 4000 miles or 6,400,000 meters (these numbers are close enough for my purposes) and we have about 9.8 meters per second squared sitting on the Earth. But on a space elevator, my thoughts went like this:

You would have the opposite forces of gravitational attraction at that altitude Vs the centripetal acceleration away from Earth, so using the numbers provided by google, GM is 3.985617248E14 and dividing that by r squared, using 4000 miles or 6,400,000 meters I get 9.73 m/sec^2. Close enough to 9.8 for me for now. So extending that as if you were sitting on a rod 22,000 miles out from the surface of the Earth, you get an altitude of 41,600,000 meters from the center of the Earth. So GM/r^2 up there gives you 0.23 meters/sec^2.

So the next step to do is to calculate the centripetal acceleration using V^2/r.

I got the circumference of 261,380,508 meters using my original numbers.

So divide that by 86,400 and I get a velocity of that orbit of 3025 meters per second. Squared=9152061 and I get the centripetal acceleration of 0.22 meters per second squared.

So the 0.23 minus 0.22 almost cancels each other out and you seem to be darn close to weightless if you were floating around in a space elevator. Does this sound correct?

And is there a better way to do this calculation? - 14 Jun '13 18:53If the space elevator is at a pole the occupant would feel some weight. If the elevator is on the equator, as you seem to be figuring, the occupant would be in free fall. An equatorial geosynchronous satellite hovering constantly a few meters away is obviously in free fall, and there is no reason to think that the occupant of the elevator would behave any differently just because he is enclosed in some kind of cage.

If the equatorial elevator puts the occupant up even higher, he is going to have to plant his feet on the ceiling. - 14 Jun '13 20:56

Yes, you've calculated the force on someone whose motion matches a satellite in a geostationary orbit, it had to match the orbital condition on the satellite.*Originally posted by sonhouse***So I was wondering what kind of acceleration you would feel if you were on a space elevator at geo synchronous orbit, call it 22,000 miles up or 35,200,000 meters high. That plus the radius of the Earth, call it 4000 miles or 6,400,000 meters (these numbers are close enough for my purposes) and we have about 9.8 meters per second squared sitting on the Eart ...[text shortened]... a space elevator. Does this sound correct?**

And is there a better way to do this calculation? - 14 Jun '13 21:56 / 2 edits

One thing weird to me about the concept of the space elevator is this:*Originally posted by DeepThought***Yes, you've calculated the force on someone whose motion matches a satellite in a geostationary orbit, it had to match the orbital condition on the satellite.**

If you could imagine instead of a cable, you have a building 22,000 miles high.

The building can take the huge winds and so forth at the bottom and it is super strong. So there you are on top of this building and you have this combination of Earth attractive force and slinging away force and it looks like they more or less balance out.

But with a space elevator they say you want a much longer dangler to balance out the forces or something. The thing I can't figure out is how do you make an object be in a stable orbit if the one big bang station is at geo altitude and the other end of the balancing act is 60,000 miles high. If you could picture a building 22,000 miles high and then one 90,000 miles high and super strong the outer end would certainly have a huge centripetal force, haven't done the arithmetic on that one yet. The thing about that imaginary object is it is all one piece and flies around the Earth as a giant building so it hangs together.

Now look at the elevator concept. Here you have the big bang at geo altitude and then another 60,000 miles of dangling line with a big weight at the end. How could it possibly not get tangled up? There is no horizontal stability in a cord and if you put something another 60,000 miles out on a cable and the inner section is quite happy going one revolution per day, the outer end could not have the same orbital velocity and would drag behind. So I can see this thing dragging behind so far the cable could theoretically slam into the elevator station at the geo altitude. What could keep the outer end in sync so it doesn't do that? I have imagined some kind of permanent rocket powered by solar energy, like a Vasimir motor or something that would force the thing to keep up with the lower station but without that how could the thing possibly be stable?

If you read about space elevators they always talk about a stabilizing object something like 100,000 miles up or some shyte. - 15 Jun '13 14:26

A building that tall would melt the ground under it, anchoring a cable would also be a problem. Neglecting atmospheric, solar wind and inductance effects from the earth's magnetic field on the wire...*Originally posted by sonhouse***One thing weird to me about the concept of the space elevator is this:**

If you could imagine instead of a cable, you have a building 22,000 miles high.

The building can take the huge winds and so forth at the bottom and it is super strong. So there you are on top of this building and you have this combination of Earth attractive force and slinging away ...[text shortened]... rs they always talk about a stabilizing object something like 100,000 miles up or some shyte.

The centripetal force is balanced by the tension in the cable as well as gravity, consider an element of wire the tension pulling upwards at the top of the wire is T+dT and down at the bottom T. The linear density is D and the radial coordinate r, r=0 is the centre of the earth. w is angular velocity, and is in radians per second, it is the same all along the wire and the instantaneous linear velocity v=wr:

Dw^2rdr + dT = GMDdr/r^2

So

dT/dr = D(GM - w^2r^3)/r^2

. = 0 at the geostationary point

d^2T/dr^2 = -D(2GM/r^3 + w^2) < 0

So the tension is maximum at the point of the geostationary orbit. I could integrate that lot to give a formula for the tension, but it wouldn't tell us much as we'd need a breaking strain and linear density for a material that doesn't exist. My formula is for a steady state, if we give the bottom a tug then since anglular momentum has to be conserved along the wire we'd get an increase in linear speed so it would move outwards again, but then you'd get oscillations along the wire which could accumulate to snap the wire... - 15 Jun '13 14:51

Thanks for the analysis. I was just using a theoretical building to show the idea of a line in space able to stay fixed over the surface. Do you know anything about the idea that keeps popping up in regards to space elevators about the extra line added many thousands of miles long to somehow stabilize the whole affair?*Originally posted by DeepThought***A building that tall would melt the ground under it, anchoring a cable would also be a problem. Neglecting atmospheric, solar wind and inductance effects from the earth's magnetic field on the wire...**

The centripetal force is balanced by the tension in the cable as well as gravity, consider an element of wire the tension pulling upwards at the top of ...[text shortened]... but then you'd get oscillations along the wire which could accumulate to snap the wire...

One problem I have is the idea that if you use a conducting cable you will have problems with the generation of electricity through the action of a conductor moving relative to a magnetic field, in this case, the Earth's field. If you are generating energy from that interaction and can utilize it that means the orbital velocity HAS to suffer and you have to make up that loss somewhere like rockets or something to keep the upper end in place or else the orbit would HAVE to start to decay. Now way around that if the cable is conductive. So it seems smart to have a non-conductive cable. I think people even TOUT the idea of using a conductor to generate energy and I have no doubt that could happen but for how long? That energy comes from somewhere and that somewhere is the kinetic energy of the cable itself interacting with the magnetic field of the Earth and the laws of physics state pretty clearly you can't get something for nothing and this nothing in this case is the loss of orbital energy which would eventually bring the whole thing crashing down to Earth.