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I understand that a space elevator should be at least as long as a geosynchronous orbit (22,236 mi), because only beyond that is the centrifugal force from the Earth's rotation is greater than the effects of its gravity.

At the same time, astronauts in a space station in low Earth orbit experience weightlessness even though gravity is about 90% of the strength as on the surface. I understand that this is because they are in free-fall, but remain in orbit due to the high speed in which the space station was initially placed.

Let's say you have a 40,000 mi space elevator at the equator, and it has a climber wherein occupants are strapped into chairs. If I understand correctly, if you were to stop the climber and hold it there or apply the brakes, then at...

  • 4,000 mi (roughly the Earth's radius), the occupants would feel about 1/4 of the effects of Earth's gravity. So if they were holding an object and let go, it would fall, but more slowly than on Earth.

  • 22,236 mi (GEO), the occupants would feel weightless. If they were holding an object and let go, it would float (in theory).

  • Greater than 22,236 mi, the occupants would feel drawn toward the ceiling. If they let go of an object, it would rise and hit the ceiling, at some speed proportional to how much greater than GEO they are stopped at.

Do I have this correct?

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  • $\begingroup$ Edited. Now it's just a physics question, right? $\endgroup$ – Zero Cool Jul 8 '17 at 22:45
  • $\begingroup$ Everything you say is correct, except that I got about 26,200 miles for the point at which you would be weightless on a space elevator. It isn't at the point where you are in geosynchronous orbit. But by coincidence, it isn't that far away. $\endgroup$ – mmesser314 Jul 9 '17 at 2:37
  • $\begingroup$ No, that isn't right. Did I make an arithmetic mistake? $\endgroup$ – mmesser314 Jul 9 '17 at 2:44
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I agree with your numbers. I'm guessing you already know the following:

$Gravity = \mu /r^2$
where $\mu$ is the gravitational constant times earth's mass.

$Centrifugal Force = \omega ^2 * r$
Where $\omega$ is an angular velocity of 2 pi radians per sidereal day.

$Net Acceleration = (\mu /r^2) - (\omega ^2 * r)$

If r is greater than 42,161,150 meters then the $\omega ^2 * r$ exceeds $\mu /r^2$ and net acceleration is away from the earth.

Here's a screen shot from a spreadsheet I whomped up:

enter image description here

Feel free to play with this spread sheet.

Over at the Space Stack Exchange I drew the various orbits payloads would follow when released from different points on the elevator. I did the same illustrations in more detail at General Template for Space Elevators

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A space elevator must reach considerably farther than geostationary orbit. The reason is, the structure can only be supported by tension. There is no remotely plausible way to build a tower tall enough to reach even a low-Earth orbit that would be capable of supporting its own weight.

The space elevator basically is a big rock, tied to the Earth by a "cable", and orbiting the Earth once per day, but somewhere out beyond geostationary height. Centrifugal force wants to fling the rock away, but the cable stops it.

A practical space elevator system would have a station, anchored to the cable at Geostationary height from which spacecraft could be conveniently launched, and it would have some number of "crawlers" that could climb up and down the cable.

It's a long trip, so the crawlers would have to "crawl" at hundreds to thousands of miles per hour once above the atmosphere.

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  • $\begingroup$ How is it plausible to support a tower that big with tension? Space elevators are a nice idea, but not practical. $\endgroup$ – mmesser314 Jul 9 '17 at 2:36
  • $\begingroup$ It's not a "tower" (i.e., it does not support its own weight). It's a rope that keeps a big rock (i.e., a captured asteroid) from flying off into space. I am not claiming that anybody could actually finance it. But we do know how to make materials that might almost be strong enough to make it work. On the other hand, we do not have a ****ing clue how to make any material strong enough under compression to support the weight of a free-standing tower even a mile high, let alone hundreds or thousands or tens of thousands of miles high. $\endgroup$ – Solomon Slow Jul 9 '17 at 21:49

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