Le Jun/11/2020 Ã* 19:41, Alain Fournier a écritÂ*:
On Jun/10/2020 at 14:58, Jeff Findley wrote :
In article , says...

I haven't done any research in this area. Does anyone know of any
studies of micro-gravity inside the cab of a space elevator?

Remember to work, the entire system has to be under elastic tension. The
designs I've seen discussed use a big counter-mass at the far space end
of the cable to hold the system in place above the anchorpoint on
Earth's equator.

The trivial case is when the cab is down on the Earth side. Obviously
we're at 1G on the surface. I've presumed as the cab rises the effect of
Earth's gravity goes down as inverse square (Universal Gravitation):

https://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

Are objects inside the cab of the space elevator near the "space" end
undergoing any form of microgravity? The system really isn't in free
fall because of the counter-mass suspended above it and the cable
running below. Does the tensive forces provide any form of microgravity
inside the cabin or are the occupants fully in 'free fall'? That doesn't
seem quite correct either. Only if the cabin were in orbit without any
connective cable. The counter-mass *is* appling force to the system to
hold it stable. Maybe the effect of any 'artificial gravity' are too
small to be consequential? If you were to suspend a cabin above the
counter-mass would you end up with an artificial gravity in the vector
direction of 180 degrees opposite the Earth's surface? i.e. the 'floor'
of the cabin becomes the surface of the cabin opposite the Earth,
alongside empty space?

I haven't studied this question at all. Any cites to any studies on this
appreciated.

See "Apparent gravitational field" he

https://en.wikipedia.org/wiki/Space_elevator

A rough approximation is that there is one point on the length of the
space elevator where the cabin crawling up the elevator is in "free
fall".Â* This altitude is the geostationary orbital altitude.Â* Anything
below that, and the earth's gravity is greater than centripetal force.
Anything above that, and the centripetal force is greater the earth's
gravity (so "down" is away from the earth!).

This also has a huge impact on what happens to any mass released from
the elevator.Â* From the Wikipedia entry:

Â*Â*Â* Any object released from the cable below that level would initially
Â*Â*Â* accelerate downward along the cable. Then gradually it would
Â*Â*Â* deflect eastward from the cable. On the cable above the level of
Â*Â*Â* stationary orbit, upward centrifugal force would be greater than
Â*Â*Â* downward gravity, so the apparent gravity would pull objects
Â*Â*Â* attached to the cable upward. Any object released from the cable
Â*Â*Â* above the geosynchronous level would initially accelerate upward
Â*Â*Â* along the cable. Then gradually it would deflect westward from
Â*Â*Â* the cable.

I think that what happens above geostationary altitude on an elevator is
often overlooked. People talk about putting a counter weight to keep the
cable taught. I think you want to have 30,000 km of cable above
geostationary altitude and you don't need to put a big massive object at
the end of those 30,000 km. What you put at the end of the cable is
another cable, this one spinning. All those cables are your counter
weight, but you can also use them to go away in the solar system. Just
going to 30,000 km above geostationary gives you enough angular momentum
to escape Earth. The spinning cable gives you more umpf, but it also
lets you go outside of Earth's equatorial plane.

I forgot to mention also that once you are past geostationary altitude,
you no longer have to figure out how to power your cabin. The cabin is
pulled out by the centrifugal force. Now you have to figure out what you
are going to do with the electricity you generate while controlling your
speed and/or slowing down. So using the cable after geostationary
altitude to go out in the solar system is really a free ride.


Alain Fournier