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All-orbital space station


Daymare17

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Imagine if you have a gigantic circular space station that is tethered with numerous space elevators, is concentric with the earth and spans the entire geocentric orbit (I think A.C.Clarke played with this idea in one of his early works). My question is, at what altitude would it need to be tethered to provide 1g of centrifugal force in the "roof" of the space station, so that people can walk around up there in a normal way?

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Some numbers that might help, which i found from a NASA study but the interconnection of which I do not understand since I do not know the maths of centrifugal force. Can anyone who does, help me?

 

1 rotation per minute (rpm) gives 1g at the inner circumference of a "circle" that is about 1600 meters across. 1g is achieved at 3 rpm with a 200 meters diameter, and a 4 rpm rate gives 1g if the circle is 110 meter across. There are 1440 minutes in one day, which is the rotational time of a space elevator/station, thus it would rotate at 1/1440 rpm. Earth's equatorial radius is 6,378 km. LiftPort corporation, which are currently planning to build an elevator within a few decades, calculates a length of 100,000 km.

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I remember reading A.C. Clarke's novel, about the space elevator but can't remember which one it is. I think that the premise is that the mid point of the lift has to be in a geostationary orbit at all times. Therefore you have to build both towards the Earth and away from the earth at all times to maintain the orbit. Once you have the first lift in operation you can extend around the equator at the geostationary orbit to build your Earth ringing space station, building further lifts as necessary.

 

The fact that this earth ringing space station is in a geostationary orbit means that it would be effectively in free fall i.e. no effective rotation and no centrifugal force i.e. no artificial gravity.

 

Now once you have this geostationary Earth ringing space station, effectively tethered to the Earth. then do you propose to build further concentric rings around it. I would guess that the stresses involved in keeping any further rings tethered would be much too great to make them practical.

However making concentric rings both above and below the geostationary orbit might balance things out.

(Not that I have thought too hard about it, someone may provide a better explanation):embarass:

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The fact that this earth ringing space station is in a geostationary orbit means that it would be effectively in free fall i.e. no effective rotation and no centrifugal force i.e. no artificial gravity.

 

I think you are mistaken there. The centrifugal force would be the same for a earth-ringing space station as for a single space elevator. The centrifugal force comes not from the rotation of the station relative to the Earth but from its rotation parallel to the Earth. My question can essentially be rephrased as: How long must a single space elevator be for the centrifugal force at its end not only to nullify the Earth's gravity but to mirror it, in effect creating -1G?

 

Thanks for the links although I already read them.

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First, not to nitpick, but there is no such thing as centrifugal force. Second, in order to create centripetal acceleration without constantly firing a rocket, it would be necessary to tether the elevator to the earth. This is not going to be possible for a very, very, very, very long time, if at all. Space elevators themselves are viable as of right now, but the thing is, they ARE NOT reliant on their tether to the Earth. The "anchor" so to speak is in space. And for that matter, if you keep the center of balance at geostationary orbit (which is a vital necessity), then the entire thing will spin at the same rate. Effectively, everything will be in the same equilibrium that exists at the geostationary point in orbit. The reactionary push to the centripetal acceleration of the elevator's center of mass will cancel out gravity. The only difference between orbit and the ground (in terms of gravity) is that the speed of the object is such that the centripetal accelleration is supplied completely by gravity, with no excess or shortage, resulting in a near-perfect state of equilibrium. It is the exact same thing as being in an airplane that is accelerating downwards at exactly [math]9.8 m/s^2[/math]. The people inside will feel weightless, when in reality, they are not - it just depends on your reference point.

 

That last bit might have been kind of vague. Long story short, IIRC, it is impossible to create artificial gravity like you are trying to do.

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For this ringstation to have one g on the surface facing the Earth, the centripetal acceleration would have to equal the gravitational acceleration, plus 1g acceleration (~9.81ms^-2), so...

 

[math]\frac{4\pi^2r}{T^2} = \frac{Gm}{r^2}+9.81[/math]

 

Errr...

 

Hm, I'm too tired to rearrange that properly now, but I'm pretty sure if you work out r from that (T=86400 (the time for one revolution in seconds), G=6.673*10^-11 (Gravitational constant), m=5.9742*10^24 (mass of the Earth)), then subtract the radius of the Earth from it (6378100 metres, according to Google), you'd get, in metres, the distance from the surface of the Earth to which this anchored planet-girdling space station would reach.

 

You'd then have 1g of gravity on the inside of the space station.

 

 

I'm pretty sure you wouldn't be able to get a ring to orbit something, but it will be much further out than the geostationary height anyway because if it was at that point the interior would be a 0g environment. As it is, the ring will still be extremely unstable and you'll either need to dynamically stabilize it (with thrusters, and such) or anchor it to the surface with (probably) unrealistically strong materials. That said, the ring will probably need to be made of unrealistically strong materials just to hold itself together.

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That was my point. If it were required to anchor the ring to the surface, even if it was anchored down miles under the surface, even if materials strong enough to compose such an anchor existed, they would still have such strain on them that they would rip themselves out from the surface, ripping out a good chunk of land - we're probably talking cubic kilometers - in the process.

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Thanks a lot for your help Xyph. I solved the equation and got the solution of 189681995,4m or 189692 km, which is about half of the distance to the moon's perigee and generally sounds reasonable. You can really ignore the part about a circumference-parallel space station - the essence of the question was how long one beanstalk must be for the centripetal acceleration (thanks for that term calbiterol :)) to provide "-1G" at its end.

 

Calbiterol, I see your point but I don't see why the feat would be impossible, if you approach it not from the standpoint of an all-orbital space station but a single space station tethered to a single elevator. The elevator doesn't have to be physically tethered to the Earth - why can't it be tethered by that part of its own mass which is within the geostationary orbit, i.e. within the area where the gravity outweighs the centripetal acceleration? The "anchor" would still be in geostationary orbit. From the standpoint of material tensile strength, the process (in essence, nothing more than the doubling or tripling or so of the cable length and the increase of the mass of the lower part of it) doesn't sound that much more difficult than keeping the huge fricking cable upright in the first place. Keep in mind that CNT's were sci-fi until 15 years ago.

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Calbiterol, I see your point but I don't see why the feat would be impossible, if you approach it not from the standpoint of an all-orbital space station but a single space station tethered to a single elevator. The elevator doesn't have to be physically tethered to the Earth - why can't it be tethered by that part of its own mass which is within the geostationary orbit, i.e. within the area where the gravity outweighs the centripetal acceleration?

That's not what I'm trying to say. I'm trying two things. The first is that if you put something like this - something that has the capability to spin itself and spans around the entire orbit - in geostationary orbit, then you're going to have to spin it in order to produce gravity at one G. Not too terribly much faster than it is already travelling, mind you, but fast enough to cause extreme strain on the materials of the ring. And then, by spinning it, you're defeating the point of putting it in geostationary orbit anyway, because you'll no longer be continually over the same point on the ground. In theory, this wouldn't matter if you had something with an enormously large tensile strength, because you could just slow down a ring placed in a higher orbit (and therefore able to produce 1g with a lower speed). The force of gravity would just succeed in centering the ring because the fact that the ring has a larger diameter than the earth would prevent the ring from coming crashing down to the ground - assuming the material it was made out of is strong enough to stand such an immense force. However, to the best of my knowledge, this is currently impossible, even with further development in carbon nanotubes.

 

One last hair to split - gravity isn't overpowering the centripetal acceleration, it's balancing it, or equal and opposite.

 

The "anchor" would still be in geostationary orbit. From the standpoint of material tensile strength, the process (in essence, nothing more than the doubling or tripling or so of the cable length and the increase of the mass of the lower part of it) doesn't sound that much more difficult than keeping the huge fricking cable upright in the first place.
You're right, a space elevator would not necessarilly need to be tethered to the ground, but simply hanging it from your ring would require an incredibly large tensile strength out of the materials, something that doesn't currently exist. Keeping the cable itself upright is just a (relatively ;)) simple task of counterweighting the other end of the cable - let gravity provide centripetal acceleration, while the inertia of the counterweight provides the reactionary push (aka "centrifugal force" :P), which will hold up the weight of the cable, and anything pulling itself up the cable. In order to remain at one point on earth, the system would have to have its center of mass at exactly geostationary orbit (hence the counterweight). It's doable, but a bit annoying to implement in the construction phase. That counterweight, by the by, is the anchor I was referring to, not the center at geostationary orbit. Alternatively, by putting its center of mass anywhere BUT geostationary orbit, while technically possible, would result in a cable that spins around the world. Trying to catch the "train" (cable) while it's moving, as I'm sure you would agree, is not the best way of doing things. :P

 

 

Long story short, this IS possible, maybe even with modern-day technology and materials - although we're quite a bit short on quantity of the materials for an entire orbital ring. The way you'd do it is by counterweighting the ring itself, towards the earth. By locating the ring's center of mass in geostationary orbit, while placing the "walking surface" of the station in a higher orbit (but still attached to the other part), the outside will have a large enough radius to produce one g at the velocity of geostationary orbit, and not require a tensile strength that is immeasurably high. Setting up the system like this, you could have as many elevators as you wanted, as long as the center of gravity (IIRC this is interchangable with center of mass) of the ring remains unchanged. Hopefully that last bit made sense. If it didn't, let me know, and I'll make a drawing for you.

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I think we have a confusion resulting from my misunderstanding of the term geostationary orbit. I took it to mean an orbit that went parallel with a definite point on the Earth's equator, regardless of altitude. Meaning, I stupidly considered each point of the length of a space elevator to be in geostationary orbit, which is obviously not true - for one, the base is not.

 

To clear things up: If you have a 190,000 km long space elevator with a space station at the end (forget the ring already), wherein people can walk in the roof at 1G, will it of necessity fly off into outer space ("tearing itself out of the ground")?

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To clear things up: If you have a 190' date='000 km long space elevator with a space station at the end (forget the ring already), wherein people can walk in the roof at 1G, will it of necessity fly off into outer space ("tearing itself out of the ground")?[/quote']

Short answer - not if it's centre of gravity is at the geostationary orbit point.

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Reading this post has just reminded me of another type of lift.

 

Get your head around this one if you can:- :eek:

 

Instead of having a stationary lift "shaft" you get the shaft to rotate around it's centre of gravity, at the geostationary orbit point.

The rotation is such that when the ends of the shaft touch the Earth they are traveling with the same rotational speed as the earth's rotation, and in the same direction.

So you can just step on to the end of the lift shaft when it touches the Earth and be whisked off up into space.

The great thing about this one is that if you "let go" of the shaft after it has whisked you off the surface then you would have an interplanetary velocity and don't need rockets for space travel.

 

It took me a while to figure this one out when I read about it, and I can't remember which book it was from - anyone else out there know?

 

Come to think of it Daymare17 the same applies, for space travel, to your 190,000 km long space elevator, don't have a space station on the end of it and just let the carriage slip off the end after accelerating up the shaft past the geostationary orbit point. You still need to expend the energy to get from the surface to past the geostationary point though.

 

Yes I know I just read too much SF.:embarass:

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Unless I've misunderstood what you're saying...

 

You can't get something to orbit an object that's in a geostationary orbit around the Earth... The Earth would have a hugely destablizing influence, and it's likely that this lift would just crash into the Earth, tearing the geostationary object it's tethered to out of orbit as it went, rather than continue to swing out to space and back again in a stable manner.

 

Also, even if you did manage to have something rotating around a point in geostationary orbit, you're not going to be able to have it matching the speed of the rotation of the Earth, because the Earth, from the perspective of the geostationary orbit, won't be rotating...

 

So if you wanted to supply 1g of gravity to the interior of this spinning thing, it would be whirling through the atmosphere at near 20km/s - stepping onto it wouldn't be very safe, to say the least, but probably no-one would even get the chance to do that since it would almost certainly burn up very quickly. The gravity would also vary greatly, since if you wanted to supply 1g to the inside surface, you'd end up with 2g at the point that touched the Earth because the centripetal acceleration would be added to the Earth's gravity.

 

 

 

To clear things up: If you have a 190,000 km long space elevator with a space station at the end (forget the ring already), wherein people can walk in the roof at 1G, will it of necessity fly off into outer space ("tearing itself out of the ground")?
Yes, most likely. If it's center of gravity is at the geostationary orbital height, and it's made of rigid materials, I suppose there's a chance it wouldn't... But the stress on such an object would probably be far too much, so even if it didn't tear itself out of the ground it might well just tear itself apart or snap.
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You can't get something to orbit an object that's in a geostationary orbit around the Earth
I think that I did not make myself clear here - the structure has a powered rotation, not an orbit around it's centre of gravity.

 

Also, even if you did manage to have something rotating around a point in geostationary orbit, you're not going to be able to have it matching the speed of the rotation of the Earth, because the Earth, from the perspective of the geostationary orbit, won't be rotating...

 

Obviously not, OK I did not think it through, or understand the concept that well, let's revise it a bit.

 

The centre of gravity is not in geostationary orbit, but it's in an orbit that would make it possible to allow the speed of the arm to match the rotation of the Earth. Would that have to be a retrograde orbit then?

 

Also I did not say anything about the amount of gravity to the interior of the spinning thing, but I take your point that "just stepping onto it" is going to be hazardous unless you can balance the accelerations as you do step on to it somehow.

However the fact that the end of the arm is stationary with respect to the spinning Earth when it touches the earth, if it is in the appropriate orbit and spinning at the appropriate speed, would mean that, at the point of stepping onto it, all you would be experiencing would be the gravitational attraction of the Earth, wouldn't it?

 

What happens after stepping on to it though ...hmm? could be a hell of a ride.:eek:

 

Hell I don't know if it's possible or not, I did not make it up, I just thought at the time I read it that I understood the concept and thought that it sounded cool.

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