Space elevator question: gravity conditions in space station

If I’m reading the Wikipedia article correctly, a space station tethered to the Earth and in geosynchronous orbit would experience free fall or micro-gravity conditions.

During the trip up the tether from Earth to the space station, travelers would initially feel Earth gravity. Launch acceleration would make the apparent gravity more intense(?), and then the further they were from Earth, the less gravitational pull they would feel until they reached the station and were in free fall. If someone were to follow the tether past the space station to the counterweight (a captured asteroid, another space station, or just a length of tether equal to the mass necessary to act as counterweight), they would feel a pull in the opposite direction due to centripetal/centrifugal (seriously, which one is it?) force?

That’s my understanding. Clarification and/or confirmation is greatly welcome. Now, here’s the questions where I’m completely ignorant.

  • given appropriate technology, would it be possible to generate apparent gravity on the geosynchronous space station by having it rotate on an axis perpendicular to its orbit (does that make sense?)?

  • what the hell would the Earth, Sun, and Moon appear to be doing if you were on a rotating geosynchronous station, looking out a window on the rotation axis? What if you were on the distal end and had a window which looked completely away from the Earth?

  • if you traveled to the end of the counterweight (distal side of the asteroid/space station or very end of the tether past the geosynchronous station) and “jumped” off, would you be launched away from the Earth? Would this be a reasonable way to launch vehicles to the Moon or other destinations?

Connected questions:

  • we know long term exposure to free fall does nasty things to the bones and muscles. But what about fractional gravity? If a person lived at 80% G or 50% G, any idea what the advantages and disadvantages are?

  • could you have a space elevator at a latitude other than the equator? How about directly above or below the axis of the Earth (and would the tilt of the axis matter?)?

Have at it, guys! (And thanks! :smiley: )

I’m sure much smarter folks than me will be along shortly with better answers. :slight_smile:

I don’t know that I would call it ‘launch’. The car would be electrically powered to climb up the cable. It would start rising up the cable and keep accelerating until it reached the desired speed. It doesn’t need to reach escape velocity like a rocket does, and I don’t think there would be any need to go fast except that you’d want to limit transit time. If it only rose as quickly as the elevator in an office building it’s take months to get to the top. :slight_smile: Also, I don’t know if there’s a provision to allow ascending and descending elevator cars to pass each other, so if t takes a long time for a car to go one way I guess it’s holding up any cars from going the other direction.

Yes, I think it would be possible to have a ringed rotating station attached to the tether. It might make the engineering a bit more complicated, but compared to being able to built a 20,000+ mile cable I don’t think it’s all that big a deal.

You wouldn’t need to jump off, just let go. Yes, It’s been suggested that vehicles could be launched to the Moon and Mars, etc. by sending them up the cable past the point of geostationary orbit.

Other than the time the Apollo astronauts spent on the Moon, people don’t really have experience living in lower than Earth gravity. There are various physical problems caused by spending long periods in zero gravity, and I’d assume they’d be lessened by living at say 50% or 80% gravity, but I don’t know how much research into it that has actually occurred.

The problem would be that it wouldn’t go straight up but bend so that the top was above the equator.

No, that won’t work. It’s the spin of the Earth that’s holding the thing up. you don’t get any spin if you build it at the top or bottom of the spinning planet.

I’m sure there will be more knowledgeable 'dopers along soon to answer, but:

I don’t think so. As long as the elevator was accelerating I think you’d have ‘apparent gravity’. When it started to slow down (presumably to approach the station) that would certainly decrease.

Sure, afaik. You could spin the thing in any direction you wanted, and if you spun it quickly enough and maintained that rotation you’d have ‘apparent gravity’. You might make the folks on board sick if you have a window and you spin the station perpendicularly to it’s orbit, since the Earth and horizon would be constantly changing.

If I’m understanding the question correctly, the Earth would appear to be rapidly through the window a lot like the sun coming up in the east and setting in the west…but very rapidly. It would be a bit disorienting I should think. Depending on where you were and the angle, the Sun and Moon would be similar I should think.

If it’s rotating (which I assume it would be) then that would impart some energy to you if you jumped off, yes. I guess it would depend on the amount of energy due to the rotation and the vector you jumped off in.

I am not sure about this one, but I think that even micro-gravity would help. Not sure what the level of loss would be compared to free fall though.

My understanding is that such an elevator has to be in geosynchronous orbit to work. The station is in effect holding up the cable and the elevator.
Huge grain of salt on all of the above, as I probably have many or most of the details wrong. Hopefully someone like Stranger will be along…and hopefully he’s in a good mood and won’t be all grouchy and such about the reality of a space elevator.


Excellent! Thank you.

That’s correct, and the proper term here is “centrifugal”. Some pedantic but ill-informed physics teachers will insist that centrifugal force doesn’t exist and that one should always speak of centripetal force instead, but centrifugal force is exactly as real a force as gravity. Which one is correct to speak of depends on which reference frame you’re in.

If you’re in an inertial (which implies non-rotating, among other things) reference frame, then you see an object moving in a curved path, and deduce that there must be some force acting on it causing the path to curve. This can be any sort of force at all; in the case of the counterweight of a space elevator it’s mostly the tension in the cable. Whatever the force is that’s causing the path to curve, that force is called the centripetal force, and it points toward the center of curvature.

If, however, you’re in the rotating reference frame (as you would be if you’re standing in a space station at the counterweight), then this explanation doesn’t work. You can’t say that your path is curving, because in your reference frame, you’re not moving along a path at all. But you can still observe that the cable is exerting a significant force on the space station, pulling it towards the Earth. To explain the observation that the station is not (in your reference frame) accelerating towards the Earth, you need to invoke some other force that’s counteracting the force from the cable. This force is called the centrifugal force, and always points away from the center of curvature. It’s an odd sort of force, since there is no other object which is exerting the force on you; the existence of the force is just an artifact of the reference frame you’re in. Forces like this are sometimes called “fictitious forces”, and this is why some pedants insist they don’t exist, but if you happen to be in a non-inertial reference frame, they work just fine.

I can’t add much to what’s already been written either by the OP or the earlier responses, but here is my take on the questions asked:

  • The centripetal acceleration goes up in proportion to the distance to the center of the earth as the elevator rises; this makes it feel like there is a centrifugal force pushing you out with a strength proportional to the distance from the center. The acceleration due to gravity goes down reciprocally with the square of distance. So, if you go up very slowly, you’ll feel the full force of earth’s gravity at the start and this will gradually be reduced to zero at the top. However, to make it it the top in a reasonable time, you’ll probably want to accelerate upward at several g’s (let’s say 5 g’s) for the first part of your flight and accelerate downward at several g’s at the end of your flight. This acceleration/deceleration will feel like another force that has to be added to the force you will feel for a slow trip, so your flight will start at 6 g’s (1+5), coast for a while at something below 1 g, then slow down at -4 g’s (1-5), then stop at 0 g.

  • Going slightly beyond the geosynchronous point, you would have a centripetal acceleration a bit larger than the local acceleration due to gravity, so it would feel like a centrifugal force pushing you out with a force small compared to gravity.

  • Yes, you can generate artificial gravity in any direction and others have described what it would look like out the window.

  • If you stepped off the distal end, you would be in an elliptical orbit around the earth with the perigee (closest approach) equal to the distance when you stepped off. The apogee (furthest distance) would be slightly greater and occur half way around the orbit from the point you stepped off.

  • This wouldn’t lauch something to the moon or elsewhere but it would help immensely, since in geosynchronous orbit you would have climbed out of a substantial fraction of the earth’s gravity well. The ability to put something into a high orbit without having to build rockets that are vastly larger than their payloads is the reason for all the interest in a space elevator.

  • Your speculation is as good as mine about the long term physiological effects of fractional gravity.

  • No, you can’t do this off the equator unless (as someone else pointed out) you’d like to design an elevator that is not vertical. A stationary geosynchronous anchor point has to be above the equator. If the anchor point were slightly above the equator, but still geosynchronous, it would follow an orbit that goes just as far below the equator as it does above. And, no, this won’t work over the poles either.

Chronos wrote his answer while I was writing, but I am glad to see that he and I are on the same wavelength. It seems to be mostly high school physics teachers who make a big deal about there being no such thing as centrifugal force. Physicists talk about it all the time.

How can you guys talk about this without citing one of the all-time great xkcd comics? Centrifugal Force

That depends on how far past geosynch you extend the cable. Extend it far enough, and something dropped off the end will escape the Earth (after which it’s mostly just a matter of aiming right, to get to the Moon, Mars, or any other given destination). Plus, by making the cable longer, you decrease the required mass for the counterweight.

Oh, and for the Earth, an equatorial cable would be most practical, not only because the cable would be straight, but also for protection from storms: You get all sorts of storms in the tropics, but right at the equator itself, there are very few, since hurricanes or other cyclones can’t cross the equator. Once you’ve got a cable on Earth, though, it’s a relatively small step to also put similar systems on other planets, and I’ve seen studies that suggest that on Mars, the top of Olympus Mons would be more practical than any equatorial spot, due to being above more of the atmosphere.

The biggest problem (well, leaving aside building the thing) is…what happens if something cuts the cable? Ouchy…


Not the worst problem on the way up if the cable is cut below you. But considering the strength required it’s going to take some mighty big bolt cutters.

The portion of the cable above the cut “falls” up into a higher orbit, and the portion below the cut falls back down to Earth and mostly burns up in the process. It’d be a pretty significant act of vandalism, but you could probably even reclaim the upper portion and reattach it. It’s no bigger a worry than “the problem with a skyscraper is that someone might fly a plane into it”.


Tensile strength does not imply shear strength.

Yes, but there will be plenty of shear forces that have to be accounted for, at least within the atmosphere.

Obviously, not a good thing for any cars currently in transit to/from GEO but as far as crashing into the Earth catastrophically, it is important to keep in mind that the cable would by necessity be made of extremely low density material. It wouldn’t be like the Golden Gate Bridge falling out of the sky but more like well, a light weight cable or ribbon.

Wow, this is great. And the xkcd strip made me laugh out loud. :smiley:

I’m still wondering though, if the geosync station were built to rotate around its long axis, perpendicular to its orbit:

  • how large would the Earth appear?
  • as the station rotates, what would the Sun appear to do?
  • as the station rotates, what would the Moon appear to do?

I’m guessing that the Earth would still appear fairly large, at 22,000 miles up, but wouldn’t necessarily dominate the view. I’m also guessing that the station would have apparent sunrise before the tether point at the equator and sunset after, because it’s so much further from the center of the Earth.

In turning on its own long axis, depending on how long the rotation cycle was (which is a function of how big the station is, in order to spin it fast enough to achieve apparent full G at a chosen point in the station), viewers not at either end will see the sun rise and set in half(?) the time it takes to complete a full rotation, if the full rotation is a short enough period. If it’s a long period - visualizing this is a biatch - then, the sun’s apparent path is a curve determined by a proportionate combination of the Earth’s rotation and the station’s? So, it would look something like a half a parabola (had to look that one up)? Maybe more than half?

Would the moon’s apparent path be a variation on that? Or are there even more factors to calculate?

I’m assuming that you mean to put the axis of rotation along the cable. By sunrise and sunset, do you mean the Sun coming up over the edge of the Earth, or over the edge of the station? That high up, the Earth would appear small enough that most “nights”, it wouldn’t even cover the Sun at all, and even when it did, it’d only be for a short time. From the perspective of the folks on the station, it would be more like an eclipse than like night. The Earth, meanwhile, wouldn’t “rise” or “set” at all: If you could see the Earth looking in a particular direction out a particular window, it’d always be at that spot in that window. But it would appear to rotate in place with whatever period the station’s rotating (how much that would be would depend on the size of the station and how much gravity you wanted), and it would also show phases that would vary over the course of 24 hours (so you could see what time it was by seeing the phase of the Earth).

Yes, by sunrise, I mean the Sun appearing over the horizon of the Earth, though from what you say, it sounds like there won’t be much horizon.

How large would the Earth appear from geosynchronous orbit?

And yes, I did mean rotation on the same axis as the tether/cable. The idea of night being an eclipse, perhaps an annular eclipse is pretty damn wild.

By phase you mean the change in illumination from the sun?

Well, geosynchronous orbit is at a height of about 3.3 times the diameter of the Earth, so find something 3 feet across, and stand 10 feet away from it. That’s about how big Earth would look. This is still well larger than the apparent size of the Sun, so you wouldn’t get annular eclipses.

I’m geeking out with happiness right now. :smiley:

Last question, promise:

How does one determine how fast an object needs to rotate in order to achieve a centripetal force equal to Earth’s gravitational pull? There must be an equation, right? Because the longer the radius, the faster it moves at its edge, right? (And I know I’m using 2D language for a 3D problem, but the brain is doing the best it can.)