What would be the gravity at the top of a "space elevator"?

Factual Questions
1

The concept of a “space elevator” or “beanstalk” is one I’ve sometimes encountered in science fiction, and a last year “Popular Mechanics” had a cover story on the idea. In sum, you tow an asteroid into geosynchronous orbit around the Earth (or whatever planet you’re starting from), string a set of cables made of some as-yet-uninvented material from the asteroid to the Earth’s surface, anchor the cables securely at both ends, and now you’ve got a way to lift things into orbit without rockets! All you need is some kind of traction motor for moving the “elevator cars” along the cable. This drastically reduces the cost of getting things out of the planet’s gravity well, and once you’re at the top of the beanstalk, you’re halfway to anywhere else in the solar system.

Astonishingly excellent idea. What it would cost, and what would have to be invented before it could be tried, I have no idea. But there’s nothing about it that is flatly impossible in view of our current knowledge.

One point, however, I’ve never seen directly addressed: If you’re in the station at the top of the beanstalk, what local gravity do you feel? On the one hand you’re on an asteroid in orbit, which means that, apart from the asteroid’s own negligible gravity field, you’re in free fall. On the other hand – you’re at the top of this long cable which, notionally, might as well be a very very high mountain – that is, if a mountain that high existed, in theory you could climb to the top of it, and at the top what you would feel would be full-strength gravity, somewhat reduced by the additional distance from the planet’s center of gravity – right?

(This puzzle reminds me of the thought experiment that led Galileo to question Aristotelian assumptions about gravity: If heavier objects fall faster than lighter objects, and if you tie two stones of different weights together and drop them – does the lighter stone drag on the heavier stone and slow it down? Or do the two weights sum so the whole thing falls even faster than either stone would by itself?)

One other point – would the beanstalk’s surface anchor have to be at the equator? Or could it be anywhere?

2

http://science.nasa.gov/headlines/y2000/ast07sep_1.htm
http://news.bbc.co.uk/2/hi/technology/2188107.stm

My guess is that the balance of the so-called ‘centrifugal force’ pulling outward exactly balancing the gravity pulling inward provides you the answer. The force pulling you ‘down’ at the asteroid would likely be 1G, but of course not due to gravity, and its towards the asteroid, as long as you’re on the side that the cable attaches to.

For the concept to work, the whole ‘string’ has to be under tension, remember.

At geosynchronous orbit you’d be at freefal, but that’s only partway to the end of a practical space elevator.

There’ve been other threads about space elevators here. None of them talk about the forces felt?

3

What do you define as the top?
The space dock would be best situated at the zero-gee point, and from everywhere else on the beanstalk this is in the up direction, so that must be the top-
and the answer in that case is zero.

But the anchor stone is much closer to the zero gee point than the earth end, so is not at 1 gee, but at a much smaller aceleration.

4

A space elevator is 72,000 miles long. The space terminal would be at the midpoint, geoynchronus orbit. There ships and people can just float away from it. The counter weight, to keep the elevator from collapsing, must be large enough to counter the weight of all vehicles on the cable. (the two equal lengths of cable cancel each other out.)

Vehicles could go past the midpoint various distances and let go, and be flung all over the solar system.

You can use a shorter and shorter cable and a larger and larger counter weight if you want instead.

This ignores tapering of the cable and things.

5

The ‘zero-G’ point can only be at geosynchronous orbit because it is only at this point that eternal free-fall from a combination of gravity and velocity of orbit coincide above a fixed point on the planet.

There would likely be a ‘dock’ here, as eburacum45 says, but this is most definitely not where the elevator ends. Well, conceivably the ride could end there, but the structure would almost certainly extend well past this point. This is because you have to have some mass at the other end of the line to counterbalance the enormous mass of the elevator and keep it from crashing down to Earth. This mass must be beyond geosynchronous orbit!

If you have a fairly long hunk of the elevator structure beyond geo orbit (which you must, by definition of keeping the line under tension and not collapsing to Earth), then why not extend the ride past that, for those loads that are destined to be flung outward from Earth towards other planetary bodies or other orbits?

So I disagree that a space-dock at zero-g is likely to be the ‘last stop’ on the ride.

Further, while the force of gravity due to the asteroid is negligible, and the asteroid may be closer to the geosynchronous orbit point along the ride than Earth is, the force you would feel there would not be negligible!

Consider: The asteroid, even if its fairly big, has to be attached some distance outside of geosynchronous orbit to keep the line under tension. If the line is under tension, then its whipping around the Earth at a speed that is much much faster than its distance from the Earth would indicate if it were untethered. It ‘wants’ to go flinging away from the Earth but the line is keeping it held back… just like swinging a bucket above your head horizontally on a line. Earth’s gravity isn’t enough to keep it at that distance constantly. Its too weak, by definition of it being outside of geo orbit. Therefore, the line is pulling back on it. But this means that anything on the inner surface of the asteroid (on the tethered side) is also experiencing this pull.

Same principle as being on the inside of a rapidly rotating wheel space station. And I think the force on you, if you were near the anchor point, would be considerable.

If you were on the other side of the asteroid, don’t let go…

6

yeah- That is what I was assuming- if the counterweight was twice as heavy as the earthside beanstalk, it would be in 0.5 gee.

You are right about getting a kick off the end of the counterweight- always good to get a bit of aceleration for nothing- but it might cause an uncomfortable reaction in the beanstalk if you do it all the time.
What you need in that case is a rotating tether.

7

Let’s clarify some things – my physics is (obviously) a bit rusty.

What IS a geosynchronous or geostationary orbit? I mean, I know it means
that the orbiting object is always over the same spot on the Earth’s surface.
As a space elevator would have to be. But is this something that is
possible only at a certain distance, or within a range of distances, from
the Earth’s center of gravity? And, again, does it have to be over the
equator? Or is a polar geosynchronous orbit possible?

8

Geosynchronous or geostationary orbit was first suggested, AFAIK, by Arthur Clark in 1945 as a good place to locate communication satellites.
It is the only orbit where a body in freefall orbits the Earth in 24 hours, and so moves at the same speed as the earth’s surface beneath it, assuming the earth is rotating in the same direction. You will find that the speed and angle of the earh’s rotation allows for a geostationary orbit directly over the equator, where the height of such a satellite is 35,800km.

9

You could not use an asteroid at the top of the elevator. Tidal forces would probably destroy an asteroid. Tidal forces cause earthquakes on the moon.

A space station could do it, though. Here, mechanical forces would be stronger than the tidal forces of gravity, and the space station would stay together. Either way its going to be mega-bucks to get this contraption to work.

10

Interesting- I wonder if you could get a free lunch out of the tidal forces- you can sometimes generate electricity by using rotating tethers- never let a good erg go to waste, even if you slow down the Earth in the process.

http://www.orionsarm.com/

11

Now I’m confused. trustno1 says “a space elevator is 72,000 miles long” – that’s 115,200 km, right? (As I remember, 1 mile = 1.6 km.) But eburacum45 says a geostationary satellite would have to be at a height of 35,800 km. Is there some reason why a space elevator would have to extend to three times the height of a geosynchronous orbit?

12

Some very simple orbital mechanics (since I’m incapable at present of working out the math):

  1. Given circular orbits, if the orbit of a mass is inside of geo orbit, then the angular rate is faster than Earth’s rotation rate (meaning the mass moves ahead of a spot on the Earth). If the orbit is outside geo orbit, then the angular rate is slower than Earth’s rotation rate (so the mass falls behind).

  2. Geo orbit is exactly the point where the angular rate of the orbit of a mass exactly matches the rotation rate of the Earth.

  3. If a mass at a given instant is moving at too fast or too slow an angular rate for its height, it will have an elliptical orbit. Important to the discussion is that if you try to have a mass inside of geo orbit with the same angular rate as the Earth, that’s too slow to orbit circularly, so the orbit is elliptical. There’s a point at which (as you lower the height) the orbit becomes so elliptical it intersects with the planet.

Given this, if you only had the elevator go up to geo orbit and for an instant had it dead straight, almost the entire elevator is inside of geo orbit, so isn’t orbiting fast enough to maintain its height given its angular rate. So the tower would collapse under its own weight, quite spectacularly. (Even if we could design the tower to support itself at that massive height under compression, there’d be a kind of orbital shear that’d bend it in two like a twig… the elevator’s really narrow with respect to its length).

Instead of it being under compression, and trying to deal with shear, if you design it so that it’s always under tension like a violin string or rope bridge that’s a lot easier to construct (I’m guessing). To do this, you need something that has the effect of ‘pulling’ on both ends of the elevator. It needs to be under tension and as straight as possible.

Gravity’s got one end, down at the Earth, and all the way up to geo orbit (because anything below this has that annoying elliptical orbit, or has to be moving ahead or falling behind the dock on Earth).

What’s pulling on the other end? Nothing, if the elevator ends at geo orbit.

If you have a big hunk of the elevator, or a fairly large mass/counterweight outside of geo orbit, something interesting happens.

At a given instant, if the part of the elevator beyond geo orbit is moving at the same angular rate as a spot on the Earth, then its moving faster than a circular orbit at that height would allow. The tendency is for that part of the elevator to go into an elliptical orbit that takes it further away from the Earth (rather than closer to the Earth like the case inside geo orbit for a fixed spot above the planet). That part of the elevator seems to be being ‘forced outward’ (it isn’t really, it’s just mass following the gravity well, but moving faster than a circular orbit would require).

But if we instead pull back on it, we can keep it in a circular orbit, even though it has too much angular rate around the Earth to ordinarily do so. Kinda like a bob on a string being spun above your head.

But we are pulling back on it, with the mass of the elevator inside of geo orbit.

So if you have enough ‘pull’ by the portion of the elevator outside of geo orbit, you can counterbalance the mass inside of geo orbit that is wanting to orbit inward towards Earth.

But if you have enough mass outside of geo orbit ‘pulling out’ to counterbalance the mass inside geo orbit pulling in, then the space elevator must be under tension and essentially straight. So even though different parts of the elevator want to follow different elliptical orbits given that the elevator is initially straight and all above one point on the Earth, the tension provided by the mass outside geo orbit helps keep it that way.
I think I got most of this right. :slight_smile:

13

I’m having a tough time parsing some of the ideas on this page, but I’m seeing a lot of stuff that’s not right.

For starters, in order to get this to work you’d have to have the center of mass of this system at a height of about 35000 km. This mass includes the asteroid and the cable, so the asteroid would most likely be “higher” than that, depending on how massive the Magic Rope turns out to be.

Secondly, you wouldn’t get “flung off” the other side or have any acceleration to speak of. Think of it this way - the Earth is in orbit around the sun right now. If you took a string and connected the two, what would happen? Answer: Nothin’. The “centrifugal force” from the Earth’s motion balances out the sun’s gravity at all points of our orbit (which is the definition of an orbit), so we only have to deal with “local” gravitational effects.

And as for the question about ground traces for Geostationary orbits, you can have a geostationary orbit anywhere you want to - if you’re circular and at that altitude, you’ll go around the earth once per day. Keep in mind, though, that if you’re anywhere other than over the equator, your ground trace will move throughout the orbit (if you start out inclined 30 degrees, say, you’ll trace out a figure 8 over the course of a day). In order to stay directly above a point on the Earth, you’ll need this orbit to be equitorial.

14

If you had the Earth and Sun connected by a rope but not attached to either one, the rope would fall inward. Most of the rope would have insufficient velocity to maintain a circular orbit. Different parts of the rope would try to orbit at different rates, but given its all one rope, I suspect it would establish an orbit somewhere in between its inner and outer distance from the sun. The orbit would be elliptical, and I think settle somewhere inside of Earth’s orbit.

We’re not talking about orbital ‘centrifugal force’. We’re talking about rotational ‘centrifugal force’. The Earth rotates much faster than its orbit (as evidenced by the fact that we’re not tidal locked, and actually have days and nights). But all this is completely moot. This would work even if the Earth were tidally locked. You’d just have the space elevator on the side of the Earth facing away from the sun, and the elevator would have to be much much longer.

The critical point here is the difference in angular orbital rates to maintain a circular orbit. It is precisely that difference that would collapse the tower if it were entirely inside geo orbit, and keeps it under tension and upright if a good deal of the tower is outside geo orbit.

Lastly if you place the groundstation anywhere but the equator, then you’d better have a damned stretchy elevator as that figure 8 occurs. By definition of the groundstation and end of the elevator not staying right above one another you’d have the elevator constantly trying to change length.

15

Yeah- about 1.5 million kilometres long…

To have a geostationary space elevator on any slow-rotating body means a very long cable, so the design works best on fast rotation planets like Mars and the Earth.
Fear not- the high tech science fiction answer is at hand- the spaceport end has to be supported in a non orbital location, so has to be subject to acceleration…
if you build a Dynamic Orbital Ring, supported by mass particle streams, you can hang any number of space elevators underneath… this is not Star Trek physics, but a good old Linear Accelerator (SF Link)

16

That’s… a freaking long elevator ride…

17

Tidal forces cause moonquakes because of lunar libration. If the asteroid was not librating, you’d have no changing stress, unless it was stressed to the failure point.

An asteroid could be just as strong as a space station. Depends upon its makeup.

18

Asteroids are a bit crumbly in their native state- better make a properly designed load bearing construct.
And yes, it will take more than megabucks- probably terabucks when the final bill comes in.
Which is why it will be a few decades/centuries before the first one is built.