# Space Elevator Question: If You're At The "Top," Are You In Zero-G? **URL:** \ **Category:** Factual Questions\ **Created:** [September 13, 2017, 5:27pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254 "2017-09-13T17:27:24Z")\ **Posts on this page:** 20\ **Page:** 3
**Author:** ![iamthewalrus\_3](https://avatars.discourse-cdn.com/v4/letter/i/258eb7/32.png) [@iamthewalrus\_3](https://boards.straightdope.com/u/iamthewalrus_3)\ **Post date:** [September 14, 2017, 5:28pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/41 "2017-09-14T17:28:17Z")
> [@scr4](#): > > In that post, I was talking about a **400 km** tall tower, in case what the OP meant by “space elevator” was just a tall tower, tall enough that the top of it was in space. > > A 400 km tall tower is just a tower. It’s just supported by compressive strength of the material. The top of the tower is nowhere near orbital speed for that height. If you dropped something from it, it would drop almost straight down. My mistake. I took the OP’s “400km” as a wild-ass guess about the proper height for an actual geosynchronous elevator, and was responding as such, but you’re correct, a 400km tower is just a tall tower. > [@Darren\_Garrison](#): > > Nope, a rigid tower can’t be at proper orbital velocity at every height. Look at [this](http://www.calctool.org/CALC/phys/astronomy/earth_orbit) calculator. At a point 50 km up, orbital period is 1.42 hours. At 100 km, 1.44 hours. At 200 km, 1.47 hours. At 400 km, 1.54 hours. (All figures rounded.) But if it is a rigid tower rising off of Earth, each section would _have_ to have an orbital period of 24 hours. Sorry, I meant “rotational velocity”, not orbital velocity. It sounded like **scr4** was saying that the top of the tower wouldn’t already be rotating at the proper speed (to remain a tower, not to be in orbit). My point was that if you are building a rigid tower, at no point do you have to do something special to accelerate the higher bits, because they are accelerated as you raise them up to build the tower. But I used some pretty imprecise language to get there. ---
**Author:** ![Elendil\_s\_Heir](https://avatars.discourse-cdn.com/v4/letter/e/7cd45c/32.png) [@Elendil\_s\_Heir](https://boards.straightdope.com/u/Elendil_s_Heir)\ **Post date:** [September 14, 2017, 6:03pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/42 "2017-09-14T18:03:36Z")
> [@carnivorousplant](#): > > …Was Clarke’s tethered in [the Fountains of Paradise](https://en.wikipedia.org/wiki/Sequoioideae)? Fascinating book - I just reread it a year or so ago. _Star Trek: Voyager_ had a space-elevator episode, too: [Rise (episode) | Memory Alpha | Fandom](http://memory-alpha.wikia.com/wiki/Rise_(episode)) ---
**Author:** ![HopDavid](https://avatars.discourse-cdn.com/v4/letter/h/a88e4f/32.png) [@HopDavid](https://boards.straightdope.com/u/HopDavid)\ **Post date:** [September 14, 2017, 8:55pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/43 "2017-09-14T20:55:06Z")
> [@HeyHomie](#): > > Let’s say the prefect the Space Elevator. You get on it and ride it up to the top, 400km above the surface of the Earth. Will you be “weightless” in the sense that astronauts aboard the ISS are? Gravity scales with 1/r^2 and so called centrifugal force scales with r. Earth’s radius is about 6378 kilometers. If you’re a 150 lb guy, climbing to the top of a 400 km tower on the equator would lose you about 3 lbs. Centrifugal force doesn’t cancel gravity until you reach an altitude of about 36,000 km. This is the altitude of geosynchronous satellites. To reach an orbit with a perigee above earth’s surface you need to climb to an altitude of around 30,000 kilometers. [Here](http://hopsblog-hop.blogspot.com/2016/09/general-template-for-space-elevators.html) I have some drawings of the paths payloads would follow if released from different parts of an elevator. ---
**Author:** ![k9bfriender](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/k9bfriender/32/3283_2.png) [@k9bfriender](https://boards.straightdope.com/u/k9bfriender)\ **Post date:** [September 14, 2017, 9:03pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/44 "2017-09-14T21:03:03Z")
> [@HopDavid](#): > > If you’re a 150 lb guy, climbing to the top of a 400 km tower on the equator would lose you about 3 lbs. I think I’d lose more than just 3 lbs if I climbed a 400km tower. 😃 ---
**Author:** ![HopDavid](https://avatars.discourse-cdn.com/v4/letter/h/a88e4f/32.png) [@HopDavid](https://boards.straightdope.com/u/HopDavid)\ **Post date:** [September 14, 2017, 9:04pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/45 "2017-09-14T21:04:29Z")
> [@sbunny8](#): > > Geosynchronous orbit is about 35,800 km above Earth’s surface. The space elevator needs to be about twice that tall, so that its center of mass is at geosynchronous orbit. As you go outward gravity falls faster than centrifugal force climbs. So the acceleration gradient isn’t symmetrical about geosynchronous orbit. To balance and maintain tension, the tether above geosynchronous would need to be around twice as long as the tether below geosynchronous orbit. ---
**Author:** ![carnivorousplant](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/carnivorousplant/32/3563_2.png) [@carnivorousplant](https://boards.straightdope.com/u/carnivorousplant)\ **Post date:** [September 14, 2017, 9:20pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/46 "2017-09-14T21:20:00Z")
🙂 > [@k9bfriender](#): > > I think I’d lose more than just 3 lbs if I climbed a 400km tower. 😃 ---
**Author:** ![K364](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/k364/32/5_2.png) [@K364](https://boards.straightdope.com/u/K364)\ **Post date:** [September 14, 2017, 11:26pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/47 "2017-09-14T23:26:20Z")
> [@HopDavid](#): > > As you go outward gravity falls faster than centrifugal force climbs. So the acceleration gradient isn’t symmetrical about geosynchronous orbit. > > To balance and maintain tension, the tether above geosynchronous would need to be around twice as long as the tether below geosynchronous orbit. I’m thinking that below geosynchronous the tower wants to drop, and above geosynchronous it wants to fly away to a higher orbit. So, centripetal force depends on the speed and mass. In an unconstrained orbit these are equal to gravity. In the tower above geosynchronous it is more than gravity and pulls the tower up. So, you don’t need any special length for the upper tether, just more mass. BTW, in the category of “Possible, just needs more technology than we have today” I think this engineering will never happen. The scale of this thing is immense and any defect or mishap will be catastrophic. ---
**Author:** ![HopDavid](https://avatars.discourse-cdn.com/v4/letter/h/a88e4f/32.png) [@HopDavid](https://boards.straightdope.com/u/HopDavid)\ **Post date:** [September 15, 2017, 1:57am UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/48 "2017-09-15T01:57:47Z")
> [@K364](#): > > I’m thinking that below geosynchronous the tower wants to drop, and above geosynchronous it wants to fly away to a higher orbit. Correct. Gravity is GM/r^2 . And so called centrifugal force is ω^2 r. ω is angular speed in radians. Net acceleration is GM/r^2 - ω^2 r. If r \> than geosynchronous, centrifugal force is greater and net acceleration is up. If r is below geosynchronous, gravity is greater and net acceleration is down. > [@K364](#): > > So, you don’t need any special length for the upper tether, just more mass. Net acceleration is small if you’re close to geosynchronous orbit. So a counterweight near geosynchronous would provide only a small amount of upward newtons unless it was many tonnes. Having the tether length extend to an altitude of about 144,000 kilometers would be the minimum mass way to counterbalance the downward newtons. And if your tower is that tall, it could fling payloads most the way to Neptune. ---
**Author:** ![eburacum45](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/eburacum45/32/8690_2.png) [@eburacum45](https://boards.straightdope.com/u/eburacum45)\ **Post date:** [September 15, 2017, 12:41pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/49 "2017-09-15T12:41:21Z")
I am pretty sure that you could use a space elevator to leave the Solar System if you aimed the payload just right- using a flyby manoeuvre at Jupiter, for instance, like Voyager. It is a remarkable thing that once you get past geostationary orbit all that acceleration would come from the rotation of the Earth. But launching payloads towards the outer solar system would have an effect - there ain’t no such thing as a free lunch. It would cause a drag on the Earth, and slow it down, That’s why you need to attach the elevator firmly at the bottom - otherwise the bottom end of elevator would career off through the atmosphere, and extract angular momentum by friction. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [September 15, 2017, 1:06pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/50 "2017-09-15T13:06:21Z")
On the other hand, you could also set up an exchange of material: Say, put another elevator on Mars, and ship equal tonnages of Earthly seawater to Mars, and Martian iron ore to Earth. In that case, you’d have no net effect on the rotation or orbit of Earth or of Mars. And you could set it up so the energy cost per ton (both ways) was arbitrarily small. (nitpicking myself: You couldn’t actually do _arbitrarily_ small, since you’d need computers to calculate the proper trajectories, and it costs energy to do those calculations. But that’s so small compared to the energies involved in the movement of the matter that it might as well be zero.) ---
**Author:** ![HopDavid](https://avatars.discourse-cdn.com/v4/letter/h/a88e4f/32.png) [@HopDavid](https://boards.straightdope.com/u/HopDavid)\ **Post date:** [September 15, 2017, 1:10pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/51 "2017-09-15T13:10:00Z")
> [@eburacum45](#): > > I am pretty sure that you could use a space elevator to leave the Solar System if you aimed the payload just right- using a flyby manoeuvre at Jupiter, for instance, like Voyager. > > It is a remarkable thing that once you get past geostationary orbit all that acceleration would come from the rotation of the Earth. But launching payloads towards the outer solar system would have an effect - there ain’t no such thing as a free lunch. It would cause a drag on the Earth, and slow it down, That’s why you need to attach the elevator firmly at the bottom - otherwise the bottom end of elevator would career off through the atmosphere, and extract angular momentum by friction. Yes, it is the upper part of elevators that I find most interesting. They can act as a sling hurling payloads in any direction in the orbital plane of the elevator. As you say, an assist from a gas giant could send an elevator flung payload out of the solar system. Or just make the elevator a little longer and the elevator could provide all the velocity. The acceleration gradient grows much shallower as you move outward up the slopes of earth’s gravity well. So it would be possible to extend the elevator without adding a horrendous quantity of upward newtons. Not that I regard an conventional Clarke style beanstalk as plausible, mind you. At least not on earth. ---
**Author:** ![Darren\_Garrison](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/darren_garrison/32/92_2.png) [@Darren\_Garrison](https://boards.straightdope.com/u/Darren_Garrison)\ **Post date:** [September 15, 2017, 1:23pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/52 "2017-09-15T13:23:00Z")
If I’m doing my math right, if you extend the trailing portion of the space elevator out to a little over 2.5 billion miles, you can exceed the speed of light. So let’s get started on building that sucker to launch starships! ---
**Author:** ![Marvin\_the\_Martian](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/marvin_the_martian/32/2898_2.png) [@Marvin\_the\_Martian](https://boards.straightdope.com/u/Marvin_the_Martian)\ **Post date:** [September 15, 2017, 1:43pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/53 "2017-09-15T13:43:16Z")
> [@eburacum45](#): > > I am pretty sure that you could use a space elevator to leave the Solar System if you aimed the payload just right- using a flyby manoeuvre at Jupiter, for instance, like Voyager. > > It is a remarkable thing that once you get past geostationary orbit all that acceleration would come from the rotation of the Earth. But launching payloads towards the outer solar system would have an effect - there ain’t no such thing as a free lunch. It would cause a drag on the Earth, and slow it down, That’s why you need to attach the elevator firmly at the bottom - otherwise the bottom end of elevator would career off through the atmosphere, and extract angular momentum by friction. That raises an interesting point. Any materials discussion regarding a space elevator that I have seen concentrates on the tensile strength of the structure, but if you launch a payload from the top of the elevator that would produce a large torque about the center of mass and a shearing force at the anchorage point. How strong are carbon nanotubes to shear force? Sent from my iPad using Tapatalk ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [September 15, 2017, 3:46pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/54 "2017-09-15T15:46:01Z")
You’ll only get a shearing force until the cable bends and turns it back into a tensile force. And for a cable that long, the amount of bending needed to do that is minuscule. ---
**Author:** ![RedSwinglineOne](https://avatars.discourse-cdn.com/v4/letter/r/a9adbd/32.png) [@RedSwinglineOne](https://boards.straightdope.com/u/RedSwinglineOne)\ **Post date:** [September 15, 2017, 10:25pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/55 "2017-09-15T22:25:47Z")
So my intuition tells me that that as weight is pulled up the cable, the weight “tries” to fall behind the cable because as the weight goes up, it is being pulled up to a height for which it does not yet have sufficient speed. This would cause the weight to act as drag on the cable. The cable would need to not only raise the weight, but pull it forward to increase the weight’s speed. Does the cable have a way to compensate for the drag or do we assume the mass of the weight being lifted is insignificant when compared to the weight of the elevator? ---
**Author:** ![carnivorousplant](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/carnivorousplant/32/3563_2.png) [@carnivorousplant](https://boards.straightdope.com/u/carnivorousplant)\ **Post date:** [September 15, 2017, 11:48pm UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/56 "2017-09-15T23:48:29Z")
> [@RedSwinglineOne](#): > > Does the cable have a way to compensate for the drag or do we assume the mass of the weight being lifted is insignificant when compared to the weight of the elevator? I am an engineer, and of instrumentation, not a physicist, but I should think that a prerequisite would be that the mass of the cable car and it’s load would be insignificant compared to the mass at geosynchronous orbit. ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [September 16, 2017, 12:49am UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/57 "2017-09-16T00:49:03Z")
If the elevator has two tracks, with cars going up one track and down the other at the same time, the lateral forces should cancel out. Assuming the separation between cars is reasonably short, and that the amount of cargo coming down is the same as going up. ---
**Author:** ![eburacum45](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/eburacum45/32/8690_2.png) [@eburacum45](https://boards.straightdope.com/u/eburacum45)\ **Post date:** [September 16, 2017, 1:03am UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/58 "2017-09-16T01:03:28Z")
I’ve seem some detailed analyses that suggest that the forces on ascending and descending cars are going to be complex and unusual, and it won’t be a simple matter of just cancelling the forces out. To reduce the forces to manageable levels you could raise and lower the cars slowly, but that means a transit time measured in days or weeks. You weren’t in a hurry, were you? ---
**Author:** ![HopDavid](https://avatars.discourse-cdn.com/v4/letter/h/a88e4f/32.png) [@HopDavid](https://boards.straightdope.com/u/HopDavid)\ **Post date:** [September 16, 2017, 1:13am UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/59 "2017-09-16T01:13:33Z")
> [@RedSwinglineOne](#): > > So my intuition tells me that that as weight is pulled up the cable, the weight “tries” to fall behind the cable because as the weight goes up, it is being pulled up to a height for which it does not yet have sufficient speed. This would cause the weight to act as drag on the cable. The cable would need to not only raise the weight, but pull it forward to increase the weight’s speed. Does the cable have a way to compensate for the drag or do we assume the mass of the weight being lifted is insignificant when compared to the weight of the elevator? What you referring to is called Coriolis force. An ascending payload would exert a Coriolis force pushing west which would induce an oscillation. When the oscillation reaches earth, it would be absorbed and give earth a tiny shove that would have negligible effect on earth. An payload descending the elevator would exert an eastward Coriolis force which would also induce oscillations. However oscillations in the elevator aren’t desirable. But oscillations can be damped by timing when elevator cars would ascend or descend. * * * Earth anchored Clarke Towers aren’t remotely plausible. Not only are we unable to make Bucky tubes of sufficient length, there is also an excessive amount of debris and satellites in low earth orbit. Impacts would cut the elevator. Also with an elevator tens of thousands kilometers in length, throughput would be excruciating. Isaac Kuo has correctly called it drinking a very thick milkshake with a very long, thin straw. It is doubtful the throughput would even be adequate to maintain the elevator. But smaller elevators are plausible. Orbital vertical tethers would be stabilized by a gravity gradient. If each tether handled just 1 km/s delta V, stress would be manageable. It could be made from conventional materials like Zylon. Orbital tethers could also be placed in regions that are free of debris and satellites thus avoiding likelihood of impact. I talk about the possibility of such elevators at [Trans Cislunar Railroad](http://hopsblog-hop.blogspot.com/2016/08/tran-cislunar-railroad.html). * * * A [Mars elevator](http://hopsblog-hop.blogspot.com/2017/06/zylon-mars-elevator.html) could be built from Zylon if you use a safety factor of 1. However no sane entity would risk human lives or valuable payloads on such a razor thin safety margin. The slightest scrape or nick would cause the tether to break. Given a more sensible safety factor of 3, Zylon Mars elevators are impractical. However Zylon Mars’ orbital tethers anchored at [Deimos](http://hopsblog-hop.blogspot.com/2016/01/deimos-tether.html) and [Phobos](http://hopsblog-hop.blogspot.com/2016/01/upper-phobos-tether.html) are doable. ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [September 16, 2017, 2:35am UTC](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254/60 "2017-09-16T02:35:17Z")
> [@Darren\_Garrison](#): > > If I’m doing my math right, if you extend the trailing portion of the space elevator out to a little over 2.5 billion miles, you can exceed the speed of light. So let’s get started on building that sucker to launch starships! We could probably make the project easier (or at least cheaper) by instead expending the effort to spin the Earth up to having just a 12 hour rotational period instead of the traditional 24. Then we could save 1.25 billion miles of tether material. I’m pretty sure that’s a money saver. Even considering the cost of all the rest of our infrastructure we’d need to update. All new clocks, new school books, 4 hour workdays, etc. Still cheaper than 1.25 billion miles of tether & the power systems to traverse it. 😃 [Previous page](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254.md?page=2) [Next page](https://boards.straightdope.com/t/space-elevator-question-if-youre-at-the-top-are-you-in-zero-g/796254.md?page=4)