What would it be like to ride a Space Elevator?
January 28, 2020 12:37 PM   Subscribe

Assuming it travels at a constant speed for most of the trip up or down, what would it feel like to be a passenger inside of a space elevator car?

Googling around suggests that you'd gradually transition from your planet's normal gravity at the bottom to feeling subjectively "weightless" at the top, but this is coming from places like Quora where I have no idea as to whether people know what they're talking about. Can you point me to a reputable source that clearly explains how this would work?

I specifically want to know about the subjective experience of a passenger, not so much the larger engineering problems involved.

(This is for a book I'm writing...please help me avoid lying to children!)
posted by Narrative Priorities to Science & Nature (19 answers total) 2 users marked this as a favorite
 
Weightlessness from being in orbit comes from traveling sideways quickly - quickly enough that the force due to gravity is balanced out. If you were on the end of a long pole attached to the earth at the height of the International Space Station's orbit, you would still feel something like 95% of the force of gravity that you'd feel on the ground. I'd bet most people wouldn't notice this change. My guess is that this would feel exactly like riding a normal elevator, except for the great views.
posted by jimmytransistor at 12:50 PM on January 28


Also, the force felt due to gravity reduces as the square of the distance between the two objects increases. There is a height where you would be weightless at the end of the pole (where the tip of the pole is in a stable orbit). This is called a geostationary orbit, and it's much higher than the ISS orbit (ISS is 250 miles, geostationary is 26,000 miles).
posted by jimmytransistor at 12:57 PM on January 28 [1 favorite]


Clarke's "The Fountains of Paradise" is a novel about the construction of a space elevator, and IIRC a character rides the thing at the end. It definitely talks in practice about how to construct it, describing the arrival of the tube that just hangs from the sky, not quite reaching the ground, as they build it in orbit and lower it down.

Considering it's a crawl straight up for at least 22,236 miles (the geostationary orbital altitude that Clarke himself is credited with first calculating), it takes days of travel, so the elevator is basically a hotel that drives up a long pole, or tape or cable, however you want to define it.

jimmytransistor, I'm not sure what you mean about weightlessness being connected to geostationary orbit -- geostationary orbit is just the orbital height that matches the rotation of the Earth, so you stay over one spot (or over a line of constant longitude, often) on Earth, and is unrelated to the force of gravity anywhere.
posted by Sunburnt at 1:01 PM on January 28 [2 favorites]


I'm not a astrophysicist but I play one on Kerbal Space Program. My understanding is that the weightlessness you feel it orbit is from constantly falling. You just happen to be going at a speed that you fall past the "side" of the source of gravity. Nothing to do with canceling or balancing the force of gravity.

But I do agree with jimmytransistor that it would feel like a normal elevator, in that its upward motion creates an artificial gravity. Which is only a sense of gravity and not actual gravity as it pertains to orbital mechanics.
posted by humboldt32 at 1:01 PM on January 28


Boredom.

If I understand the physics involved they won't shoot you up at the speed of a rocket because the whole point of the structure is to save costs and not pollute by using counterweights. This means you're not going to shoot up there incredibly fast or you increase costs by needing to both speed up and then slow down. Generally things that run slowly put less stress on the structure because things that run fast will introduce vibrations and impact and such. The space elevator is going to be one of those things you try not to break as breaking it could be catastrophic.

My expectation is that it's going to chug you up there nice and slowly and bring you back nice and slowly. I'd be surprised if it didn't take a few hours to get to the space station above, especially if the farthest point is not merely in low earth orbit where most of our space stations and satellites hang out. Low earth orbit is actually not high enough up to get your completely out of the atmosphere. Our space stations stay close to the skin of the planet to take advantage of the protection of its environment. If there are people on board they don't want them to die from exposure to solar radiation. But if the space elevator is actually the first way point to actual space travel such as we would need to colonize Mars and mine the asteroids it's going to be sending us beyond the atmosphere and the problem of shielding the occupants from solar radiation will have to be solved, or else there's no point going up there at all.

So you're going to be sitting there being bored for a long while. And if there is solar radiation there may not be anything to look at either, since it is easier to protect people with solid walls than with glass ones. I suspect you'll be watching a movie not looking out the windows.
posted by Jane the Brown at 1:10 PM on January 28 [1 favorite]


From this not-a-physicist: this is essentially correct. As you travel away from the surface you will experience less and less gravity over time. In addition, at geosynchronous height (assuming that's where you get off), you will also be in orbit, thus totally cancelling out the remaining gravity that you would otherwise feel. As others have mentioned, expect this to take a long time (days at car-like speeds). The amount of gravity you will experience is 9.81 * 6357^2 / (6357 + your_altitude_in_km)^2, where 6357 is the earth's radius in km. This doesn't consider the rotation of the elevator.

Another fun thing you could do (thanks Kim Stanley Robinson) is keep going up your space elevator above that height. The rotational speed will now force you away from the earth, and you could launch vehicles towards the moon / other destinations from there at some less fuel cost. Note that I'm assuming that the destination of your elevator is geosynchronous orbit. In addition to balance out the bottom section that lets you get there, you will definitely want a top half, of the same mass.
posted by Phredward at 1:15 PM on January 28


The results in this result on Physics StackExchange are, I'm pretty sure, correct (as is the discussion on Wikipedia.)

Effectively, you would "feel" your weight gradually decrease as you ascended the elevator. This is because the elevator car is, by virtue of going in a circle around the Earth once a day, accelerating towards the center of the Earth. You, in the car, will also be accelerating towards the center of the Earth at the same rate, and there must be a force on you to prevent this. To get the "right amount" of force on you so that you move in a circle with the elevator, some of the inward force of gravity pulling you down must be cancelled out by an outward force from the floor of the elevator. It's this force from the floor that you "feel" as your weight.

As you ascend, the inward force of gravity decreases (due to Newton's Law of Universal Gravitation), while the acceleration you need to stay at rest with respect to the elevator car increases. Thus, the upward force exerted on you by the floor must decrease, and so your apparent weight will decrease. Geosynchronous orbit is, by definition, the elevation at which the gravitational force on any object is "just right" to make it go around in a circle once a day; so when you get to geosynchronous orbit, you would feel weightless, with no forces on you from the elevator at all. And if you climbed up past geosynchronous orbit, you would see things "fall up" towards the ceiling, since beyond geosynchronous orbit there's not enough gravitational force to keep you moving in a circle; the ceiling of the elevator would have to start pushing you inwards.
posted by Johnny Assay at 1:16 PM on January 28


I'd be surprised if it didn't take a few hours to get to the space station above, especially if the farthest point is not merely in low earth orbit where most of our space stations and satellites hang out.

So, the center of mass of a space elevator must be in geostationary orbit around the earth, or else the whole thing is subject to some pretty extreme unbalanced forces that will tear it apart. Most of the designs I'm familiar with have a big space station at the end, somewhere just beyond the geostationary altitude, to serve as essentially a giant counterweight against the mass of the extremely long strand of the "elevator" per se.
posted by tobascodagama at 1:23 PM on January 28 [1 favorite]


I figured out my mistake, above, regarding gravity at the station in geosynchronous orbit. Oops.

How to transition from elevator gravity, which can be a combination of earth's diminishing pull and the acceleration of the elevator, to station zero-G is an interesting engineering problem.
posted by Sunburnt at 1:51 PM on January 28


Sunburnt, your rotational velocity is constrained by the magic pole which you are attached to, and which is anchored to the earth (the elevator). Since you can't change your rotational velocity, the only altitude that you can feel weightless is the altitude of geostationary orbit. Below this altitude, you will feel some pull of the earth's gravity that increases as you get closer to the ground. Above this altitude, you will feel an increasing sense of being flung off into space.
posted by jimmytransistor at 1:57 PM on January 28 [1 favorite]


jimmytransistor Yup, you're 100% right-- I had something else in my head at the time. Even if the geosync station is where it is because of earth's rotation, an orbit is an orbit, and orbits are weightless. Sorry I called you out; mea culpa. I'll take a point off my space nerd license.
posted by Sunburnt at 2:58 PM on January 28 [1 favorite]


There is also a slight sideways acceleration you would feel, depending on how fast you ascend. Your rotational velocity at the surface of the earth is about 1000 miles per hour and at geosynchronous orbit it is about 4300 miles per hour. You would feel deceleration from the opposite side on descent.
posted by JackFlash at 3:22 PM on January 28 [1 favorite]


You might like Issac Arthur's Upward Bound: Space Elevators video. He does a ton of Space Futurism videos about orbitals, colonization, space elevators, sky hooks, etc. There's little actually describing the trip up/down, but really that could be anything...

You basically have free energy by the point of space elevators, energy can be recovered from things going from orbit to ground, and once you're doing geostationary space stations you have basically unlimited access to solar power.

If the tether is strong enough, you could take a seat and accelerate at 1-2g about half the way and then rotate the passenger car and deccelerate the same until you arrived at the orbital and stopped into 0g. Or you could just accelerate for a bit and cruise along with just enough acceleration to maintain earth gravity for most of the trip and then slowly decelerate and gently crank down to 0g.

The video goes into a few things like the car basically being a spacecraft that could pop-off and re-enter the atmosphere in an emergency or on purpose.
posted by zengargoyle at 5:22 PM on January 28


Just wanted to add that the space elevator on Mars in KSR's trilogy had an interesting complication: the areostationary orbital altitude is 10,000-something miles, but the orbit of Phobos, the smaller irregular captive moon of Mars, orbits in the 9,000-something range of altitudes, meaning that the Martian space elevator had to undulate in a calculated, deliberate fashion in order to dodge Phobos speeding by, and that's in addition to the undulations by which the tailing end, the end above the areostationary orbit, would capture incoming spacecraft or throwing forth outgoing craft.

Such a thing might be necessary depending on what planet you're putting your beanstalk upon.
posted by Sunburnt at 2:20 AM on January 29 [2 favorites]


Here's a PDF from NASA circa 2000: Space Elevators: An Advanced Earth-Space Infrastructure for the New Millennium.

Being NASA and all serious, like we could do this now if we really tried just maybe... The actual journey bits boiled down to:
If an energy system were used equivalent to the ISS solar arrays producing ≈60 W/kg, it would take ≈12 days to climb the structure at an average speed of 125 km/hr.
...
To reduce the travel time to ≈10 hr, a power level of ≈1 kW/kg would be needed, resulting in speeds of ≈2,000 km/hr.
Those are slow-ish numbers compared with things like bullet trains (~400 km/hr) so boil down to pulling some minor g's at the start and end and traveling most of the way at constant velocity in a gradually reducing gravitational field.

At constant speed, you'll go light rather quickly. Ignoring the coriolis force and orbital velocity and such... the orbital is 35,768 km above the equator, choose your speed. If we ignore the start/stop bits and fudge a bit... going from 0/10ths the distance (time) to 10/10ths (arrival), gravity goes like this:

100
41.0583781392469
22.22361397984224
13.90340048125929
9.511257315578597
6.913796029150277
5.2514343002058315
4.123778644729888
3.32387171712367
2.7359934690171888
2.291328804442763

That's % (and that final 2.3% is the zero-g if you're also going sideways and orbiting). At 1/10th the trip you're already less than 1/2 your weight. By 2/10ths you're less weight than on the moon. By half-way you're at 7%. You're all bouncey floaty most of the trip. If you go the 12 days *or* the 10 hours, the whole trip is gravity being different every time you think about it.

Constant velocity should probably be avoided if possible... either accelerate/decelerate at 1g and flip at the right spot, or keep 1g and then 0g. Depending on your destination and training....

If you're heading to the moon, keep it at that 1/3 gravity for as long as possible for practice, same for mars. If the orbital is rotating, acclimate to the gravity of the orbital. If you're still heading out to a zero-g situation... do that.

Story-wise, it depends on the culture/technology. If it's early and they're still gritty astronauts... it's a space-capsule and just like the Apollo missions coasting to the moon. If it's mature culture/technology... It's a couple or three hours at mostly 1g with a couple of zero-g transitions between starting/ending gravity.
posted by zengargoyle at 10:35 AM on January 29 [1 favorite]


There is also a slight sideways acceleration you would feel, depending on how fast you ascend. Your rotational velocity at the surface of the earth is about 1000 miles per hour and at geosynchronous orbit it is about 4300 miles per hour. You would feel deceleration from the opposite side on descent.

FYI, this sideways force is also explainable as due to the Coriolis force in a frame rotating with the Earth.
posted by Johnny Assay at 8:42 AM on January 30 [1 favorite]


The coriolis effect is handled by creative engineering. You just need to make the internal structure of the transport capsule like a pendulum (hopefully damped a bit) and it will autocorrect for the forces to keep a true down. The up and sideways field vectors sum to a single vector defining up/down and subject only to differences in field strength between the bottom and top. Then it's down to to toilets in the northern hemisphere and the southern hemisphere of earth rotate in different directions.
posted by zengargoyle at 2:03 AM on January 31




THANK YOU, everyone!! This has been incredibly helpful!!!!

(Also, since I posted this question, the project it relates to was announced!)
posted by Narrative Priorities at 5:55 PM on February 10 [2 favorites]


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