4
$\begingroup$

On Earth, from how much of a planet's surface will the Space Elevator be visible, as in "a prominent object in the sky"?

The simple geometry seems straightforward: If we assume the elevator originates from the north pole (which it wouldn't since it should be built on equator) than it should be visible from all of the according hemisphere, since it's much higher than the planet's atmosphere.

But what will the people on the equator actually see in that case? The elevator is pretty much aligned with the horizon for them.

And what about the atmosphere and lighting? During the day we don't see the stars - will we able to notice the SE? Or only in the night? Will we be able to notice all of the length, or only the part in atmosphere? But the string's upper part is lit by the sun almost all the time - since it's much longer than the planet's diameter and the planet doesn't obstruct much. So is there a reason not to see it in daylight?

Also, if it's got a diameter of ten miles, for example, will it really be visible for so far as from thousand miles?

Relation to world building: I am considering a post-apocalyptic civilization, who doesn't navigate by stars but orients itself around a derelict SE that is visible from all the discovered lands.

$\endgroup$
  • $\begingroup$ Related meta discussion. $\endgroup$ – HDE 226868 Nov 1 '15 at 16:17
  • 2
    $\begingroup$ Space elevator for supergiants, perhaps? 1 mile is huge for a space elevator! $\endgroup$ – Loren Pechtel Nov 1 '15 at 20:41
  • $\begingroup$ Just to be pedantic, you'd have the space elevator at the equator not the north pole as it has to have a geostationary orbit... $\endgroup$ – colmde Nov 2 '15 at 12:44
  • 2
    $\begingroup$ Yeah, I would reduce the diameter of your elevator. In Sci-Fi literature, the SE's on various planets are so thin, most of the planet will only notice a glint off the sun, if even, unless you're very close to it. $\endgroup$ – Mikey Nov 2 '15 at 18:39
  • 1
    $\begingroup$ @HDE 226868 , no, I don't think it is. There is no need to see the base of the cable. If it's visible at all - it's a vertical line on the horizon, so seeing only an upper half works as well. $\endgroup$ – Nox Nov 3 '15 at 9:56
7
$\begingroup$

Part 1: Initial assumptions

If we were to assume that the elevator cable is visible from any distance (assuming no obstacles), then this is actually a simple exercise in geometry.

Here's a quick diagram I drew (note: elevator length not to scale):

The black circle is the planet, the blue line is the elevator, and the red lines are imaginary lines tangent to points on the planet's surface.

At any point on the planet's surface, we can draw a line tangent to it that will intersect the elevator cable at some point. We can then draw a line from that point to the center of the planet (this line will be perpendicular to the tangent line):

The computation for any length is simple. At some height $h$, we have $$\text{Length of tangent line}=\sqrt{R_{\text{planet}}^2+(R_{\text{planet}}+h)^2}$$ We also have $$\text{Angle between point on surface and horizontal}=\arccos\left(\frac{R_{\text{planet}}}{R_{\text{planet}}+h}\right)$$ You should be able to find the area from this, which I leave as an exercise for the reader. Hint: Use the arc length of the central cross-section of the planet.

This whole approach fails, however, if we consider that the Earth is not a perfect sphere, and is not smooth. A better approximation is to treat it as an oblate spheroid. We can describe it in Cartesian coordinates by the equation $$\frac{x^2}{6,378,137^2}+\frac{y^2}{6,378,137^2}+\frac{z^2}{6,356,752^2}=1$$ Using spherical coordinates, we then have $$\frac{r^2\sin^2\theta\cos^2\varphi}{6,378,137^2}+\frac{r^2\sin^2\theta\sin^2\varphi}{6,378,137^2}+\frac{r^2\cos^2\theta}{6,356,752^2}=1$$ Doing some algebra, we have $$r=r(\theta,\varphi)=\left(\sqrt{\frac{\sin^2\theta\cos^2\varphi}{6,378,137^2}+\frac{\sin^2\theta\sin^2\varphi}{6,378,137^2}+\frac{\cos^2\theta}{6,356,752^2}}\right)^{-1}$$ We can then set some $(\theta_0,\varphi_0)$ to be the location of the elevator, and make the changes $$\theta\to\theta-\theta_0,\qquad\varphi\to\varphi-\varphi_0$$ At this point, you can use a similar method to the one used in the spherical approximation part.

Part 2: Accounting for visibility problems

Here, we need to discuss angular diameter. The angular diameter of an object $d$ meters wide at a distance of $D$ meters is $$\delta=2\arctan\left(\frac{d}{2D}\right)$$ In your case, $d=$10 mile$=s$16,000 meters. According to Wikipedia, the minimum angular resolution the human eye can see is about 60 arcseconds, or 1/60 of a degree. Plugging this in, we have $$D=\frac{1}{2}\left(\frac{16,000}{\tan(1/120)}\right)$$ I find a distance of about 55,000 kilometers.

[Under construction]

$\endgroup$
  • 1
    $\begingroup$ This also presumes a completely smooth earth surface, which it's not ... from my understanding it's not a perfect sphere, either, but I believe you are as close as it's going to get a real answer here. $\endgroup$ – Pᴀᴜʟsᴛᴇʀ2 Nov 1 '15 at 0:00
  • 1
    $\begingroup$ @Paulster2 Good point. I hadn't considered that. I suppose my method is only an approximation then. :P Darn geology and rotation. $\endgroup$ – HDE 226868 Nov 1 '15 at 0:01
  • $\begingroup$ Part 2 is wrong. Your 5,500 kilometers is eyeball to cable--but at anything like 5,500 kilometers the base of the cable is well below the horizon. Your red lines in Part 1 are a reasonable approximation of what you can actually see. $\endgroup$ – Loren Pechtel Nov 1 '15 at 20:40
  • 1
    $\begingroup$ @LorenPechtel I'm talking about seeing any point on the elevator cable. $\endgroup$ – HDE 226868 Nov 1 '15 at 20:44
  • 1
    $\begingroup$ @LorenPechtel I think there's perhaps some miscommunication. I'm saying that at any one of the points on the surface, there are lines of site from an observer to points on the cable of the elevator. These lines of site can have arbitrarily long distances. $\endgroup$ – HDE 226868 Nov 1 '15 at 22:38
4
$\begingroup$

Regardless of the size of the tower, the important thing to note is the maximum visibility given by the atmosphere. In some places, the very, very best visibility you can expect to get (looking at something unlit, which I assume your tower is as it's derelict) is 150km, but it's generally much, much lower. Fog, clouds, rain, any number of pollutants and even pollen will all contribute to your survivors being unable to see the tower, with the mile wide structure fading into invisibility in the distance like a skyscraper in a fog bank.

Of course, you've had an apocalypse. Whether this will decrease the visibility (extra pollutants/particulates in the air) or increase it (fewer humans pumping soot into the air), will have a big effect. Another important note is air humidity. Water vapour will obscure vision at a distance, so a drier place (lets say the Atacama Desert, where there are several high fidelity telescopes for this very reason) will increase visibility. Your chosen location at the north pole will suffer a lot from snow, as even a light snowfall will rapidly reduce the visibility distance.

If you're lucky, the day is clear and the sun is bouncing off the tower at the right angle then HDE 226868's answer about the limits of the human eye and world geometry applies.

The other consideration is sunshine at nighttime. This is a fairly obvious effect if you look for the ISS at the right time, so sunlight glinting off the structure could theoretically make it much easier to spot. Sadly this only works if you're stood on the right side of the tower (if you're in its 'shadow' you can't use this trick) the clouds are clear enough for you to have 360 degree visibility, and you catch it at the right time. At the north pole the 'right time' pretty much encompasses half of the year, thanks to the day/night cycle there, but nearer the equator this will diminish to a short time around dusk/dawn as the tower is eclipsed by the earth.

TL:DR: Less than 100km most days, up to 5000 on clear days and if you're staring in the right direction at nighttime.

All in all: You're better off looking for the Big Dipper. That isn't going anywhere.

$\endgroup$
  • $\begingroup$ It might be a reasonable assumption to say that the counterweight or upper station would be solar powered and still have working lights visible from the ground. How would that change your answer? $\endgroup$ – Samuel Nov 2 '15 at 19:19
  • $\begingroup$ It makes it a lot more visible at night, but the essence of the answer remains the same: atmospheric effects will almost always dominate and make the tower unreliable for navigation past the hundreds of km's. $\endgroup$ – Joe Bloggs Nov 3 '15 at 9:28
  • $\begingroup$ For a more distant view of the cable you'll be looking out of the atmosphere for most of the path. While clouds or fog will certainly be an issue at times I don't think that's generally going to be the limiting factor. $\endgroup$ – Loren Pechtel Nov 4 '15 at 4:04
  • $\begingroup$ Very good point. I'd forgotten that the atmosphere is comparatively tiny, but in daytime you're going to be looking for a (very) low contrast line high in the sky, and at night it'll be subject to exactly the same (and possibly greater if the lights are off) visibility constraints as the stars, so you may as well use those instead. $\endgroup$ – Joe Bloggs Nov 4 '15 at 9:25
3
$\begingroup$

A real space elevator cable will be invisible to the human eye. The hyper filaments (nanotubes or whatever else is being used) will have to be as small and light as possible, with just enough strength to hold the weight of the cable below them, the climber(s) and a safety factor to account for vibration, possible severing of strands by micrometers and so on.

The cable itself will be tapered so that it is thickest at geostationary orbit (it is supporting the entire mass of the cable below and the pull of the entire cable+counterweight assembly above) and gradually tapering down to the ends. Liftport, a company which hopes too build the real thing, is experimenting with ribbons of material 5cm wide by 13mm thick. The cable will be invisible to the naked eye even when relatively close (you are looking at something essentially like a strip of standard duct tape rising vertically into the atmosphere.

The astute viewer may be able to see evidence of the cable as climbers ascend or descend, especially if (using Liftport's design) the climbers are externally powered by photovoltaic cells energized by ground based lasers. The panels should reflect some of the light and be visible as rising or falling bright "sparks", and should be visible from up to 5000km away, as HDE suggests. As well, if the cable has intermediate stations attached, they should reflect the sun and be visible much the way satellites are visible in orbit, with the strange twist that anything attached to the cable will be fixed in space, not moving through the sky like a conventional satellite.

$\endgroup$
  • $\begingroup$ While I agree that the individual filaments may be invisible, they will likely be bundled in a manner to provide the ability to share load between fibers - such as how metal cables are braided. But even with the equivalent of a "braided" bundle of carbon nanotube fibers feet in diameter, they likely will be too thin to resolved at a distance of 1000 miles. $\endgroup$ – Jim2B Nov 2 '15 at 5:12
  • $\begingroup$ Isn't Liftport is trying to make a Lunar space elevator? The specifications of the cable would be quite different for an Earth based system. $\endgroup$ – Samuel Nov 2 '15 at 19:30
  • $\begingroup$ Liftport wants to build both a space and a lunar elevator. Physics supports a lunar elevator; it can be built using current materials like Kevlar, Spectra 100 or M5, while an Earth based geostationary elevator requires materials that are on the edge of theoretically possible. $\endgroup$ – Thucydides Nov 3 '15 at 1:00
3
$\begingroup$

Static discharge off the cable might aid visibility. Experiments with long cables in orbit have produced interesting results.

http://www-spof.gsfc.nasa.gov/Education/wtether.html

It won't be moving through the magnetic field like a low orbit does, but it should still be interacting with winds and clouds.

Lightning can be visible for over a hundred miles.

$\endgroup$
  • $\begingroup$ That would be a really, really great visual. And also really good for the storyline. Can you tell us more about this effect (e.g. - about the experiments that produced results, etc.) $\endgroup$ – Mikey Nov 2 '15 at 21:46
  • $\begingroup$ Updated with a tether experiment link. $\endgroup$ – Zan Lynx Nov 2 '15 at 22:27
  • $\begingroup$ I think you've got a really good idea to 'spark' interest in the OP's story for making the cable visible. $\endgroup$ – Mikey Nov 2 '15 at 22:38
2
$\begingroup$

Descriptions of the tangential views are important here (because of the premise of the story), but I have a description that is anecdotal, that should be taken into consideration.

When approaching Dubai from inland, the stretch of Sheikh Zayed Highway (a highway lined on either side with hundreds of skyscrapers) from Jumeirah to Umm Sequim is about 10 miles. The slight elevation increase and the unobstructed view of this thin stretch of skyscrapers can be seen on a clear day coming in on E-66 once you get about 15 miles closer to it. This is a clear, dry, sunny atmosphere as well. It isn't that the buildings are under the horizon, either, it's just that they're far away (or the atmosphere obscures it).

So my guess is a 10 mile wide "tower" (or space elevator) would not materialize into your vision until you are 15 miles++ away from it, maybe a bit more because of it's "solid" nature in the sky.

If you illuminate your Space Elevator somehow very, very brightly, then the dynamics of the horizon are the only thing important.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.