FYS: I probably should have tried to figure this out before.

[jeriendhal]

So, given that the Ring is in geosynchronous orbit by necessity, since it's connected to the Earth via the beanstalks, what exactly is the local surface gravity? It can't be Earth normal, and there's no magi-tech gravity generation in this setting.

Um, is there a rocket scientist in the house?

Comments

[allah-sulu.livejournal.com]

A geostationary orbit can only be achieved at an altitude very close to 35,786 km (22,236 mi), and directly above the Equator. This equates to an orbital velocity of 3.07 km/s (1.91 mi/s) or a period of 1,436 minutes, which equates to almost exactly one sidereal day or 23.934461223 hours. Given that the diameter of the Earth is 12,742 km (7,918 miles), then a person/being on the Ring will be 42,157 km (26,195 mi) above the center of the Earth. That converts to 138,309,600 feet. Also, the rotational period of one sidereal day converts to approximately 86,164 seconds. We'll use those values for the formulae given on this site: Suppose that 'R' is the distance in feet between the axis of rotation of the station, and your position on the inside of the rotating shell: the floor. Let 'T' be the time it takes to rotate once around the axis. Then, your velocity will be simply V = 2πR / T = 2 × 3.1415926535 × 138309600 / 86164 ≈ 10085.7 ft / sec The acceleration would be V² / R = (10085.7)² / 138309600 ≈ 0.7354612 ft / sec² As the acceleration of Earth's gravity (1g) is 32 ft / sec², the gravity on the Ring would be 0.02298316g or about 2.3% of the gravity on the Earth's surface. That is, if I'm doing this right. (Neither Skylab nor Mir nor the ISS were in geosynchronous orbit, granted, but I don't recall any of them having any noticeable gravity strictly from their speed orbiting the Earth. Being within the Ring would be similar to being within a space station at the same altitude, so one would expect an environment of microgravity.) You could add habitation blocks that rotate around the Ring to generate higher gravity within those areas, and use the low-gravity center of the Ring itself for transport of people and equipment between the rotating blocks. Safety permitting, there could also be low-gravity parks in the Ring for humans to have closely-monitored fun within, for limited amounts of time, before they were returned to their Earth-standard gravity habitations (as extended periods in low gravity are not healthy for humans).

[allah-sulu.livejournal.com]

I just recalled that I did read a novel recently which had a beanstalk to geosynchronous orbit and beyond. There was a station at geosynch, and the beanstalk extended an equal distance beyond that to a second, counterweight station. I believe the station at geosynch was described as being low/micro-gravity, with higher gravity as you moved up toward the counterweight station.

I wonder how fast Ringworld is rotating about its star...

[harvey-rrit.livejournal.com]

That's 770 miles per second. Gravity remained constant as it drifted off-center.

An object that displays centrifugal force on its inner side cannot be orbiting; it would be in free fall. The Ringworld needed attitude jets to keep station.

I surmise the Ring to be at the counterweight end of the Beanstalks; geosynchronous orbit is a Beanstalk's midpoint. That being the case, inner surface gravity would be the same as planetary surface gravity.

[jeriendhal.livejournal.com]

Which implies that the Ring must have an attitude control system (preferably with beefed up security so idiots a million years from now won't be borrowing the rockets).

[harvey-rrit.livejournal.com]

A million? You must know something about people I don't. I'd have said fifty, tops.

(Studies show optimists are happier, but pessimists have more fingers.)

[harvey-rrit.livejournal.com]

Brain fart over now. Your Ring is anchored in place by the Beanstalks. Ve's goot.

[jeriendhal.livejournal.com]

It's likely got one anyway. The Groupmind is a belt and suspenders and duct tape sorta AI. I can't see the beanstalks serving any structural purpose in holding the Ring in place. At minimum there has to be a lot of give in their anchors to account for flexing in the main Ring structure.

[allah-sulu.livejournal.com]

Not to mention flexing in the surface of the Earth to which the other ends of the beanstalks are tethered: earthquakes, continental drift, tides...

[jeriendhal.livejournal.com]

Well crap. The Ring is definitely going to have to be rotating faster than that. Nanny Groupmind isn't going to let anyone develop weak bone structure due to microgravity effects. Obviously either the beanstalks don't connect directly to the ring, or they might be on a separate connecting track on the 'roof' of the Ring.

Note that I'm assuming whatever Unobtanium the Ring is constructed of, if it's strong enough to support the Ring as a solid structure, it's probably strong enough to survive the effects of a 1g rotation for X number of millennia.

I do like the idea of the roof area having microgravity fun zones for the inhabitants. Well, however much fun you can have once you're wearing the required padding...

[allah-sulu.livejournal.com]

Sorry about that.

On an unrelated note, I bought most of your books a while back and I'm just getting caught up on some of them on my Kindle now. Loved The Dragon's Companion, and now I'm halfway through Triumvirate.

[jeriendhal.livejournal.com]

Oh, thank you! Hopefully Triumvirate holds up for you. It got a little rushed towards the end IMO.

[siderea]

That converts to 138,309,600 feet.

WHY?! What did SI ever do to you????

As the acceleration of Earth's gravity (1g) is 32 ft / sec², the gravity on the Ring would be 0.02298316g or about 2.3% of the gravity on the Earth's surface.

Well, the centrifugal pseudo-gravity on the Ring.

Not that I expect it to be other than negligible, but I do wonder just how much the Ring masses and whether it has enough suck of its own to be felt. After all, it's much, much bigger than most space ship scenarios usually discussed.

This all just begs the question of just what it's made of, anyway. What's the molar mass of Handwavium, again?

Okay, fine: 4.2157x10^7m above the center of the Earth, 2πR = 2 × 3.1416 × 4.2157x10^7m = 2.649x10^8m of circumference.

Steel's average density is 7.85 g/cm^3 (I read it on the internet, so it must be true), which would be 7.85x10^9 g/m^3. So if the Ring were a band of steel one km wide (1x10^3m), and 1cm thick (1x10^-3m) and 2.649x10^8m in circumference, mass is:

M = 7.85x10^9 g/m^3 × 1x10^3m × 1x10^-3m × 2.649x10^8m = 2.077x10^18g = 2.077x10^15kg.

The Earth masses 5.97x10^24 kg, according to google, and the Moon masses 7.35×10^22 kg. A Ring of steel 1cm thick and 1km wide would mass about one 10-millionth of a Moon.

So, not very much as gravity-field generating celestial objects go, but I still wouldn't want to have to push it out of my driveway. Also, because the center of mass of the Ring is the center of the Earth, most of the Ring's gravitational attraction will not be pulling your boot soles against the surface of the Ring (what we dirtballs used to call "up"), but towards the center of the Earth (the direction formerly known as "down"). However, the actual gravitational force on you would be a product of all the points of the Ring acting on you, to the inverse square of their distance from you. I'm not up to figuring out the integral that would express that relationship, much less solving it.

Anyway. We do not know if the Ring is more or less massive than a 1cm thick x 1km wide ribbon of steel.

I have to say, this all raises all sorts of questions for me: Which planet did they dismantle for the raw materials? What are the tidal forces on the Earth of wearing a heavy belt like that? How much of the Ring is pulling you more toward the Ring and how much of the Ring is pulling you toward the center of the earth (that would be an awesome visualisation, if someone does figure out the integral)?

(ETA: Somebody should check my math. I am very, very, very out of practice at this sort of thing, and not feeling to sharp at the moment due to tireds.)

(ETA2: Goddamn. My brain just announced that the expression for the force of gravity of the Ring on someone standing on its inner surface is the integral of 1/(2*4.2157x10^7m*sinθ)^2 dθ. Is that right?)

[allah-sulu.livejournal.com]

I converted to feet in the instance above because feet were used in this site that I'd consulted. I used meters below.

I have to say, this all raises all sorts of questions for me: Which planet did they dismantle for the raw materials?

I assume they would start by using any asteroid, comet, or other body whose orbit crosses the Earth's at any point in the foreseeable future, since they're all about the safety of Earth's (and the Ring's) inhabitants. In fact, they'd probably sweep the asteroid belt, Kuiper belt, Oort cloud, and the Trojans clean; leaving just the planets and their moons (unless they still needed even more materials).

Somebody should check my math. I am very, very, very out of practice at this sort of thing

You're not the only one.

[siderea]

No, wait, I need the mass in there. The integral of 2.077x10^18g /(2*4.2157x10^7m*sinθ)^2 dθ maybe?

[allah-sulu.livejournal.com]

I'm willing to stop at the line "A Ring of steel 1cm thick and 1km wide would mass about one 10-millionth of a Moon." However we crunch it, we're getting fairly negligible results. In order to get anything approaching Earth-normal gravity, something has to change. The mass of the Ring has to be significantly higher and/or it has to rotate a lot faster (normally, orbital speed is fixed as a function of the orbit's radius; that's why we have a defined geosynchronous orbit in the first place) and/or there have to be additional constructions rotating around the Ring.

It's a pity the Groupmind didn't just Terraform Mars for humans to live on. They're already Terraforming one planet (Earth); why not two?

[jeriendhal.livejournal.com]

Eyes cross. Okay, now I know why the Doctor was always saying "I'll explain later".

Thanks. I hadn't considered the mass of the Ring's primary material. That's going to be something to consider.

Also, Pluto was never a real planet anyway. :)

[sraun]

My memory from freshman physics is that the net gravitational pull at any point inside an ideal hollow sphere is zero. I would assume a ring would be the same.

[selenite.livejournal.com]

Yep. If you're outside the ring it acts equal to a point source at its center. So adding a bit of upside rounding error to the Earth's attraction.

[natf.livejournal.com]

Fascinating! I am also intrigued as to how you added a lovely quadrille paper bacground to your comment. Straight HTML?

[allah-sulu.livejournal.com]

Straight HTML (a <table> with that pattern as the background). LJ doesn't let you use any of that fancy-shmancy newfangled CSS in comments anymore.

To be more precise, this is the block of code that I placed before my text…

<table background="http://DammitJa.net/lj/bg/xkcd-graph-tile.png" cellpadding="8"><tbody><tr><td><div align="justify">

…and this is the block that goes after the text:

</div></td></tr></tbody></table>

[natf.livejournal.com]

Ah hah! Thank you! Do you mind if I use that elsewhere?

[allah-sulu.livejournal.com]

Go right ahead! Depending upon where you're using it, you might want to copy the image file there rather than hotlinking it but I don't think that would be a problem with most sites. Other paper options:

Paper - Graph http://DammitJa.net/lj/bg/xkcd-graph-tile.png

Paper - Lined http://DammitJa.net/lj/bg/notepaper-bg.gif

Paper - Loose Leaf http://DammitJa.net/graphics/page.gif

Paper - Spiral Bound http://DammitJa.net/lj/bg/a-spiralbound.gif

Paper - Triangles http://DammitJa.net/lj/bg/trigp1.jpg

Paper - Wrinkled http://DammitJa.net/lj/bg/paperbackground2.gif

[natf.livejournal.com]

Thank you for these! *copies files*

[allah-sulu.livejournal.com]

No prob. If you want more, or need other HTML tricks, just ask.

[mckenzee.livejournal.com]

What's the height?

[allah-sulu.livejournal.com]

35,786 km above sea level (which I assume is the zero point on that chart).

Using the formula from that Wiki page, since that chart doesn't go out far enough:

gh = go (	re	)	²
	re + h

g_o = 9.8 m/sec² (Earth's gravitational constant)

r_e = 6371000 m (Earth's mean radius)

h = 35786000 m (height above sea level)

∴ g_h = 0.22382155 m/sec² = 0.022838933g

That's almost exactly what I got above (within the margin of error). However, since the centrifugal force vector is directly away from the center of the Earth, and the gravitational vector is directly toward the center of the Earth, the two would pretty much cancel out.

[selenite.livejournal.com]

Certified rocket engineer speaking. Standard consulting fees waived.

Yep, as you proceed up along your Beanstalk the apparent gravity dwindles until you're in free fall at geosynchronous altitude*. Keep going and centrifugal force overwhelms gravity and you stand with your head now pointed toward Earth.

A moving ring sliding on the geosynch one can spin hard enough to give you your desired local acceleration. Walter Jon Williams' Dread Empire's Fall space-opera series has some large scale examples of this.

I can shoot you a copy of a Pyramid article I wrote for SJG on this once upon a time.


* Exact altitude varies with irregularities of Earth's gravitation field.

PS. I will happily field any questions you have, I was just answering the OP.

[jeriendhal.livejournal.com]

Thanks. If you could send a copy of that article to my sirtalen gmail account I'd really appreciate it. I"m sure I'll have more questions for you later.