Am I the only one who doesn't want ringworlds to be buildable?

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Peko?

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That's not how that works, the star would feel nothing as the circular ring world would apply a gravitational force equally from every direction on every point of the star. Consequently no points on the star would feel a net force and it is like the ringworld didn't even exist.
Except that pesky divide by distance squared thing. A ringworld should exert a small outward pull on the star.
 

Kat Tsun

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Not really. I think Jupiter would probably be sufficient and it's not the only gas giant in the solar system. In fact it's not even a big gas giant compared to some in the galaxy.
Also this is mass not volume. Volume really doesn't matter that much but you need to keep the actual mass of the ringworld down or it wills tart causing problems.

I'm sorry how are you not lifting matter from the star? Also you can't control what a star fuses and to what. You're much better of using artificial fusion. Also you can't get gold from sustained fusion, fusing anything heavier than iron requires that you actually spend energy. In nature it only happens in supernovas and such.

Unfortunately P. Birch doesn't go into much detail in his little paper, but he discusses neutron star mining as a means of creating planets of pure elements with a "clever mass spectrometer arrangement". The figure is missing so I can't really say how this would look.

To mine a neutron star (Fig 9.1) we use the powerful magnetic fields of the star itself and an energy beam spearing down to the magnetic pole. Now we can guide the near-relativistic jet of neutron star material through a clever mass spectrometer arrangement to yield streams of pure elements. By tuning the beam, we can obtain any element we want. it becomes as easy to build a planet of gold as a planet of muck. And if you want a miniaturised personalised, private star — just assemble a moon of uranium and stand back!

You are lifting matter from the star, in a sense (the neutrons presumably are allowed to decay into however many protons and electrons are required for forming spheres of gold/muck/uranium), but you're not actually trying to pull particles of iron or whatever from its atmosphere (or break the iron shell off with a celestial pickaxe?) with a big magnetic scoop on a space truck or something. That would be too slow. You use the star itself to do that instead.

It's the only thing that I can think of, off the top of my head, that would supply sufficient raw elements to allow for something approaching the economics needed for megascale engineering. You can hand wave it as being arbitrarily quick if you want I suppose. Almost certainly quicker than sucking up a gas giant and converting it into carbon and waste heat.

Besides, you need Jupiter. Not only is he cute, cuddly, and yandere, but he keeps the inner system free from comet and meteor strikes. Or he does the opposite and gave us water by being a space bully. Either way, a good lad all the same. Would definitely buy him a pint. Seems a bit ingrate to want to disassemble him after all the good work he has done over the millennia, too.

OTOH, no one needs a neighbourhood Space Hitler who likes to flash innocent youth with his gamma and x-rays and literally melt their eyeballs.
 
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I think "wonders" you come across, like ringworlds, would become less awe-inspiring if you could build them yourself, so I have to vote no.
 
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TheDungen

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That's not how that works, the star would feel nothing as the circular ring world would apply a gravitational force equally from every direction on every point of the star. Consequently no points on the star would feel a net force and it is like the ringworld didn't even exist.

Also in regards to construction, you need to have more than 2 meters of 'ground' as that would be insufficient to insulate the side facing the sun to your prefered climate. I have no idea how much you actually need, but I'd hypothesis 10-20 meters or probably sufficient. Keep in mind that only needs to exist where you plan on having people and 10-20 meters also gives you sufficient room to work with whatever underground projects you need, say a sewer system or a subway.

Now it just occurred to me, that you would probably need a rather large amoumt of depth so that your oceans would be sufficiently deep. Otherwise your oceans would be very warm thus causes alot of hurricanes/cyclones or other similar things on the planet. Additionally the surface temperature would be very high if the oceans were 1.5 meters deep as was your example. Again, I'm not sure how deep your ocean needs to be but I'd hazard to guess between 400 and 2000 meters would be fine.
You're talking about displacement I'm talking about deformation. Yes the ring would not displace the star but the equilibrium between the outward push from the centre of the fusion, and the inward crunch towards the centre of the gravity would be affected. The gravity of the ring world would give the outward forces a small additonal push.

Think polar coordinates with the origin at the stars centre of gravity.
 
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Kat Tsun

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Not only that, but if you get rid of him, everyone goes in or out. Cat-ass-trophy. Orbits and masses are not to be messed with

It is the year 2.1205427e7 AD.

The U.S. Senate Committee on Environment and Celestial Works hears a proposal from the United States Space Navy (USSN) to construct a megascale ringworld for housing the new Starkiller-class ultra graser, which channels energy directly from the Sun (i.e. solar power) to generate massively powerful, narrowly directed gamma ray beams to nearby star systems. The Navy believes that this weapon will be valuable in the event of war with the Denebians, provided they haven't already fired such a weapon a thousand years ago. Two methods of construction are proposed:

- Deep star mining where the United States Space Mining Corps (USSMC) will construct massive magnetic satellites around Calvera to produce sufficient quantities of carbon for the diamandoid skeleton.

- Disassembly of the major gas giants of the Outer System.

- A third option, considered "legally questionable", regards the use of extra-solar burning stellar masses and Super Jovian gas giants.

Use of ultra-large exoplanets (ULEs) and Super Jovians for Sol's defense purposes was successfully challenged by the Luhman Commonwealth (a protectorate, territory, voting constituency, and Senate candidate for Vela [the 207th state of the Union] of the United States 204816th Congress [although Luhman's unique status as both a United States Territory and United States Permanent Resident had given some pause, he was eventually allowed to be represented on the ballot]) in Luhman v. Luhman, et. al., a landmark Supreme Court case regarding the property rights of sentient gas giants.

After several hours of subconscious simulation and deliberation, the Senate Committee on Celestial Works requested that the Navy and Senate Solar Environmental Protection Commission conduct separate studies into the effects of such megascale architecture on the natural gravity of the Solar System and its effect on human habitation and natural plant/animal life of the celestial bodies.

Eight years later they have yet to reach a firm conclusion, but the final publishing of the initial studies is expected to occur by the end of the decade.
 

Otto of england

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Except that pesky divide by distance squared thing. A ringworld should exert a small outward pull on the star.

Ill give you that I abstracted a bit, but given the mass, which is very small since only section very close together can be counted for the gravitational force at any point, we are looking at the distance (1 AU) of the ringworld, and given that its circular you get a very small force, basically negligible force.

If we assume a Star of the Suns mass (not unreasonable since we are using a AU distance ) then a simplified representation of the force is:

Fg1 = GmM/r^2
Where M = 1 solar mass r = 1 AU G = the gravitational constant

Fg1 = (6.67408 × 10-11 m3 kg-1 s-2)(m kg)(1.989 × 10^30 kg)/(149 597 870 700 m)^2
Fg1 = (0.00593 * m) N

If we assume only the opposite side of the ringworld has any counter balance effect, then the maximum possible additional distance is 1 391 400 000 m. This gives:

Fg2 = (6.67408 × 10-11 m3 kg-1 s-2)(m kg)(1.989 × 10^30 kg)/(149 597 870 700 m + 1 391 400 000 m)^2
Fg2 = (0.00582 * m ) N

Thus we get 0.00011 * m N of net force at the furthest possible distance from the star.

If we assume that a section, m, is 0.00017453 radians (0.01 degree) then we get the arc length of a section to be 26 109 875.4 m, 12 000 000 m tall and say 500 m deep, for a volume of 1.56659*10^17 m^3. If we then assume this has a density equivalent to 1/10 of earth (since we are smart and not using solid rock to build this) then we get 8.63819*10^20 kg of section. This then gives us a net force of 9.50201 *10^16 N (assuming I haven't made a mistake). While this seems very large, when compared to the earths gravitational pull, 3.54162 * 10^22 N, to the sun we see its rather small being ~1 000 000 times smaller.

Now considering that we are ignoring the other 359.98 degrees of this ringworld it is safe the say the net force on a point on the suns surface is even less, likely approaching zero. Even assuming it does not get further reduced (which it does), considering that the rotation of the Earth and other planets does not cause destructive events on the sun, I would hypothesis that a Synthetic sun with only a ringworld orbiting it would experience no more solar weirdness than Earth currently does.

Note: If we use theDungeons 14 m depth we get a force ~10,000,000x smaller.
Note2: The section length I used is probably too large, but I abstracted it by assuming only points within 0.01 degree were distances, R, from point, p. I expect +/- 10% error as a result.



Disclaimer the above math is more probably right, but owning to abstractions we should keep in mind it is not a perfect model of what happens.
 
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Dëzaël

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It is the year 2.1205427e7 AD.....

"Save our gas giants! Spheric farts have the right to live!"
And a decade later, citizen rights were given to any gas bubble weighting more than a ton. A great victory for democracy indeed! :cool:
 

TheDungen

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Ill give you that I abstracted a bit, but given the mass, which is very small since only section very close together can be counted for the gravitational force at any point, we are looking at the distance (1 AU) of the ringworld, and given that its circular you get a very small force, basically negligible force.

If we assume a Star of the Suns mass (not unreasonable since we are using a AU distance ) then a simplified representation of the force is:

Fg1 = GmM/r^2
Where M = 1 solar mass r = 1 AU G = the gravitational constant

Fg1 = (6.67408 × 10-11 m3 kg-1 s-2)(m kg)(1.989 × 10^30 kg)/(149 597 870 700 m)^2
Fg1 = (0.00593 * m) N

If we assume only the opposite side of the ringworld has any counter balance effect, then the maximum possible additional distance is 1 391 400 000 m. This gives:

Fg2 = (6.67408 × 10-11 m3 kg-1 s-2)(m kg)(1.989 × 10^30 kg)/(149 597 870 700 m + 1 391 400 000 m)^2
Fg2 = (0.00582 * m ) N

Thus we get 0.00011 * m N of net force at the furthest possible distance from the star.

If we assume that a section, m, is 0.00017453 radians (0.01 degree) then we get the arc length of a section to be 26 109 875.4 m, 12 000 000 m tall and say 500 m deep, for a volume of 1.56659*10^17 m^3. If we then assume this has a density equivalent to 1/10 of earth (since we are smart and not using solid rock to build this) then we get 8.63819*10^20 kg of section. This then gives us a net force of 9.50201 *10^16 N (assuming I haven't made a mistake). While this seems very large, when compared to the earths gravitational pull, 3.54162 * 10^22 N, to the sun we see its rather small being ~1 000 000 times smaller.

Now considering that we are ignoring the other 359.98 degrees of this ringworld it is safe the say the net force on a point on the suns surface is even less, likely approaching zero. Even assuming it does not get further reduced (which it does), considering that the rotation of the Earth and other planets does not cause destructive events on the sun, I would hypothesis that a Synthetic sun with only a ringworld orbiting it would experience no more solar weirdness than Earth currently does.

Note: If we use theDungeons 14 m depth we get a force ~10,000,000x smaller.
Note2: The section length I used is probably too large, but I abstracted it by assuming only points within 0.01 degree were distances, R, from point, p. I expect +/- 10% error as a result.



Disclaimer the above math is more probably right, but owning to abstractions we should keep in mind it is not a perfect model of what happens.
You are counting displacement not outwards inwards force from the centre.

That's not how you co it, you integrate over in interval from 0 to 2pi from as I said earlier a point of origin at the starts centre of gravity. You do the same for the stars inward gravity, and then set up a force equation between the outward force the explosion and the outwards pull of the ring and the inward gravity of the star. We should be able to forget about the differences in distance between different parts of the sun considering the distances involved are two orders of magnitude larger than the internal distances in the body. And obviously the internal distances of the ring are even more negligible.

anyway you set up a equitation for the pull from a infinitesimally thin sliver of your ringworld and then you sum those up using integration so you get the total outwards pull.

Let's assume that every such sliver is roughly a rectangular block of height same as the height of the ringworld, depth same as the depth of the ringworld and a width delta W that approaches zero as we let the number of such blocks approach infinity (yes it would not be perfectly rectangular but the difference is negligible since they are so thin). The volume of said block times the density, let's say the density of the earth is the mass of said block, and the second mass is the sun. the distance between them is 1 au. just plug it into newtons gravity equation and integrate it from 0 to 2pi and you should get the total outward force of the rings gravity.

for the crunching force you'd have to to a two dimensional ingeral where you integrate the suns own density in the formula, over the intervals 0 to 2pi and r from 0 to the suns radius.

And the last equation is the outward force due to the sustained fusion.

Many of these factors depend on the star in question.

The crunch and the outward force are always relatively balanced, if the outward force is to great then the star goes supernova (or less dramatically start growing toward being a giant), if the inwards force is the great it collapses to a super dense state.
 
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Otto of england

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You are counting displacement not outwards inwards force from the centre.

That's not how you co it, you integrate over in interval from 0 to 2pi from as I said earlier a point of origin at the starts centre of gravity. You do the same for the stars inward gravity, and then set up a force equation between the outward force the explosion and the outwards pull of the ring and the inward gravity of the star. We should be able to forget about the differences in distance between different parts of the sun considering the distances involved are two orders of magnitude larger than the internal distances in the body. And obviously the internal distances of the ring are even more negligible.

anyway you set up a equitation for the pull from a infinitesimally thin sliver of your ringworld and then you sum those up using integration so you get the total outwards pull.

Let's assume that every such sliver is roughly a rectangular block of height same as the height of the ringworld, depth same as the depth of the ringworld and a width delta W that approaches zero as we let the number of such blocks approach infinity (yes it would not be perfectly rectangular but the difference is negligible since they are so thin). The volume of said block times the density, let's say the density of the earth is the mass of said block, and the second mass is the sun. the distance between them is 1 au. just plug it into newtons gravity equation and integrate it from 0 to 2pi and you should get the total outward force of the rings gravity.

for the crunching force you'd have to to a two dimensional ingeral where you integrate the suns own density in the formula, over the intervals 0 to 2pi and r from 0 to the suns radius.

And the last equation is the outward force due to the sustained fusion.

Many of these factors depend on the star in question.

The crunch and the outward force are always relatively balanced, if the outward force is to great then the star goes supernova (or less dramatically start growing toward being a giant), if the inwards force is the great it collapses to a super dense state.

I agree my method was a very simplified look at an individual point on the star, because A) I didn't feel like doing the added calculations of the integration and B) it is good enough to outline my point, which was the outward pull from one side of the ringworld is dampened enough by the outward pull from the other side of the ring world that its effect is insignificant.

If we take my previous abstraction, the net force at the center of the star is 0 from the ringworld as every additional gravitational pull is perfectly cancelled out by the ringworld from the opposite side of the ring. As you move towards the outside that grows, and in a simplified 1-d perspective it grows to 0.00011*m which is very small. In a 2-d perspective it changes slightly but, the net force at point, P, on the circle circumference is now <0.00011*m as there are more points to the right of P than on its left. As you move to 3-d it is again less than what it calculated in 2-d as again there are more points to the right of P than their are to its left.

When considering though, that a regular star system such as ours has numerous asteroids, planets, moons, and other random small objects all pulling gravitationally on the Sun. I estimate that the average net outward pull of the Solar system on the sun is => the artificial ring world system. Therefore, the ringworld is inconsequential on the stability on said star.
 
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Person012345

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You don't experience gravity inside a sphere, is the same not true for a ring? It doesn't matter where you are within the sphere, whether you're 5ft from one of the surfaces or 5000 miles, the gravitational forces cancel out completely. So I'm not really understanding where you're getting this phantom "pull" from. Unless the rules are different for rings.
 
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LordMagus

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While in the millions of years, only a few civilizations have made ringworlds, don't forget the story of Yuht. They were around for ages, and NEVER EVEN DEVELOPED FTL, and were defeated by a new kid on the block. Imagine how crazy the game would be if new civilizations could pop up, and not start always start at 'pathetic' technology compared to you, and developed ringworlds in record time. That'd spice up the game a bit, and new custom ringworlds allowing you to specialise your resource generation (and allow for more things to spend the resources on) would be kinda cool.
 

Kat Tsun

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"Save our gas giants! Spheric farts have the right to live!"
And a decade later, citizen rights were given to any gas bubble weighting more than a ton. A great victory for democracy indeed! :cool:

The metaphysics of the universe meant that each particle of antimatter harvested by the US Space Navy retained a connection to Senator Luhman, however slight, and the accidental (the review board's term, which was challenged by The Honorable Luhman 16 in a closed door session of the Senate Committee of the Armed Forces, but ultimately accepted) use of antimatter, intended for propulsion, as atomic triggers for Mark 67 fusion bombs during a raid on Gilese terror camps preceded a "great consternation" among Senator Luhman's constituent sentient molecules. This manifested as a high energy radiation burst and shedding of several thin layers of Senator Luhman's atmosphere (Senate Astrophysicians compared it to a starquake). The Senator's response was that he fel "as if millions of voices suddenly cried out in terror and were suddenly silenced".

The Senator's appreciation of "the classics" was completely lost on the entire committee and military officers present. ):
 
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Jabby

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Everyone is arguing about the materials but remember this: the ringworlds are made of LIVING metal which implies that it can grow in size.
 

LordMagus

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the ringworlds are made of LIVING metal which implies that it can grow in size.

The living metal would still require some material in order to grow, unless it can turn solar energy into matter (why the hell not, it's sci-fi, and we know it's theoretically possible, even if requiring loads of energy).
 

TheDungen

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I agree my method was a very simplified look at an individual point on the star, because A) I didn't feel like doing the added calculations of the integration and B) it is good enough to outline my point, which was the outward pull from one side of the ringworld is dampened enough by the outward pull from the other side of the ring world that its effect is insignificant.

If we take my previous abstraction, the net force at the center of the star is 0 from the ringworld as every additional gravitational pull is perfectly cancelled out by the ringworld from the opposite side of the ring. As you move towards the outside that grows, and in a simplified 1-d perspective it grows to 0.00011*m which is very small. In a 2-d perspective it changes slightly but, the net force at point, P, on the circle circumference is now <0.00011*m as there are more points to the right of P than on its left. As you move to 3-d it is again less than what it calculated in 2-d as again there are more points to the right of P than their are to its left.

When considering though, that a regular star system such as ours has numerous asteroids, planets, moons, and other random small objects all pulling gravitationally on the Sun. I estimate that the average net outward pull of the Solar system on the sun is => the artificial ring world system. Therefore, the ringworld is inconsequential on the stability on said star.
Except the ringworld they suggested had the mass of a thousand earths remember? Not to mention they placed it all at 1 au, which is considerably closer than most of the mass of the solar system, and remember distance is even more important than mass.
And you could make the same argument for the crunch on any single particle inside of the star, but the difference does matter. And I'm not talking about the gravity of the ringworld overcoming the gravity of the star on it's own just weakening it enough so that the outwards push of the fusion wins out.
And to know the force is there we actually don't need to count, for every action there is an equal an opposite reaction remember? The star pulls on the ringworld and subsequently the ringworld also pulls on the star.

You don't experience gravity inside a sphere, is the same not true for a ring? It doesn't matter where you are within the sphere, whether you're 5ft from one of the surfaces or 5000 miles, the gravitational forces cancel out completely. So I'm not really understanding where you're getting this phantom "pull" from. Unless the rules are different for rings.
With that logic there should be no pressure at the centre of the earth because it all cancels out unfortunately that only applies if you're looking at displacement. It won't move you but it will deform you.
 
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BlackUmbrellas

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You don't experience gravity inside a sphere, is the same not true for a ring? It doesn't matter where you are within the sphere, whether you're 5ft from one of the surfaces or 5000 miles, the gravitational forces cancel out completely. So I'm not really understanding where you're getting this phantom "pull" from. Unless the rules are different for rings.
...what?

I mean, yeah, you would absolutely experience gravity within a hollow sphere- gravity fades with distance, so unless you're in the exact centre of the sphere, you would experience a slightly higher gravitational pull in the direction of the closest surface.

The strength of that pull might be negligible depending on the density of the sphere, of course, but it would absolutely still be there- same for a ring.

This is actually one of the greatest engineering problems of such stellar super-constructs as I understand it- stellar fluctuations and even tiny amounts of drift risk destabilizing the entire thing. This was even a major plot point for Niven's Ringworld; it had been without maintenance for so long that the correctional thrusters meant to keep it properly "balanced" had failed and placed the entire structure in danger.
 
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TheDungen

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Everyone is arguing about the materials but remember this: the ringworlds are made of LIVING metal which implies that it can grow in size.
The only problem I have with living metal is that it's metal. The only reason mankind had built with metal up until now is that it's easy to produce. In all other ways it's inferior to fibrous materials. It took the bacteria of the world millions of years to figure out how to break down lignin for an example.

I'm hoping we can make Iserlohn with living metal.

Just a big sphere that has like 10 Titan lasers and 20 Tachyon lances.
wouldn't be a ring world though

...what?

I mean, yeah, you would absolutely experience gravity within a hollow sphere- gravity fades with distance, so unless you're in the exact centre of the sphere, you would experience a slightly higher gravitational pull in the direction of the closest surface.

The strength of that pull might be negligible depending on the density of the sphere, of course, but it would absolutely still be there- same for a ring.

This is actually one of the greatest engineering problems of such stellar super-constructs as I understand it- stellar fluctuations and even tiny amounts of drift risk destabilizing the entire thing. This was even a major plot point for Niven's Ringworld; it had been without maintenance for so long that the correctional thrusters meant to keep it properly "balanced" had failed and placed the entire structure in danger.
Goes further than that, since any object except a point mass is actually made up of a huge number of particles very few of which are at the actual centre of the circle/sphere.
 

Person012345

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...what?

I mean, yeah, you would absolutely experience gravity within a hollow sphere- gravity fades with distance, so unless you're in the exact centre of the sphere, you would experience a slightly higher gravitational pull in the direction of the closest surface.

The strength of that pull might be negligible depending on the density of the sphere, of course, but it would absolutely still be there- same for a ring.

This is actually one of the greatest engineering problems of such stellar super-constructs as I understand it- stellar fluctuations and even tiny amounts of drift risk destabilizing the entire thing. This was even a major plot point for Niven's Ringworld; it had been without maintenance for so long that the correctional thrusters meant to keep it properly "balanced" had failed and placed the entire structure in danger.
No, you're incorrect. Inside a hollow sphere you feel no gravity no matter where you are, the forces all balance out.

https://en.wikipedia.org/wiki/Shell_theorem

I don't know about the last paragraph, the ringworld would still feel gravitational effects from the star, but the star would not feel gravitational effects from the ringworld (gravity wouldn't exert an "outward force" on the particles inside it, including the star) - assuming the effect holds true within a ring.
 
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