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Alain wrote:
>>>I have a BIG problem with Dyson spheres! Inside an hollow sphere, there
>>>is NO gravity, if the sphere is built around a star, everything not held
>>>in place will fall in the star. If you make it spin, all the air will
>>>collect at the equator, untill the sphere collapses unto itself. The
>>>equatorial part goes flying away and the poles plunging into the star.
Wasn't it Tim Cook who wrote:
>>Why is there no gravity? If you take a sun-sized star and build a
>>sphere of Earths around it at 1 AU (dunno where you'd get that many
>>Earths), does the now hollow sphere not have gravity on either its
>>inside or outside surface?
Mike Williams wrote:
> There's gravity on the outside, but not on the inside. It just so
> happens that the strong contributions from the small regions under your
> feet are exactly cancelled out by the weaker contributions from the much
> larger regions over your head. I don't think there's any simple way to
> demonstrate that, you have to calculate the integral like Newton did.
There's a good lot of discussion on Wikipedia. Search for Dyson Sphere
and then go to the discussion page (tab at the top). Talks about the
calculations et. al.
There'd be neutral gravity on the inside, but if you spin the sphere
fast enough, then momentum-in-a-circular-restriction (aka centrifical
force) will stick you to the sphere. Something like 32km/s (or 320 ..
it's been a while) for a 1AU sphere.
(And the gravity on the outside would be horrendous, the gases from the
sun would kill you, there's not enough matter to make it .. yada yada ..
great concept for SciFi, and ray tracing, but not real life!)
Cheers!
Rick Measham
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Wasn't it Rick Measham who wrote:
>
>There'd be neutral gravity on the inside, but if you spin the sphere
>fast enough, then momentum-in-a-circular-restriction (aka centrifical
>force) will stick you to the sphere. Something like 32km/s (or 320 ..
>it's been a while) for a 1AU sphere.
But only the people standing on the equator would get vertical gravity
at full strength.
The strength of the force varies with cos(latitude) and the angle to the
local vertical is proportional to latitude. So people standing near the
poles get very little force, and what force there is would be almost
horizontal.
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Mike Williams wrote:
> The strength of the force varies with cos(latitude) and the angle to the
> local vertical is proportional to latitude. So people standing near the
> poles get very little force, and what force there is would be almost
> horizontal.
Which is where the original argument came from (and I agree) .. the
poles would implode, thus degrading the structure and causing the
equator to explode.
One poster (somewhere, not here) suggested a band rather than a sphere.
The 'band' would be the equatorial region and would (somehow) hold
together as it spun.
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Anthony D. Baye nous apporta ses lumieres en ce 2005-10-12 20:50:
> Mike Williams wrote:
>
>> Apart from the absence of the plasma tube, I'd say it was Thistledown.
>>
>> Where does your hollow asteroid get its light from?
>>
> Wow. Someone else reads what I read.
>
> A.D.B.
>
> P.S. Alain, Gravity is based on mass. A sphere the size of earth's
> orbit with a shell thick enough to withstand impacts would naturally
> have a reasonable amount of gravity on it's inside surface.
The gravitational pull from the relatively small but very near region under your feet
is exactly
countered by that of the extremely large and distant part over your head.
That leave you with ONLY the gravity from the sun witch is straight up... NOT a good
thing.
--
Alain
-------------------------------------------------
'hAS ANYONE SEEN MY cAPSLOCK KEY?'
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Karl Anders wrote:
> at least if the sphere is homogeneuos
<English swot>
Homogenuous!
</English swot>
See you in Khyberspace!
Yadgar
Now playing: Starship Trooper (Yes)
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Alain wrote:
> Anthony D. Baye nous apporta ses lumieres en ce 2005-10-12 20:50:
>
>> Mike Williams wrote:
>>
>>> Apart from the absence of the plasma tube, I'd say it was Thistledown.
>>>
>>> Where does your hollow asteroid get its light from?
>>>
>> Wow. Someone else reads what I read.
>>
>> A.D.B.
>>
>> P.S. Alain, Gravity is based on mass. A sphere the size of earth's
>> orbit with a shell thick enough to withstand impacts would naturally
>> have a reasonable amount of gravity on it's inside surface.
>
>
> The gravitational pull from the relatively small but very near region
> under your feet is exactly countered by that of the extremely large and
> distant part over your head.
> That leave you with ONLY the gravity from the sun witch is straight
> up... NOT a good thing.
>
Far be it from me to dislike being proved wrong.
but an interesting factoid: Larry Niven's Ringworld, which was
conceived as an intermediate step toward a dyson sphere, spun at 770
mi/h. It was determined that, in order to withstand the shearing forces
from the spin, the base material which measured 1000' thick would have
had to have had a tensile strength on the order of the force which holds
the nucleus of an atom together.
A.D.B.
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Alain wrote:
> The gravitational pull from the relatively small but very near region
> under your feet is exactly countered by that of the extremely large and
> distant part over your head.
> That leave you with ONLY the gravity from the sun witch is straight
> up... NOT a good thing.
Let me get this straight. If you put a hypothetical mass, say, ten
times the mass of the earth, in an orbit exactly opposite the earth, the
gravity under your feet will be countered by a certain amount by the
extremely large and distant mass over your head? In other words, to
simplify things, without extra masses...you weigh less at noon?
--
Tim Cook
http://home.bellsouth.net/p/PWP-empyrean
-----BEGIN GEEK CODE BLOCK-----
Version: 3.12
GFA dpu- s: a?-- C++(++++) U P? L E--- W++(+++)>$
N++ o? K- w(+) O? M-(--) V? PS+(+++) PE(--) Y(--)
PGP-(--) t* 5++>+++++ X+ R* tv+ b++(+++) DI
D++(---) G(++) e*>++ h+ !r--- !y--
------END GEEK CODE BLOCK------
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news:434e9f3b$1@news.povray.org...
In other words, to
> simplify things, without extra masses...you weigh less at noon?
>
Don't tides work this way?
Marc
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Anthony D. Baye wrote:
> Alain wrote:
>
>> Anthony D. Baye nous apporta ses lumieres en ce 2005-10-12 20:50:
>>
>>> Mike Williams wrote:
>>>
>>>> Apart from the absence of the plasma tube, I'd say it was Thistledown.
>>>>
>>>> Where does your hollow asteroid get its light from?
>>>>
>>> Wow. Someone else reads what I read.
>>>
>>> A.D.B.
>>>
>>> P.S. Alain, Gravity is based on mass. A sphere the size of earth's
>>> orbit with a shell thick enough to withstand impacts would naturally
>>> have a reasonable amount of gravity on it's inside surface.
>>
>>
>>
>> The gravitational pull from the relatively small but very near region
>> under your feet is exactly countered by that of the extremely large
>> and distant part over your head.
>> That leave you with ONLY the gravity from the sun witch is straight
>> up... NOT a good thing.
>>
> Far be it from me to dislike being proved wrong.
>
> but an interesting factoid: Larry Niven's Ringworld, which was
> conceived as an intermediate step toward a dyson sphere, spun at 770
> mi/h. It was determined that, in order to withstand the shearing forces
> from the spin, the base material which measured 1000' thick would have
> had to have had a tensile strength on the order of the force which holds
> the nucleus of an atom together.
>
> A.D.B.
Still, if it were spun up for gravitational effects, and only the area
around the equator were used for habitation, and the rest was used for
energy collection, storage, life support and suchlike, then there would
still be a massive amount of livable area. f/ex. if the sphere were the
size of earth's orbit, and assuming that the livable area were a million
miles wide, then the total inside surface area (Not accounting for
variations in topography) would be 2(pi)(9.28e6)(1e6) mi^2 that's
several thousand times the surface area of earth.
A.D.B.
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Tim Cook <z99### [at] bellsouth net> wrote:
> Alain wrote:
> > I have a BIG problem with Dyson spheres! Inside an hollow sphere, there
> > is NO gravity, if the sphere is built around a star, everything not held
> > in place will fall in the star. If you make it spin, all the air will
> > collect at the equator, untill the sphere collapses unto itself. The
> > equatorial part goes flying away and the poles plunging into the star.
>
> Why is there no gravity? If you take a sun-sized star and build a
> sphere of Earths around it at 1 AU (dunno where you'd get that many
> Earths), does the now hollow sphere not have gravity on either its
> inside or outside surface?
A spherical shell exerts no gravity on anything within. If you do the math
it turns out that no matter where you are within the shell, the gravity
from the material behind you exactly cancels the gravity from the material
in front of you.
Outside the shell, gravity works like you'd expect.
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