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Alderson Disk general solution

  1. Mar 3, 2008 #1
    Is there a general solution for gravity force in Alderson Disk system?

    Given: Alderson Disk system, disk inner radius R1, outer radius R2, thickness T, density Rho, star mass M (irrelevant due to superposition?).
    [​IMG]
    How to find amount and direction of gravity force at some point in the system, including points in-disk, in the solar gap and out in space (preferably in a real-time calculable way)?
     
  2. jcsd
  3. Mar 5, 2008 #2
    As the lack of replies suggests, either there is no such solution, no one know it, or i didn't ask as supposed.

    Is there a way to find the gravity acceleration acting on a space ship approaching above system, at lease an acceptably imprecise method?
     
  4. Jun 5, 2008 #3
    The question still stands...
     
  5. Oct 9, 2008 #4
    And no replies in half a year.
    Maybe someone could at least tell me why is that?
     
  6. Dec 16, 2010 #5
    > Given: Alderson Disk system, disk inner radius R1, outer radius R2, thickness T, density Rho,
    > star mass M (irrelevant due to superposition?).

    > How to find amount and direction of gravity force at some point in the system, including
    > points in-disk, in the solar gap and out in space (preferably in a real-time calculable way)?

    g exerted by an alderson disk can be approximated by that of an infinite plane, which is 2*pi*G*T*Rho (using your variables). This is uniform near the disk itself, away from the edges. At a density 1/2 that of steel (assuming you have internal, pressurized structures within the disk) or 6 g/m3, you'd need a disk 4,000 km thick to approximate 1 earth gravity. The theoreticial habitible zone around most main-sequence stars is considerably larger than the diamater of the earth, so I assume one extends the disk the diamater of the earth inside and outside of the hab zone in order to avoid edge effects (not detailed here... complex math required, and not usually relevent).

    the influence of the star is calculated normally, G*M/r^2. I assumed a 4x sol brightness star, which would put r1 at about 1 AU and r2 at about 3 AU (minus and plus diamiter earth, as explained above). since a 4x sol star is only marginally more massive than sol (about 2x), the direct effects of it's gravity would be minor, and could be compensated by having the disk rotate (perhaps in seperate sections... haven't worked this all out yet). This would also aid the material strength issues encountered in construction (which could be further relieved by masses in evacuated tubes traveling at superorbital velocities applying supportive force to the structure and accelerated via a dyson "bubble" of solar sails exterior to the radius of the disk itself, similar to the "space fountain" idea... but that's another story.)

    within the disk, since infinite parallel planes of equal density have NO gravitational influence on objects within, the math is easy. double the fraction of the depth of the object and multiply that by the surface gravity to find the gravity at that point. ex: an object 1/8th of the way through the disk (in the above example, 500 km "down") would experience 1/4 earth G.

    within the solar gap, and sufficently distant from the edge, I'm guessing gravitational effects from the disk itself would be neglagible within the thickness of the disk itself (see the Shell Theorm), and a slight attraction towards the plane of the disk otherwise (again, steller attraction is calculated normally). I have not done a rigerous proof of this.

    "Above" the disk, gravity would remain constant and only diminish once the distance from the surface became a significant percentage of the distance to the edge. At a sufficent distance, of course, the mass of the disk could just be treated as part of the system itself.

    I realize this is not all of what you requested, but I hope it helps. Also, I hope that you're still working on it... just started looking into this recently.
     
  7. Dec 18, 2010 #6
    minor miscalculation. Density of steel is about 7 g/cm3, so a structure with internal features could have an avarage internal density of about 3 g/cm3, which puts the thickness of about 8,000 km.

    worked out the grav effect on a non-rotating disk from the twice-solar-mass star, and it starts at about .0012g at 1 au away, and ends at 1/9th that at the exterior. So, for the most part, neglagable. For a general solution, use the surface gravity formula, with distance from the star as r expressed in earth radii (1 AU = 23,010.5 re, approx) and mass of the star expressed in earth masses (sol = approx 333,000 me). g = m/r^2

    near the edge, gravity towards the disk would go down and become angular towards the disk itself. On the exterior surface of the disk, there would be minor but noticible gravity towards the center of the system (somewhere between .5 and .3 ... haven't done the math on it exactly).

    The biggest issue that I've encountered on designing the Alderson Disk is the accumulation of extra atmosphere. The gravity of the disk would extend a significant distance from the surface (at least a quarter of an AU, possibly more; again, I haven't tackled the math) without significant diminishment, and would thus accumulate a significant portion of the steller wind. You'd want to limit the total atmospheric accumulation, so as to limit the atmospheric pressure. My personal solution would be "vaccume towers", approx 20km tall (a bit more than the height of the troposphere), that suck the hydrogen (steller wind would eventually recombine into elemental or ionic hydrogen, poss some helium) from the upper atmosphere into the core of the disk to power fusion generators and provide raw materials for manufacturing (assuming sufficently advanced fusion technology, iron would be the most likely ultimate product).
     
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