1. Subscribersonhouse
    Fast and Curious
    slatington, pa, usa
    Joined
    28 Dec '04
    Moves
    53223
    01 Apr '06 18:233 edits
    Originally posted by sonhouse
    I was looking at the formula for circular orbits, its
    P(orbital period in seconds)=SQRT (4 (PI^2)* R^3)/GM
    R = distance of orbit from center of earth, and the 4PI^2/GM for the planet earth is just a constant, 9.904 E-14 (to 4 places) so its really easy, R is in meters, which is the radius of the earth in meters added to the altitude of the orbit in meter ...[text shortened]... about 2 ish Km/s.
    How can two masses connected by a line ever be stable in such an arrangement?
    I made a mistake on the presentation of the orbital period formula, it should be P=((4*(PI^2)*R^3)/GM))^0.5. Same thing basically but I did not include the GM part in the sqrt. For anyone who is interested.
    But the 4PI^2/GM is still a constant for the earth, which is 9.904E-14
    so its still just (9.904E-14*R^3)^1/2=P. A lot simpler.
    Just remember if you want to do a few yourself, R= radius of earth plus the height of the orbit. Units are in meters and seconds. P=seconds per orbit. R = radius of earth in meters plus height of orbit in meters above the earth. There is a nice derivation of the earths radius at this link: http://en.wikipedia.org/wiki/Earth_radius
    It gives the mean radius as 6370.998 or 6370998 meters, which will probably give better circular orbit times. So going back to the original formula and solving for R^3, calling the 9.9etc as C, its R=(P^2/C)^1/3 so to find the geosync orbit, now its just R=((86400)^2/C)^1/3
    or (7464960000/9.904E14)^1/3 =7.537318256E22^1/3 =
    42241462 Meters minus 6370998 (radius of earth)=35870464.5 Meters or 35,870.565 Km / 1.609 = 22,293.6 Miles for an orbital period of 86400 seconds or one day. Thats close enough for our purposes, the C is actually slightly more than 9.904E-14, I just rounded it to that # so I didn't have to keep typing in all those digits tied to PI.
  2. Subscribersonhouse
    Fast and Curious
    slatington, pa, usa
    Joined
    28 Dec '04
    Moves
    53223
    13 Aug '06 05:44
    Originally posted by sonhouse
    I made a mistake on the presentation of the orbital period formula, it should be P=((4*(PI^2)*R^3)/GM))^0.5. Same thing basically but I did not include the GM part in the sqrt. For anyone who is interested.
    But the 4PI^2/GM is still a constant for the earth, which is 9.904E-14
    so its still just (9.904E-14*R^3)^1/2=P. A lot simpler.
    Just remember if you w ...[text shortened]... 14, I just rounded it to that # so I didn't have to keep typing in all those digits tied to PI.
    I think the answer is this: If you have a large mass at Geosync then the mass at 100,000 Km has to be forced to hold things taut, so it has to be purposely sped up by something, rocket or whatever to a velocity of about 12 Km/s if I remember right. And it would have to have monitors to check for movements straying away from taut and motors to keep the whole thing in a straight line. I think if left to itself it would rather quickly become unstable.
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