01 Apr '06 18:23>3 edits
Originally posted by sonhouseI 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.
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?
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.