# [geocentrism] Zero gravity

*From*: Bernard Brauer <bbrauer777@xxxxxxxxx>*To*: geocentrism@xxxxxxxxxxxxx*Date*: Mon, 12 Mar 2007 14:07:46 -0700 (PDT)

A satellite company wanted to deliver it's satellite to a point above the equator of the Earth where the effects of gravity would be zero, so that...................IT WOULD NOT FALL! So they hire a math scientist and he says 22,200 miles high. So the satellite company hires a rocket company to deliver the satellite to that level. So what is all the fuss about satellites? Heliocentrists keep asking, "why doesn't the satellite fall"? Because it's not supposed to at zero gravity. That was the whole point of putting it there. Sheesh! The Earth itself "hangs on nothing" and the same with the satellite - it "hangs on nothing." Bernie ----------------------------- Found on internet: Question: "I know that an object in geosynchronous orbit effectively "hovers" over a specific point on earth, due to it actually orbiting the planet at the same relative speed as the earth is revolving. Yet the object does not actually "fall". So how high up does an object need to be before it can do this without 'falling'?" Answer: So how do we determine this velocity and height? F=ma Ft (tangential force, the component of our force causing acceleration required for geosynchronous orbit) = R*alpha (the radius from the point mass center of earth's gravity multiplied with the angular acceleration {the rate at which the object speeds up or slows down}) If you followed all that, then you should guess that for our geosynchronous orbit the "alpha" must be zero (the object doesn't really change speed as it "hovers" over a point on earth), which means the tangential component of the force is zero! Fc (centripedal force, the component of the force that causes the object to constantly change direction, i.e., "fall" towards earth at just the correct rate that it keeps falling downward at just the right speed that as it hurtles forward it never really hits the earth) = m*v^2/R (the object's mass times it's forward velocity squared divided by the radius at which it's orbiting). But what *causes* this m*v^2/R force? F=G*m1*m2/R^2 (or in intuitive terms, the force due to gravity is proportional to the masses of the objects in question and inversely proportional the the square of the distance between their centers of gravity). So we'll call m2 Earth, and replace it with Me and we already refered to the mass of the object as m, so m1=m. So the duelling Newtonian determinations of the force on the object can be set on separate sides of the equation: m*v^2/R = G*m*Me/R^2 And calling in for algebraic backup, we can simplify things to: v = sqrt(G*Me/R) G is a constant and Me is something that we can look up in a Physics book. So what we've got is a generic equation relation between v and R. What we haven't yet used in our equation is the fact that we're specifically after geosynchronous motion. Or the v=R*omega (velocity at some radius is equal to the radius times the rotational speed). omega is one we can figure out ourselves, as it turns out the earth rotates once every... (wait for it)... 23 hours and 56 minutes! Okay, just kidding. Once every 24 hours. In terms friendly to calculations that means one revolution (2*pi radians) per (24 hours/day * 60 minutes/hour * 60 seconds/minute) 86400 seconds. So now we've got two equations with two unknowns which we can solve a couple of different ways. I'm not going to go through the bother of describing the actual mechanics (the process is to set one equation such that one unknown is isolated and then plug that equation into the other), but what you end up with is v = sqrt(G*Me*omega/v) v^2 = G*Me*omega/v v^3 = G*Me*omega v = [(6.67 x 10^-11 N*m^2/kg^2) * (5.98 x 10^24 kg) * (2*pi/86400 s)]^1/3 v = 3070 m/s = 6870 mph we can then put that into either of our equations to determine R: R = v/omega = 42.3 x 10^6 m (the distance from the center of the earth) or, subtracting off the radius of the earth (say at the equator), d = R - Re = 35.8 x 10^6 m = 35800 km or 22200 miles above the equator. --------------------------------- TV dinner still cooling? Check out "Tonight's Picks" on Yahoo! TV.