Your "weight" on a "Space Elevator"? I'm not sure the term "Space Elevator" is widely used, and outside of those who've read Clarke's THE FOUNTAINS OF PARADISE I'm not sure the term would mean much to many, so by way of an overly simplified description of the term... You put a big rock in geosynchronous orbit, drop a rope from it, and tie off the rope to a spot on Earth's equator. Now you run an elevator car up and down the rope to get in and out of orbit, and you have your "Space Elevator". For a MUCH better explanation (not to mention a nifty "artist's concept" pic) check out this article over on FIRST SCIENCE.COM: http://www.firstscience.com/site/articles/elevator.asp For the sake of discussion, OUR "Space Elevator" consists of 4 parts: 1) A "counter weight" (perhaps a large asteroid) slightly father out in its orbit than geosynchronous (so that the c.g. of the entire "Space Elevator" as a system is what orbits at the g.s.o. point as opposed to just the "counter weight"). 2) A "tether" that runs from the "counter weight" down to within 300 miles of Earth's surface where the "tether" is attached to... 3) A Penn Station of New York sized "Railroad terminal" in orbit that remains stationary 300 miles above the Earth's equator, from which dangles... 4) A large number of "tracks" that make their way down from the "Railroad Terminal" 300 miles up, to anchor points on Earth's equator. It's an old Idea, I know, but here's DA QUESTION... If you weighed 100pounds at Earth's sea level, how much would you weigh once you made your way up from Earth's surface to the "Railroad Terminal" 300 miles above Earth's surface? I can't imagine where you'd be weightless, even though you're in orbit. In fact the only place I'd imagine you'd be weightless is at the point, somewhere along the "tether", where you're actually at the g.s.o. point. Until you got that far out I'd expect you'd still have weight, and beyond the g.s.o. point (getting closer to the "counter weight" that's slightly farther out than g.s.o.) I'd expect your "weight" would start to go slightly negative (i.e. once past g.s.o. you'd have a tendency to start drifting toward the ceiling of your elevator car). First - Am I right about that. Secondly - how quickly would your weight drop as you ascended the elevator? Would it drop in a linear fashion from 100 pounds on Earth's surface, to zero at the g.s.o. point, to a slightly negative value past g.s.o.? How would you calculate your weight at 300 miles up under these conditions?