Tucson_Tim

2009-Apr-19, 11:05 PM

Let's assume a fully operational space elevator, attached at the equator, where the center of mass is at the geosynchronous orbit altitude (approx 22,000 miles), with a large counterbalancing mass (space station?) at the top. For my question, let's also ignore the Earth's atmosphere and its effects. If you like, the space elevator can be on the Moon or Mars. Please see the following link:

http://en.wikipedia.org/wiki/Space_elevator

My question evolves from reading the Clarke novel Fountains of Paradise:

http://en.wikipedia.org/wiki/Fountains_of_Paradise

Question: Let's say on our trip up the elevator we stop every 20 miles or so and drop a ball from a "window". We don't give the ball any other velocity -- we just drop it straight down. Now, in Clarke's novel, Morgan is able to free an auxillary battery and it drops. He states that at the current altitude, which is 405 km, that it will land 10 km directly east of the base of the elevator. I believe the dropped balls (at every 20 miles of altitude) will follow parabolic curves to the ground. We continue to do this till we get to the geosnychronous altititude of 22,000 km. Now at 22,000 miles, the ball will not fall but will just float along next to the elevator -- in geosynch orbit. I have several questions now:

1) For balls dropped from under the geosynch point, are the falling paths indeed parabolic? The balls don't have enough radial speed to be in these orbits.

2) Are there points nearing the geosynch point where the balls will fall but miss the Earth and end up in some elliptical orbit?

3) What will be the path of balls dropped from above the 22,000 mile geosych orbit? Say from the space station at the top? The balls have too much radial speed to be in these orbits.

http://en.wikipedia.org/wiki/Space_elevator

My question evolves from reading the Clarke novel Fountains of Paradise:

http://en.wikipedia.org/wiki/Fountains_of_Paradise

Question: Let's say on our trip up the elevator we stop every 20 miles or so and drop a ball from a "window". We don't give the ball any other velocity -- we just drop it straight down. Now, in Clarke's novel, Morgan is able to free an auxillary battery and it drops. He states that at the current altitude, which is 405 km, that it will land 10 km directly east of the base of the elevator. I believe the dropped balls (at every 20 miles of altitude) will follow parabolic curves to the ground. We continue to do this till we get to the geosnychronous altititude of 22,000 km. Now at 22,000 miles, the ball will not fall but will just float along next to the elevator -- in geosynch orbit. I have several questions now:

1) For balls dropped from under the geosynch point, are the falling paths indeed parabolic? The balls don't have enough radial speed to be in these orbits.

2) Are there points nearing the geosynch point where the balls will fall but miss the Earth and end up in some elliptical orbit?

3) What will be the path of balls dropped from above the 22,000 mile geosych orbit? Say from the space station at the top? The balls have too much radial speed to be in these orbits.