If the Moon or Mars will we . . .

> Another way to think of this is, how much farther is
> 80,000 ft. from the center of Earth? Earth's radius
> is about 20,718,750 feet, so 80,000 ft. is less than
> 1/2 of 1 percent, which is not much compared to all
> of the trouble of trying to lift a big rocket?
>
Another way to compare is looking at the escape
velocity from these two points.

(all numbers from http://en2.wikipedia.org/wiki/Earth)

Using equatorial diameter of earth = 1.27563 * 10 ^7 m
Adding 80,000 feet or 24384 meters you get so that
puts you at 1.27807 m.

Using Ve = sqrt(G*M/r)
Ve(earth)=sqrt((6.67 * 10^-11)*(5.97 * 10^24)/(1.27563
* 10 ^7 m)) = 5587 m/s or 5.587 km/s
Ve(80k feet up))=sqrt((6.67 * 10^-11)*(5.97 *
10^24)/(1.27807 * 10 ^7 m)) = 5581 m/s or 5.581 km/s

So there you have it, you shave roughly 6 m/s off the
escape velocity from launching a bit higher up or
about 0.09% of the total velocity.

On a similar note, I recently found a neat-o link at
nasa for atmospheric modeling for earth and mars which
is related to this discussion called the Interactive
Atmosphere Simulator:
http://www.grc.nasa.gov/WWW/K-12/airplane/atmosi.html

Ryan Z