Re: Launch Rails..Was:Re: Orbital Towers.

From: Adrian Tymes (
Date: Fri Apr 14 2000 - 23:12:30 MDT

"Michael S. Lorrey" wrote:
> OK, new/old idea to talk about: mass drivers as launch system.

Just so you know, they've got a model showcased on the back page of
this month's Wired. (Or is that what inspired you to ask this?)

> Practical questions:
> How long does the rail need to be to accelerate up to 2,000, 4,000, 6,000 and
> 8,000 mph respectively?

If you mean "miles per hour" (usual meaning of mph), it's usually better
(and safer) to keep all discussion of measures in the metric system when
dealing with space systems. As to the question itself, it fortunately
comes down to basic math.

L = length of launch rail in meters
A = acceleration in meters per second per second (assumed constant)
V = exit velocity in meters per second
T = time from start to exit in seconds

L = VT/2 (since initial velocity is zero)
V = AT
T = V/A
L = VV/2A

So, with an acceleration of 40 m/ss (just over 4 Gs), and a desired
exit velocity of 3,600 km/h = 1,000 m/s...

L = 1,000 * 1,000 / 2 * 40
  = 12,500 meters

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