wrote:
 > Saying it operated in the vacuum of space. is there a speed limit to
 > how fast an electromagnetic repulsion accelerator can launch an
 > object?

 You don't seem to have gotten a completely straight answer,
 so I'll throw my $.02 in.

 > Naturally, there is the speed of light barrier, but I don't
 > think that's possible ... or is it?

 No. But you could get so close to c
Relative to what?
that it would be hard
 to measure the difference. The thing about the relativistic speed
 limit is that it takes more and more energy as you approach
 c.
Relative to what?
It makes more sense to talk about your energy, or your
 relativistic "gamma" factor, than it does to talk about speed
 for objects very close to c.
Relative to what?

 Thus, there are accelerators with gamma factors of 100, and
 others with gamma = 1000. In both cases the velocity is just
 a tiny bit different from c,
Relative to what?
 but the energy of a particle with
 gamma = 1000 is 10 times the energy of the same particle
 with gamma = 100, and it has 10 times the momentum
 also.

 You can never reach v = c
Relative to what?
(gamma>oo as v>c
Relative to what?
), but there's
 no limit to high gamma can be. An incredibly highenergy
 particle detected a few years ago, nicknamed the "Ohmy
 god particle", had a gamma somewhere in the neighborhood
 of 10^20 as I recall. But still a speed v < c.
Relative to what?

 > And whatever speed limit such an accelerator had, could it be reached
 > if the accelerator was 60,000 miles long?

 You could use work = Force * distance to figure out how
 much energy you could get on such an accelerator.

 Let's say you can accelerate a mass of 1 kg at 1000 g's,
 i.e. you can exert a force of 9800 N on that mass. Then
 over 60000 miles (96500 km) you would give the mass
 9800 N * 96,500,000 m = 9.46*10^11 J.

 But actually, this isn't a relativistic speed. Let's check it
 with the Newtonian formula, which is OK at low speeds:
 KE = 0.5*mv^2

 v = sqrt(2*9.46*10^11/m) = 1.38 *10^6 m/sec, or about
 0.0046 c.
Relative to what?

 That gives a gamma of 1/sqrt(1  v^2/c^2) = 1.00001
Relative to what?
 > If so, what would the rate
 > of acceleration from a standing start to achieving that speed at the
 > end of the 60,000 miles?

 Not a sensible question since you can't achieve c.
Relative to what?
Not a sensible answer,
"But the ray moves relatively to the initial point of k, when measured in
the stationary system, with the velocity cv".  Einstein.