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Re: starship-design: Re: FTL travel
> In a message dated 4/6/00 11:18:40 PM Pacific Daylight Time, firstname.lastname@example.org
> > K = m * ((1 / sqrt(1 - v^2/c^2)) - 1)
> > K = kinetic energy
> > m = object mass
> > v = object velocity
> > c = speed of light
> > K = 1/2 * m * v^2 is only a low-speed approximation.
> 1/10 C is low speed so the K calculated is near exact. The relativistic
> effects are insignificant though not zero.
You're claiming that a smaller mass with an equal amount of kinetic
energy will go faster than c. That's not what will happen; putting that
amount of kinetic energy into the smaller mass makes it go closer to,
but not faster than, c.
> > Relativistic
> > kinetic energy grows without bound as an object approaches the speed of
> > light.
> I agree totally. Real kinetic energy does not.
Relativistic kinetic energy _is_ real. Newtonian physics is merely a
low-speed approximation to relativistic physics; the approximation
breaks down when you start dealing with velocities that are significant
fractions of c. If you want calculations that are accurate at any
velocity and not just small ones, you need to use the relativistic
> To be more precise you cannot see the rocket traveling at twice light speed
> from the below viewpoint of earth. Such velocities relativistic are limited
> by the equation you give to below C. The velocity I calculated was velocity
> real as an observer at rest on earth was not part of the problem and I did
> not ask what velocity relative to earth was obtained.
No observer sees the rocket traveling faster than c, not even one on the