Re: starship-design: Re: FTL travel

[Steve VanDevender <stevev@efn.org>]

STAR1SHIP@aol.com writes:
 > Taking for example the general equation MVof exhaust=MVof payload.
 > to accelerate a given payload mass of 5 tons twice light speed requires I 
 > accelerate 100 tons of exhaust to 1/10 light speed (at 1 g acceleration).
 > Using the equation E(kinetic)=1/2 MV we get the energy required.
 > using the equation E=Mc^2 we get the mass that needs to be converted 
 > providing the energy.

That equation isn't relativistically correct.

In the situation where your 5 ton payload _instantaneously_ reacts 100
tons of fuel which is ejected in one direction at 0.1 c, then we would
have:

Mp * Vp / sqrt(1 - Vp^2) = Mf * Vf / sqrt(1 - Vf^2)

Solving for Vp we would get about 0.89 c.  And this is still
oversimplified.

If you carry 100 tons of fuel and accelerate at 1 g the resulting
velocity is much lower because you aren't just pushing 5 tons of payload
but all the remaining fuel at that point in the trip.

If your fuel is capable of reacting to produce exhaust products that
travel at 0.1 c, then you're also converting a nonnegligible amount of
fuel mass into energy, and that needs to be accounted for too.

 > Machines like partical accelerators that use the general instantaneous 
 > acceleration equation of E'=1/2 MV gamma, reguire the Energy relativistic 
 > equation to be used for the energy is delivered intantaneously (small 
 > increment of time)from the relativistic(observer or rest) frame instead of 
 > the frame of moving(inertial) object and is therefore limited to below light 
 > speed relativistic so that momentum is conserved.

The same laws of physics apply to rockets as to particles in particle
accelerators.  You don't get to pick and choose which laws of physics
you want to apply in which situations.