Re: Velocity limits on Starships...

On Mon, 7 Apr 1997, Joachim Heck - SunSoft wrote:

>   F = ma = mv/t, so force is directly proportional to velocity.  

Acceleration, not velocity. Not quite the same thing.

>		2
>   E = 1/2 * mv, so energy is proportional to the square of the
> velocity.  That means the faster you go the more energy you need to
> provide the same force (thrust in FT terms).	So that's why an engine
> would become less capable at higher speeds.

Well... either an engine produces a set force just like that (a 
reactionless drive, decidedly unrealistic anyway), or it chucks out 
something from the rear end.

We'll assume this something masses m, which is small but non-zero.

The drive system imparts a kinetic energy of 1/2mv^2 on the chunk.
We're not interested in the exact v, but rather the force it takes,
since 
the opposite reaction will kick us in the direction we want to go.

The chunk has momentum mv. It used to have 0, so it just had an impulse 
of mv push it. We could speculate the length of the impulse and
calculate 
the exact force, but that's not necessary.

The total momentum of the system must remain constant (Newton's 3rd).
Therefore, the ship's momentum increases by mv in the opposite
direction, 
which translates to a velocity increase of:

 delta-v = mv/ship's mass

Thus, for a constant energy E, you can accelerate by a constant amount.
(Actually, a little more since you're getting lighter all the time).

Not to worry, it's a common mistake to apply conservation of energy when

one needs to use the conservation of momentum. 

I won't speculate reactionless drives, since they're impossible as we 
understand physics. If they are possible, they probably won't follow any

traditional laws of physics anyway.

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