Reactionless Redux
- From: "Knobby" <knobbynobbs@xxxxxxxxxxx>
- Date: 21 Mar 2007 16:03:42 -0700
Hello to all the regulars and irregulars from this humble newbie.
I was prompted to sign up and post here from the discussion on
reactionless drives and energy costs thereof that was linked to from
the excellent Atomic Rocket site.
Sorry if this dead horse has been beaten so much as to have been
reduced to it's constituent atoms, but if I may be indulged, I'd like
to re-visit the question of what the hypothetical energy cost of a
reactionless drive might be.
I would also like to apologize in advance if this has been asked and
answered elsewhere, and similarly would like to thank any who deign to
chime in for putting straight my admittedly meagre knowledge /
understanding of physics.
Anyhoo, on with the show...
It is my understanding that, regardless of frame of reference,
relativistic effects on a mass are the same. IOW, if two space craft
moving at relativistic velocities are pacing each other side-by-side,
looking from one to the other, one will be able to observe a length
contraction in the other, while not noticing seeing it in one's self.
So, if I've got this right, and relativistic effects on a body are the
same regardless of frame, why not use a spacecraft's relativistic mass
(the extra) and the grand ol' equation of E=mc^2 to figure out the
energy costs associated with having moved that spacecraft?
For instance, if the spacecraft's velocity is such that due to
relativistic effects it's mass (in kg) has increased by 1%, the energy
it would have taken to achieve that velocity would be (at 100%
efficiency) 0.01*mass*c^2 joules.
Okay, that's it. Hack away.
.
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