[TML] Thrust Distribution Field

[Craig Berry]

> Date: Wed, 01 Aug 2007 17:12:24 -0700
> From: Anthony Jackson <ajackson at iii.com>
>
> > It's worse than that; the needed energy depends on how fast you're
> > going relative to a particular observer, or vice versa.
>
> No, it really doesn't. As long as you're pushing on something, the total
> energy requirement is constant, though where the energy winds up varies
> depending on reference frame.

Yes, it really does. :>  Let's assume you're looking at such a "remote
push" ship from a frame in which both the ship is moving at 10m/s
relative to the mass it plans to push on.  Make the ship mass 10kg and
the pushed-on reaction mass 1kg, as in my previous example.  The ship
will accelerate by 1m/s, again as before.

First, analyze this in the initial rest frame of the pushed body.  The
initial kinetic energy of the system is then all in the ship, and is
0.5(10kg)(10m/s)^2 = 500J.  The final KE, with the reaction mass now
moving 10m/s and the ship 11m/s, is 0.5(1kg)(10m/s)^2 +
0.5(10kg)(11m/s)^2 = 50J + 605J = 655J.  The change in KE is thus
155J.

Now, start in the pre-acceleration rest frame of the ship.  The
initial KE is all in the reaction mass: 0.5(1kg)(10m/s)^2 = 50J.  The
final KE, with the reaction mass moving 11m/s and the ship 1m/s, is
0.5(1kg)(11m/s)^2 + 0.5(10kg)(1m/s)^2 = 60.5J + 50J = 110.5J  The
change in KE is thus 60.5J.

Again, the same maneuver takes different amounts of energy depending
on the frame in which it's being observed, which breaks physics.

-- 
Craig Berry - http://www.cine.net/~cberry/
"The road of excess leads to the palace of wisdom." - William Blake
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