Re: Propellantless propulsion fun 3 (recirculative propelant)

Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
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Greg Neill wrote:
"Spaceman" <spaceman@xxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:7padnbhwn6puiBvVnZ2dnUVZ_szinZ2d@xxxxxxxxxxx
Greg Neill wrote:
Sorry, I don't get your point. KE is simply (1/2)*m*v^2
where m is whatever mass is involved. What is
calculating the kinetic energy going to accomplish
here?

It is going to show you that it will take more
kinetic energy to stop the larger mass then it
took to "move" the larger mass.

Since kinetic energy isn't conserved, it would not
surprise me. In fact, you could employ nearly any
amount of kinetic energy you wish to stop the
larger mass.

How? Well kinetic energy is proportional to the
mass and the square of the velocity. Say that your
mass in motion is M travelling at velocity V. You
want to bring it to a halt, but you want to use
more kinetic energy than M has to do it. So you
fire a smaller mass at it with a much higher
velocity.

Say we choose a mass m = M/10.

In order to bring the mass to a halt the smaller mass m
travelling at velocity v must be carrying exactly the
same momentum as the larger mass M. So

M*V = m*v

v = V*M/m

but m = M/10 so

v = V*10

We fire the smaller mass at ten times the velocity as the
larger mass is travelling. That gives it a kinetic energy
of

KE = (1/2)*m*v^2

= (1/2)*(M/10)*(V*10)^2

= 10*(1/2)*M*V^2

It's kinetic energy is ten times that of the mass M.



Anyways, if you want the kinetic energy of the launched
ball it's (1/2)*m*v^2. If you want the kinetic energy
of the ball and object after impact it's

(1/2)*(M + m)*v2^2
= (1/2)*(M + m)*[v*m/(M + m)]^2
= (1/2)*m^2*v^2*/(M + m)

So it's smaller than the kinetic energy of the ball alone
was by a factor of m/(M + m). Again, kinetic energy is
*not* a conserved quantity in general.

So now you think the ball getting stuck in the large mass
will violate the conservation of energy.

Nope. Kinetic energy is not conserved. It trades off
with other forms of energy to keep the sum total conserved
(liek it does with potential energy in orbits) but kinetic
energy by itslef is *not* a conserved quantity.

And you typed all that *** and still can not grasp
That the same amount of force to move the larger
object, will not stop the larger object once in motion because the
larger object now has a greater KE than it had when
it was "at rest".

That's an absurd contention. Kinetic energy is
frame dependent.

Suppose you have an observer at rest with respect to
a mass M. In his frame of reference he applies a
force F to the mass M for period of t seconds. That
is, he causes the mass to accelerate with an
acceleration of a = F/M for a time duration of t.

At the end of that one second the mass is travelling
at a speed v = a*t = F/M*t.

Now, a second observer happens to have been coasting
by when this happend. By coincidence he happend to
be travelling along at the speed v with respect to
the first observer, so what he sees from his point
of view is the mass M starting off with a velocity
of -v in his frame, shoved by the other observer
(decellerated) and coming to rest in his frame.

So both observers witnessed a change of velocity of
magnitude v for the same mass M using the same force
F for the same length of time t. One saw the mass
accelerate to speed v, the other saw it decelerate
from speed v to rest in his frame.

The situations are entirely symmetrical. You can
reverse the roles of the observers, starting with
the mass at rest in the second observer's frame
and him shoving it to cause it to come to rest in
first's frame. In each case the same force F is
applied for time t.

Now Greg pulls the frame jumping bull*** that I am
never allowed to do for his relativity crap, so he can yet
once more, ignore and not admit that the large mass
"will" in fact move according to the outside the box
frame.
LOL

--
James M Driscoll Jr
Spaceman




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