Re: Konstantin Tsiolokvsky was wrong?!


"The Ghost In The Machine" <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:hj27d3-fug.ln1@xxxxxxxxxxxxxxxxxxxxxxxxxx
In sci.physics, Spaceman
<Realspace@xxxxxxxxxxx>
wrote
on Sat, 25 Feb 2006 15:44:45 -0500
<CvGdnTfYu-YmXJ3ZRVn-tg@xxxxxxxxxxx>:

"The Ghost In The Machine" <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote in
message
news:2ap6d3-j3g.ln1@xxxxxxxxxxxxxxxxxxxxxxxxxx
As you wish.

In order to derive Newtonian/Tsiokovski, one must have
conservation of momentum. If M = M(t) is the mass of
the rocket, v = v(t) the velocity, e = e(t) the exhaust
speed (which is NOT constant relative to the rocket,
in this derivation), and t of course is time (which for
purposes of this derivation is assumed absolute throughout
the Universe), one gets

M*v = (M - dM) * (v + dv) + (v-e)*dM

since the rocket is shooting mass backwards.

We now get

M*v = (M - dM) * (v + dv) + (v-e)*dM
= M*v - v*dM + dv*M - dM*dv + v*dM - e*dM

and therefore

0 = -v*dM + dv*M + v*dM - e*dM
= dv*M - e*dM

which shows that the two forms -- my more complicated expression and
the one mentioned in the Wikipedia -- are in fact equivalent.

So

dv*M = e*dM

We now have to get M into the right place. Dividing by M,

dv = e/M*dM

and therefore

v = v(0) + integral(t=0,t) (e(t)/M(t) M' dt)

where the integration constant is replaced by v(0).

If we assume constant fuel flow f, then M' is itself a constant, and we
can replace M' by a constant (-f) and pull it out, giving


M' is not a constant,
again you have assumed to far.

I said IF.




v = v(0) + integral(t=0,t) (e(t)/M(t) (-f) dt)
= v(0) + integral(t=0,t) (e(t)/(M - tf) (-f) dt)
= v(0) - f*integral(t=0,t) ( (e(t) / (M/f - t) dt) )

I cannot go further without some knowledge of e(t). If e(t) is
in fact assumed constant (i.e., the exhaust velocity *relative to
the rocket* is constant), then of course one ultimately gets

v = v(0) + e*f*eval(t=0,t) log(M/f - t)
= v(0) + e*f*(log(M/f) - log(M/f - t))

which is in more or less the usual form. (Yes, the signs are correct;
integral 1/(k-t) dt = -log(k-t) + C.)

As for stopping -- what stops the propellant? Certainly not
anything related to the rocket. Once it leaves the rocket the
spent propellant pursues its own lonely existence in orbit
around the Sun, or, if the rocket's exhaust is going fast enough,
it will shoot out of the solar system entirely.

If the propellant hits a planet the planet will be nudged a bit,
but the rocket won't care.

You keep creating "constants" from variables that are never
actually constant.
The only constant about speeds of a rocket or a fuel is .... nothing.
Each and every part of the mass of the fuel, mass of the rocket,
velocities etc.. are never "constant".
C,mon Ghost,
how are you falling for those "variable" constants?
:)


I have shown how it is possible for a rocket to exceed its
exhaust speed. Is this sufficient or not?

No,
You have created a constant that does not exist,
and worked upon such a non existing fact.

By your use of such,
You are basically stating that someone can push
a merry go round faster than they can run without using
their arms for an extra push while running speed.
:)
It is crazy you can even accept such crap.

So if e = 500 meters per second exhaust speed
The rockets will be crazy fast and will go much
faster than 500 meters per second.
(care to put that e into place and come up with rocket speed?)
:)



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