Re: Konstantin Tsiolokvsky was wrong?!


"The Ghost In The Machine" <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:6ld5d3-km2.ln1@xxxxxxxxxxxxxxxxxxxxxxxxxx
This is going to annoy the hell out of me, but hopefully
someone can point out what must be an obvious (in hindsight) error.

A rocket is initially motionless in space, next to a space
buoy (for purposes of this illustration the origin of our
coordinate system). It is full of propellant.
Define M(t) = M as the mass of the rocket at time t.
Obviously M(t) < M(0) if t > 0, since the rocket will be
releasing propellant. Also, v = v(t) is the velocity of
the rocket.

In a Newtonian space, if the rocket releases a piece of propellant
of mass dM at velocity -e (since it's coming out of the back
of the rocket), momentum must be conserved at all times. Therefore

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

where the notation 'du' is a corruption of 'delta-u', and for the
sake of ASCII simplicity I use the former; you might also be
familiar with Leibnitz notation -- dM/dt=f, the flow rate of
the propellant.

If we assume e is constant, this is a differential equation with
two variables v and M. With some further manipulation we get

Why do you asume e is constant?
What is e standing for in reality to make it constant?




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

If we divide by M * (v-e):

0 = dv/(v-e) - dM/M - (dv * dM)/(M * (v-e))

The third term simply vanishes (I know, it looks like magic),
giving us

dv/(v-e) = dM/M

log(v-e) = K + log(M)

where K is the obligatory integration constant.

At time 0, I will set v(0) = 0 for simplicity; therefore

log(e) = K + log(M(0))
K = log(e) - log(M(0))

and

log(v-e) = log(e) + log(M) - log(M(0))

v-e = exp(log(e) + log(M) - log(M(0)))
= exp(log(e)) * exp(log(M)) / exp(log(M(0)))
= e * M/M(0)

This is not the form given in the back of the book (or in the Wikipedia),
which gives the much simpler momentum equation

M dv = - e dM

or

-dv/e = dM/M

-v/e = K + log(M)

and K = -log(M(0)) since I'm assuming v(0) = 0; therefore

v = -e * log(M) / log(M(0))

Can someone point out where I goofed? I can't believe
such an elementary result misled everyone for more than
a century... :-) Nor am I convinced that my magic didn't
blow up in my face, despite the issue that dv * dM/ dt = 0.

Help....

--
#191, ewill3@xxxxxxxxxxxxx
It's still legal to go .sigless.


.

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