Re: Physics Answer to a Question...

["A. R. Marlow" <marlow@beta.loyno.edu>]

On Sun, 9 Feb 1997, John Mallinckrodt wrote:

> ...
> I disagree.  No need for any limits (although in this case it doesn't
> matter.)
>
> Let's apply conservation of momentum to the nth spit in the frame of the
> earth:
>
>     Momentum before = Mv
>
> where M = mass of the old man before the nth spit [(50 - (n-1)*0.001) kg]
> and v = speed of the old man before the nth spit
>
>     Momentum after = (M - m)(v + dv) - m(V - v - dv)
> ...

Here is where the problem lies  --  you are using V - v - dv as the speed
of the ejected tobacco after the spit.  But by definition V is the
difference between the speed of the rocket (old man) and the speed of
the ejected material in the given frame, and so the correct expression
for the speed of the ejected material in your formula above is simply
V-v.  (See, for example, Marion & Thornton, Classical Dynamics of
Particles and Systems, Fourth Edition, Saunders, 1995, p. 89).  By
including the extra dv you get a cancelation of the earlier mdv term, but
that cancelation in fact does not actually occur.  In the standard 
derivations of the rocket formula, the mdv term is dropped because it
is a product of two infinitesimals (specifically noted in the text of
Marion & Thornton cited above), but we cannot argue that here, and
hence the term remains and contributes to the motion of the old man.
 
 
> where m is the spit mass [0.001 kg], dv is the old man's added speed (not
> infinitesimal), and V is the speed of the spit relative to the old man 
> (5 m/s)
> ...

A. R. Marlow                          E-MAIL:  marlow@beta.loyno.edu
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