Re: Rocket action

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

Hugh Haskell wrote:

> But if the exhaust gases that were emitted at first are still moving
> (at high speed relative to the initial point of firing) they are a
> long way from the cm by this time and thus provide the
> "counterbalance" that can keep the cm fixed. I haven't tried this
> calculation, so my impression may be wrong, but ....

Several days ago Roger Freedman wrote:

> The final velocity of the rocket will be greater in magnitude
> (and is often much greater) than the relative speed v_ex
>  if ln (m_0/m) >1, that is, if m_0/m > e = 2.71828

 m_0 is the initial mass of the rocket + fuel
 m is the final mass of the rocket
 **********************************************

Here is a numerical model in which the escaping fuel is
represented by "bullets" of equal mass dm, fired one after
another. The speed of each bullet is v0 with respect to the
rocket. Its speed with respect to the laboratory frame,
vB=v0-vR, decreases as the mass of the rocket, mR
decreases and as the rocket's speed, vR, grows. It turns
out that vR=0.98*v0 when mR=37 kg and vR=1.01*v0
when mR=36 kg, as theoretically shown by Roger. (For
the initial mass of 100 kg). The listing of the code and its
partial output are shown below.

program rocket
let mR=100      ! initial total mass kg
let vR=0        ! rocket initially at rest
let dm=1        ! mass of each "bullet" kg
let v0=1000     ! m/s (bullet wrt to rocket)
DO                      ! one bullet per loop
print mR,vR,v0-vR
let mR=mR-dm            ! remaining rocket mass
let vR=vR+(v0-vR)*dm/mR ! new rocket speed
if mR<2*dm then stop
LOOP                    ! next bullet
end
************************************
rocket mass      rocket speed
     kg            m/s (lab)
-----------------------------------
     99             10.
     98             20.101
     97             30.3051
     96             40.6144
..........................
     40            908.834
     39            933.834
     38            959.476
     37            985.791
     36           1012.82
     35           1040.6
..........................
      6           2737.38
      5           2904.04
      4           3104.04
      3           3354.04
      2           3687.38
      1           4187.38

In this case the payload of 1 kg used 99 kg
of fuel. To get a larger vR one would use a
more massive rocket (more fuel).

Ludwik Kowalski
********************************************************
To understand is to find a satisfactory causal relation.
To explain is to express that understanding.
To teach is to promote understanding.
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