🚀 go-pugleaf

In article <4576@cruzio.santa-cruz.ca.us> snarfy@cruzio.santa-cruz.ca.us writes:

>    According to conventional "THEORIES" of your SO CALLED "science"  ,  the
>   Moon (Luna) has only 1/6th of the Earth's gravity. 

	Also according to the landing gear (and takeoff capability) of
the Apollo moon missions.  So far, the theory and the experiment are
in agreement.

> Newton formulated his
>   Law  of Unversal Gravitation in 1666 which led to this conclusion.  This
>   conclusion was based on assumptions that (1) The density and  composition
>   of  the  moon  was  similar  to  the  earth's,  and (2.) This density and
>   composition was fairly uniform throughout the lunar globe.

	Neither assumption is necessary.  The size of the moon, its
distance from the Earth, and the mass ratio of Earth and Luna are
the relevant numbers; none of these is 'assumed', all are measurable.

>    These  assumptions  and  observations,  taken together , lead one to the
>   conclusion that the moon has "1/6" of the earth's gravitational  pull  at
>   it's  surface.  This  figure  has been accepted for centuries. 

>               a(moon)/a(earth) = 1/81.3/(.273)^2 = .165 ( @ 1/6 )
>
>               This calculation is apparently based on the observed distance
>   of  2903  miles of the earth's distance from the earth-moon barycenter as
>   that distance is divided into the observed distance to  the  moon.   This
>   calculation  yields  a distance to the moon of 81.3 times the distance of
>   earth's center to the barycenter, locating this  barycenter  well  inside
>   the earth.

	81.3 is the ratio of the Earth's mass and Luna's; this is both
the currently accepted figure and the figure in my 1935-edition CRC
Handbook of Chemistry and Physics.

>    Throughout the fifties and sixties  ,  the  so  called  "neutral  point"
>   where  the  gravitational  forces  of  the earth-moon system balance,was
>   given over and over again as being between 22,100 and 25,200  miles  from
>   the surface of the moon in the direction of the earth. 

	Given by who?  It doesn't follow from anything I see in
the data, nor do you give any attribution.

>   presumed to represent logical guesses from trained scientists.  However ,
>   it  is  acknowledged  in  Abell's  book that only by observing falling or
>   orbiting bodies in the moon's vicinity could  the  actual  neutral  point
>   distance, hence the moon's true gravity , be determined.

	Earth is an orbiting body in the Moon's vicinity.  

>     Using   the   above   neutral  point  and  assumed  gravity  for  their
>   calculations, the U.S. and the Soviet Union began to send space probes to
>   the moon in the late fifties.  They  met  with  miserable  failure.  

	Not 'miserable' failure; 'initial' failure, followed by 
'inevitable' success.

>
>    Missions  became  more succesfull after this, this possibly attributable
>   to the notion that the  Soviets  and  the  Americans  had  been  able  to
>   recalculate  the  neutral  point and the correct gravity of the moon 

	Nope.  No one calculates an orbit using any mythical 'neutral
point' number; you use the positions and masses of celestial objects
(Earth, Sol, Luna, etc.).  None of those are hard to locate, and 
there was never any great uncertainty about the masses, either
(due in part to Kepler's laws... the orbital periods give this 
information pretty accurately).


>    The  July  25th  ,  1969  issue of TIME magazine stated that the Neutral
>    Point was 43,495 miles from the center of  the moon.  Werner  von  Braun
>    stated  in  the  1969  edition of "History of Rocketry and Space Travel"
>    that the neutral point was 43,495 miles the center of  the  moon  .  The
>    pre-Apollo  distances  were  given  as  20,000  to 25,000 miles from the
>    center of the moon. 

	Given by whom?  Either there is a second definition involved, or
SOMEONE made an error.  Several someones, if the range of 20,000 to 25,000
is to be believed.  Again, I see no source for the questionable numbers.
Let me guess: 43,495 miles from the moon (in the direction of
the Earth) the ratio of Earth's gravity to Luna's gravity (absolute
values of local acceleration) is roughly


	(40,000)^2 *81/(240,000 - 40,000)^2   ~=  4

so this distance corresponds to something rather different
from 'neutral gravitational acceleration'.  How about
looking at the ratio of gravitational potential energy for
a test object at this position?

	(40,000) *81/(240,000 - 40,000)   ~=  1


	It looks to me like someone has introduced more than one
concept under the name 'neutral point', and has caused some confusion.

	John Whitmore