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<DIV><FONT size=2 face=Arial>An important gravitation lesson was given to me in
late 1963 as a<BR>First year Physics student. Just two weeks earlier I had
never even<BR>seen an Integral Calculus symbol, and now I wound up having to
solve<BR>multiple Integral Calculus problems as homework. The Professor
was<BR>teaching us about Newton and his Fluxions (which we call
Integral<BR>Calculus now). He had just taught us Newton's standard
equation of<BR>gravitational force (for the surface of the Earth).</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>The far more interesting lesson was to be
next. Clearly the Professor<BR>had decided to demonstrate how brilliant
Isaac Newton was, even regarding<BR>the Fluxions that he had just
invented. The Professor then<BR>decided to do the calculation (of
Newton's) for a location halfway down<BR>inside the Earth. He stated a
requirement that the Earth must be<BR>uniformly symmetric, although density
increase with depth is allowed.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>The Professor then decided to use a Polar
coordinate system, and then<BR>he set up a triple Integral Calculus problem in
that coordinate<BR>system.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>We students (me just barely 17 years old) were
given the actual math as<BR>homework. The Professor had given us some
suggestions. Possibly the<BR>most important of those was that to calculate
the Integral Vector Calculus<BR>total for a very thin spherical shell centered
on the center of the Earth, but<BR>for such a shell which was at a radius which
was greater than the radius<BR>of the chosen location inside the
Earth.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>The homework problem was initially just a double
Vector Integral, for all locations<BR>everywhere on that shell, which had
delta-r (miniscule) shell thickness. I encourage<BR>all readers of this to
repeat that homework assignment. It turns out<BR>that an incremental
location right 'behind' you is close so the inverse-square<BR>law applies, but
there is such an enormous area 'across' the shell, that<BR>the net attractive
force from way over there EXACTLY EQUALS the attractive<BR>force from that tiny
area 'behind' you, which is a Force Vector which is in<BR>the exact opposite
direction. When you do the complete Vector Integrals for<BR>that specific
shell, Newton found that the Calculus Vector Integral is<BR>EXACTLY
ZERO.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>The next part of that homework was to do a Third
Integral, for the series<BR>of shells from your location within the Earth up to
the surface of the<BR>Earth. It was obvious to us students that the Vector
Integral of a bunch<BR>of zero-amplitude Vectors is zero.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>What the Professor had taught us, of the Fluxions
math which Newton had<BR>done three hundred years ago is that, the strength of
the gravitational<BR>field at ANY location inside the (symmetric) Earth was ONLY
due to the Mass<BR>of the Earth which happened to be in the Core part of the
Earth which<BR>was the size of the sphere of the Earth which was the size of the
location<BR>you happened to be at. For the homework problem the Professor
gave<BR>us, of half the radius, that smaller ball which actually was
providing<BR>a gravitational field for you, was only 1/8 the volume of the
entire<BR>Earth. The density of the Core of the Earth was much higher than
our<BR>Crustal rocks, but gravitation acts at an inverse-square law. For
the<BR>student, the net effective mass of t he Earth acts as though it is at
the<BR>Center of the Earth, so the effective gravitational field would have<BR>a
1/4 factor compared to at the surface.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>Therefore, there are four effects which must be
calculated, (1) the<BR>ignoring of all the mass of the Earth which was at
greater radius than<BR>you are at, (2) the average density of that portion of
the Earth which<BR>is within that radius (around 2.5 times as great), (3) the
net mass of<BR>that portion of the Earth, that is, density times volume or (2.5
* 1/8),<BR>and (4) the inverse-square distance of you from the center of the
Earth<BR>(which is 4 times greater..</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>For the specific homework problem, the Professor
noted that the Core of<BR>the Earth is considerably more dense than our Crust,
and so the net<BR>measured effect of all these factors would be a SLIGHT
INCREASE in<BR>local gravitational field (1/8 * 4 * 2.5), as you did an entire
trip<BR>"Journey to the Center of the Earth" as Jules Verne wrote long
ago.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>I see an entirely different important example of
Newton's mathematical analysis.<BR>Consider the Sun or any other star. We
usually describe it as though the whole<BR>WEIGHT of the entire Sun is pressing
down onto the very Core of the Sun,<BR>which we then say creates spectacular
temperatures of billions of<BR>degrees Kelvin, which is then what we claim
causes Hydrogen nuclei to<BR>FUSE into each other to form Helium atoms and which
releases the<BR>spectacular amounts of energy that our Sun radiates away and
which<BR>keeps us all alive.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>But consider Newton's reasoning and math
above. Consider the very Core<BR>of the Sun, maybe a space the size of the
Earth. There is NOT the mass<BR>of 330,000 times the Earth pressing down
on that Core (which has always<BR>been totally accepted as logical for creating
the enormous pressure and<BR>therefore temperature). Per Newton's Integral
Calculus, that simply cannot be true. That portion of<BR>the Sun, its very
Core, REALLY, only has roughly the weight of ONE<BR>Earth gravitationally
pressing down on it. How in the world could THAT<BR>create enough pressure
and therefore temperature of billions of degrees<BR>Kelvin, to initiate and
maintain nuclear fusion?</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>Logic being what it is, this reasoning and math
seems airtight. There MUST<BR>BE some other explanation for how the Sun
(and all other stars) create<BR>sufficient temperature for Fusion. We
necessarily must be very wrong in our<BR>understanding of even this basic
idea.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>Newton WAS right, and a whole lot of Physics
students did the homework<BR>problems to confirm it.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>I realize that these comments involve some Integral
Vector Calculus which some of you may n ot be familiar. I only felt it
important to present this logic of Newton for some background for your
group. Whether in the Earth or within any star, Newton proved that all of
the mass which happened to be at greater radial distance, has no net effect
regarding the Vector Force of gravitation. Personally, I findn this most
troubling inside of stars, as the enormous "gravitational weight" which alleged
presses down to create enormous pressure and therefore the billions of degrees
of Kelvin temperature in the very Core of every star, simply is not possibly
true.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT size=2 face=Arial>Carl Johnson</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
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