Solar Eruption and Electrostatic Gravity

The Solar eruption and its effects on the earth may be partly due
to the electrostatic nature of the gravitational field of the
earth. That is electrostatic dipoles inside each atomic nucleus
on the spinning orbiting earth transverse to these motions can
explain the gravitational field (and the magnetic field ) of the
earth.

Crapsnip


I was just thinking the loons are quiet tonight, they must all be
outside watching the strange lights in the sky while they chant 'Nancy!
Nancy!' And masturbate until the sky falls down, but no! They've sent in
a premature ejaculator to keep us from getting lonely.....

--
DT
Replace nospam with the antithesis of hills
*******************************************

If anyone would like to discuss these matters without going
through the unpleasantness and lunacy of most of the people out
there please write .

I assume that the spinning and orbiting motion of planets and
moons and suns is associated with charge polarization in their
nuclei transverse to these motions and that the attractions in a
radial direction account for the gravitational force of these
objects.
The Sun is spinning counterclockwise as one might view it
looking down on the roughly planar solar system. The force
producing this motion would also produce a dipole with the center
of negative charge on the side of each nucleus closest to the
center of the Sun. The received wisdom is that the planets were
formed from outer material of the Sun that was spun off and
coalesced later into the form of the planets. This protoplanetary
material on the surface of the Sun would have dipoles with the
positive pole facing the Sun's center and when it was spun off,
the same orientation of this dipole would persist; although the
spinning of the planet would produce another dipole that would
have a positive pole on the outside of the planet; On the night
side of a planet like the Earth, the dipole associated with the
Earth's spin would have the positive pole on the outside but the
dipole associated with the Earth's motion about the Sun would
have its negative pole on the outside; the accumulation of
negatively charged particles on the surface of the Earth and the
similar potential gradient of the atmosphere; if this was the
case then the outer pole of each of the Sun's dipoles is
negative. Thus the outer positive pole of the Earth's atomic
dipoles are attracted to the negative outer pole of the Sun's
atomic dipoles.
At a greater distance from the planet, the dipoles associated
with the spin of the planet and facing the Sun may be
substantially weakened by oppositely directed such dipoles on the
dark side of the planet. That is the dipoles in the Earth's
nuclei on the opposite side of the Earth from the Sun are
repelled by the Sun. This demands that we add a solar dipole
component in the planet's atomic nuclei of a size that is similar
to the spin dipole component oriented along the planet's radii
and that the solar component dipole in each atomic nucleus
changes orientation as the planet changes its position with
respect to the Sun, just as the spin component dipole changes
orientation as the Earth's radius on which it is situated changes
direction as the Earth spins.
Also the orbital component dipole may be larger than the spin
component dipole because the orbital speed is about seventy times
greater than the spin speed. The dipole in each atom is caused in
part by this velocity but also by the distance between
the dipole and other dipoles. Hence, the just neighboring
dipoles exert a stronger influence than the more distant dipoles;
the influence is such that the greater the distance between
dipoles, the greater the size of the dipole associated with a
given velocity. If there are many nearby dipoles, this has the
effect of limiting the size of the dipole produced by the spin
velocity and size of the dipole produced by the orbital velocity
and so of their difference.
Still the component factor associated with the orbital
velocity should be greater than the component factor associated
with the spin velocity. This would imply a greater attraction
between objects on a line toward the Sun than on a line toward
the center of the Earth, if the lines contained as many atoms in
both cases- which they don't.
Also these orbital component dipoles are rotated by the
spinning motion of the Earth so that they are made to line up
with the spin component dipoles and add to the spin component
dipoles forming attractive dipoles along radii of the Earth and
on radii facing the Sun along a line toward the Sun.
The need for such an added dipole is that it would help to
explain why the Earth does not fall apart under the influence of
the Sun's attraction of one side and its repulsion of the others
oppositely oriented dipoles. That is the side of the Earth
nearest the Sun is more attracted to the Sun but also because of
the added dipole in the atomic nuclei, the atoms of the Earth
nearest the Sun are more attracted to each other when compared to
atoms on the dark side of the Earth. Both of these effects
largely cancel so that the net gravitational force on the Sunny
side of the Earth is the same as that on the dark side except for
the observed tidal effects. Similar considerations apply for
dipoles in the atomic nuclei of the the Earth, other planets and
the Sun tracking the center of the galaxy.
Now the largest distance between atomic nuclear dipoles on the
Earth implicitly determining the maximal size of the dipoles is
about 106.8 meters whereas the distances for planets to the Sun
is (5.79)(1010) for Mercury, (1.49)(1011)for Earth to 5.9(1012)
meters for Pluto and for the Sun to the galactic center 104
parsecs = (3)(1020) meters. Lets see what the atomic nuclear
dipoles in the Sun and Earth must be to give the observed
gravitational force between them and if they are small enough to
be consistent with the known distances between atoms at various
temperatures etc..
That is we must be able to write the total dipoles as keRs
and KeRS where k and K are functions of the relative influence of
the total dipoles on each other etc; the total dipoles here are
proportional to the masses (note the planet masses are .22, 4.87,
5.97, .64, 1899.7, 568.8, 86.9, 103.0, and .013 times 1024kg vs
the Sun's(2)(1030)kg.); that is, to the number of protons plus
neutrons, denoted, protons+neutrons, in each mass.
Since the Sun is .75H+.25He so that 1.75kg of Sun contains
6.02 times 1026 molecules each of which contains on average 1.75
protons+neutrons so 1kg of the gaseous Sun contains 6.02 times
1026 protons+neutrons in a volume that is larger of course than
that of 1 kg of a solid planet; but 1kg of any planet or the Sun
contains the same number of protons+neutrons. There are about
2(1030) kg in the Sun. Hence the Sun contains 6.02 times 1026
times M or 12 times 1056 and the Earth contains 6.02 times 1026
times m or 3.59 times 1051 unit dipoles in the Earth. The total
dipoles a 1.2(1057)k(s)RS* and 3.59(1051)K(S)Rs*.
Hence GmM/R2 = 9(109)mM[6.02)(1026)]2 times kK times s*S*
times (N)(2.56) times 10-38 divided by R2. If N=1, this implies
kKs*S*=(.0079)10(-61-11+38) = 10-36 approximately. Now RkS* and
RKs* are the magnitudes of the dipoles associated with the Sun
and planet respectively where R is about 1010 to 1013 meters.
But we also know that the Earth's dipoles cannot be much larger
than atomic nuclei about
10-15 =RKs* that Ks*=10-26 which implies kS*=10-10 and also
RkS*= 10(-10+11) so the dipoles on the Sun are 10 meters in
length.
This sounds impossible. Perhaps the charge of the dipole could
be somehow larger so that instead of the Sun's dipoles being eS*
etc., it could be e*S* where e* is the charge on say 1000
electrons or more and S* could be that much smaller. After all
at the high temperatures (T=5.77(103) to 1.5(107) degrees Kelvin
of the Sun the average kinetic energy is .5mv2=
(1.5)(1.38)(10-23)T Joules where 1.602(10-19) Joules =1eV and
9.1(10-31)kg times v2 gives the speed of an electron at this
temperature; that is about (10-20)Joules /(10-30) at the low
5770 degree value of T suggesting v=105 meters per second for
this temperature; but below the Sun's surface then with much
greater temperatures, v is far in excess of the 106 meter/sec
velocity of the electron around the hydrogen or helium nuclei.
This suggests that dipoles much larger than those proposed for
atomic nuclei are possible within plasmas between groups of
electrons and groups of ions, protons or helium nuclei separated
by distances that can still be many orders of magnitude smaller
than ten meters.
Similar reasoning could explain the dipole attraction between
the solar system and the center of the galaxy. But what about the
moon 3.84 times 108 meters away which suggests that if RKS* =
(108)KS*=10-15 say, that (108)ks*=10(-36+15) suggesting that
Rks*=10-13 meters. Could atomic nuclei on the moon be larger than
those on Earth? Perhaps this is a problem or perhaps the tidal
effects of the moon on the Earth and vice versa and perhaps the
amount of charge polarized inside the Earth's atomic nuclei are
larger than we first considered; that is, e*s instead of es
where, e*, is greater than, e.
What is the relationship of gravity to the net spin of the
planet, satellite, star, galaxy etc. and to the number of atoms
contained in each? Clearly as in Newtonian gravity theory, the
gravitational attractive force of a planet etc is proportional to
the number of atoms. Is it then proportional to the angular
momentum and if the angular velocity was increased and the mass
was decreased so that the angular momentum remained the same
would the attractive force remain the same?
Blackett suggests such a possibility and a correlation between
magnetic field and gravitational field in the May 1947 issue of
Nature regarding the planets, the Sun, and a few stars. An
extension of this idea is that a primordial electrostatic force
produced a linear momentum of galaxies or clusters of galaxies
which was partitioned first into the angular momenta of the
spinning galaxies and then into the spinning stars and then into
the spinning planets and their satellites. That is, the strength
of the magnetic field is a function of the total of the angular
momentum components and the linear momentum component and the
number of protons-neutrons in the mass considered.
The total force may also be evident in each of these objects
down to the planetary satellites.If for example the total force
produces charge polarization inside atomic nuclei and electrons
initially in a high temperature plasma state, the effect of the
assumed linear force on charge polarized nuclei and plasmas would
be to cause a torque on individual nuclei but also on large
clumps of electrons and nuclei. This mechanism could provide a
rationale for the approximate covariation of gravity with angular
momentum that Blackett, Wilson and others had observed and an
explanation of why the relationship might not be more exact.
Thus any accelerated object, eg a bullet, a rocket, a plane,
a car, a frisbee, a skidding or spinning billiard ball etc has
electrostatic dipoles produced in its atomic nuclei transverse
to and proportional to the accelerating force which even if
mechanical is still ultimately electrostatic; The tendency of
linearly propelled atomic nuclei to then rotate may add to the
aerodynamic efficiency of spinning projectiles. The resulting
dipole field may or may not be self sustaining against thermal
disturbances as in the dipole chain model of ferroelectrics
(Feynman v2p5-5, 11-10).
In the above mentioned ferroelectric model the dipoles are
assumed to be composed of poles, concentrations of charge, that
are fairly constant over time unlike our model of charge
polarization inside atomic nuclei which changes rapidly with the
position of the orbiting charged particle(s) inside the nuclei
but which averaged over the orbital time period represents a
displacement of centers of negative and positive charge in a
specific direction. In both models the dipole-dipole interaction
is the same but the interaction of one dipole with a single pole
of the other is different in the two models.
In our model the action of one dipole on the single pole of
another is to produce a transverse elliptical motion of the
single pole, rather than as in the ferroelectric model to produce
a motion of the pole only in the direction of the dipole field
and thereby to sustain a dipole field.

"ralph sansbury"
wrote in message ...
The Solar eruption and its effects on the earth may be partly
due
to the electrostatic nature of the gravitational field of the
earth. That is electrostatic dipoles inside each atomic nucleus
on the spinning orbiting earth transverse to these motions can
explain the gravitational field (and the magnetic field ) of
the
earth.

Bigcrapsnip


Your first error, Ralph, was to begin your entire diatribe with 'I
assume...'. Your second was to refer to 'looking down on the solar
system' without defining your frame of reference. There is no 'down'. At
this point I lost interest.
Read more widely, and be more critical of others assertions, value
evidence more than opinion, learn mathematics.
These things will improve your ability to reject crap.
Best wishes,
--
DT
Replace nospam with the antithesis of hills
*******************************************

"DT" wrote in message
...
Bigcrapsnip


Your first error, Ralph, was to begin your entire diatribe with 'I
assume...'. Your second was to refer to 'looking down on the solar
system' without defining your frame of reference. There is no 'down'. At
this point I lost interest.

That's a bit of a low blow IMVHO. There's a general convention (that annoys
the pants off antipodeans (a term which is itself based on a general
assumption about which way is "up") about which way is "up") which is that
North is up. True, there is no "up" in space, but in a vernacular sense most
people would consider looking "down" on the solar system to indicate looking
at the ecliptic plane in plan view from a vantage point above one pole of
the sun, approximately, that pole being the one that points in the
approximate same direction as the North pole of the Earth. That being for
purely conventional reasons, because earth maps, by arbitrary convention,
generally have North at the top.

He's not drawing some detailed conclusion from this choice of reference
frame (as he would be say in a discussion on SR), he's just setting the
scene. If, for instance, one said "looking down on America, one can see the
Grand Canyon" one doesn't need to specify precisely where in space one is. A
reasonable reader can easily understand what the writer means.

Read more widely, and be more critical of others assertions, value
evidence more than opinion, learn mathematics.
These things will improve your ability to reject crap.
Best wishes,
--
DT
Replace nospam with the antithesis of hills
*******************************************

Valleys? Holes? Pits? Plains? Please be specific

Ian

In message , Ian Bland
writes
"DT" wrote in message
...
Bigcrapsnip


Your first error, Ralph, was to begin your entire diatribe with 'I
assume...'. Your second was to refer to 'looking down on the solar
system' without defining your frame of reference. There is no 'down'. At
this point I lost interest.

That's a bit of a low blow IMVHO. There's a general convention (that annoys
the pants off antipodeans (a term which is itself based on a general
assumption about which way is "up") about which way is "up") which is that
North is up. True, there is no "up" in space, but in a vernacular sense most
people would consider looking "down" on the solar system to indicate looking
at the ecliptic plane in plan view from a vantage point above one pole of
the sun, approximately, that pole being the one that points in the
approximate same direction as the North pole of the Earth. That being for
purely conventional reasons, because earth maps, by arbitrary convention,
generally have North at the top.

He's not drawing some detailed conclusion from this choice of reference
frame (as he would be say in a discussion on SR), he's just setting the
scene. If, for instance, one said "looking down on America, one can see the
Grand Canyon" one doesn't need to specify precisely where in space one is. A
reasonable reader can easily understand what the writer means.


Snip my own condescending crap
*******************************************


Valleys? Holes? Pits? Plains? Please be specific

Ian


I don't feel it was a low blow at all, and I'll tell you why.
I find astronomy a fascinating subject, and as a middle-aged engineer I
am also naturally interested in the associated physics. Because of my
training and background perhaps, I have developed an aversion to
metaphorical constructs that bear no relation to perceived reality. As I
am a learner in this field (astronomy/astrophysics), I am always looking
for information that will help me towards a clearer picture.
Most discussions that I've seen on sci.phy.rel are either beyond me at
the moment or are such poorly constructed metaphors that they quickly
degenerate into woolly philosophical arguments that ultimately lead to
'I think the universe exists, therefore it does'
Ralph started his metaphor on exactly these lines, by defining a false
scenario. Everything that followed must (quite reasonably in my view!)
be dismissed as irreconcilable with reality.
Verbal metaphors for mathematical constructs should be very carefully
constructed so that ambiguity is minimised. When they're not, I get
irritable sometimes.
This little diatribe is in a sense a compliment to the groups as I do
learn a lot here, living the lowly life of a lurker, occasionally
popping his head up to laugh as naked kings go by.
In answer to your final point, I've no intention of being more specific,
however you can test your results by asking 'what's the antithesis of a
plain' etc.
As I only have a rudimentary understanding of relativity, I shan't
cross-post again.

Regards, Denis
--
DT
Replace nospam with the antithesis of hills
*******************************************

ralph sansbury:
If anyone would like to discuss these matters without going
through the unpleasantness and lunacy of most of the people out
there please write .

I assume that the spinning and orbiting motion of planets and
moons and suns is associated with charge polarization in their
nuclei transverse to these motions and that the attractions in a
radial direction account for the gravitational force of these
objects.

You assume a lot of things that are blantly silly. So what?

"ralph sansbury" wrote in message ...
[snip]

I assume that the spinning and orbiting motion of planets and
moons and suns is associated with charge polarization in their
nuclei transverse to these motions and that the attractions in a
radial direction account for the gravitational force of these
objects.

Well then how do you account for the Cavendish experiment results?

----- Original Message -----
From: "luke"
Newsgroups: sci.astro,sci.physics.relativity
Sent: Thursday, October 30, 2003 11:11 AM
Subject: Solar Eruption and Electrostatic Gravity


"ralph sansbury" wrote in message
...
[snip]

I assume that the spinning and orbiting motion of planets
and
moons and suns is associated with charge polarization in
their
nuclei transverse to these motions and that the attractions
in a
radial direction account for the gravitational force of these
objects.


Well then how do you account for the Cavendish experiment
results?
The horizontal projection or the radial force as mentioned in
the part
you snipped
see also www.bestweb.net/~sansbury



The proposed polarization of charge in atomic nuclei and
electrons and mesons etc overlaps the concept of spin.
And so perhaps also the attempted explanation of forces as
being associated with and determined in some mystical way by the
exchange of photons or short lived charged particles in CERN
'pictures'
of collisions of protons etc.
So the gravitational field may be due to radially oriented
electrostatic dipoles. This would explain the radial
attraction of objects toward the center of the earth and of
objects as in Cavendish's experiment. In the latter case the
horizontal force between the objects may be the projection of the
radial force in the horizontal direction.
(see www.bestweb.net/~sansbury)

"ralph sansbury" wrote in message ...
----- Original Message -----
From: "luke"
Newsgroups: sci.astro,sci.physics.relativity
Sent: Thursday, October 30, 2003 11:11 AM
Subject: Solar Eruption and Electrostatic Gravity


"ralph sansbury" wrote in message
...
[snip]

I assume that the spinning and orbiting motion of planets
and
moons and suns is associated with charge polarization in
their
nuclei transverse to these motions and that the attractions
in a
radial direction account for the gravitational force of these
objects.


Well then how do you account for the Cavendish experiment
results?

The horizontal projection or the radial force as mentioned in
the part
you snipped
see also www.bestweb.net/~sansbury



Am I to understand that in your theory the cavendish force is directly
proportional to the spin of the Earth? So a Cavendish experiment
performed in an intertial frame would yield a null result?

Sorry if I didn't read your material thoroughly but I imagine you
would have interesting things to say about the mass and density of
Mars as it has a much smaller rotation rate..

To be more general the currently formulated orbital mechanics works
for a central force and is independent of angular momentum.. wouldn't
a theory such as yours destroy this symmetry and change the orbit
equation? For example an elliptical orbit has more centrifugal force
at perihelion, and so should have stronger dipole moments, and more
gravity than 1/r^2..



The proposed polarization of charge in atomic nuclei and
electrons and mesons etc overlaps the concept of spin.
And so perhaps also the attempted explanation of forces as
being associated with and determined in some mystical way by the
exchange of photons or short lived charged particles in CERN
'pictures'
of collisions of protons etc.
So the gravitational field may be due to radially oriented
electrostatic dipoles. This would explain the radial
attraction of objects toward the center of the earth and of
objects as in Cavendish's experiment. In the latter case the
horizontal force between the objects may be the projection of the
radial force in the horizontal direction.
(see www.bestweb.net/~sansbury)