How many protons/electrons in Sun and Earth to replace mass-gravity with EM-gravity #166 New Physics #275 ATOM TOTALITY 5th ed

[Archimedes Plutonium]

Alright, no-one in physics has ever proposed that all of Newtonian
gravity and General Relativity are fakes and replaced by the EM force
of Faraday's law.

So what I am doing is beginning a compilation of a table in which we
can easily go from Newtonian gravity as first approximation and then
go to the Maxwell Equations for replacing the two masses by charge in
the Faraday law.

In a recent post I started this process of compilation and want to
correct some errors of calculation such as the nails without use of
Avogadro's number.

The physicist that is not brilliant, looks at Saturn Rings or barred
spiral galaxies 
and sees rigid body rotation and then due to his lack
of physics 
skills invokes dark matter and dark energy. The brilliant
and true physicist seeing 
rigid body rotation 
says "gravity can no
longer be mass-gravity but must be EM-gravity."

The Cavendish Experiment is a fool's experiment for it cannot
eliminate the
EM bonds of the materials used in the experiment which those EM bonds
are 10^40 stronger than mass-gravity and so I tiny number of charges
in uneven distribution overwhelms a measurement of mass-gravity and
calls
into question whether mass-gravity is a fake force. The Earth and Sun
are round globes, not because of the force of gravity, but due to the
force of EM.

The mass of the Sun is 2x10^30 kg and the surface area is 6x10^12
km^2
The mass of Earth is 6x10^24 kg and its surface area is 5x10^8 km^2
The mass of a single proton is 1.6x10^-27 kg
The force coupling strength of EM versus mass-gravity is approx 10^40
The number of protons that make up the Sun is 10^57 and for Earth it
is 10^51

Since the coupling strength of EM to gravity-mass is 10^40 then the
number of 
protons needed for Earth is 10^51/10^40 is 10^11
The 10^11 protons is about the number of protons in a tiny speck of
iron filing.

The number of electrons needed for the sun replaced of its mass-
gravity with 
EM gravity is 10^57/10^40 = 10^17 which is about the
number of protons
in a small needle.

We have the molar mass/gram of iron as 55 and the mass of iron sewing
needle I measured is about 1/10 g. So we need an iron needle that is
even smaller by a factor of about 10^-3 and so if I cut that needle
into 1,000 pieces and divide by 55 and then multiply by Avogadro's
number at 6*10^23, I get 10^17 protons in that tiny sliver of a piece
of a needle.

So as you can see, if the EM force was gravity in the form of
Faraday's law
then the entire Sun is replaced by a tiny needle of 10^17 electrons
(or protons) and the Earth is replaced by a smaller iron sliver of
10^11 protons or electrons.

The diameter of the hydrogen atom is about 10^-13 km 
The surface area
of Sun is 6x10^12 km^2 so that for the surface of the sun
would have
6x10^38 protons. And all we need is a surface area of 10^17 of
electrons or protons. Now I wonder if the Sun Spots near the equator
are about 10^17 in area versus the total area of the Sun 10^38 (and
also why Jupiter has its Red Spot)?

So all it takes for the Sun to hold Earth in a bonded condition of EM
force is an "order of magnitude" of the size of needles
of ions as a
net overall charge and for Earth to have a net overall charge of the
protons in a sliver of a needle. Is it not in the Bible of words to
the effect-- "to see the world in the eye of a needle."

It is remarkable that we all know the Sun has a huge number of ions
that do not cancel out but leave a net charge. And the number of
cosmic electron rays (or proton rays) that hit Earth every minute, is
remarkable to
think that Earth would not have a net charge the amount
of protons in a sliver of a needle.

The concept of Net Overall Charge is perhaps too new, too
revolutionary 
for all those old minds in physics who have been
brainwashed into
thinking that mass, just plain mass attracts other
mass.

Now many are going to complain as to why and how the Sun is charged
opposite to the charge of a planet to have them attract. And my answer
is that in Faraday's law, not Coulomb law that the Faraday's law needs
a bar magnet, not isolated charges and
that a net charge differential
make the entire Sun and a planet be a
magnet.

Now the above may still have computation errors, but I plan to keep
and improve the above for it basically tells the entire picture of EM-
gravity. That rigid body rotation does not allow us to invoke the
silly and stupid dark matter and dark energy, but rather instead,
forces us to realize that mass-gravity is a fake and that EM-gravity
is the true force.

Archimedes Plutonium http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies

[Archimedes Plutonium]

On Feb 3, 2:53 am, Archimedes Plutonium

Now in this compilation, I hope to show why the only rigid body
rotation display in
our Solar System are the Rings of Saturn and some of the other gas
giants. I have
a suspicion it is due to the fact that ice globules that is the prime
composition of those
Rings, is that ice is a polar bonded molecule and thus uneven in
charge distribution, so that the EM-force of Saturn is so much
stronger on the ice globules.

Now I have to review the three laws of Kepler on the motion of
planets, and infuse the
EM-gravity. Now I do recall that several planets were supposed to be
void of a magnetic field
but out satellites surprised us with magnetic fields. In EM-gravity,
all planets have a magnetic field and it takes but a tiny magnetic
field to be EM-bound to the Sun.

Also, in EM-gravity, that acceleration in a flyby is larger than
expected from Newtonian gravity or General Relativity. And in
Faraday's law, we expect inordinately more acceleration in a flyby due
to the moving bar magnet.

Now I have not given up on exploring the Cavendish Experiment, because
I suspect also, that
the Cavendish Experiment makes no allowance for the radio waves in the
air. During 1797 when this experiment was first performed the air
waves had not the abundance of radio transmission that today's air
waves have. Of course, in 1797 there were Jupiter radio signals as
well as Solar radio signals, and the Cavendish Experiment is probably
more of a antenna than it is a experiment to measure mass-gravity. Now
I do not know how much of a quantity of radio waves equals that of the
EM energy of a single electron or single proton? Let us say that a
quantity of radio wave transmission down that wires of the Cavendish
Experiment of 1797 is sufficient a quantity that would give the torque
on those wires to fool the experimenter into thinking the torque was
due to mass attracting mass, when in effect, the torque was simply due
to ambient radio wave transmission. And one way of confirming this is
to see if a modern day Cavendish set-up yields more of a deflection
than what Cavendish reported, due to the increased flood of radio
waves in modern time. I do not know how good or how bad is the
Cavendish set-up to take in radio waves as an antenna.