In article <epr12a$c2...@skeeter.ucdavis.edu>,

carlip-nos...@physics.ucdavis.edu wrote:

frankli...@yahoo.com wrote:

[...]

I was looking for something like "Why gravity can't be the
electrostatic force", but I find almost nothing in the literature
or the web.

The simplest reason is this: in gravity, all masses attract, while
in electrostatics, like charges repel.

For example, the Earth ands the Moon attract, so in an electrostatic
model, they must have opposite charges.

***{That's seems rather too strong, don't you think? In the particular
electrostatic model you are discussing, they do, in fact, have opposite
charges; but if you intend to claim that must be so in all possible
electrostatic models of gravity, then I must disagree. For example,
suppose that the attraction between unlike charges is greater than the
repulsion between like charges, in the very slight amount necessary to
account for the observed strength of gravity. In such an electrostatic
model, the Earth and Luna (the Moon) attract electrostatically with
gravitational force, yet have the same charges--to wit: essentially
zero.

Let me elaborate a bit.

According to Coulomb's law, the magnitude of the electrostatic force is
equal to a constant of proportionality times the product of the charges,
divided by the square of the distance between their centers.

By such a rule, the electrostatic force between Earth and Luna would
have four components:

(1) If K is the Coulomb constant, e1 is the charge of the electrons in
the Earth, e2 is the charge of the electrons in Luna, and r is the
distance from the center of the Earth to the center of Luna, we have:

F1 = Ke1e2/r^2

(2) If p1 is the charge of the protons in the Earth and p2 is the charge
of the protons in Luna, we have:

F2 = Kp1p2/r^2

(3) If k is the ratio, slightly greater than 1, of the absolute
magnitude of the attraction between like charges to that of the
repulsion between unlike charges, then we have:

F3 = Kke1p2/r^2

(4) And, similarly, we have:

F4 = Kke2p1/r^2

Therefore the force of gravity, Fg, would be such that

Fg = F1 + F2 + F3 + F4

Fg = Ke1e2/r^2 + Kp1p2/r^2 + Kke1p2/r^2 + Kke2p1/r^2

If M is the mass of Earth, m is the mass of Luna, and <e> is the mass of
the electron, then:

e1 = -(M/1836)/<e> = -M/1836<e>

p1 = -e1 = M/1836<e>

e2 = -(m/1836)/<e> = -m/1836<e>

p2 = -e2 = m/1836<e>

Substitution into the last force equation, above, gives:

Fg = K(-M/1836<e>)(-m/1836<e>)/r^2

+ K(M/1836<e>)(m/1836<e>)/r^2

+ Kk(-M/1836<e>)(m/1836<e>)/r^2

+ Kk(-m/1836<e>)(M/1836<e>)/r^2

= [2K/(1836<e>)^2][Mm/r^2] - k[2K/(1836<e>)^2][Mm/r^2]

= [2K/(1836<e>)^2][1 - k][Mm/r^2]

For simplicity, let z = 1 - k, so that the above becomes

Fg = [2zK/(1836<e>)^2][Mm/r^2]

By Newton's law of gravitation we have

Fg = GMm/r^2

And so we conclude that

G = [2zK/(1836<e>)^2]

Of course, 1836<e> = <p>, the mass of the proton, in the notation I'm
using, so we obtain

G = 2zK/<p>^2

In SI units, G = 6.6742x10^-11 Nm^2/kg^2, K = 8.988x10^9 Nm^2/C^2, and
<e> = 9.109x10^-31 kg. Hence <p> = (1836)(9.109x10^-31) = 1.672x10^-27.
Since in this problem we want to distinguish between attractive and
repulsive forces, and since the attractive ones traditionally get a
negative sign, I'm going to treat the value of G as a negative number
when I plug it in below. (Mass and radius are clearly positive, so G is
the only palusible way to enforce that distinction.) And so we have:

z = (-6.6742x10^-11)/{2(8.988x10^9)/[(1.672x10^-27)]^2}

z = -1.038x10^-74

What the above argument accomplishes, hopefully, is to demonstrate the
theoretical possibility that gravity is electromagnetic in nature, and
arises out of Coulomb's law. That's not to say that it would be
convenient to do gravitational calculations that way, of course. Such a
demonstration would be useful in the same sense that it is useful to
note that the relativistic kinetic energy formula, Ek = m0[1/(1 -
v^2/c^2)^.5 -1]c^2, could in principle be used to calculate the kinetic
energy involved in ordinary automobile collisions, despite the practical
inconvenience of doing so. The idea is to reveal a possible
connectedness that would in most circumstances remain hidden from view.

There are lots of counterarguments that can be directed at the above
idea, of course--so many, in fact, that I don't offer it as a necessary
truth, but merely as something worthy of consideration. I have been
thinking along these lines off and on for a long time, and there is
another theory of gravitation that I feel has more explanatory power
than this one, even though the issue between them is not settled by any
means.

What would be the point of trying to unify electromagnetics and
gravitation, after all? :-)

Any errors in the above are of course mine, but the credit for the basic
idea, for what it's worth, goes to Prof. Thomas Barnes. (See his book,
Space Medium: the key to unified physics, published in 1986.)

--Mitchell Jones}***

The Earth and the Sun
attract, so they, too, must have opposite charges. But that would
make the charge of the Moon the same sign as the charge of the Sun,
so they would repel. This does not happen -- the Moon's orbit
very clearly shows that it is attracted by both the Earth and the
Sun.

(Or for a more dramatic example, what "charge" would you ascribe to
the Apollo 11 lunar lander? It was attracted to the Earth -- they
didn't have to scrape it off the ceiling during assembly -- so it
must have had the opposite charge from the Earth. But the Moon is
also attracted to the Earth, and must also have the opposite charge
from the Earth. Why, then, wasn't the lunar lander repelled from
the Moon?)

Steve Carlip

*****************************************************************
If I seem to be ignoring you, consider the possibility
that you are in my killfile. --MJ- Hide quoted text -

- Show quoted text -

In article <epr12a$c2...@skeeter.ucdavis.edu>,

carlip-nos...@physics.ucdavis.edu wrote:

frankli...@yahoo.com wrote:

[...]

I was looking for something like "Why gravity can't be the
electrostatic force", but I find almost nothing in the literature
or the web.

The simplest reason is this: in gravity, all masses attract, while
in electrostatics, like charges repel.

For example, the Earth ands the Moon attract, so in an electrostatic
model, they must have opposite charges.

***{That's seems rather too strong, don't you think? In the particular
electrostatic model you are discussing, they do, in fact, have opposite
charges; but if you intend to claim that must be so in all possible
electrostatic models of gravity, then I must disagree. For example,
suppose that the attraction between unlike charges is greater than the
repulsion between like charges, in the very slight amount necessary to
account for the observed strength of gravity. In such an electrostatic
model, the Earth and Luna (the Moon) attract electrostatically with
gravitational force, yet have the same charges--to wit: essentially
zero.

Let me elaborate a bit.

According to Coulomb's law, the magnitude of the electrostatic force is
equal to a constant of proportionality times the product of the charges,
divided by the square of the distance between their centers.

By such a rule, the electrostatic force between Earth and Luna would
have four components:

(1) If K is the Coulomb constant, e1 is the charge of the electrons in
the Earth, e2 is the charge of the electrons in Luna, and r is the
distance from the center of the Earth to the center of Luna, we have:

F1 = Ke1e2/r^2

(2) If p1 is the charge of the protons in the Earth and p2 is the charge
of the protons in Luna, we have:

F2 = Kp1p2/r^2

(3) If k is the ratio, slightly greater than 1, of the absolute
magnitude of the attraction between like charges to that of the
repulsion between unlike charges, then we have:

F3 = Kke1p2/r^2

(4) And, similarly, we have:

F4 = Kke2p1/r^2

Therefore the force of gravity, Fg, would be such that

Fg = F1 + F2 + F3 + F4

Fg = Ke1e2/r^2 + Kp1p2/r^2 + Kke1p2/r^2 + Kke2p1/r^2

If M is the mass of Earth, m is the mass of Luna, and <e> is the mass of
the electron, then:

e1 = -(M/1836)/<e> = -M/1836<e>

p1 = -e1 = M/1836<e>

e2 = -(m/1836)/<e> = -m/1836<e>

p2 = -e2 = m/1836<e>

Substitution into the last force equation, above, gives:

Fg = K(-M/1836<e>)(-m/1836<e>)/r^2

+ K(M/1836<e>)(m/1836<e>)/r^2

+ Kk(-M/1836<e>)(m/1836<e>)/r^2

+ Kk(-m/1836<e>)(M/1836<e>)/r^2

= [2K/(1836<e>)^2][Mm/r^2] - k[2K/(1836<e>)^2][Mm/r^2]

= [2K/(1836<e>)^2][1 - k][Mm/r^2]

For simplicity, let z = 1 - k, so that the above becomes

Fg = [2zK/(1836<e>)^2][Mm/r^2]

By Newton's law of gravitation we have

Fg = GMm/r^2

And so we conclude that

G = [2zK/(1836<e>)^2]

Of course, 1836<e> = <p>, the mass of the proton, in the notation I'm
using, so we obtain

G = 2zK/<p>^2

In SI units, G = 6.6742x10^-11 Nm^2/kg^2, K = 8.988x10^9 Nm^2/C^2, and
<e> = 9.109x10^-31 kg. Hence <p> = (1836)(9.109x10^-31) = 1.672x10^-27.
Since in this problem we want to distinguish between attractive and
repulsive forces, and since the attractive ones traditionally get a
negative sign, I'm going to treat the value of G as a negative number
when I plug it in below. (Mass and radius are clearly positive, so G is
the only palusible way to enforce that distinction.) And so we have:

z = (-6.6742x10^-11)/{2(8.988x10^9)/[(1.672x10^-27)]^2}

z = -1.038x10^-74

What the above argument accomplishes, hopefully, is to demonstrate the
theoretical possibility that gravity is electromagnetic in nature, and
arises out of Coulomb's law. That's not to say that it would be
convenient to do gravitational calculations that way, of course. Such a
demonstration would be useful in the same sense that it is useful to
note that the relativistic kinetic energy formula, Ek = m0[1/(1 -
v^2/c^2)^.5 -1]c^2, could in principle be used to calculate the kinetic
energy involved in ordinary automobile collisions, despite the practical
inconvenience of doing so. The idea is to reveal a possible
connectedness that would in most circumstances remain hidden from view.

There are lots of counterarguments that can be directed at the above
idea, of course--so many, in fact, that I don't offer it as a necessary
truth, but merely as something worthy of consideration. I have been
thinking along these lines off and on for a long time, and there is
another theory of gravitation that I feel has more explanatory power
than this one, even though the issue between them is not settled by any
means.

What would be the point of trying to unify electromagnetics and
gravitation, after all? :-)

Any errors in the above are of course mine, but the credit for the basic
idea, for what it's worth, goes to Prof. Thomas Barnes. (See his book,
Space Medium: the key to unified physics, published in 1986.)

--Mitchell Jones}***

The Earth and the Sun
attract, so they, too, must have opposite charges. But that would
make the charge of the Moon the same sign as the charge of the Sun,
so they would repel. This does not happen -- the Moon's orbit
very clearly shows that it is attracted by both the Earth and the
Sun.

(Or for a more dramatic example, what "charge" would you ascribe to
the Apollo 11 lunar lander? It was attracted to the Earth -- they
didn't have to scrape it off the ceiling during assembly -- so it
must have had the opposite charge from the Earth. But the Moon is
also attracted to the Earth, and must also have the opposite charge
from the Earth. Why, then, wasn't the lunar lander repelled from
the Moon?)

Steve Carlip

*****************************************************************
If I seem to be ignoring you, consider the possibility
that you are in my killfile. --MJ- Hide quoted text -

- Show quoted text -