# Atomic Model Of The Solar System Theory

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This theory is based on the assumption that a fluorine atom is a small-scale model of our solar system. The nine planets that orbit the sun, are the nine electrons that orbit the nucleus of the fluorine atom.

An electron has mass, and a point mass creates a radial gravitational field. Therefore an electron can attract particles that have mass that is much smaller than electonic mass. These “electron moons” are rotating about the electron on the same way the moons rotate about the planets.

The total mass of our solar system is the product of the atomic mass of the fluorine atom and a constant. Where the mass of the sun is the product of the mass of the fluorine nucleus and the constant. Hence, the total mass of the planets orbiting the sun is the product of the total mass of the electrons around the fluorine nucleus and the constant.

Planet/ Star....Mass (×10^24kg)

sun...............1989100

mercury.........0.3302

venus............4.8685

earth.............5.9736

mars.............. 0.64185

jupiter............1898.6

saturn............568.46

uranus...........86.832

neptune........102.43

pluto.............0.0125

Total mass of the solar system, without moons of the planets as they will have a negligible effect, is approximately 1.992×10^30 kg

The atomic mass of fluorine is 18.9984u, which is 3.154×10^-26 kg

Therefore,

3.154×10^-26 × k = 1.992×10^30

where k is a constant and has a value of 6.316×10^55

The mass of any star can therefore be calculated by finding out how many planets orbit that star and referring to the periodic table.

© Alexey 2004

Contact: [email protected]

An electron has mass, and a point mass creates a radial gravitational field. Therefore an electron can attract particles that have mass that is much smaller than electonic mass. These “electron moons” are rotating about the electron on the same way the moons rotate about the planets.

The total mass of our solar system is the product of the atomic mass of the fluorine atom and a constant. Where the mass of the sun is the product of the mass of the fluorine nucleus and the constant. Hence, the total mass of the planets orbiting the sun is the product of the total mass of the electrons around the fluorine nucleus and the constant.

Planet/ Star....Mass (×10^24kg)

sun...............1989100

mercury.........0.3302

venus............4.8685

earth.............5.9736

mars.............. 0.64185

jupiter............1898.6

saturn............568.46

uranus...........86.832

neptune........102.43

pluto.............0.0125

Total mass of the solar system, without moons of the planets as they will have a negligible effect, is approximately 1.992×10^30 kg

The atomic mass of fluorine is 18.9984u, which is 3.154×10^-26 kg

Therefore,

3.154×10^-26 × k = 1.992×10^30

where k is a constant and has a value of 6.316×10^55

The mass of any star can therefore be calculated by finding out how many planets orbit that star and referring to the periodic table.

© Alexey 2004

Contact: [email protected]

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#2

(Original post by

This theory is based on the assumption that a fluorine atom is a small-scale model of our solar system. The nine planets that orbit the sun, are the nine electrons that orbit the nucleus of the fluorine atom.

An electron has mass, and a point mass creates a radial gravitational field. Therefore an electron can attract particles that have mass that is much smaller than electonic mass. These “electron moons” are rotating about the electron on the same way the moons rotate about the planets.

The total mass of our solar system is the product of the atomic mass of the fluorine atom and a constant. Where the mass of the sun is the product of the mass of the fluorine nucleus and the constant. Hence, the total mass of the planets orbiting the sun is the product of the total mass of the electrons around the fluorine nucleus and the constant.

Planet/ Star....Mass (×10^24kg)

sun...............1989100

mercury.........0.3302

venus............4.8685

earth.............5.9736

mars.............. 0.64185

jupiter............1898.6

saturn............568.46

uranus...........86.832

neptune........102.43

pluto.............0.0125

Total mass of the solar system, without moons of the planets as they will have a negligible effect, is approximately 1.992×10^30 kg

The atomic mass of fluorine is 18.9984u, which is 3.154×10^-26 kg

Therefore,

3.154×10^-26 × k = 1.992×10^30

where k is a constant and has a value of 6.316×10^55

The mass of any star can therefore be calculated by finding out how many planets orbit that star and referring to the periodic table.

© Alexey 2004

Contact: [email protected]

**RussianDude**)This theory is based on the assumption that a fluorine atom is a small-scale model of our solar system. The nine planets that orbit the sun, are the nine electrons that orbit the nucleus of the fluorine atom.

An electron has mass, and a point mass creates a radial gravitational field. Therefore an electron can attract particles that have mass that is much smaller than electonic mass. These “electron moons” are rotating about the electron on the same way the moons rotate about the planets.

The total mass of our solar system is the product of the atomic mass of the fluorine atom and a constant. Where the mass of the sun is the product of the mass of the fluorine nucleus and the constant. Hence, the total mass of the planets orbiting the sun is the product of the total mass of the electrons around the fluorine nucleus and the constant.

Planet/ Star....Mass (×10^24kg)

sun...............1989100

mercury.........0.3302

venus............4.8685

earth.............5.9736

mars.............. 0.64185

jupiter............1898.6

saturn............568.46

uranus...........86.832

neptune........102.43

pluto.............0.0125

Total mass of the solar system, without moons of the planets as they will have a negligible effect, is approximately 1.992×10^30 kg

The atomic mass of fluorine is 18.9984u, which is 3.154×10^-26 kg

Therefore,

3.154×10^-26 × k = 1.992×10^30

where k is a constant and has a value of 6.316×10^55

The mass of any star can therefore be calculated by finding out how many planets orbit that star and referring to the periodic table.

© Alexey 2004

Contact: [email protected]

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#3

**RussianDude**)

This theory is based on the assumption that a fluorine atom is a small-scale model of our solar system. The nine planets that orbit the sun, are the nine electrons that orbit the nucleus of the fluorine atom.

An electron has mass, and a point mass creates a radial gravitational field. Therefore an electron can attract particles that have mass that is much smaller than electonic mass. These “electron moons” are rotating about the electron on the same way the moons rotate about the planets.

The total mass of our solar system is the product of the atomic mass of the fluorine atom and a constant. Where the mass of the sun is the product of the mass of the fluorine nucleus and the constant. Hence, the total mass of the planets orbiting the sun is the product of the total mass of the electrons around the fluorine nucleus and the constant.

Planet/ Star....Mass (×10^24kg)

sun...............1989100

mercury.........0.3302

venus............4.8685

earth.............5.9736

mars.............. 0.64185

jupiter............1898.6

saturn............568.46

uranus...........86.832

neptune........102.43

pluto.............0.0125

Total mass of the solar system, without moons of the planets as they will have a negligible effect, is approximately 1.992×10^30 kg

The atomic mass of fluorine is 18.9984u, which is 3.154×10^-26 kg

Therefore,

3.154×10^-26 × k = 1.992×10^30

where k is a constant and has a value of 6.316×10^55

The mass of any star can therefore be calculated by finding out how many planets orbit that star and referring to the periodic table.

© Alexey 2004

Contact: [email protected]

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#6

(Original post by

Surely that constant only applies to

**mik1a**)Surely that constant only applies to

*our*solar system?
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(Original post by

let alone the fat that unlike electrons, the planets have different masses. oh, and they're not attracted electrically to the sun, nor are tehy stuck in discrete quantum states.

**elpaw**)let alone the fat that unlike electrons, the planets have different masses. oh, and they're not attracted electrically to the sun, nor are tehy stuck in discrete quantum states.

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#9

(Original post by

Surely that constant only applies to

**mik1a**)Surely that constant only applies to

*our*solar system?
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(Original post by

HA HA HA HA HA HA made me laff

**Lee86**)HA HA HA HA HA HA made me laff

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#11

(Original post by

nor are tehy stuck in discrete quantum states.

**elpaw**)nor are tehy stuck in discrete quantum states.

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#12

There's always going to be dodgy areas when you compare an atom to a star system, namely that while the celestial model relies only on a gravitational force, the microscopic model needs a repulsive force (electrostatic) to prevent the electrons falling into the the nucleus. Aside from this, there are still flaws in your theory. The mass of a star is actually determined by the density of matter in the dust from which the star was formed - and has nothing to do with the number of planets. In fact, what is a planet? Does Pluto count as one, its mass having virtually no effect on the movement of the Sun? Does the moon count as a planet, since it has about the same mass as Pluto? What about other satellites of similar masses? Should the asteriod belt between Mars and Jupiter count? What about the solar systems with no planets- is the star then massless? And what about multiple star systems? The biggest problem about testing this theory though is that outside our Solar System only about 10 planets have been confirmed, so actually counting the number of planets in the first place will be the biggest problem. Nice try, but it's probably best to stick to our primitive ways of calculating a star's mass for now!

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#13

(Original post by

There's always going to be dodgy areas when you compare an atom to a star system, namely that while the celestial model relies only on a gravitational force, the microscopic model needs a repulsive force (electrostatic) to prevent the electrons falling into the the nucleus.

**el GaZZa**)There's always going to be dodgy areas when you compare an atom to a star system, namely that while the celestial model relies only on a gravitational force, the microscopic model needs a repulsive force (electrostatic) to prevent the electrons falling into the the nucleus.

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#14

(Original post by

the microscopic model needs a repulsive force (electrostatic) to prevent the electrons falling into the the nucleus.

**el GaZZa**)the microscopic model needs a repulsive force (electrostatic) to prevent the electrons falling into the the nucleus.

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2 el GaZZa,

According to my theory the mass of the sun is the product of the mass of the fluorine nucleus and the constant. You are right, it has nothing to do with a number of planets.

Everything should count, i just didnt include all of that in the calculation above.

if there are no planets we have to look at the nucleus that doesnt have any electrons that orbit it

I havent thought about multiple star systems, hmmmm

According to my theory the mass of the sun is the product of the mass of the fluorine nucleus and the constant. You are right, it has nothing to do with a number of planets.

(Original post by

In fact, what is a planet? Does Pluto count as one, its mass having virtually no effect on the movement of the Sun? Does the moon count as a planet, since it has about the same mass as Pluto? What about other satellites of similar masses? Should the asteriod belt between Mars and Jupiter count?

**el GaZZa**)In fact, what is a planet? Does Pluto count as one, its mass having virtually no effect on the movement of the Sun? Does the moon count as a planet, since it has about the same mass as Pluto? What about other satellites of similar masses? Should the asteriod belt between Mars and Jupiter count?

(Original post by

What about the solar systems with no planets- is the star then massless? And what about multiple star systems?

**el GaZZa**)What about the solar systems with no planets- is the star then massless? And what about multiple star systems?

I havent thought about multiple star systems, hmmmm

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#17

Why a fluorine atom? Why not a hydrogen atom with a different value of k? Surely the reason you've used a fluorine atom is because you've decided there are 9 'planets' but as el gaZZa points out you could easily make a case for the solar system having anything between 7 and about 15 'planets'.

In your original post you state 'the masses of moons are disregarded as these are small'. - why include pluto (mass 1.25x10^22kg) and exclude Ganymede (mass 14.8x10^22kg), Titan (13.4x10^22kg), Callisto (10.8x10^22kg), Io (8.93x10^22kg), the moon (7.35x 10^22 kg), Europa (4.8x10^22kg) and Triton (2.15x10^22kg)?

Hmmmmmm right so you want to look at a hydrogen (+) ion? Or perhaps an alpha particle? Which one? Whichever one you choose surely by implication all stars with no planets will have the same mass (k*atomic mass of whatever ion you choose). This is clearly bollox. In fact you're saying that stars must have discrete masses dependant on their number of planets. This is at complete odds with observations which show stellar masses are non-discrete. You clearly know nothing about star formation and I can't believe you're taking this theory seriously.

I accept that there are some major problems with the current theories of planetary formation but I'm far happier with them than I am with this.

In your original post you state 'the masses of moons are disregarded as these are small'. - why include pluto (mass 1.25x10^22kg) and exclude Ganymede (mass 14.8x10^22kg), Titan (13.4x10^22kg), Callisto (10.8x10^22kg), Io (8.93x10^22kg), the moon (7.35x 10^22 kg), Europa (4.8x10^22kg) and Triton (2.15x10^22kg)?

(Original post by

if there are no planets we have to look at the nucleus that doesnt have any electrons that orbit it

**RussianDude**)if there are no planets we have to look at the nucleus that doesnt have any electrons that orbit it

I accept that there are some major problems with the current theories of planetary formation but I'm far happier with them than I am with this.

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#18

(Original post by

in reality i think the masses of electrons are not the same, and there were no words "attracted electrically" in the theory

**RussianDude**)in reality i think the masses of electrons are not the same, and there were no words "attracted electrically" in the theory

Also all electrons have the same charge, but the solar system is held together by gravity so a difference in mass implies varying charges in your comparison.

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In your original post you state 'the masses of moons are disregarded as these are small'. - why include pluto (mass 1.25x10^22kg) and exclude Ganymede (mass 14.8x10^22kg), Titan (13.4x10^22kg), Callisto (10.8x10^22kg), Io (8.93x10^22kg), the moon (7.35x 10^22 kg), Europa (4.8x10^22kg) and Triton (2.15x10^22kg)?

In fact you're saying that stars must have discrete masses dependant on their number of planets.

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#20

(Original post by

no, i dont know how you came up with he fact that masses are discrete

**RussianDude**)no, i dont know how you came up with he fact that masses are discrete

1) Count the # of planets

2) look up the mass of an atom with that # of electrons

3) The mass of the star is given by the mass of that atom times k

If that's the case then stars can only have masses equal to the masses of atoms times k and since the masses of atoms are discrete, masses of stars are discrete. I must have misunderstood.

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