# Atomic Model Of The Solar System Theory

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#1
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.

Contact: [email protected]
0
15 years ago
#2
(Original post by 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.

Contact: [email protected]
Thank you very much, certainly interesting for the prospective physicists out there.
0
15 years ago
#3
(Original post by 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.

Contact: [email protected]
Erm. There's so many holes to be picked out of this.... Maybe I'm missing a joke....
0
15 years ago
#4
[Dr Evil]Riiiiiiiiiiiiiiiiigh....t[/Dr Evil]
0
15 years ago
#5
Surely that constant only applies to our solar system?
0
15 years ago
#6
(Original post by mik1a)
Surely that constant only applies to our solar system?
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.
0
15 years ago
#7
HA HA HA HA HA HA made me laff
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#8
(Original post by 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.
in reality i think the masses of electrons are not the same, and there were no words "attracted electrically" in the theory
0
15 years ago
#9
(Original post by mik1a)
Surely that constant only applies to our solar system?
Or any that has a similar structure to ours?
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#10
(Original post by Lee86)
HA HA HA HA HA HA made me laff
Einsteins theory of relativity would probably make you laugh as well, if you were around at that time.
0
15 years ago
#11
(Original post by elpaw)
nor are tehy stuck in discrete quantum states.
Technically they are. Not in terms of their electrical energy, but their kinetic energy. It would require a certain minimum (work function) of Kinetic applied energy to remove a planet from its orbit. At least I am seeing that connection.
0
15 years ago
#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!
0
15 years ago
#13
(Original post by 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.
i think you've got a bit muddled up on your atomic theory.
0
15 years ago
#14
(Original post by el GaZZa)
the microscopic model needs a repulsive force (electrostatic) to prevent the electrons falling into the the nucleus.
last time i checked, the electrostatic force is attractive.
0
15 years ago
#15
Bah. I stand corrected.
0
#16
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.

(Original post by 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?
Everything should count, i just didnt include all of that in the calculation above.

(Original post by el GaZZa)
What about the solar systems with no planets- is the star then massless? And what about multiple star systems?
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
0
15 years ago
#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)?

(Original post by RussianDude)

if there are no planets we have to look at the nucleus that doesnt have any electrons that orbit it
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.
0
15 years ago
#18
(Original post by RussianDude)
in reality i think the masses of electrons are not the same, and there were no words "attracted electrically" in the theory
Not the same by the factor that Jupiter is more massive than pluto?

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.
0
#19
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)?
Everything should count, i just didnt include all of that in the calculation above, my bad.

In fact you're saying that stars must have discrete masses dependant on their number of planets.
no, i dont know how you came up with he fact that masses are discrete
0
15 years ago
#20
(Original post by RussianDude)
no, i dont know how you came up with he fact that masses are discrete
Ok in that case I've no idea what your theory is saying. I understood it as:
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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