acsinuk Posted January 7, 2016 Posted January 7, 2016 By using the inverse square law we know that at balance point mass of sun will be ten thousand times that of earth.. Our planet weights 6x10^24 kg so the sun weighs 6x10^28 kg. I am confused as text books state 2x10^30 kg? Keplers 3rd law will keep an object at the L1 position but it will not set that balance position.
EdEarl Posted January 7, 2016 Posted January 7, 2016 Wikipedia says Sun L1 is 1.5x106 from earth and 1.5x108 from the Sun. Perhaps showing your calculations would help.
swansont Posted January 7, 2016 Posted January 7, 2016 By using the inverse square law we know that at balance point mass of sun will be ten thousand times that of earth.. Our planet weights 6x10^24 kg so the sun weighs 6x10^28 kg. I am confused as text books state 2x10^30 kg? Keplers 3rd law will keep an object at the L1 position but it will not set that balance position. That assumes that the forces are equal at that point, but that's not what L1 is. The force at L1 is not zero. https://en.wikipedia.org/wiki/Lagrangian_point
imatfaal Posted January 7, 2016 Posted January 7, 2016 That assumes that the forces are equal at that point, but that's not what L1 is. The force at L1 is not zero. https://en.wikipedia.org/wiki/Lagrangian_point I had to re-read this but upon a bit of thought it makes sense. If the force was zero then any object would continue in a straight linear fashion at a constant velocity - whereas at a Lagrange point the object can maintain a stable position with respect to the two major object (ie an orbit) which is a state which requires an centripetal force 2
swansont Posted January 7, 2016 Posted January 7, 2016 I had to re-read this but upon a bit of thought it makes sense. If the force was zero then any object would continue in a straight linear fashion at a constant velocity - whereas at a Lagrange point the object can maintain a stable position with respect to the two major object (ie an orbit) which is a state which requires an centripetal force Exactly. From an inertial view, the point is not stationary, even though it looks that way from the earth. It's a special orbit, meaning there still has to be a force.
studiot Posted January 7, 2016 Posted January 7, 2016 I had to re-read this but upon a bit of thought it makes sense. If the force was zero then any object would continue in a straight linear fashion at a constant velocity - whereas at a Lagrange point the object can maintain a stable position with respect to the two major object (ie an orbit) which is a state which requires an centripetal force Isn't is a good feeling when you puzzle something out for yourself and find you have puzzled it correctly? +1
acsinuk Posted January 8, 2016 Author Posted January 8, 2016 There could be another explanation. If the force of gravity, is assisted by another dark energy force that is just over 20 times stronger then the maths is correct. So mass attraction force is 5% and dark energy force 95%. IMHO that dark energy force will turn out to be electro-magnetic! -1
swansont Posted January 8, 2016 Posted January 8, 2016 There could be another explanation. If the force of gravity, is assisted by another dark energy force that is just over 20 times stronger then the maths is correct. So mass attraction force is 5% and dark energy force 95%. IMHO that dark energy force will turn out to be electro-magnetic! No, not really. And your opinion doesn't matter. It's what you can show with models and data. 1
imatfaal Posted January 8, 2016 Posted January 8, 2016 There could be another explanation. If the force of gravity, is assisted by another dark energy force that is just over 20 times stronger then the maths is correct. So mass attraction force is 5% and dark energy force 95%. IMHO that dark energy force will turn out to be electro-magnetic! Clive - we don't need a different more complicated explanation which doesn't fit the facts; to advance an alternative hypothesis we really need you to be able to show where the current model fails and the proposed model succeeds. And that would need to happen in the Speculations Forum. We can predict where Lagrange points (L1 L2 L3) will lie using geometry and the results of a fifth order polynomial and L4 and 5 are more easily worked out (they are on point of equilateral triangles with the two large masses at other points - ie simple trig). We make these predictions using simple Newtonian gravity - and we have confirmed them by looking at the motion of natural objects there and even sticking satellites into place there (we have at least six in Sun-Earth L1). What is wrong with this picture which requires a new, exotic, and untested hypothesis
acsinuk Posted January 9, 2016 Author Posted January 9, 2016 What is wrong with existing model is that we have lost 95% of the forces that hold the solar system together and the 95% of the repelling force that is forcing stars and galaxies apart. This can be solved by assuming the galaxies are magnetised which can induce balanced electric charges into the universe which is witnessed by us as neutron electro-magnetic [magnoflux3D] matter. CliveS
studiot Posted January 9, 2016 Posted January 9, 2016 acsinuk Our planet weights 6x10^24 kg so the sun weighs 6x10^28 kg. Weighs? What sort of kilogramme are you talking about?
swansont Posted January 9, 2016 Posted January 9, 2016 What is wrong with existing model is that we have lost 95% of the forces that hold the solar system together and the 95% of the repelling force that is forcing stars and galaxies apart. Lost? I'm sure it's around here somewhere. What are you talking about?
Sensei Posted January 10, 2016 Posted January 10, 2016 This can be solved by assuming the galaxies are magnetised which can induce balanced electric charges into the universe which is witnessed by us as neutron electro-magnetic [magnoflux3D] matter. CliveS Magnetized objects are very easily visible, as they f.e. attract iron, bend paths of charged particles, split spectral lines of gases in plasma state etc. If star would be magnetized to serious level, it would be visible in spectrum of that star. Magnetization spreading on entire galaxy would have to have tremendous power.
acsinuk Posted January 11, 2016 Author Posted January 11, 2016 Its not so much that the stars and planets are magnetised its more to do with the deep space between them that is magnetised. Thus this magnetisation forces stars apart in space by induction. We cannot measure the EM force as we have no fixed reference points to measure from but the WMAP results mean that it is a force of around 23 times stronger than the gravity mass-mass attraction force. Now we are in speculations forum, you can see that I have speculated this in part9 of my magnoflux3D blog. I am sure there are errors in my blog which need to be addressed but basically we need a model of the universe that electro-statically balances and incorperates the magnoflux spin effect. The existing mass gravity time model is insufficient which is why I want the basic MKS dimensional units expanded to include flux-current and voltage units MKSIV.
Mordred Posted January 11, 2016 Posted January 11, 2016 (edited) That's where your wrong. We can measure plasma, and look for magnetized influence. We've done this via the CMB. Overall the universe is electromagnetic neutral. The subject of electromagnetic influence in the overall universe isn't a new one. It's been around for quite some time. As such there have been repeated tests to confirm or disprove. The tests show electromagnetic neutrality. I would imagine though there is still papers around suggesting the possibility. Edited January 11, 2016 by Mordred
Mordred Posted January 11, 2016 Posted January 11, 2016 (edited) Proof from Whom? A good reference is "Conversations on Electric and Magnetic Fields in the Cosmos" by Eegene N Parker. Though that's certainly not the only textbook that states the universe is overall electrically neutral. I'd have to wait till I get home from the field to check Dodelsons Modern Cosmology. One line that discusses it is on page 11 chapter 1. Of the Eugene Parker book above. "The Lyttleton-Bondi conjecture challenged experimentalists to look for a slight difference between the magnitudes of the electron and proton charges. For instance, one may contemplate what is implied by the decay of a free neutron into an electron and a proton if the difference in charge is nonzero. One part in [latex]2*10^{18}[/latex] is a very small difference (see discussion in appendix A), and it was not until Dylla and King (1973) that the experimental upper limit was pushed down to one part in [latex]10^{19}[/latex], showing that any difference is negligible so far as cosmology is concerned." Sorry I can't post the source for copyright reasons. Though I've provided the legal reference format (I hope lol) Edited January 11, 2016 by Mordred
acsinuk Posted January 13, 2016 Author Posted January 13, 2016 The cosmic calculator looks very impressive but the problem is it is not able to evaluate the electric charges which must nearly balance.We know that on planet earth all material is made from neutrons, protons and electrons but they do not balance as the plus charge in molecules is not as large as the negative neutron/electron charge. So if planet weighs 6x10^24 kg then the equivalent of 1.675+0.009- 1.673 = 0.011 parts in 3.357 are excess negative charge as a minimum. This means planets and all material/charge that is enclosed in electron shells are negatives.As the solar system must be electrically balances then the sun must be positively charged and its molecules although identical in composition will be enclosed in positron shells and stars will be overall positives and made of anti-matter/charge. The sun must weigh about 6x10^28 as otherwise the charges will not balance.Now the cosmic calculator should be able to sense a change in the ratio of charge difference over mass ratio which is 0.011/3.357 at present.As our galaxy is magnetised; if we were to look out from the magnetic hub in Sag.A* black hole we would see all the stars with their positive charges moving round due to magnoflux3D spin effect and note that they repel each other. We know that this electrostatic force is about 23 times stronger than the force of gravity and is ineptly called dark energy. If this ratio changed then the comic calculation would change.In the solar system we have a solar wind bringing H+ ions to us as calculated in magnoflux part1. . This attractive electrostatic force accelerates the ion across space to, say, planet Earth. We now know that the solar wind takes 11 days to reach Earth. From this time we can calculate the average speed of the wind to about 155km/sec resulting with a final velocity of around 310,000 metres per second. From this, using e/m constant for a proton H+ ion and formula: ½ mv2 = e.V (1/2x1/9.59x10^-7x310^2x10^6)we obtain the voltage per metre of about 500 volts, as the distance between the Earth and the Sun is 150 million kilometres, then the DC voltage will be around 75 million, million Volts.Somehow the cosmic calculator must also be able to record voltage differences and probable also changes to the impedance of free space 376 ohms at present. The MKS system needs to become MKSVICosΦ to enable the cosmic calculator to operate correctly.
swansont Posted January 13, 2016 Posted January 13, 2016 We know that on planet earth all material is made from neutrons, protons and electrons but they do not balance as the plus charge in molecules is not as large as the negative neutron/electron charge. What's your evidence that this is true? So if planet weighs 6x10^24 kg then the equivalent of 1.675+0.009- 1.673 = 0.011 parts in 3.357 are excess negative charge as a minimum. This means planets and all material/charge that is enclosed in electron shells are negatives. As the solar system must be electrically balances then the sun must be positively charged and its molecules although identical in composition will be enclosed in positron shells and stars will be overall positives and made of anti-matter/charge. The sun must weigh about 6x10^28 as otherwise the charges will not balance. How much negative charge is on the moon? Calculate the resulting force between the earth and the moon if they are charged as you claim.
Strange Posted January 13, 2016 Posted January 13, 2016 We know that on planet earth all material is made from neutrons, protons and electrons but they do not balance as the plus charge in molecules is not as large as the negative neutron/electron charge. 10/10 for chutzpah.
acsinuk Posted February 4, 2016 Author Posted February 4, 2016 The moon is negative charge material same as planet earth but does not repel as it is magnetically locked to the planet so only mass attraction applies. We know that if it were unlocked magnetically there would be a voltage across space that would magnoflux spin the moon which would allow us to see the backside of moon. So the moon must be magnetically locked to us. Gravity is the magnetic locking force.
Strange Posted February 4, 2016 Posted February 4, 2016 The moon is negative charge material same as planet earth What evidence do you have for this? but does not repel as it is magnetically locked to the planet so only mass attraction applies. What evidence do you have for this? We know that if it were unlocked magnetically there would be a voltage across space How do we know this? What evidence do you have for this? that would magnoflux spin the moon which would allow us to see the backside of moon. What is magnoflux spin? What evidence do you have for this? So the moon must be magnetically locked to us. This is the fallacy of "begging the question". Gravity is the magnetic locking force. Gravity is not a magnetic force. And assertions are not evidence.
Mordred Posted February 4, 2016 Posted February 4, 2016 (edited) The moon is negative charge material same as planet earth but does not repel as it is magnetically locked to the planet so only mass attraction applies. We know that if it were unlocked magnetically there would be a voltage across space that would magnoflux spin the moon which would allow us to see the backside of moon. So the moon must be magnetically locked to us. Gravity is the magnetic locking force.You know we can measure particle with a magnetic field. It's surprisingly easy to do so. I suggest you Google magnetosphere and how it interacts with the solar winds. As Strange pointed out gravity is definitely not electromagnetic... As pointed out we can easily measure plasma interactions with a magnetic field. This has already been pointed out several times this thread. A little common sense would easily tell you otherwise. "How many non magnetic objects can you name that still have weight" ? What's your explanation for that question Edited February 4, 2016 by Mordred
Phi for All Posted February 5, 2016 Posted February 5, 2016 The moon is negative charge material same as planet earth but does not repel as it is magnetically locked to the planet so only mass attraction applies. We know that if it were unlocked magnetically there would be a voltage across space that would magnoflux spin the moon which would allow us to see the backside of moon. So the moon must be magnetically locked to us. Gravity is the magnetic locking force. ! Moderator Note Two pages of claims with nothing of substance to back them up. In essence, you're preaching this idea, not supporting it. Without evidence, this thread is wasting time for everyone. Do NOT re-open the subject unless you check with staff to reassure us you have some actual evidence for your assertions. Thread closed. 1
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