Strange Posted October 6, 2014 Posted October 6, 2014 (edited) I am not quite sure what you are trying to say. You have, I assume, the gravitational force between the two masses (That should be G not g, and r2 not r2). The other equation is for the force of a charge moving in a magnetic field. There is all sorts of information missing: what is the source and orientation of the magnetic field, what is charged, what direction is the charge moving, etc. Also, the force in this case is at right angles to both the direction of movement and the magnetic field. I don't think you could contrive a case where it acts in the direction you have shown. Maybe you wanted to compare the force of gravity and the Coulomb force? http://en.wikipedia.org/wiki/Coulomb%27s_law Then you could balance the repulsive force of the charges and the attractive force of gravity. But note that it is impossible for this to be stable. The two objects will either fall towards each other under gravity or be repelled by the electric charges. However, a variation of this was used as the basis of the Millikan oil-drop experiment to measure electric charge. Edited October 6, 2014 by Strange
andrewcellini Posted October 6, 2014 Posted October 6, 2014 it would help if you discuss what is going on in your diagram. just to clarify, you have 2 charged massive objects and you're saying the electromagnetic and gravitational forces cancel out?
Relative Posted October 6, 2014 Author Posted October 6, 2014 (edited) it would help if you discuss what is going on in your diagram. just to clarify, you have 2 charged massive objects and you're saying the electromagnetic and gravitational forces cancel out? Yes exactly that, the arrows are in direction of the opposing forces, the arrows are not in their relative positions, the force is isotropic on both accounts. Edited October 6, 2014 by Relative
Relative Posted October 6, 2014 Author Posted October 6, 2014 I am not quite sure what you are trying to say. You have, I assume, the gravitational force between the two masses (That should be G not g, and r2 not r2). The other equation is for the force of a charge moving in a magnetic field. There is all sorts of information missing: what is the source and orientation of the magnetic field, what is charged, what direction is the charge moving, etc. Also, the force in this case is at right angles to both the direction of movement and the magnetic field. I don't think you could contrive a case where it acts in the direction you have shown. Maybe you wanted to compare the force of gravity and the Coulomb force? http://en.wikipedia.org/wiki/Coulomb%27s_law Then you could balance the repulsive force of the charges and the attractive force of gravity. But note that it is impossible for this to be stable. The two objects will either fall towards each other under gravity or be repelled by the electric charges. However, a variation of this was used as the basis of the Millikan oil-drop experiment to measure electric charge. Remember when i asked on a keyboard how to do all the respective symbols in place, I could not do it still, it is gravity and lorenz force What is fn or fn0? The equilibrium of force , like the ground pushes back, fn = 0
Strange Posted October 6, 2014 Posted October 6, 2014 Remember when i asked on a keyboard how to do all the respective symbols in place , I could not do it still I think you could still type G (gravitational constant) instead of g (acceleration due to gravity at the Earth's surface). it is gravity and lorenz force The Lorentz force (note spelling) depends on velocity (including direction), charge, electric field (including direction) and magnetic field (including direction). You have not said what or where q, v, E or B are.
Relative Posted October 6, 2014 Author Posted October 6, 2014 I think you could still type G (gravitational constant) instead of g (acceleration due to gravity at the Earth's surface). , The Lorentz force (note spelling) depends on velocity (including direction), charge, electric field (including direction) and magnetic field (including direction). You have not said what or where q, v, E or B are. I am sorry , I should put word in my diagrams then theres no mistake, I do not know what q,v,e or b even represents, I needed you to know this which ever way I got the message across. ''you have 2 charged massive objects and you're saying the electromagnetic and gravitational forces cancel out?'' You guys know how to do the correct maths and correct tags
Strange Posted October 6, 2014 Posted October 6, 2014 I do not know what q,v,e or b even represents Then why do you think it makes sense to use them? You might as well write random Chinese characters and expect people to understand. needed you to know this which ever way I got the message across. ''you have 2 charged massive objects and you're saying the electromagnetic and gravitational forces cancel out?'' So, you clearly want the Coulomb force, not Lorentz. (And it is much simpler) Form that you can work out what value of charge will balance a given mass. (I might do it later, if I have time).
Relative Posted October 6, 2014 Author Posted October 6, 2014 (edited) Then why do you think it makes sense to use them? You might as well write random Chinese characters and expect people to understand. So, you clearly want the Coulomb force, not Lorentz. (And it is much simpler) Form that you can work out what value of charge will balance a given mass. (I might do it later, if I have time). I used them so you would actually understand the idea I had, I am unsure what to use, I just consider gravity and consider an opposite reaction, the opposite reaction been a stopping force that stops m1 and m2 colliding, I understand the centrifugal gravity explanation of Einstein and even the curvature of space, but from day one when I looked, I thought there was something missing. And yes probably is Coulomb force , but I do not know, like I said , to me, even a beginner in science maths is a Maxwell to me. I needed a Maxwell to put in the correct forces to my diagram and see if it works or not. My idea, his or hers maths , Just tried to copy and paste the formula from wiki to my diagram, doe snot work and can not type that in sorry Edited October 6, 2014 by Relative
swansont Posted October 6, 2014 Posted October 6, 2014 What evidence/data do you have to determine the strength of the Lorentz force? And how do you reconcile this with the Cavendish experiment? http://en.wikipedia.org/wiki/Cavendish_experiment
andrewcellini Posted October 6, 2014 Posted October 6, 2014 Relative, Coulomb's Law F = k(q1q2)/r^2 = qE F is the magnitude of the electrical force k is coulombs constat 8.99e9 q is the magnitude of the charge r is the distance between charges E is the electric field due to a source, kq(source)/r^2 if you have two point charges of equal but opposite magnitude at a distance r, they exert an equal magnitude force in opposite directions. similarly if you have one charge q as as source with an electric field E at distance r from the source, you can set your other charge at this distance and find the source exerts a force on the charge.
Relative Posted October 6, 2014 Author Posted October 6, 2014 What evidence/data do you have to determine the strength of the Lorentz force? And how do you reconcile this with the Cavendish experiment? http://en.wikipedia.org/wiki/Cavendish_experiment define wire in the Cavendish experiment please? Model correction and thx. for the formula Updated the model to look better.
andrewcellini Posted October 6, 2014 Posted October 6, 2014 it would be better to learn more about the applications of the formulae you are using rather than taking my brief introduction to couolmbs law and immediately adding it to your theory.
Relative Posted October 6, 2014 Author Posted October 6, 2014 it would be better to learn more about the applications of the formulae you are using rather than taking my brief introduction to couolmbs law and immediately adding it to your theory. I understand but your explanation of the formula looked correct to what i am saying, or something very similar...
Strange Posted October 6, 2014 Posted October 6, 2014 (edited) Gravity: [latex]\displaystyle f = G \frac{m_1 m_2}{r^2}[/latex] Coulomb: [latex]\displaystyle f = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}[/latex] You want these two forces to be equal so: [latex]\displaystyle G \frac{m_1 m_2}{r^2} = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}[/latex] The distance ® cancels: [latex]\displaystyle G m_1 m_2 = \frac{1}{4 \pi \epsilon_0} q_1 q_2[/latex] If we make the two charges equal and the two masses equal: [latex]\displaystyle G m^2 = \frac{q^2}{4 \pi \epsilon_0}[/latex] From this we can work out the ratio of mass to charge for the forces to just cancel: [latex]\displaystyle \frac{m}{q} = \frac{1}{\sqrt{4 \pi \epsilon_0 G}}[/latex] If we make the charge equal to the electron we get: [latex]\displaystyle m = \frac{e}{\sqrt{4 \pi \epsilon_0 G}}[/latex] Which is about 2 micrograms http://www.wolframalpha.com/input/?i=electron+charge+%2F+\sqrt%284+\pi+\epsilon0+G%29 So, if you take a grain of sand, split in two and put an electron on each half, there will be no net force between them. I used them so you would actually understand the idea I had, I am unsure what to use, I just consider gravity and consider an opposite reaction, the opposite reaction been a stopping force that stops m1 and m2 colliding, I understand the centrifugal gravity explanation of Einstein and even the curvature of space, but from day one when I looked, I thought there was something missing. The trouble is if you were to add an extra force, as you want to do, then none of our maths would work for for calculating the orbits of planets, the paths of cricket balls, and getting rockets into orbit, etc. And if, as above, you match the gravitational force with a repulsive force, then there would be no orbits because there would be no net force. Edited October 6, 2014 by Strange
swansont Posted October 6, 2014 Posted October 6, 2014 And if, as above, you match the gravitational force with a repulsive force, then there would be no orbits because there would be no net force. Which really needs to sink in, here. The orbits we observe are consistent with Newtonian gravity and not other forces of any appreciable size. Mess with that, and nothing fits. You can't have an electrostatic force between planets and the sun augment gravity or subtract from it, because they can't all be repulsive or attractive. To me it looks like you are trying to force the equations to fit your idea. That you're learning the bare minimum so that you can adopt a formula and make things look like there's a viable model here. It doesn't work. There is no appreciable electromagnetic force at play here. In comparison to gravity, it vanishes. I'll go back to my previous question: do you have any evidence that there is either an electrostatic or magnetic force between the earth and the sun?
Relative Posted October 6, 2014 Author Posted October 6, 2014 Which really needs to sink in, here. The orbits we observe are consistent with Newtonian gravity and not other forces of any appreciable size. Mess with that, and nothing fits. You can't have an electrostatic force between planets and the sun augment gravity or subtract from it, because they can't all be repulsive or attractive. To me it looks like you are trying to force the equations to fit your idea. That you're learning the bare minimum so that you can adopt a formula and make things look like there's a viable model here. It doesn't work. There is no appreciable electromagnetic force at play here. In comparison to gravity, it vanishes. I'll go back to my previous question: do you have any evidence that there is either an electrostatic or magnetic force between the earth and the sun? I can offer you evidence from your science, I can offer you evidence from the forum today, unless you can deny the sun emits electrons and the earth emits electrons, I think I have given you something to think about. I know the net force is equal to zero, that is what i have put the fn = 0, this does not change the maths of current, and I can explain where the orbital motion comes from, I can explain Venus, but am in fear of thread closure. I am getting use to the formulas now, I pretty much, am understanding them. Still not 100% but getting there. I now know things like a lower case g is different to a capital G , so will start to be more considerate when i post formulas. I am trying please do not shut me down again, I need to go back to fn on the ground diagrams to show, that is zero, but the object is still in motion with the spin, and aslo I need to relate to another link from one of my closed threads to show magnetic motion or something like that it was called, I need to explain that the ground is attracted to the core but also attracted to the sun and thats what gives the motion. But you will close me down.
Strange Posted October 6, 2014 Posted October 6, 2014 I can offer you evidence from your science, I can offer you evidence from the forum today, unless you can deny the sun emits electrons The sun emits a mixture of particles. There is no net charge. You are just ignoring anything that contradicts your idea. I think I have given you something to think about. Nope. 1. If you work out orbits based just on gravity you get the right answer. 2. If you work out the forces due to solar wind, magnet fields, whatever else, they are approximately zero.
Relative Posted October 6, 2014 Author Posted October 6, 2014 My first evidence i offer is this link, orbital magnetic moment, the sun emits electrons and so does the earths core, but the mantle , the surface does not emit electrons making it a negative except probably iron content which can hold electrons. http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/orbmag.html The sun emits a mixture of particles. There is no net charge. You are just ignoring anything that contradicts your idea. Nope. 1. If you work out orbits based just on gravity you get the right answer. 2. If you work out the forces due to solar wind, magnet fields, whatever else, they are approximately zero. sorry what you mean no net charge , no energy? The ionosphere is a shell of electrons and electrically charged atoms forgive me if this sounds really stupid , that sounds friction to me? Is the pressure released at night? ''At night the F layer is the only layer of significant ionization present, while the ionization in the E and D layers is extremely low. During the day, the D and E layers become much more heavily ionized, as does the F layer, which develops an additional, weaker region of ionisation known as the F1 layer. The F2 layer persists by day and night and is the region mainly responsible for the refraction of radio waves.'' Does the suns magnetic field rotate with the sun?
Strange Posted October 6, 2014 Posted October 6, 2014 (edited) My first evidence i offer is this link, orbital magnetic moment, That is irrelevant. That is about the magnetism of a single electron in an atom. the sun emits electrons It also emits positively charged ions. So there is no net charge. and so does the earths core How can the core emit electrons? It is in the middle of the Earth. Where are they going to go. And no it doesn't anyway. , but the mantle , the surface does not emit electrons making it a negative Nope. the Earth is electrically neutral. sorry what you mean no net charge , no energy? No net charge. Nothing to do with energy. Charge is not energy. As you have been told repeatedly. The ionosphere is a shell of electrons and electrically charged atoms forgive me if this sounds really stupid , that sounds friction to me? I can't see how you get from electrons and ions (electrically charged atoms) to friction. You might as well say it sounds like nougat or reggae. Is the pressure released at night? What pressure? Edited October 6, 2014 by Strange
Relative Posted October 6, 2014 Author Posted October 6, 2014 (edited) I am getting confused now , there seems to be a contradictory to the earlier answers in my questions threads, the sun has a magnetic field, the earth has a magnetic field , both the shields expand outwards, towards each other, gravity wants us to join together, but the magnetic shields do not let us. 'But a magnet's field doesn't come from a large current traveling through a wire -- it comes from the movement of electrons.'' http://science.howstuffworks.com/magnet3.htm ''electrons are negatively charged and like charges repel therefore, electrons repel similarly protons repel (two +'s) but a proton and electron attract, thus causing atoms, and the universe we live in today'' https://answers.yahoo.com/question/index?qid=1006042024566 ''The other observation is much easier to understand that the already emitted positive and negative filaments do not neutralize each other in space because they are antiparallel electric currents which repulse each other stronger than they attract each other via electrostatic attraction. Electrons of a negative filament do not fly to ions to recombine it in a positive filament in a distance of only 1000km, they fly parallel to each other millions of kilometres and later diverge.'' http://www.electric-sun.info/main.html And to the diagram daytime is more compressed which suggest there is more force on the layer in the day , which shows us we are pulling towards the sun, hence blue sky. and i can draw a diagram showing you this with a basic formula using the Doppler redshift and blue shift And sorry mods this is to continue to fight that my model may have some meaning and the same subject. added to coincide with my model idea. added from forum questions from some one. ''So anyway, if we refered to an electron as having a positive charge and a proton as having a negative charge. And we changed all the equations around respectively. Would there be any real difference?'' your answers were there would be no difference, so in answer to the maths , and if you changed gravity to opposing and did the equations the results would be the same? Edited October 6, 2014 by Relative
Endy0816 Posted October 7, 2014 Posted October 7, 2014 The sky is blue due to our atmosphere scattering the light. Us not hitting the Sun has nothing to do with charge and everything to do with our velocity. How do you think satellites orbit? Same concept.
andrewcellini Posted October 7, 2014 Posted October 7, 2014 relative, i'd like to repeat that you should probably do more studying on these subjects, including the formulae and their applications, and doing problems so that you can have a stronger grasp on the subjects. it seems you are avoiding doing this so far.
Strange Posted October 7, 2014 Posted October 7, 2014 I am getting confused now , there seems to be a contradictory to the earlier answers in my questions threads To be honest, that is because you have a very vague understanding of about 10% of what you are told. The rest just confuses you. You also ignore and misinterpret anything that contradicts your idea. , the sun has a magnetic field, the earth has a magnetic field But these are two weak to have any effect of the motion of the Earth. , both the shields expand outwards, towards each other, gravity wants us to join together, but the magnetic shields do not let us. The magnetic fields are too weak to overcome gravity. Also, whether the magnetic fields attract or repel depends on their orientation - which is constantly changing. http://www.electric-sun.info/main.html This is a crackpot pseudoscience website. (Unfortunately, you don't know enough basic science to tell.) And to the diagram daytime is more compressed which suggest there is more force on the layer in the day , which shows us we are pulling towards the sun, hence blue sky. Just no. The reason for the sky being blue is well understood and nothing to do with that. so in answer to the maths , and if you changed gravity to opposing and did the equations the results would be the same No. <sigh> There are both positive and negative charges. If you were to swap their signs, it would make no difference. Gravity does not have positive and negative components so there is nothing to swap.
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