

LAGRANGE POINTS
© Lloyd

http://milesmathis.com/lagrange2.html (THE CHARGE FIELD causes LAGRANGE POINTS)
 Here I will show that the current Lagrange points are misplaced in the field, due to mathematical errors by Lagrange.
 I will rerun the 3body problem with my simple unified field equations, showing not only the new points of balance for the Moon, for a point, and for a satellite like SOHO; but also the precise places where the current math and theory fail.
 Gravitational potential energy was just energy that would be expressed kinetically [by the field] if you allowed an object to move in the field [].
 So kinetic energy and potential energy weren't really two separate things.
 One was gravity being expressed by motion, and the other was gravity about to be expressed by motion.
 So you cannot sum potential and kinetic energy in a gravitational field.
 You cannot sum the future with the present, and claim you have two different things.
 This is how Lagrange cheated.
 In the Lagrangian, the potential and kinetic energy don't resolve.
 If they did, the Lagrangian would always be zero.
 But Lagrange discovered [] that the two don't resolve, in fact.
 A celestial body has kinetic energy that can't be explained by the gravity equations or the potential.
 In other words, there is more to the field than just mass and distance.
 Once we have exhausted the potential, we still have kinetic energy left over.
 What this should have told Lagrange is that there is another mechanism at work in the field, to give us that residual kinetic energy.
 Something else is driving celestial bodies besides gravity.
 What this all means is that Lagrange had a hidden unified field, just like Newton.
 Newton's unified field was hidden in G, and Lagrange's unified field is hidden in the Lagrangian.
 According to the math of Lagrange, you can fall to the center of a gravity field and still have half your potential left.
 The reason he has that huge error in his math is that he borrowed Newton's math without analyzing it, and Newton's math already contained that huge error.
 It was already embedded in the equation a=v^2/r, and Lagrange didn't spot it.
 According to Newton's own variable assignments and math, it should have been a=v^2/2r.
 Newton made a basic calculus error.
 Lagrange has written an equation in which the charge field is the same size as the gravity field.
 I have shown that isn't physically true. Or, it is true only for objects of a certain size.
 It is true for objects that are around 1 to 10 meters in diameter.
 This is why the Lagrangian works well at the human scale.
 But for smaller and larger objects, the Lagrangian is false.
 One way that Lagrange's orbital equations are semisuccessful is in their prediction of Lagrange points.
 Jupiter's Trojans are cited as proof of this success, and that is in indeed what is happening with the Trojans.
 They are inhabiting areas where the field more or less balances.
 However, this has nothing to do with kinetic and potential energy, it has to do with gravity and charge.
 Neither [gravitational] kinetic energy nor potential energy can hold real objects at a distance, and the only way that the Trojans can be kept from moving closer to Jupiter is with some real force of exclusion.
 The Trojans must be excluded for some other reason.
 Some other field must be balancing the gravitational field here.
 The Trojans are held at bay by the charge field of Jupiter.
 We can see this most clearly if we go to Lagrange point 1, instead of 4 and 5.
 It is known that Lagrange's points 1 and 2 don't really exist where they are supposed to.
 We have tried to take satellites to the Earth's point 1, with no success.
 I mean, the satellites are there, but there is only a reduced instability, not a stability.
 I will recalculate point 1 below with my unified field, showing the errors in the current math.
 First, let's compare the Earth's points 1 and 2 to the motion of the Moon.
 It seems to me that a single Moon would try to hit those points, since it would be a great energy saver if it did.
 Action is supposed to be "least motion", which would imply an energy saving like this, but according to the current math the Moon ignores the points, orbiting well inside them.
 That is the first sign something is wrong with the equations.
 Next, let us look at the eccentricity of the Moon.
 According to current equations, the Moon should have an eccentricity of infinity.
 It should crash into the Sun.
 At New Moon the Moon is seriously out of balance, for instance, and although it corrects that, there is no physical explanation of how it corrects that.
 We find the Moon has an eccentricity of .055 [].
 I have shown that we require the Solar Wind, which is an E/M effect, to calculate it.
 If we know how the field really works, we can solve such problems without any difficult math at all.
 All we need is fractions.
 Yes, the Solar Wind at the distance of the Earth/Moon is strong enough to positively affect the Moon's orbit.
 Not only is charge an [e]ffect of the unified field, but secondary effects of charge also have to be factored in, like the Solar Wind.
 I will now recalculate Lagrange point 1 for the Earth.
 I have done similar math in my papers on weight and on the magnetosphere, showing where the two fields balance.
 According to current math, Lagrange point 1 is about 1.5 million km from the Earth.
 Not only is including the centrifugal force illogical as a piece of Newtonian mechanics, but we know from data that celestial bodies don't feel centrifugal forces.
 We have mountains of evidence that they don't, straight from the Moon.
 We can look for evidence in the crust.
 Since the Moon is in tidal lock, the forces don't travel.
 Therefore they should stack, year after year after millions of years, making the evidence obvious.
 If we had centrifugal forces, we would see their effects on the Moon.
 We would see a big tide at the front and back, and we would see shearing sideways, one direction forward and one direction back.
 What do we have? The most glaring negative data imaginable. No tide in the back, and a negative tide in the front. And no shearing.
 We also have negative data that is very easy to read from the moons of Jupiter and Saturn, including the very small moons inside the Roche limit.
 We find that at 2.586 x 10^5 km [or 258,600 km], the Earth's acceleration upon point 1 matches that of the Sun.
 That is a long way from the current Lagrange point. [But it's] in the same ballpark as the orbit of the Moon.
 Maybe the Moon really IS hitting the Lagrange points, or trying to.
 The Moon is inclined five degrees to the ecliptic, so it doesn't hit the right plane every month, but it isn't far away.
 And since the nodes travel, it will hit them occasionally. We know that from eclipses.
 At Solar eclipse, the Moon is nearest Lagrange point 1, one way or another, since it is right between Earth and Sun.
 I have shown that charge is a function of both mass and density.
 Since we seek a charge density to work with, and since mass and charge are equivalent in the field equations, we seek a mass density.
 So to calculate relative charge (charge of one body relative to another), you multiply mass times density.
 This means that the Sun has 85,063 times as much charge as the Earth.
 Therefore, if we give the Earth a charge of 1, the Sun has a charge of 85,063.
 Since the Sun's charge is moving out from center, we take the fourth root. 4√85,063 = 17.078
 But since the charge field of the Earth is actually .009545m/s2, not 1, the actual charge field of the Sun is 17.078(.009545) = .16301m/s2
 Since the Earth is 1/388 times as far away as the Sun, the Earth's relative charge at the Moon is only .000025.
 To find the total charge field at the Moon, we add eq.1 .16301 + .000025 = .163035m/s2  Now we do the gravity.
 eq.2 Gravity from Earth to Moon 9.7895/60.27 = .162427
 eq.3 Gravity from Moon to Earth 2.668/60.27 = .044267
 eq.4 Gravity from Sun to Moon 1070/23,395 = .045736
 eq.5 Gravity from Moon to Sun 2.668/23,395 = .000114
 We add eqs.4 and 5, then subtract 3 from that, then subtract that from 2, to get .16084.
 Then we subtract that from eq.1, giving us .002195m/s2.
 Since that is very close to my corrected number for the offset of the tangential velocity of the Moon (calculated in another paper) of .002208m/s2**, I think I may claim to have solved the problem.
 So I have found that the Moon is at its own Lagrange point 1, given its velocity.
 I have shown that all the accelerations and vectors balance, at a single position, without any difficult math.
 Another thing to notice is that we only have to slow the Moon down a bit to make it hit a more tightly defined Lagrange point.
 Historically, the Lagrange point hasn't been stationary in the field, of course, since if the Earth is moving, the point has to move with it.
 We would drop the Moon's velocity from 31km/s to just under 30km/s, to make it stop orbiting.
 If we could slow it instantaneously right at that position, we might make it hover in eclipse, permanently.
 What I want to do now is see if that Lagrange point is at the distance we found above, using Newton's simple equations.
 Remember that we found the number 258,600 for the Lagrange point, using Newton's math instead of Lagrange's.
 What if we put the Moon at that point with an Earthshadowing velocity of 29.75km/s?
 Would it stay there, without orbiting the Earth (ignoring other instabilities)?
 No, if we run the numbers again, we find a repulsion of .101m/s2, so we have gone way too close.
 What we find is that the correct distance for balance, with no orbit, is around 380,500km.
 We only have to move the Moon 3,500 km from its average orbital distance to achieve a nonorbiting balance at Lagrange point 1.
 Since that is already in the current range of the Moon, you can see that the forces that cause the Moon to orbit aren't very different from a nonorbiting balance.
 In other words, it wouldn't take much of a blow at eclipse to make the Moon hover in eclipse (or it wouldn't if the Moon were orbiting retrograde).
 We just slow it from about 31km/s to about 29.5km/s, relative to the Sun.
 And so, the Moon's Lagrange point 1 is at about 380,500km.
 The Moon is where it is because it is staying near its Lagrange point, which completely contradicts current math and theory.
 However, it confirms logic.
 As I said, we should have expected the Moon to hit its Lagrange point at Solar eclipse, since we know the Moon is in balance.
 If the Moon weren't in balance, it would fly off into space.
 In fact, some of the old guys like Euler and Lagrange did expect it.
 Some of them were surprised that the Moon didn't hit this balancing point at eclipse.
 [S]atellites are neither points nor bodies with much charge, so they won't go to either the Lagrange point for a point or the Lagrange point for a Moon.
 [I]f we start at my Lagrange point 1, and we expand the point by giving it both radius and mass, it will have charge also, and we will have to go closer to the Earth to keep the balance.
 The Sun will respond to the increasing charge, and will push it away.
 That is why the Moon is inside the Lagrange point proper.
 But if we increase the radius and don't increase the mass, we will have to go away from the Earth to keep the balance.
 The larger radius makes the Earth seem to push it away, as in the equations above.
 But the Sun does not respond in kind, because the charge hasn't increased.
 This is what is happening with our satellites that are supposed to be at Lagrange point 1.
