How is Lagrange point L1's location reconciled with the mass of the Sun?

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Offline acsinuk

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By using the inverse square law we can calculate that weight of sun is ten thousand times that of planet earth which weight 6x10^24  kg.  This means the sun weights about 6x10^28 kg and not 2x10^30 kg as shown in text books? 
I am confused as Keplers third law only keeps objects in the balanced position L1 not set that position.
« Last Edit: 07/01/2016 16:24:40 by chris »
A.C.Stevens

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Offline acsinuk

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The cosmic calculator looks very impressive in predicting future 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: 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.
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Offline evan_au

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Quote from: acsinuk
we can calculate that weight of sun is ten thousand times that of planet earth
Today we estimate the ratio of masses as about 330000:1.

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Keplers third law
The mass of the Sun and the Earth can be calculated from Keplers Third Law (with the knowledge of Newton's Gravitational constant).

With a tiny object orbiting a much more massive object, the mass of the central body can be calculated as follows: M=4*pi^2*a^3/(P^2*G)
Where:
M is the mass of the central body (in kg)
a is the average radius of the orbit (in meters): 1.5e11 m for Earth, 6.77e6 for the ISS
P is the period of the orbit (in seconds): 3.16e7 for Earth, 5.56e3 for the ISS

Plugging in these values we get:
Mass of the Sun: 2e30 kg
Mass of the Earth: 6e24 kg

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the balanced position L1
The L1 Lagrange point is between the Earth and the Sun, about 1.5 million km inwards from the Earth (1% of the Earth-Sun distance). It is not a stable point in space, but several space probes have been put into wide orbits around L1 (eg for Sun observation).

Applying Kepler's third law again, to find the orbital period of a satellite at L1 around the Earth (ignoring the Sun):
P=SQRT(4*pi^2*a^3/GM)

a=1.5e9 m
M=Mass of the Earth = 5.95e24 kg
P=0.58 years = 1/SQRT(3) years

So when the tug of the Sun is included, the orbital period of a satellite at the L1 point becomes 1 year (the same as the Earth).

See the derivation of the L1 position at: http://en.wikipedia.org/wiki/Lagrangian_point#L1 

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By using the inverse square law we can calculate that weight of sun is ten thousand times that of planet earth
Please explain how the inverse square law was used to deduce this result?

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Offline acsinuk

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What we are looking for is the null force point in space where the attractive gravitational force towards the earth is equal to the attractive force towards the sun. At this balance point an item will not fall either way.  If L1 is that balance point then the inverse square law will ratio sun mass to earth mass surely.
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Offline evan_au

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Quote from: acsinuk
What we are looking for is the null force point in space where the attractive gravitational force towards the earth is equal to the attractive force towards the sun
I tried working a back-of-the-envelope calculation using the conventional masses of the Earth & Sun, and the inverse square aw, as you describe.
I found the balance point of the forces to be about 260,000 km sunward from the Earth (pretty close to the Moon's orbit). This is about 6 times closer to the Earth than the conventional location of L1, which is 1.5 million km sunward.

I think the flaw in this argument is that at this balance point, the space probe will feel no net force from Sun or Earth, and so it will follow a straight line (as dictated by Newton's laws).

This means that as the Earth continues around its orbit, the space probe will approach the Earth more closely, and will no longer be at this balance point. So this balance point is very unstable (unlike the real L1 which is semi-stable).

In fact, L1 must be sunward of this force balance point, as the gravity of the Sun must exceed the attraction of the Earth, pulling the space probe into an arc that keeps it a constant distance from both Sun and Earth. See the animation here.

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the DC voltage will be around 75 million, million Volts
The Sun and Earth must be (on average) electrically neutral, or the Solar wind would quickly balance the charges, or dissipate any excess charge.

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We now know that the solar wind takes 11 days to reach Earth.
Typical speed of the Solar wind on quiet days is around 300km/s (6 days), but we know that the the solar wind is very gusty, sometimes reaching 1000km/s (2 days).
In the Carrington event, a disturbance on the Sun's surface was observed to reach Earth in just 18 hours.

I'm afraid you can't use electrostatics to explain the solar wind. Bursts are associated with rupturing magnetic fields in solar flares, or streams emitted from a hole in the Sun's corona.

See: http://www.spaceweatherlive.com/en/help/the-solar-wind

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Offline acsinuk

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It is the slowest. What we need to use is a single hydrogen ion that just lifts off from the suns surface due to electric attraction of negative charged planet.  Why the moon is mass attraction only must have something to do with being magnetically locked to the planet. If there were a voltage then the moon would rotate and we would see its outside face. The laws of electrostatic may need to be revisited to understand the implications of magnetic locking on charged particle forces which are applicable to unlocked magnetic particles only
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Offline evan_au

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Quote from: acsinuk
Why the moon is mass attraction only must have something to do with being magnetically locked to the planet.
The Earth's magnetic field is fairly weak - around 50 microTeslas. Not enough to modify the path of the massive Moon.
And since the Earth's magnetic field decays and reverses periodically, the Moon would have flung off into space long ago if it were being held by magnetic forces.
See: http://en.wikipedia.org/wiki/Earth%27s_magnetic_field

The Moon's magnetic field is far weaker than the Earth's (measured in nanoTeslas). It is not globally aligned, but pointing in different local directions. So there is nothing that the Earth's magnetic field could attract.
http://en.wikipedia.org/wiki/Magnetic_field_of_the_Moon

You can't use magnetic fields to explain the orbit of the Moon, any more than you can use electric forces to explain the orbit of the Earth around the Sun.

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Offline jeffreyH

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Dealing with any kind of assumed equilibrium point in a dynamic system is tricky. It's like a 3 body problem on steroids.

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Offline acsinuk

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The point is that if moon and planet are locked magnetically then there is no necessity for an external magnetic field link.This magnetic binding force becomes the force of mass attraction. If the magnetic fields are unlocked then an external magnetic field will be necessary, which with a voltage difference will cause spin.
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Offline evan_au

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Quote from: acsinuk
as the distance between the Earth and the Sun is 150 million kilometres, then the DC voltage will be around 75 million, million Volts.
In this New Theory, the Earth orbits the Sun because it has an electrostatic attraction.
You could extend this theory to say that the other planets also orbit the Sun because they have different (but incredibly high) voltages, with the opposite polarity to the Sun.

However, this produces the following fundamental contradiction:
  • According to this theory, Earth, Mars and Venus must have the same polarity, as they are all attracted to the Sun
  • Because Earth, Mars & Venus have the same polarity, they must repel each other.
  • Because Earth passes closer to Mars and Venus than it does to the Sun, Mars and Venus must repel the Earth more strongly than the Sun attracts the Earth.
  • In this theory, the Solar system would fly apart in under a year.

In fact, Newton's law of gravitation is extremely precise in calculating the attraction of every planet to every other planet: massive ones like Jupiter, and ones that pass closely, like Earth, Mars and Venus.

It is so accurate at adding up all these tiny disturbances that it was able to highlight a microscopic discrepancy with the perihelion of Mercury: Newton's gravity predicts 532 arc-seconds per century, while the observed value is 574. Much of the difference is due to the time dilation of general relativity in the Sun's intense gravitational field.

http://en.wikipedia.org/wiki/Tests_of_general_relativity#Perihelion_precession_of_Mercury

Similar logic applies to planets with multiple moons, like Jupiter and Saturn. If the polarity of the magnetic field is oriented to attract the moons to the planet, then it must be oriented so that all the moons repel each other. But the orbits of all these moons have been studied in enormous detail, and used to plot the path of space probes examining these moons. They do not repel each other; they attract. And the space probe itself is intended to measure magnetic fields of the planet and its moons, so it has virtually no magnetic field of its own; and yet it remains in the planned orbit.

So I am afraid that electrostatics cannot hold the Solar system together, any more than magnetism can hold together a planet and its moon(s).

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Offline acsinuk

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Yes,the answer is that the magnetic field is locked. Now, to get electromagnetic wave energy to flow in a conductor we need to synchronise the magnetic field in the Ixdirection only and lock it in space and time. We can then push 3D energy into the system. This appears to automatically happen when magnetized foreign objects approach a magnetized planet, they just align and the electrostatic field disappears leaving only mass attraction. Further, plasma gas is magnetically locked unless we use it in electric arc welding which only unlocks it for an instant. This means that TOCOMACS will need to find some way of locking the Ixfield only if they are to get EM power from their plasma engines
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