E.M ON AMAZON KINDLE

https://www.amazon.com/dp/1799165647

https://www.amazon.com/dp/B07PG44BMG

Dear  Readers,

You are welcome here to share my view, my way in scientific research.

-Where a mass is, there is a gravity field around.

-Where an electric charge is, there is an electric field around.

-Where a magnet is, there is a magnetic field around.                                                                                

Nonetheless, on the reverse wise of the above statement:

‘’Where a magnetic field is available, there is a magnet in it’’ creates a long debate; especially about Earth Magnetism. The debate does not just take a decade, it is prolonging in centuries.

I am not writing about an analogue to this: ‘’Where a mass is, there is a gravity field around’’; but basically I argue about Earth Magnetism that the field of Earth Magnetism is made out of the moves of many electric charges, then the Earth core or anything else is magnetized, reciprocally the apparent Earth Magnetic field is seen.

I trust and I am backed by a rule that ‘’Where an electric charge moves, there is a magnetic field to be induced’’, because that’s how the Earth Magnetism is built. The material in a magnetic field can be magnetized, as Faraday’s Law of Magnetic Induction did point out in history.

The way I go along is the way I share my view with you.

Although the problems and solutions in this book are quite theoretical but every assumption and solution depends on real data from many different researchers in the world. The work costs me decades.

Where are those electric charges? That is a question, and I got to find out the electric charges on the Earth, in atmosphere and in the space near our Earth.

This book ‘’The basic problems for Earth Magnetism’’ contains problems and solution to each of them.

The major theory used in this book is Biot-Savart and Ampere-Maxwell. The data used in this book are findings of NOAA, NASA and from many other data sources in the world.

The book is broken down into two major parts; the first is for problems of charges on Earth. The charge on Earth is realized long ago such as charged cloud and its opposite patch on Earth surface. I am not the first who figures out the charges, but a person who connects the area of world’s charged (thunder) cloud with air-earth current in every 24hr of our Earth rotation, even if no direct equation is given out for Earth Magnetism.

As matter of facts, the Earth is rotating around its own axis. Therefore every charge on it can induce magnetism to Earth.

Several people keep questioning me about the move of electric charge, I don’t respond directly to their question but hint that whether the speed 462 m/s of everything at Earth’s equator makes sense to them?!

The second part is for external charge such as one on Van Allen Belt and the other is on Moon. The charge on any of those above is obviously assumed, but depending on a reality that there is almost no absolutely neutral object found, which is neither negative nor positive. The principal: ‘’The more sun light strikes on an object surface, the more electrons it loses’’ did lead Albert Einstein to his invention of solar cell. The mentioned invention leads to assumption that the Moon has two opposite semi -spheres, the shined one is positive charged while the dark one is definitely negative. That is not it; the Moon is even not neutral or balanced in electric charge but is also lacking electrons. Lunar total charge may be positive sometime and keeps changing due to the change of solar radiation. My assumption does encounter a lot, certainly, critics. A predominant question is also a hint that if an object lacks of electron, its ambiance will provide electron; if the object is rich of electrons, the protons from its ambiance will come to balance it. This question actually is still in the invention of Albert Einstein.

The last major question in my debate with other people is ‘’Why do the magnetic poles stay away from Earth’s true poles?’’, although I mentioned in the preamble that I never work with rotors or anything inside the Earth. But I do propose a non-professional answer that each plate is a huge magnet in general Earth Magnetic field; the poles in apparent Earth Magnetic field are shaped by the total of every individual magnetic field of all continents and plates. Definitely, many people are not satisfied with such a non-professional answer and go forward with another question that the North magnetic pole is drifting toward Russia, and the drifting is the beginning of flipping which will happen in 1000 year time, can charge move help to stop it? Although this book is not written about the magnetic pole flipping, and I don’t believe in the conclusion that the magnetic pole drifting; but anyway I point out that the African continent is moving northward as fast as 8 inches/year. Look at our Earth again; we may realize that Americas continent is Earth’s back bone while Africa and Euro-Asia are moving northward, whereby Africa is much faster than Euro-Asia.

Certainly many people are not satisfied with my answer; they have no choice but wait to read the research result from world scientists who study that subject.

For the 1^st publishing on Amazon Kindle, I would like to give ‘’I-Biot-Sawart’’ expression for anyone who attempts to calculate how much an electric charge on Earth surface can contribute to Earth Magnetism.

In order to apply I-Biot-Sawart expression, we need to have value of electric charge (q) and a table of distance to Earth rotation axis (the ‘’r’’).

With I-Biot-Sawart (Interpolated Biot-Sawart), we can state that: if I am given with a charge and its position on our global coordinates, I can calculate right away how much that charge contributes to Earth Magnetism.

This publishing is bearing in it a great thank to the Late Professor Nguyen Dinh Noan, and PhD Vu Bang for encourage and great help during I study this subject.

Thank for your kind attention.

Ho Chi Minh City March 9, 2019 

NV Cuong

EARTH HISTORY RE-CONSIDERATION

(Geologist Edward Young, Theia, Moon, collision, impact)

The Earth core is assumed to be a ‘’globe’’ with another smaller ‘’globe’’ in it; so far they are named as ‘’outer core’’ and ‘’inner core’’. This essay is to discuss about re-consideration to the core(s) on basis of updated information about gravity from NASA and some other space researching bodies.

I-NASA GRAVITY MAP:

How was the Earth established?

Figure 1-Gravity map with data observed by GRACE (Credited to NASA, updated on May 14^th 2017)

The above Figure ‘’1’’ is re-built by the real NASA data, which are harvested by some satellites in observing the Earth gravity. The following is a quote from a title of NASA news:

‘’The joint –NASA-German Aerospace Centre Gravity Recovery and Climate Experiment (GRACE) mission has released its first science product, the most accurate map yet of Earth’s gravity field. GRACE is the newest tool for scientists working to unlock secrets of ocean circulation and its effect on climate.

Created from 111 days of selected GRACE data to help calibrate and validate the mission’s instruments, this preliminary model improves knowledge of the gravity field so much it is being released to oceanographers now, months in advance of the scheduled start of routine GRACE science operations. The data are expected to significantly improve our ability to understand ocean circulation, which strongly influences weather and climate.’’

Although the Earth has been established in almost 4.5 billion years ago, since then there are lot of changes on the Earth; nowadays it looks as a perfect sphere. But actually (by NASA gravity map in figure 1) the evidence is still like alive despite the change in billions of years. The present Earth is made out of 2 major parts: 

Figure 2-Junction of 2 parts.

Junction is un-covered.

Despite there are bundles of hypothesis about the Earth and Moon establishment, every of them claims recognition; this evidence is to confirm the giant impact  hypothesis of Geologist Edward Young. The collision between another planet and our primitive Earth left evidence behind, that is junction, let’s go further with it.  

In order to understand the ‘’why’’, the following net force or vector addition rule is employed:

Figure 3 - On-junction gravity is less than on-main-body one.

+Junction: The dark blue trace is nearly about Earth’s longitude and getting through India, China and Russia over Artic…and Atlantic is assumed to be the junction between the two mentioned parts. The Gravity at Arctic and the channel between Argentine and Antarctic continent are normal, this is another issue in the subject that requires more research.

+ On-junction gravity is less than on-main-body: The gravity observed at junction or border between the two parts (each gravity centre is belonging to a different part) is less than normal because:

Two vectors make the largest total when the angle between them is 0⁰, this is known to the world.

II-EARTH IS A PLANET OF TWEEN GRAVITY CENTRES:

What we should debate is the tween gravity-centre structure of our present Earth.

The tween gravity centre planet is made out of a combination of one and the other, the annexation is assumed to be made by collision; if it was not known to us before, it is debated today. The following is a quote from Wikipedia dictionary:

By the Moon's origin  ‘’The Moon is an astronomical body that orbits planet Earth and is Earth's only permanent natural satellite. It is the fifth-largest natural satellite in the Solar System, and the largest among planetary satellites relative to the size of the planet that it orbits (its primary). The Moon is after Jupiter's satellite LO, the second-densest satellite in the Solar System among those whose densities are known.

The Moon is thought to have formed about 4.51 billion years ago, not long after Earth. The most widely accepted explanation is that the Moon is formed from the debris left over after a giant impact between Earth and a Mars-sized body called Theia.’’

The above explanation is accepted and certainly my argument is about it.

a)-More demonstration needed:

In addition to the assumed demonstration on Wikipedia for Theia the following ‘’equation’’ can explain a bit more on term of structure and power in the process:

Figure 4-My Earth Collision Equation

‘’a’’ is a planet (Theia) or a huge comet in solar system.

‘’b’’ is our primitive Earth (in the shape of recent found comet Rosetta, (ESA Image)).

‘’E’’ is denoted for the Earth with the latest gravity data (NASA Image), mind that the Earth does not look like that without water in the oceans.

‘’M’’ is the Moon, debris after collision.

b-Non-aligned or head-on collision:

Almost collision that happened when one collider’s gravity centre locating on the same moving line with the other is assumed as head-on collision; the followings are examples:

+Two bricks on rail.

+Two wagons on a rail.

+Two globes of pure material (un-pure material globe may have gravity centre off from its geometric centre).

+Two pure cylinders standing straight upward, one moves toward the other than hit.

Otherwise, the collisions that gravity centre of one collider is not on the same moving line with the other, are assumed as non-aligned. Consequence of non-aligned collision always includes a moment or a change in spinning wise and speed.

Figure 5-Non-aligned collision

-The ‘’equation’’ in (Figure 4) is to depict a process when a planet ‘’a’’ merges with planet ‘’b’’, the products are present Earth and the Moon. The Earth after collision is not just in a new shape, but also with new structure as following illustration:

Figure 6-New structure of our Earth

The individual gravity centre of each part might be proposed as following:

Figure 7-New structure with new gravity centres

As assumption repeated more than twice, the collision between Earth and Theia is head-on type, but actually there is rare head-on collision but non-aligned one in natural reality. The reason is quite simple that there is no perfect pure globe or cylinder found in nature, or the way one moves along is always out of the way where the other moves along. On the other hand, the shape of each is never a perfect square or perfect global in nature…therefore, the head-on collision is hardly found, so every collision leads to both deformation and spinning change, which is not debated in this title.

Although we have neither image of primitive Earth nor the Theia, but have title to suppose that each of them is a certain globe, so those two globes after collision might become a combination as above illustration, where the part left by Theia on our Earth is yellow or yellow half. The original core of Theia and its gravity centre are eliminated after collision then a new gravity centre of yellow half is established.

-Evidence:

+Diamond is never produced in normal condition but very high pressure and temperature. Notwithstanding, the diamonds are found near Earth surface or even in open mines of Africa, those evidences are indicating that Africa rock and soil used to be at very deep layer and belonging to Theia planet. This argument may lead to an assumption that there is still some diamond on Moon surface, this is quite in a normal logic, but nobody can suggest where on the Moon is rich of diamonds.

At present there are several hypothesises explaining that the diamond of Africa is floated up from deep layer by impact of collision or what so ever since billions years ago, while metal is always on one way to Earth’s core without return, the diamond should be assumed to be on the move of the same wise as metal.

+The lunar rocks  and soil from Moon are stored at NASA; although experiment in lab with them is facing some difficult, I believe that they will be successful in proving that the relation between Moon and Earth is substantial in the wise that rock and soil are common somewhere. The sample rock, anyhow need to be collected from larger area on the Moon as well as Earth surface.

Conclusion

-The gravity map issued by NASA reveals a‘’junction’’ which can be like a disk that created by the collision between our Earth and another planet, that used to be named as Theia (Wikipedia for Moon).

-The gravity centre of each is still viewed as independent to the other; the junction disk is assumed as very hard to be beaten through; at least there is no evidence of breaking through.

-The Earth’s core is in the junction of the historic collision and approximately in the middle of Earth’s rotation axis. The junction is a product of ‘’metallurgy’’ during colliding, it must be at good heat conduction or heat transferring. For heat conduction, the evidence is that the temperature at either the Northern or Southern Pole never been chilled to -90⁰C while it might be -150⁰ C in space near the Earth; this fact is realized as the Poles are warmed up by the heat from Earth core.

Therefore, the Earth core is assumed as cooler than all sides around.

Hanoi 29^th June 2018

Nguyen Van Cuong

CHAPTER III-MOON'S TOTAL CHARGE

MOON's TOTAL CHARGE ON E.M

Every previous solution is solved with many assumptions. Anyhow, the results are meaningful; they do confirm that the semi-moons in every day of Moon are approximately the reverting points when the Moon’s hemisphere effect changes from negative to positive or vice versa. Hence the two semi-moons divide the Moon’s orbit into 2 different, opposite halves; the effect on one half is going against or contrarian the current E.M, while the other’s effect is supporting or backing the current E.M.

The radiation from the Sun is changing and the Moon charge therefore is changing accordingly; so there must be a sequent change in Moon’s effect on E.M, and this chapter is a discussion about the change in circumstance when Moon’s charge is changing and the total charge ‘’Qa+Qb’’ is no more a ‘’0’’.

We leave the influence of permeability of the media between Moon and Earth centre for another problem; just consider the Moon’s influence on Earth’s surface.

Let’s set and solve the following problems for Moon’s effects.

2/-The problem of Moon’s total charge for Houston:

Houston of U.S chosen as observer in this problem is rather close to Earth's equator and quite in the middle of Van Allen belt and plasma sphere; therefore, it is very hard to detect the Moon’s effect. Nonetheless, we keep working with that location as observer then sort out the issue, and following is the problem:

U.S observer location (30N, 95W), USA on 02^nd May 2017, Moon’s total charge Qm and other parameters:

-Illumination: 49.1%≈50%, this is to make sure that each of the two hemispheres is capable to eliminate the other, so the hemispheric effect can be null.

-Meridian passing: 19.31 hrs.

-Moon rise/set (excerpt from Moon’s time table):

-Distance: 377652 km from point P to Moon.

-Moon’s average radius: R=1738.1 km.

Question:

With Moon’s total charge Qm≠0, consider the amplitude of magnetism that directly induced by Moon on Houston at 19.00 hrs LT.

2/1-To understand more about the question (analysis):

-The meridian passing is on 19.31 hrs at 75.7⁰.

-From the Moon’s timetable: Moon rise/set is 12.38/(next day)02.20 hrs. or 12.38/26.20  (or Moon’s on sky within 822 minutes).

-The distance from A or B to a general P (on longitude of P) can be viewed as the same and depicted in the following figure:

  

-The given P is on land surface with eye height ‘’0’’, the horizon viewed from P is a plane with the straight line from Sunrise (A) to Sunset (B).

-Note that AP = PB, by Pythagorean theory AP=(AO²-OP²)^1/2 (Figure 4/IV), where OP=Re, the value of Earth average radius despite ‘’O’’ is not coinciding on Oe (Figure 5/IV).

-A circle with centre (O) is designed to meet AP=(AO²-OP²)^1/2 and distance from P to Moon 377652 km. This centre (O) may not be coinciding on the terrestrial geometric centre (Oe).

 2/2-Solution:

2/2/1-Step 1: to re-think about the real things.

Like any other electric current or moving charge, the Moon definitely contributes magnetism. Moon’s total charge has been assumed as neutral with 2 opposite hemispheres in previous problem.

But in reality the total charge of any object under solar wind can’t be neutral at all the time; it is actually changing, and is given a value Qm (positive) in this problem.

The questions are how magnificent is the magnetism induced by the lunar total charge? Which wise (north or south) does it direct to?

2/2/2-Step 2: Assumptions.

- P (30N, 95W) is assumed as the stance for observer with eye height ‘’0’’.

- Distance from P to A is the same as P to B or any position on the longitude of P to those 2 positions of Moon.

-The Moon is assumed as on ‘’no real move’’ throughout the time of calculation. By this condition, the Moon’s real move is not considered. Instead, we consider the Moon’s relative move or 12(+) hr. trace around the Earth.

-Moon’s total charge (Qm=Qa+Qb and Qm≠0) stays unchanged throughout this calculation.

-Distance: 377652 km is from P to the nearest position of the Moon. The distance from P to charge centre of Moon can be approximately added with ½ of Moon’s average radius 1/2*1738.1 km.

The distance d=378521 km is assumed as distance from observer to Moon’s charge centre.

3-Step 3: Solution with reliable accuracy.

-Recall Biot-Savart expression for a single charge. We add some more lines on the basic figure 12b/v to facilitate the solution.

-In the above figure:  The relative Moon’s trace (not the Moons’ orbit) in 12 hrs is demonstrated (from Moonrise till Moon set) in 2D coordinates, the larger half of the circle (from A to B through under part) is not viewed. Within this duration, the Earth completes almost a half of cycle around its own axis. We can imagine that a certain observer standing somewhere near either North or South Pole, with a restricted view she/he can observes the Moon rising like depicted in the figure (while the Moon after Moonset is not seen, and the Earth makes no move).

The problem becomes pure geometric on 2D coordinates; and the figure requires some additional assumptions and quantities:

+Moon is at a general position (M) on sky, its distance to P is ‘’r’’ which varies against time and reaches the given value of d at meridian passing;

+Moon is rising at A and setting at B;

+Moon is expecting to pass meridian at Z over P; where angle APM=angle APZ =90⁰ while Moon’s altitude reaches its max of that day 75.7⁰.

+Vector  ''v'' is perpendicular to the radius that drawn through Moon (OM);

+PD is perpendicular to OM and parallel to;

+The angles are named after each point and numbered: M1, M2 and O1,O2; or P1,P2,P3,P4;

+Parameters: d=378521*10³m (minimum distance from the given position P to Moon’s centre).

+The Moon’s angle (P2 or angle APM) is taken by a professional equipment (Sextant) (at the given time); (the ‘’altitude’’ of Moon or any object on sky is understood as the angle between the line from observer to the object and horizontal plane at observer. Therefore, a large gap between the two aforementioned concepts (altitude and P2) is found, and no chance for a mixing up or confusion). Nonetheless, we can’t deny that the Moon’s altitude depends on P2 and reaches max almost at the middle of Moon’s trace between A and B, therefore we may keep considering altitude in relation with Moon’s angle P2. Furthermore, by the laws of 3D trigonometric, we can calculate P2 from altitude or vice versa, but no discussion for that issue is encouraged in this book.

The following is a figure that depicts approximately the altitude of Moon and the Moon’s angle P2:

 Figure 9/V-Moon’s altitude

(At Moon’s meridian passing, the angle APM is obviously 90⁰ or ½ π while Moon’s altitude keeps varying (angle MPH is not always to be 90⁰ even if we measure it from Earth equator)).

Again, we refer to the basic expression of Biot-Savart:

With Biot-Savart expression, this problem is the normal geometric one in astrophysics, we can demonstrate every important parameter on 2D coordinates. The sensitive issue in this expression is the product of vector ''v''  and unit vector ''r'', a vector directs from this page toward reader, and its  non-directional value is:

v*Sin(M1)

The non-directional value of B is:

b/1- To seek for (r ):

+Consider the triangle MDP (Vector ''v'' and PD are extended for a better view with M’P’//MP):

 Cos(O2)=Cos(07.95⁰)=0.990389

+In the triangle MPO:

In the triangle ODP:

OD=Re*Cos(O2)= 6356.8*0.990389=6295.704795 km

And Sin (M1) = Sin(81.91⁰+07.95⁰)=0.99999

MD = MO-OD or MD=(Re+d)-Re*Cos(O2)=6356.8+378521-6356.8*0.990389=

=378582.0952048 Km

Thus, r=MD/Sin(M1)= 378582.0952048/0.99999 =378585.88106 km=

r=378585.88106*10³m.

b/2-Angle M1:

M1=P2+P3 (Vector v is parallel with PD, the rule for alternate angles is applied)

P2 is Sextant reading (or astronomy almanac), and the P3=O2. Thus:

Angle M1=P2+O2, with given P2=81.91⁰ and P3=07.95⁰   

 Angle M1=81.91⁰+07.95⁰=89.86⁰ = 89051’’

Final calculation:

-Replace the followings in the expression and calculate value of B(tc):

Radius r=378585.881*10³m

Speed v=21230.8834 m/s

Angle M1=89⁰51”, Sin(M1)=0.99997≈1.

As assumption of this problem, q=Qm Coulomb, every length is in meter (m), the unit of B is Tesla (T).

We have:

B(tc)=10^-7*21230.8834*378585.881^-2*10^-6*Qm=

=1.481287*10^-20*Qm Tesla

(The previous approximate calculation gives result: B(tc)=1.439480*10^-20*Qm)

Because every of (r,v,M1) is depending on observer’s position, therefore the above result cannot be used for everywhere. Nonetheless it is to confirm that the Moon exerts its static electric force on the

Earth and especially induces a magnetic field on it, the value can be calculated.

For instant, a reading on a Tesla meter is indicating that the Moon’s contribution at Houston (South U.S.A) reaches its max +5*10^-7 T (normally the meter reading about 2-3 hrs before meridian passing can indicate the average magnetic level); we can find Moon’s total charge as following:

B(tc)= 1.481287*10^-20*Qm=5*10^-7 Tesla

Qm=B(tc)/ 1.481287*10^-20=5*10^-7/ 1.481287*10^-20 =

=3.375443*10¹³ Coulombs.

Everywhere inside the loop, the magnetic vector is by right-hand law.

(Problem: (the following problem is designated for 60 N to make sure that Moon signal is detected without being totally barred by plasma sphere): 3.375443*10¹³ Coulombs is a value of Moon’s charge that we detected when semi-moon (illumination 50%) is passing meridian of a position ‘’P’’ (60N, 0.0). If in 36 hrs later, when Moon’s illumination of 40% we detect Moon’s charge to be ‘’0’’ at a certain position P’ (60N, 180.0). Calculate the Moon’s total charge components Qa & Qb? 

This theoretic problem is definitely inspired by many science students but not included in this book. After this problem, we find out more about the roles of 2 hemispheres of Moon on E.M).

Figure 13/V-Earth’s magnetic field amplitude (in mG)

The above is a real graph of magnetic amplitude observed by our correspondent at latitude 70 N on Atlantic, about new moon. The records are bold for 10 minutes before and after the Moon’s meridian passing.

CHAPTER II/p2-TWO OPPOSITE HEMISPHERES OF MOON

TWO OPPOSITE H.S OF MOON

Moon is the only satellite to Earth. In human life, the Moon merges in poet, music and children’s daily dream.

Since 50^th of last century, human being had succeeded in researching the Moon, U.S scientists even launched in there and planned to build an outpost on it, their plan is still going on.  Nonetheless we don’t discuss about their plan, but we do discuss about another issue: the charges on Moon and their influence to the Earth.

We denote magnetic compositions contributed by Moon’s hemisphere the Δ4, Δ5 and Δ6 where Δ4 is for negative hemi-sphere influence, Δ5 is for positive hemi sphere; and Δ6 is for total charge contribution.

The following is charged Moon that depicted by NASA:

The ion wind from Sun is striking on Moon’s surface and energizing the electrons on there, as NASA’s argument, the direct result is to electrify the shined part positively. The highly energized electrons flock to settle on the dark side of the Moon where they lost their energy. That’s how the Moon surface is such illustrated in the Image 1/v. In order to clarify the Moon’s hemisphere effects on Earth, we set and solve the following problem.

1-Problem 1: The lunar hemisphere effect.

Under sunlight, the Moon is a charged or an electrified object. Suppose that the nature of object is neutral and so the positive charge on shined side is balanced against the negative charge on the dark side.

Conclusion for solution 1:

The opposite hemi-spheres of Moon induce magnetism on P(Houston). As seen in the figure 6/v above, Qb is nearer to Earth; by the inverse law of electric force, Qb can influence more than Qa.

Moon is considered as stationary during the Earth is rotating, so Moon’s magnetic vector is parallel with Earth’s rotation axis and consists of X and D compositions.

Qb might be negative which induces a contribution that heads to North, against current Earth’s magnetism.

The result of calculation (6.241*10^-23*Qb Tesla) is not too tiny because Qb is still unknown and varying at all the time. If we give Qb or Qa a certain figure such as 10 6 Coulombs, we may have B(hs), but we don’t do it now as our discussion is still going on.