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.  

CHAPTER I/p2-VAN ALLEN BELT CONTRIBUTION

VAN ALLEN BELT

I-General about Van Allen belts:

I and many among my colleagues used to view Earth’s magnetic field as one that can only facilitate the magnetic compass to indicate the North magnetic direction and no more it can offer. But the Van Allen belt appears to change our mind; the Earth’s magnetic field is not very strong but good enough in facilitating the establishment of Van Allen belt which can stop any fast moving charged particle. The Van Allen belt, as matter of fact, plays an important role in protecting our planet.

(The following is almost prepared with close reference to SPACE & NASA’s documents and papers, which are open to everyone and nothing is from classified information.)

(By NASA)

1-What is Van Allen belt: a collection of charged particles, gathered by Earth’s magnetism, in a shape of belts around Earth.

2-The earliest discovery to the belts: 1958 by Explorer 1 (NASA’s Satellite).

-Belt’s structure: normally 2 belts, one is separated from the other by a large drain/gap; can split up to 3 belts or merged in one sometimes. The inner belt is almost close to Earth surface at minimum about 400 miles; while the outer belt is rather far and even stretches to 36000 miles.

The drain is discovered in 2012; a remarkable feature of the belt is that its drain can stop the ultra-fast electron in there. The scientists from NASA are still not determined about reason for the belt separation. Their hesitation is quite understandable because no mechanism of electric force or magnetic field is found linking to the wax & wane activity of the belts (which is in conjunction with Sun’s activity). The scientists likely ascribe to the particles from space or the Sun.

The unidentified particles or quarks might be gathered among the belts and create an unknown field between them, the field makes the belts split-up wide.

-How the belts work: They can wax and wane in response to incoming energy from the Sun, sometime swelling up enough to expose satellites in low-Earth orbit to damaging radiation.

-Undiscovered about the belts: With some satellites working on space at all the time; the more research to the belts are carried out, the more expect to be discovered; even the unidentified particle or quark.

-Plasmasphere: A companion to the belts is a giant plasma sphere which can be assumed as twinned to the belts. The plasma sphere is cloud of relative cool and charged particles that fills the outermost region of Earth’s atmosphere. That sphere begins approximately at 400-600 miles (rather close to Earth surface) and extends partially to outer Van Allen belt.

PROBLEM

If total charge of the Van Allen belts and plasma sphere is equivalent to a single band of +10⁶ Coulombs around Earth equator at a distance 1500 km from Earth surface, and evenly distributed on it.

How is the magnetism induced by the Band to the equator (Earth)? 

(V-A 01)

   

    

Where unit of ‘’Q’’ is in Coulomb and unit of ‘’d’’ is km.

Replace Q=10⁶ Coulombs, d=1500 km (as assumed) into the above formula:

  Δ(3)= -8.4169*10^-23 *666.666667=10^-20*5.611267 Tesla

(Note: the Belt is assumed as on a plane perpendicular to Earth’s rotating axis, the assumption is not verified and the reality definitely is not be so. Therefore, in order to calculate for a better accuracy, the belt as well as the plasma sphere can be represented by many bands (or hoops) in a larger problem).

Conclusion:

If Van Allen belt and plasma sphere are equivalent to a band of Q Coulombs (the Band) around the Earth on plane of Earth’s equator. The Band is to induce magnetism on Earth’s equator; its additional intensity is calculated as (V-A 01):

 

In reality, Van Allen belts and plasma sphere are not only changing in shape and size, but also in intensity. With such a complicated change, the above mentioned ‘’Δ3’’ demonstrates nothing but ‘’yes’’ to the question ‘’influence?’’. Furthermore, as noted above, the larger problem is required to estimate the additional magnetic intensity with better accuracy.

PART II

EXTERNALITIES OR ASTRONOMY ON E.M