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.