Gravity Control by means of Electromagnetic Field         through Gas or Plasma at Ultra-Low Pressure                     ...
2I. INTRODUCTION       It will be shown that the local                           In general, the momentum variationgravity...
3       Equation (6) can be rewritten in              Equation (9) shows that ω κ r = v . Thus,the following form         ...
4electrical    conductor     mean            has                            j lampρ << 1 Kg.m −3                and σ >> 1...
5plates, placed at distance d , is equal to                         mg1(Hg plasma)the outer diameter (max * ) of the      ...
6                                                                  V0 = 1.5V and the frequency decreasedg3 ⎧      ⎡       ...
7      Note that the gravity acceleration                sufficient to encapsulate the GCC withabove the air becomes negat...
8Thus, the activity [8] of the sample is                  At temperature of 300K, the air                                 ...
9to Vrms ≅ 23.5V the voltage necessary to              Thus, the local inertia is just thereach χ air ≅ −1 .              ...
10       It was shown that, when the                                      Fgj = M g (imaginary) g ′j =gravitational mass o...
11       Now consider the GCCs presented                        rotor in order to become negative thein Fig. 8 (a). Note t...
12                                                                    (Novel superconducting magnets areThen Eq. (40) give...
13and furthermore reveals that the inertial               spacecraft is subjected to a gravityeffects upon a body can be s...
14Thrusters can also provide the horizontal                                        The gravitational force dF12 thatdispla...
15                                            −8If mi1 ≅ mi 2 = ρ air V1 ≅ ρ air V2 ≅ 10 kg , and            aspect of the...
16mgV = p 1−V c and m′ V ′ = p 1 − V ′ c into                  2   2                           2   2            Obviously ...
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
Fran de aquino   gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p
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Fran de aquino gravity control by means of electromagnetic field through gas or plasma at ultra-low pressure, 2009, 44p

  1. 1. Gravity Control by means of Electromagnetic Field through Gas or Plasma at Ultra-Low Pressure Fran De Aquino Maranhao State University, Physics Department, S.Luis/MA, Brazil. Copyright © 2007-2009 by Fran De Aquino. All Rights ReservedIt is shown that the gravity acceleration just above a chamber filled with gas or plasma at ultra-lowpressure can be strongly reduced by applying an Extra Low-Frequency (ELF) electromagnetic fieldacross the gas or the plasma. This Gravitational Shielding Effect is related to recent discovery ofquantum correlation between gravitational mass and inertial mass. According to the theory sampleshung above the gas or the plasma should exhibit a weight decrease when the frequency of theelectromagnetic field is decreased or when the intensity of the electromagnetic field is increased. ThisGravitational Shielding Effect is unprecedented in the literature and can not be understood in theframework of the General Relativity. From the technical point of view, there are several applications forthis discovery; possibly it will change the paradigms of energy generation, transportation andtelecommunications.Key words: Phenomenology of quantum gravity, Experimental Tests of Gravitational Theories,Vacuum Chambers, Plasmas devices. PACs: 04.60.Bc, 04.80.Cc, 07.30.Kf, 52.75.-d.CONTENTSI. INTRODUCTION 02II. THEORY 02Gravity Control Cells (GCC) 07III. CONSEQUENCES 09Gravitational Motor using GCC 11Gravitational Spacecraft 12Decreasing of inertial forces on the Gravitational Spacecraft 13Gravity Control inside the Gravitational Spacecraft 13Gravitational Thrusters 14Artificial Atmosphere surrounds the Gravitational Spacecraft. 15Gravitational Lifter 15High Power Electromagnetic Bomb (A new type of E-bomb). 16Gravitational Press of Ultra-High Pressure 16Generation and Detection of Gravitational Radiation 17Quantum Gravitational Antennas. Quantum Transceivers 18Instantaneous Interstellar Communications 18Wireless Electric Power Transmission, by using Quantum Gravitational Antennas. 18Method and Device using GCCs for obtaining images of Imaginary Bodies 19IV. CONCLUSION 20APPENDIX 40References 44
  2. 2. 2I. INTRODUCTION It will be shown that the local In general, the momentum variationgravity acceleration can be controlled by Δp is expressed by Δp = FΔt where Fmeans of a device called Gravity Control is the applied force during a timeCell (GCC) which is basically a recipient interval Δt . Note that there is nofilled with gas or plasma where is applied restriction concerning the nature of thean electromagnetic field. According to force F , i.e., it can be mechanical,the theory samples hung above the gas electromagnetic, etc.or plasma should exhibit a weight For example, we can look on thedecrease when the frequency of the momentum variation Δp as due toelectromagnetic field is decreased or absorption or emission of electromagneticwhen the intensity of the electromagnetic energy by the particle.field is increased. The electrical In the case of radiation, Δp can beconductivity and the density of the gas or obtained as follows: It is known that theplasma are also highly relevant in this radiation pressure, dP , upon an areaprocess. dA = dxdy of a volume d V = dxdydz of With a GCC it is possible toconvert the gravitational energy into a particle ( the incident radiation normalrotational mechanical energy by means to the surface dA )is equal to theof the Gravitational Motor. In addition, a energy dU absorbed per unit volumenew concept of spacecraft (the (dU dV ) .i.e.,Gravitational Spacecraft) and aerospace dP = dU = dU = dU (2)flight is presented here based on the dV dxdydz dAdzpossibility of gravity control. We will alsosee that the gravity control will be very Substitution of dz = vdt ( v is the speedimportant to Telecommunication. of radiation) into the equation above givesII. THEORY dU (dU dAdt ) dD It was shown [1] that the relativistic dP = = = (3) dV v vgravitational mass M g = m g 1 − V 2 c2and the relativistic inertial mass Since dPdA = dF we can write:M i = mi 0 1 − V 2 c 2 are quantized, and dFdt= dU (4) vgiven by M g = n g mi 0(min ) , 2 M i = ni2 mi 0(min ) However we know that dF = dp dt , thenwhere n g and ni are respectively, the dp = dU (5)gravitational quantum number and the vinertial quantum number ; From this equation it follows that mi 0(min ) = ±3.9 × 10 kg is the elementary −73 U ⎛c⎞ U Δp = ⎜ ⎟ = nrquantum of inertial mass. The masses v ⎝c⎠ cm g and mi 0 are correlated by means of Substitution into Eq. (1) yieldsthe following expression: ⎧ ⎡ 2 ⎤⎫ ⎪ ⎛ U ⎞ mg = ⎨1 − 2⎢ 1+ ⎜ ⎟ − 1⎥⎪mi0 (6) ⎡ ⎤ ⎢ ⎜ m c 2 nr ⎟ ⎥⎬ 2 mg = mi0 − 2⎢ 1 + ⎛ Δp ⎞ − 1⎥mi0 . ⎜ ⎟ (1) ⎪ ⎩ ⎣ ⎝ i0 ⎠ ⎪ ⎦⎭ ⎢ ⎜m c⎟ ⎥ ⎝ i ⎠ ⎣ ⎦ Where U , is the electromagnetic energyWhere Δp is the momentum variation on absorbed by the particle; nr is the indexthe particle and mi 0 is the inertial mass of refraction.at rest.
  3. 3. 3 Equation (6) can be rewritten in Equation (9) shows that ω κ r = v . Thus,the following form E B = ω k r = v , i.e., E = vB = vμH . ⎧ ⎡ 2 ⎤⎫ ⎢ 1 + ⎛ W n ⎞ − 1⎥⎪m ⎪ Then, Eq. (8) can be rewritten in the mg = ⎨1 − 2 ⎢ ⎜ ⎟ ⎜ ρ c2 r ⎟ ⎥⎬ i 0 (7) following form: ⎪ ⎩ ⎢ ⎣ ⎝ ⎠ ⎥⎪ ⎦⎭ W = 1 (ε v2μ)μH 2 + 1 μH 2 2 2 (12)Where W = U V is the density of For σ << ωε , Eq. (9) reduces toelectromagnetic energy and ρ = mi 0 V c v=is the density of inertial mass. ε r μr The Eq. (7) is the expression of the Then, Eq. (12) givesquantum correlation between the ⎡ ⎛ c2 ⎞ ⎤ 2 1 2gravitational mass and the inertial mass 2 ⎜ ε μ ⎟μ⎥μH + 2 μH = μH W = 1 ⎢ε ⎜ ⎟ 2 (13) ⎢ ⎝ r r⎠ ⎥ ⎣ ⎦as a function of the density ofelectromagnetic energy. This is also the This equation can be rewritten in theexpression of correlation between following forms: B2gravitation and electromagnetism. W = (14) The density of electromagnetic μenergy in an electromagnetic field can be ordeduced from Maxwell’s equations [2] W = ε E2 (15)and has the following expression W = 1 ε E 2 + 1 μH 2 (8) For σ >> ωε , Eq. (9) gives 2 2 2ωIt is known that B = μH , E B = ω k r [3] v= (16 ) μσand dz ω Then, from Eq. (12) we getv= = = c (9) ⎡ ⎛ 2ω ⎞ ⎤ ⎛ ωε ⎞ dt κ r ε r μr ⎛ W = 1 ⎢ε⎜ ⎟μ⎥μH2 + 1 μH2 = ⎜ ⎟μH2 + 1 μH2 ≅ ⎜ 1 + (σ ωε ) + 1⎞⎟ ⎜ μσ ⎟ 2 ⎝σ ⎠ 2 2 2 2 ⎝ ⎠ ⎣⎝ ⎠ ⎦Where kr is the real part of the ≅ 1 μH2 2 (17) rpropagation vector k (also called phase Since E = vB = vμH , we can rewrite (17) r in the following forms:constant [4]); k = k = k r + iki ; ε , μ and σ, B2are the electromagnetic characteristics of W ≅ (18) 2μthe medium in which the incident (oremitted) radiation is propagating or ⎛σ ⎞ 2( ε = ε r ε 0 where ε r is the relative W ≅⎜ ⎟E (19 ) ⎝ 4ω ⎠dielectric permittivity and ε0 = 8.854×10−12F/ m By comparing equations (14) (15) (18); μ = μrμ0 where μ r is the relative and (19) we see that Eq. (19) shows thatmagnetic permeability and μ0 = 4π ×10−7 H / m; the better way to obtain a strong value of W in practice is by applying an Extraσ is the electrical conductivity). It is Low-Frequency (ELF) electric fieldknown that for free-space σ = 0 andε r = μ r = 1 then Eq. (9) gives (w = 2πf << 1Hz ) through a mean with v=c (10) high electrical conductivity. Substitution of Eq. (19) into Eq. From (9) we see that the index of (7), givesrefraction nr = c v will be given by ⎧ ⎡ 3 ⎤⎫ ⎪ ⎢ 1 + μ ⎛ σ ⎞ E − 1⎥⎪mi0 4 c εμ = r r ⎛ 1 + (σ ωε) + 1⎞ (11) mg = ⎨1 − 2 ⎜ ⎟ ⎬ (20) nr = ⎜ ⎟ 4c 2 ⎜ 4πf ⎟ ρ 2 ⎥⎪ 2 ⎪ ⎢ ⎝ ⎠ v 2 ⎝ ⎠ ⎩ ⎣ ⎦⎭ This equation shows clearly that if an
  4. 4. 4electrical conductor mean has j lampρ << 1 Kg.m −3 and σ >> 1 , then it is σ Hg plasma = = 3.419 S .m −1 (23) E lamppossible obtain strong changes in its Substitution of (22) and (23) into (20)gravitational mass, with a relatively small yieldsELF electric field. An electrical conductormean with ρ << 1 Kg.m−3 is obviously a mg(Hg plasma ⎧ ⎡ ) ⎪ E 4 ⎤⎫ ⎪ = ⎨1 − 2 ⎢ 1 + 1.909×10−17 −1⎥⎬ (24) mi(Hg plasma ⎪ ⎢ ) f 3 ⎥⎪plasma. ⎩ ⎣ ⎦⎭ There is a very simple way to test Thus, if an Extra Low-Frequency electricEq. (20). It is known that inside a field E ELF with the followingfluorescent lamp lit there is low-pressureMercury plasma. Consider a 20W characteristics: E ELF ≈ 100V .m −1 andT-12 fluorescent lamp (80044– f < 1mHZ is applied through theF20T12/C50/ECO GE, Ecolux® T12), Mercury plasma then a strong decreasewhose characteristics and dimensions in the gravitational mass of the Hgare well-known [5]. At around plasma will be produced.T ≅ 318.15 K , an optimum mercury 0 It was shown [1] that there is anvapor pressure of P = 6 ×10−3Torr= 0.8N.m−2 additional effect of gravitational shieldingis obtained, which is required for produced by a substance under thesemaintenance of high luminous efficacy conditions. Above the substance thethroughout life. Under these conditions, gravity acceleration g 1 is reduced at thethe mass density of the Hg plasma can same ratio χ = m g mi 0 , i.e., g1 = χ g ,be calculated by means of the well- ( g is the gravity acceleration under theknown Equation of State substance). Therefore, due to the ρ= PM 0 (21) gravitational shielding effect produced by ZRT the decrease of m g (Hg plasma ) in the regionWhere M 0 = 0.2006 kg.mol −1 is the where the ELF electric field E ELF ismolecular mass of the Hg; Z ≅ 1 is thecompressibility factor for the Hg plasma; applied, the gravity acceleration just R = 8.314 joule.mol −1 . 0 K −1 is the gases above this region will be given by m g (Hg plasma)universal constant. Thus we get g1 = χ (Hg plasma) g = g= ρ Hg plasma ≅ 6.067 × 10 −5 kg.m −3 (22) mi (Hg plasma)The electrical conductivity of the Hg ⎧ ⎪ ⎡ E4 ⎤⎫ ⎪plasma can be deduced from the = ⎨1 − 2⎢ 1 + 1.909 × 10 −17 ELF − 1⎥ ⎬ g (25) r r ⎪ ⎢ 3 f ELF ⎥⎪continuum form of Ohms Law j = σE , ⎩ ⎣ ⎦⎭since the operating current through the The trajectories of thelamp and the current density are well- electrons/ions through the lamp areknown and respectively given by determined by the electric field E lamp alongi = 0.35A [5] and jlamp = i S = i π φint , where 4 2 the lamp. If the ELF electric field acrossφint = 36.1mm is the inner diameter of the the lamp E ELF is much greater than E lamp ,lamp. The voltage drop across the the current through the lamp can beelectrodes of the lamp is 57V [5] and the interrupted. However, if EELF <<Elamp, thesedistance between them l = 570mm . Then trajectories will be only slightly modified.the electrical field along the lamp E lamp is Since here Elamp = 100 V .m−1 , then we cangiven by Elamp = 57V 0.570m = 100 V .m −1 . arbitrarily choose E ELF ≅ 33 V . m −1 . This maxThus, we have means that the maximum voltage drop, which can be applied across the metallic
  5. 5. 5plates, placed at distance d , is equal to mg1(Hg plasma)the outer diameter (max * ) of the χ1(Hg plasma) = = max mi1(Hg plasma)bulb φlamp of the 20W T-12 Fluorescent ⎧ ⎡ E4 ⎤⎫lamp, is given by ⎪ ⎢ 1 + 1.909×10−17 ELF(1) − 1⎥⎪ Vmax = E ELF φlamp ≅ 1.5 V max max = ⎨1 − 2 ⎬ (27) ⎪ ⎢ f ELF(1) ⎥⎪ 3 ⎩ ⎣ ⎦⎭Since φlamp = 40.3mm [5]. max Then, above the second lamp, the Substitution of EELF ≅ 33 V.m−1 into max gravity acceleration becomes(25) yields r r r mg (Hg plasma) g 2 = χ 2(Hg plasma) g1 = χ 2(Hg plasma) χ1(Hg plasma) g (28)g1 = χ (Hg plasma) g = g= where mi (Hg plasma) mg 2(Hg plasma) ⎧ χ 2(Hg plasma) = = ⎪ ⎡ 2.264× 10−11 ⎤⎫ ⎪ = ⎨1 − 2⎢ 1 + 3 − 1⎥⎬g (26) mi 2(Hg plasma) ⎪ ⎩ ⎢ ⎣ f ELF ⎥⎪ ⎦⎭ ⎧ ⎡ 4 ⎤⎫ ⎪ E ELF(2 ) ⎪Note that, for f < 1mHz = 10 −3 Hz , the = ⎨1 − 2⎢ 1 + 1.909 × 10 −17 − 1⎥⎬ (29) ⎪ ⎢ 3 f ELF(2) ⎥⎪gravity acceleration can be strongly ⎩ ⎣ ⎦⎭reduced. These conclusions show that Then, results g2 ⎧ ⎡ EELF(1) ⎤⎫the ELF Voltage Source of the set-up 4 ⎪ ⎪shown in Fig.1 should have the following = ⎨1 − 2⎢ 1 + 1.909×10−17 3 − 1⎥⎬ ×characteristics: g ⎪ ⎢ f ELF(1) ⎥⎪ ⎩ ⎣ ⎦⎭ - Voltage range: 0 – 1.5 V ⎧ ⎪ ⎡ EELF(2) ⎤⎫ 4 ⎪ - Frequency range: 10-4Hz – 10-3Hz × ⎨1 − 2⎢ 1 + 1.909×10−17 − 1⎥⎬ (30) ⎪ ⎢ f ELF(2) ⎥⎪ 3 ⎩ ⎣ ⎦⎭ In the experimental arrangementshown in Fig.1, an ELF electric field with From Eq. (28), we then conclude that ifintensity E ELF = V d crosses the χ1(Hg plasma ) < 0 and also χ 2(Hg plasma ) < 0 ,fluorescent lamp; V is the voltage drop then g 2 will have the same directionacross the metallic plates of the of g . This way it is possible to intensifycapacitor and d = φlamp = 40.3mm . max several times the gravity in the directionWhen the ELF electric field is applied, r of g . On the other hand, if χ1(Hg plasma ) < 0the gravity acceleration just above the rlamp (inside the dotted box) decreases and χ 2(Hg plasma ) > 0 the direction of g 2 willaccording to (25) and the changes can r be contrary to direction of g . In this casebe measured by means of the system will be possible to intensify andbalance/sphere presented on the top of r r become g 2 repulsive in respect to g .Figure 1. In Fig. 2 is presented an If we put a lamp above the secondexperimental arrangement with two lamp, the gravity acceleration above thefluorescent lamps in order to test the third lamp becomes r rgravity acceleration above the second g 3 = χ 3(Hg plasma) g 2 = rlamp. Since gravity acceleration above = χ 3(Hg plasma) χ 2(Hg plasma) χ1(Hg plasma) g (31)the first lamp is given by r r or g1 = χ1(Hg plasma ) g , where* After heating.
  6. 6. 6 V0 = 1.5V and the frequency decreasedg3 ⎧ ⎡ EELF(1) ⎤⎫ 4 ⎪ ⎪ = ⎨1 − 2⎢ 1 + 1.909×10−17 3 − 1⎥⎬ × in the above mentioned sequence.g ⎪ ⎢ f ELF(1) ⎥⎪ ⎩ ⎣ ⎦⎭ Table1 presents the theoretical ⎧ values for g 1 and g 2 , calculated ⎪ ⎡ EELF(2) ⎤⎫ 4 ⎪ × ⎨1 − 2⎢ 1 + 1.909×10−17 − 1⎥⎬ × respectively by means of (25) and ⎪ ⎢ f ELF(2) ⎥⎪ 3 (30).They are also plotted on Figures 5, ⎩ ⎣ ⎦⎭ 6 and 7 as a function of the ⎧ ⎡ EELF(3) ⎤⎫ 4 ⎪ ⎪ frequency f ELF . × ⎨1 − 2⎢ 1 + 1.909×10−17 3 − 1⎥⎬ (32) Now consider a chamber filled ⎪ ⎢ f ELF(3) ⎥⎪ ⎩ ⎣ ⎦⎭ with Air at 3 × 10 −12 torr and 300K asIf f ELF (1) = f ELF (2 ) = f ELF (3 ) = f and shown in Figure 8 (a). Under these circumstances, the mass density of the E ELF (1) = E ELF (2 ) = E ELF (3 ) = V φ = air inside the chamber, according to Eq. (21) is ρ air ≅ 4.94 × 10 −15 kg.m −3 . = V0 sin ωt 40.3mm = If the frequency of the magnetic = 24.814V0 sin 2πft. field, B , through the air is f = 60 Hz thenThen, for t = T 4 we get ωε = 2πfε ≅ 3 × 10 −9 S / m . Assuming that E ELF (1) = E ELF (2 ) = E ELF (3 ) = 24.814V0 . the electric conductivity of the air insideThus, Eq. (32) gives the chamber, σ (air ) is much less than ωε , 3 g3 ⎧⎪ ⎡ V 4 ⎤⎫ ⎪ i.e., σ (air ) << ωε (The atmospheric air = ⎨1 − 2⎢ 1 + 7.237×10−12 03 − 1⎥⎬ (33) conductivity is of the order of g ⎪ ⎩ ⎢ ⎣ f ⎥⎪ ⎦⎭ 2 − 100 × 10 −15 S .m −1 [6, 7]) then we canFor V0 = 1.5V and f = 0.2mHz rewritten the Eq. (11) as follows(t = T 4 = 1250s = 20.83min) the gravity racceleration g 3 above the third lamp will nr(air) ≅ ε r μr ≅ 1 (34)be given by r r g 3 = −5.126 g From Eqs. (7), (14) and (34) we thusAbove the second lamp, the gravity obtainacceleration given by (30), is ⎧ ⎡ ⎤⎫ r r 2 ⎪ ⎢ 1+ ⎛ B ⎞ ⎥⎪ 2 g 2 = +2.972g . ⎜ ⎟ mg(air) = ⎨1 − 2 ⎢ ⎜ μ ρ c2 nr(air) ⎟ − 1⎥⎬mi(air) = ⎪ ⎝ air air ⎠ ⎦⎪According to (27) the gravity acceleration ⎩ ⎣ ⎭ { [ ]}above the first lamp is r r g1 = -1,724g = 1 − 2 1 + 3.2 ×106 B4 −1 mi(air) (35)Note that, by this process an racceleration g can be increased several Therefore, due to the gravitational rtimes in the direction of g or in the shielding effect produced by theopposite direction. decreasing of m g (air ) , the gravity In the experiment proposed in Fig. acceleration above the air inside the1, we can start with ELF voltage chamber will be given bysinusoidal wave of amplitude V0 = 1.0V m g (air )and frequency 1mHz . Next, the frequency g ′ = χ air g = g = m i (air ) { [ ]}will be progressively decreased downto 0.8mHz , 0.6mHz , 0.4mHz and = 1 − 2 1 + 3 . 2 × 10 6 B 4 − 1 g0.2mHz . Afterwards, the amplitude of thevoltage wave must be increased to
  7. 7. 7 Note that the gravity acceleration sufficient to encapsulate the GCC withabove the air becomes negative epoxy. The alpha particles generated byfor B > 2.5 × 10 −2 T . the americium ionize the oxygen and For B = 0.1T the gravity nitrogen atoms of the air in the ionizationacceleration above the air becomes chamber (See Fig. 3(c)) increasing the electrical conductivity of the air inside the g ′ ≅ −32.8 g chamber. The high-speed alpha particles hit molecules in the air and knock offTherefore the ultra-low pressure air electrons to form ions, according to theinside the chamber, such as the Hg following expressionsplasma inside the fluorescent lamp,works like a Gravitational Shield that in O2 + H e+ + → O2 + e − + H e+ + +practice, may be used to build Gravity N 2 + H e+ + → N 2 + e − + H e+ + +Control Cells (GCC) for several practical It is known that the electricalapplications. conductivity is proportional to both the Consider for example the GCCs of concentration and the mobility of the ionsPlasma presented in Fig.3. The and the free electrons, and is expressedionization of the plasma can be made of byseveral manners. For example, by σ = ρ e μe + ρi μimeans of an electric field between theelectrodes (Fig. 3(a)) or by means of a Where ρ e and ρ i express respectivelyRF signal (Fig. 3(b)). In the first case the ( ) the concentrations C m 3 of electronsELF electric field and the ionizing electric and ions; μ e and μ i are respectively thefield can be the same. Figure 3(c) shows a GCC filled mobilities of the electrons and the ions.with air (at ambient temperature and 1 In order to calculate the electricalatm) strongly ionized by means of alpha conductivity of the air inside theparticles emitted from 36 radioactive ions ionization chamber, we first need tosources (a very small quantity of calculate the concentrations ρ e and ρ i .Americium 241 † ). The radioactive We start calculating the disintegrationelement Americium has a half-life of 432 constant, λ , for the Am 241 :years, and emits alpha particles and low 0.693 0.693 λ= = = 5.1 × 10 −11 s −1energy gamma rays (≈ 60 KeV ) . In order T2 1 ( 432 3.15 × 10 s 7 )to shield the alpha particles and gamma 1 Where T = 432 years is the half-life of 2rays emitted from the Americium 241 it is the Am 241.† One kmole of an isotope has mass The radioactive element Americium (Am-241) is equal to atomic mass of the isotopewidely used in ionization smoke detectors. Thistype of smoke detector is more common because expressed in kilograms. Therefore, 1g ofit is inexpensive and better at detecting the Am 241 hassmaller amounts of smoke produced by flaming 10 −3 kgfires. Inside an ionization detector there is a small = 4.15 × 10 −6 kmolesamount (perhaps 1/5000th of a gram) of 241 kg kmoleamericium-241. The Americium is present in One kmole of any isotope contains theoxide form (AmO2) in the detector. The cost ofthe AmO2 is US$ 1,500 per gram. The amount of Avogadro’s number of atoms. Thereforeradiation in a smoke detector is extremely small. 1g of Am 241 hasIt is also predominantly alpha radiation. Alpharadiation cannot penetrate a sheet of paper, andit is blocked by several centimeters of air. The N = 4.15 × 10−6 kmoles×americium in the smoke detector could only pose × 6.025 × 1026 atoms kmole = 2.50 × 1021 atomsa danger if inhaled.
  8. 8. 8Thus, the activity [8] of the sample is At temperature of 300K, the air density inside the GCC, is R = λN = 1.3 × 1011 disintegrations/s. ρ air = 1.1452kg.m . Thus, for d = 2cm , −3 σ air ≈ 10 3 S .m −1 and f = 60 Hz Eq. (20)However, we will use 36 ionization givessources each one with 1/5000th of a mg (air)gram of Am 241. Therefore we will only χ air = =use 7.2 × 10 −3 g of Am 241. Thus, R mi(air)reduces to: ⎧ ⎡ ⎤⎫ ⎢ 1 + μ ⎛ σ air ⎞ Vrms − 1⎥⎪ = 3 ⎪ 4 = ⎨1 − 2 ⎜ ⎟ ⎬ R = λN ≅ 10 9 disintegrations/s ⎪ ⎢ 4c 2 ⎜ 4πf ⎟ d 4 ρair ⎥⎪ ⎝ ⎠ 2 ⎩ ⎣ ⎦⎭This means that at one second, about { [ = 1 − 2 1 + 3.10×10−16Vrms − 1 4 ]}10 9 α particles hit molecules in the air Note that, for Vrms ≅ 7.96KV , we obtain:and knock off electrons to form ions χ (air ) ≅ 0 . Therefore, if the voltages + +O2 and N 2 inside the ionization chamber. range of this GCC is: 0 − 10KV then it isAssuming that each alpha particle yields possible to reach χ air ≅ −1 whenone ion at each 1 10 9 second then the Vrms ≅ 10KV .total number of ions produced in onesecond will be Ni ≅ 1018 ions. This It is interesting to note that σ air cancorresponds to an ions concentration be strongly increased by increasing the ρ i = eN i V ≈ 0.1 V (C m 3 ) amount of Am 241. For example, by using 0.1g of Am 241 the value of RWhere V is the volume of the ionization increases to:chamber. Obviously, the concentration ofelectrons will be the same, i.e., ρ e = ρ i . R = λN ≅ 1010 disintegrations/sFor d = 2cm and φ = 20cm (See Fig.3(c))we obtain This means Ni ≅ 1020 ions that yieldV = π (0.20) (2 × 10 −2 ) = 6.28 × 10 −4 m 3 The ρ i = eN i V ≈ 10 V (C m 3 ) 2 4n we get: Then, by reducing, d and φ respectively, to 5mm and to 11.5cm, the ρ e = ρ i ≈ 10 2 C m 3 volume of the ionization chamber reduces to:This corresponds to the minimum V = π (0 .115 ) (5 × 10 −3 ) = 5 .19 × 10 −5 m 3 2 4concentration level in the case ofconducting materials. For these Consequently, we get:materials, at temperature of 300K, themobilities μ e and μ i vary from 10 up ρ e = ρ i ≈ 10 5 C m 3to 100 m 2V −1 s −1 [9]. Then we can assume Assuming that μ e = μi ≈ 10 m 2V −1 s −1 ,that μe = μi ≈ 10 m2V −1s −1 . (minimum then the electrical conductivity of the airmobility level for conducting materials). inside the ionization chamber becomesUnder these conditions, the electricalconductivity of the air inside theionization chamber is σ air = ρ e μ e + ρ i μ i ≈ 10 6 S .m −1 σ air = ρ e μ e + ρ i μ i ≈ 10 3 S .m −1 This reduces for Vrms ≅ 18.8V the voltage necessary to yield χ(air) ≅ 0 and reduces
  9. 9. 9to Vrms ≅ 23.5V the voltage necessary to Thus, the local inertia is just thereach χ air ≅ −1 . gravitational influence of the rest of matter existing in the Universe. If the outer surface of a metallic Consequently, if we reduce thesphere with radius a is covered with a gravitational interactions between aradioactive element (for example Am spacecraft and the rest of the Universe,241), then the electrical conductivity of then the inertial properties of thethe air (very close to the sphere) can be spacecraft will be also reduced. Thisstrongly increased (for example up effect leads to a new concept ofto σ air ≅ 10 6 s.m −1 ). By applying a low- spacecraft and space flight.frequency electrical potential Vrms to the Since χ air is given bysphere, in order to produce an electricfield E rms starting from the outer surface ⎧ ⎡ 3 ⎤⎫ mg(air) ⎪ ⎢ μ0 ⎛ σair ⎞ Vrms ⎥⎪ 4of the sphere, then very close to the χair = = ⎨1 − 2 1 + 2 ⎜ ⎟ 4 2 −1 ⎬sphere the low-frequency electromagnetic mi0(air) ⎪ ⎢ 4c ⎜ 4πf ⎟ a ρair ⎥⎪ ⎝ ⎠ ⎩ ⎣ ⎦⎭field is E rms = Vrms a , and according toEq. (20), the gravitational mass of the air Then, for σ air ≅ 106 s.m −1 , f = 6Hz , a = 5m,in this region expressed by ρair ≅ 1Kg.m−3 and Vrms = 3.35 KV we get ⎧ ⎡ μ0 ⎛σair ⎞ Vrms 3 ⎤⎫ ⎪ ⎢ ⎪ 4 mg(air) = ⎨1− 2 1+ 2 ⎜ ⎟ 4 2 −1⎥⎬mi0(air) , ⎜ 4πf ⎟ a ρ ⎪ ⎢ 4c ⎝ ⎠ ⎥⎪ χ air ≅ 0 ⎩ ⎣ air ⎦⎭can be easily reduced, making possible Under these conditions, the gravitationalto produce a controlled Gravitational forces upon the spacecraft becomeShielding (similar to a GCC) surround approximately nulls and consequently,the sphere. the spacecraft practically loses its inertial This becomes possible to build a properties.spacecraft to work with a gravitational Out of the terrestrial atmosphere,shielding as shown in Fig. 4. the gravity acceleration upon the The gravity accelerations on the spacecraft is negligible and therefore thespacecraft (due to the rest of the gravitational shielding is not necessary.Universe. See Fig.4) is given by However, if the spacecraft is in the outer space and we want to use the g i′ = χ air g i i = 1, 2, 3 … n gravitational shielding then, χ air must be replaced by χ vac whereWhere χ air = m g (air ) mi 0 (air ) . Thus, thegravitational forces acting on the ⎧ ⎡ 3 μ0 ⎛ σvac ⎞ Vrms ⎤⎫ ⎪ ⎢ ⎥⎪ 4spacecraft are given by mg(vac) χvac = = ⎨1− 2 1+ 2 ⎜ ⎜ 4πf ⎟ a4 ρ 2 −1⎥⎬ ⎟ Fis = M g g i′ = M g (χ air g i ) mi0(vac) ⎪ ⎢ ⎩ ⎣ 4c ⎝ ⎠ vac ⎦⎪ ⎭By reducing the value of χ air , these The electrical conductivity of the ionized outer space (very close to theforces can be reduced. spacecraft) is small; however, its density According to the Mach’s principle; ( is remarkably small << 10 −16 Kg.m −3 , in ) “The local inertial forces are such a manner that the smaller value ofdetermined by the gravitational the factor σ vac ρ vac can be easily 3 2interactions of the local system with the compensated by the increase of Vrms .distribution of the cosmic masses”.
  10. 10. 10 It was shown that, when the Fgj = M g (imaginary) g ′j =gravitational mass of a particle isreduced to the gravitational mass ( ) = M g (imaginary) − χGmgj (imaginary) r j2 =ranging between + 0.159M i to ( ) = M g i − χGmgj i r j2 = + χGM g mgj r j2 . − 0.159M i , it becomes imaginary [1], i.e., Note that these forces are real. Remindthe gravitational and the inertial masses that, the Mach’s principle says that theof the particle become imaginary. inertial effects upon a particle areConsequently, the particle disappears consequence of the gravitationalfrom our ordinary space-time. However, interaction of the particle with the rest ofthe factor χ = M g (imaginary ) M i (imaginary ) the Universe. Then we can conclude that the inertial forces upon an imaginaryremains real because spacecraft are also real. Consequently, it M g (imaginary ) M gi Mg can travel in the imaginary space-time χ = = = = real M i (imaginary ) M ii Mi using its thrusters.Thus, if the gravitational mass of the It was shown that, imaginaryparticle is reduced by means of particles can have infinite speed in theabsorption of an amount of imaginary space-time [1] . Therefore, thiselectromagnetic energy U , for example, is also the speed upper limit for thewe have spacecraft in the imaginary space-time. Since the gravitational spacecraft χ= Mg ⎧ ⎢ ( = ⎨1 − 2⎡ 1 + U mi0 c 2 − 1⎤⎬ 2 ⎥⎭ ⎫ ) can use its thrusters after to becoming Mi ⎩ ⎣ ⎦ an imaginary body, then if the thrusters This shows that the energy U of the produce a total thrust F = 1000kN andelectromagnetic field remains acting on the gravitational mass of the spacecraftthe imaginary particle. In practice, this is reduced from M g = M i = 10 5 kg downmeans that electromagnetic fields act onimaginary particles. Therefore, the to M g ≅ 10 −6 kg , the acceleration of theelectromagnetic field of a GCC remains spacecraft will be, a = F Mg ≅ 10 m.s−2 . 12acting on the particles inside the GCC With this acceleration the spacecrafteven when their gravitational masses crosses the “visible” Universereach the gravitational mass ranging ( diameter= d ≈ 10 m ) in a time interval 26between + 0.159 M i to − 0.159M i and Δt = 2d a ≅ 1.4 × 107 m.s −1 ≅ 5.5 monthsthey become imaginary particles. This isvery important because it means that the Since the inertial effects upon theGCCs of a gravitational spacecraft keep spacecraft are reduced by −11on working when the spacecraft M g M i ≅ 10 then, in spite of thebecomes imaginary. effective spacecraft acceleration be Under these conditions, the gravity a = 1012 m. s −1 , the effects for the crewaccelerations on the imaginary and for the spacecraft will be equivalentspacecraft particle (due to the rest of the to an acceleration a′ given byimaginary Universe) are given by Mg a′ = a ≈ 10m.s −1 g ′j = χ g j j = 1,2,3,..., n. Mi This is the order of magnitude of the acceleration upon of a commercial jetWhere χ = M g (imaginary ) M i (imaginary ) aircraft.and g j = − Gmgj (imaginary) r . 2 j Thus, the On the other hand, the travel in thegravitational forces acting on the imaginary space-time can be very safe,spacecraft are given by because there won’t any material body along the trajectory of the spacecraft.
  11. 11. 11 Now consider the GCCs presented rotor in order to become negative thein Fig. 8 (a). Note that below and above acceleration of gravity inside half of thethe air are the bottom and the top of thechamber. Therefore the choice of the ( 2 ) rotor g ′ = (χ steel ) χ air g ≅ χ air g = − ng . Obviously this causes a torquematerial of the chamber is highlyrelevant. If the chamber is made of steel, T = (− F ′ + F )r and the rotor spins withfor example, and the gravity acceleration angular velocity ω . The averagebelow the chamber is g then at the power, P , of the motor is given bybottom of the chamber, the gravitybecomes g ′ = χ steel g ; in the air, the P = Tω = [(− F ′ + F )r ]ω (36) Wheregravity is g′′ = χairg′ = χairχsteelg . At the top g′′′ = χsteelg′′ = (χsteel) χairg . 2of the chamber, F ′ = 1 mg g ′ 2 F = 1 mg g 2Thus, out of the chamber (close to thetop) the gravity acceleration becomes and m g ≅ mi ( mass of the rotor ). Thus, g ′′′ . (See Fig. 8 (a)). However, for the Eq. (36) givessteel at B < 300T and f = 1 × 10 −6 Hz , we mi gω rhave P = (n + 1) (37) 2 mg (steel) ⎧ ⎡ ⎪ ⎢ σ (steel) B 4 ⎤⎫ ⎪ On the other hand, we have thatχ steel = = ⎨1− 2 1+ −1⎥⎬ ≅ 1 − g′ + g = ω 2r (38) mi(steel) ⎪ ⎢ 4πfμρ(steel) c 2 2 ⎥⎪ ⎩ ⎣ ⎦⎭ Therefore the angular speed of the rotorSince ρ steel = 1.1 × 10 6 S .m −1 , μ r = 300 and is given byρ (steel ) = 7800k .m −3 . ω= (n + 1)g (39) Thus, due to χ steel ≅ 1 it follows rthat By substituting (39) into (37) we obtain g ′′′ ≅ g ′′ = χ air g ′ ≅ χ air g the expression of the average power of If instead of one GCC we have the gravitational motor, i.e., P = 1 mi (n + 1) g 3 r (40)three GCC, all with steel box (Fig. 8(b)), 3then the gravity acceleration above the 2second GCC, g 2 will be given by Now consider an electric generator coupling to the gravitational motor in g 2 ≅ χ air g1 ≅ χ air χ air g order to produce electric energy.and the gravity acceleration above the Since ω = 2πf then for f = 60 Hzthird GCC, g 3 will be expressed by we have ω = 120 πrad . s − 1 = 3600 rpm . g 3 ≅ χ air g ′′ ≅ χ air g 3 Therefore for ω = 120πrad .s −1 and n = 788 (B ≅ 0.22T ) the Eq. (40) tell us that we must haveIII. CONSEQUENCES r= (n + 1)g = 0.0545m These results point to the 2 ωpossibility to convert gravitational energy Since r = R 3 and mi = ρπR 2 h where ρ ,into rotational mechanical energy. R and h are respectively the massConsider for example the system density, the radius and the height of thepresented in Fig. 9. Basically it is a motor rotor then for h = 0.5m andwith massive iron rotor and a box filled −3 ρ = 7800 Kg .m (iron) we obtainwith gas or plasma at ultra-low pressure(Gravity Control Cell-GCC) as shown inFig. 9. The GCC is placed below the mi = 327.05kg
  12. 12. 12 (Novel superconducting magnets areThen Eq. (40) gives able to produce up to 14.7T [10, 11]). Then the gravity acceleration inP ≅ 2.19 × 105 watts ≅ 219 KW ≅ 294HP (41) any direction inside the spacecraft, g l′ , will be reduced and given byThis shows that the gravitational motorcan be used to yield electric energy at mg ( Al )large scale. g l′ = g l = χ Al g l ≅ −10−9 g l l = 1,2,..,n mi ( Al ) The possibility of gravity controlleads to a new concept of spacecraftwhich is presented in Fig. 10. Due to the Where g l is the external gravity in theMeissner effect, the magnetic field B is direction l . We thus conclude that theexpelled from the superconducting shell. gravity acceleration inside the spacecraftThe Eq. (35) shows that a magneticfield, B , through the aluminum shell of becomes negligible if g l << 10 9 m .s −2 .the spacecraft reduces its gravitational This means that the aluminum shell,mass according to the following under these conditions, works like aexpression: gravity shielding. ⎧ ⎡ 2 ⎤⎫ Consequently, the gravitational ⎪ ⎛ B2 ⎞ mg ( Al) = ⎨1 − 2⎢ 1+ ⎜ ⎟ − 1⎥⎪mi( Al) (42) forces between anyone point inside the ⎥⎬ nr ( Al) ⎢ ⎜ μc2 ρ ⎟ spacecraft with gravitational mass, m gj , ⎪ ⎢ ⎝ ( Al) ⎠ ⎥⎪ ⎩ ⎣ ⎦⎭ and another external to the spacecraftIf the frequency of the magnetic field is (gravitational mass m gk ) are given by f = 10 −4 Hz then we have that r r m gj m gkσ ( Al ) >> ωε since the electric F j = − Fk = −G 2 μ ˆ r jkconductivity of the aluminum −1 where m gk ≅ mik and m gj = χ Al mij .is σ ( Al ) = 3.82 × 10 S .m . In this case, the 7 Therefore we can rewrite equation aboveEq. (11) tell us that in the following form μc 2σ ( Al ) nr ( Al ) = (43) 4πf r r mij mik F j = − Fk = − χ Al G 2 μ ˆSubstitution of (43) into (42) yields r jk Note that when B = 0 the initial ⎧ ⎡ σ ( Al ) B 4 ⎤⎫ ⎪ ⎪ gravitational forces aremg ( Al ) = ⎨1 − 2⎢ 1+ − 1⎥⎬mi ( Al ) (44) r r ⎪ ⎢ 4πfμρ( Al ) c 2 2 ⎥⎪ F j = − Fk = −G mij mik μ ⎩ ⎣ ⎦⎭ 2 ˆ r jkSince the mass density of the Aluminum Thus, if χ Al ≅ −10 −9 then the initialis ρ ( Al ) = 2700 kg .m −3 then the Eq. (44) gravitational forces are reduced from 109can be rewritten in the following form: times and become repulsives. According to the new expression r r χ Al = mg( Al) mi( Al) { [ = 1 − 2 1 + 3.68×10−8 B4 −1]} (45) for the inertial forces [1], F = m g a , we see that these forces have origin in the gravitational interaction between a particle and the others of the Universe,In practice it is possible to adjust B in just as Mach’s principle predicts. Henceorder to become, for example, mentioned expression incorporates the χ Al ≅ 10 . This occurs to B ≅ 76 .3T . −9 Mach’s principle into Gravitation Theory,
  13. 13. 13and furthermore reveals that the inertial spacecraft is subjected to a gravityeffects upon a body can be strongly acceleration given byreduced by means of the decreasing ofits gravitational mass. r Mg aman = ngMˆ = −χ airG 2 ˆ μ μ (55) Consequently, we conclude that if rthe gravitational forces upon thespacecraft are reduced from 109 times Inside the GCC we have, mg (air) ⎧ ⎡ ⎤⎫then also the inertial forces upon the ⎪ σ(air)B4 ⎪spacecraft will be reduced from 109 times χair = = ⎨1 − 2⎢ 1 + − 1⎥⎬ (56) mi(air) ⎪ ⎢ 4πfμρ(air)c 2 2 ⎥⎪when χ Al ≅ −10 −9 . Under these ⎩ ⎣ ⎦⎭conditions, the inertial effects on the By ionizing the air inside the GCCcrew would be strongly decreased. (Fig. 10), for example, by means of aObviously this leads to a new concept of radioactive material, it is possible toaerospace flight. increase the air conductivity inside the Inside the spacecraft the GCC up to σ (air) ≅ 106 S .m−1 . Thengravitational forces between thedielectric with gravitational mass, M g for f = 10 Hz ; ρ (air ) = 4.94 × 10−15 kg.m −3and the man (gravitational mass, m g ), (Air at 3 ×10-12 torr, 300K) and we obtainwhen B = 0 are r r M g mg {[ ] } χ air = 2 1 + 2.8 × 1021 B 4 − 1 − 1 (57) Fm = − FM = −G ˆ μ (46) r2or For B = BGCC = 0.1T (note that, due to r Mg the Meissner effect, the magnetic field Fm = −G 2 m gˆ = − m g g M ˆ μ μ (47 ) BGCC stay confined inside the r superconducting box) the Eq. (57) yields r mg FM = +G 2 M gˆ = + M g g mˆ μ μ (48) r χ air ≅ −10 9If the superconducting box under M g Since there is no magnetic field(Fig. 10) is filled with air at ultra-low through the dielectric presented in Fig.10pressure (3×10-12 torr, 300K for example) then, Mg ≅ Mi . Therefore if M g ≅ Mi = 100Kgthen, when B ≠ 0 , the gravitational mass and r = r0 ≅ 1m the gravity accelerationof the air will be reduced according to(35). Consequently, we have upon the man, according to Eq. (55), is g ′ = (χ steel ) χ air g M ≅ χ air g M (49) 2 a man ≅ 10m .s −1 M g m = (χ steel ) χ air g m ≅ χ air g m ′ (50) 2 r rThen the forces Fm and FM become Consequently it is easy to see that this r Fm = −m g (χ air g M )ˆ μ (51) system is ideal to yield artificial gravity r inside the spacecraft in the case of inter- FM = + M g (χ air g m )ˆ μ (52) stellar travel, when the gravityTherefore if χ air = −n we will have acceleration out of the spacecraft - due r to the Universe - becomes negligible. Fm = +nmg g M ˆ μ (53) The vertical displacement of the r FM = −nM g g mˆ μ (54) spacecraft can be produced by means of r r Gravitational Thrusters. A schematicThus, Fm and FM become repulsive. diagram of a Gravitational Thruster isConsequently, the man inside the shown in Fig.11. The Gravitational
  14. 14. 14Thrusters can also provide the horizontal The gravitational force dF12 thatdisplacement of the spacecraft. dm g1 exerts upon dm g 2 , and the The concept of GravitationalThruster results from the theory of the gravitational force dF21 that dm g 2 exertsGravity Control Battery, showed in Fig. 8 upon dm g1 are given by(b). Note that the number of GCCincreases the thrust of the thruster. For r r dmg2 dmg1example, if the thruster has three GCCs dF12 = dF21 = −G ˆ μ (58)then the gravity acceleration upon the r2gas sprayed inside the thruster will berepulsive in respect to M g (See Fig. 11) Thus, the gravitational forces between the air layer 1, gravitational mass m g1 ,and given by Mg and the air layer 2, gravitational mass a gas = (χ air ) (χ steel ) g ≅ −(χ air ) G 3 4 3 m g 2 , around the spacecraft are r02 r r G mg1 mg 2Thus, if inside the GCCs, χair ≅ −109 F12 = −F21 = − 2 ∫ ∫ dmg1dmg 2ˆ = μ r 0 0(See Eq. 56 and 57) then the equation mg1mg 2 mmabove gives = −G 2 ˆ = −χ air χ airG i1 2 i 2 ˆ μ μ (59) Mi r r a gas ≅ +10 G 27 r02 At 100km altitude the air pressure isFor M i ≅ 10kg , r0 ≅ 1m and m gas ≅ 10 −12 kg 5.691 10−3 torr and ρ(air) =5.998 10−6 kg.m−3 [12]. × ×the thrust is By ionizing the air surround the F = m gas a gas ≅ 10 5 N spacecraft, for example, by means of anThus, the Gravitational Thrusters are oscillating electric field, E osc , startingable to produce strong thrusts. from the surface of the spacecraft ( See It is important to note that if F is Fig. 13) it is possible to increase the airforce produced by a thruster then the conductivity near the spacecraft up tospacecraft acquires acceleration a spacecraft given by [1] σ (air) ≅ 106 S .m−1 . Since f = 1Hz and, in this case σ (air ) >> ωε , then, according toa spacecraft = F = F Eq. (11), nr = μσ(air)c 2 4πf . From M g (spacecraft) χ Al M i (inside) + mi ( Al ) Eq.(56) we thus obtain ⎧ ⎡ σ B4 ⎤⎫ m ⎪ ⎪Therefore if χ Al ≅ 10 −9 ; M i(inside) = 104 Kg χair = g(air) = ⎨1 − 2⎢ 1 + (air) 2 2 −1⎥⎬ (60) mi(air) ⎪ ⎢ 4πfμ0ρ(air)c ⎥⎪and mi ( Al ) = 100 Kg (inertial mass of the ⎩ ⎣ ⎦⎭aluminum shell) then it will be necessary Then for B = 763T the Eq. (60) givesF = 10kN to produce a spacecraft = 100m .s −2 { [ ]} χ air = 1 − 2 1+ ~ 104 B 4 − 1 ≅ −108 (61)Note that the concept of Gravitational By substitution of χ air ≅ −108 into Eq.,Thrusters leads directly to the (59) we getGravitational Turbo Motor concept (SeeFig. 12). r r Let us now calculate the m m F12 = −F21 = −1016 G i1 2 i 2 ˆ μ (62) rgravitational forces between two veryclose thin layers of the air around thespacecraft. (See Fig. 13).
  15. 15. 15 −8If mi1 ≅ mi 2 = ρ air V1 ≅ ρ air V2 ≅ 10 kg , and aspect of the flight dynamics of a Gravitational Spacecraft.r = 10 −3 m we obtain Before starting the flight, the r r gravitational mass of the spacecraft, M g , F12 = −F21 ≅ −10−4 N (63) must be strongly reduced, by means of a gravity control system, in order toThese forces are much more intense r produce – with a weak thrust F , a strongthan the inter-atomic forces (the forces r acceleration, a , given by [1]which maintain joined atoms, and r r Fmolecules that make the solids and a=liquids) whose intensities, according to Mgthe Coulomb’s law, is of the order of In this way, the spacecraft could be1-1000×10-8N. strongly accelerated and quickly to reach Consequently, the air around the very high speeds near speed of light.spacecraft will be strongly compressed If the gravity control system of theupon their surface, making an “air shell” spacecraft is suddenly turned off, thethat will accompany the spacecraft gravitational mass of the spacecraftduring its displacement and will protect becomes immediately equal to its inertialthe aluminum shell of the direct attrition mass, M i , (M g = M i ) and the velocity ′with the Earth’s atmosphere. r r In this way, during the flight, the V becomes equal to V ′ . According toattrition would occur just between the “air the Momentum Conservation Principle,shell” and the atmospheric air around we have thather. Thus, the spacecraft would stay free M gV = M gV ′ ′of the thermal effects that would be Supposing that the spacecraft wasproduced by the direct attrition of the traveling in space with speed V ≈ c , andaluminum shell with the Earth’s that its gravitational mass it wasatmosphere. M g = 1Kg and M i = 10 4 Kg then the Another interesting effect produced velocity of the spacecraft is reduced toby the magnetic field B of the Mg Mgspacecraft is the possibility of to lift a V′ = V= V ≈ 10−4 cbody from the surface of the Earth to the ′ Mg Mispacecraft as shown in Fig. 14. By Initially, when the velocity of theionizing the air surround the spacecraft, r spacecraft is V , its kinetic energy isby means of an oscillating electric field, E osc , the air conductivity near the Ek = (Mg −m )c2. Where Mg = mg 1 − V 2 c2 . gspacecraft can reach, for example, At the instant in which the gravity controlσ (air ) ≅ 106 S .m −1 . Then for f = 1Hz ; system of the spacecraft is turned off, the kinetic energy becomesB = 40.8T and ρ(air) ≅ 1.2kg.m−3 (300K and Ek = (Mg − m′ )c . Where Mg = m′ 1 − V ′2 c2 . ′ ′ g 2 ′ g1 atm) the Eq. (56) yields We can rewritten the expressions of χ air = ⎧1 − 2⎡ 1 + 4.9 ×10−7 B4 − 1⎤⎫ ≅ −0.1 ⎨ ⎢ ⎥⎬ ′ E k and E k in the following form ⎩ ⎣ ⎦⎭ Ek = (MgV − mgV )Thus, the weight of the body becomes c2 Pbody = mg (body) g = χ air mi (body) g = mi(body) g ′ V Ek = (M gV ′ − m′ V ′) c2Consequently, the body will be lifted on ′ ′ V′ gthe direction of the spacecraft withacceleration Substitution of M gV = M gV ′ = p , ′ g ′ = χ air g ≅ +0.98m.s −1 Let us now consider an important
  16. 16. 16mgV = p 1−V c and m′ V ′ = p 1 − V ′ c into 2 2 2 2 Obviously this electromagnetic g pulse (EMP) will induce heavy currents ′the equations of E k and E k gives in all electronic equipment that mainly ( ) pc contains semiconducting and conducting 2 Ek = 1 − 1 − V 2 c 2 materials. This produces immense heat V E′ = ( − c ) 2 that melts the circuitry inside. As such, pc 1 1 − V ′2 2 while not being directly responsible for V′ k the loss of lives, these EMP are capableSince V ≈ c then follows that of disabling electric/electronic systems. Therefore, we possibly have a new type E k ≈ pc of electromagnetic bomb. An electromagnetic bomb or E-bomb is aOn the other hand, since V ′ << c we get well-known weapon designed to disable ( ) pc′ electric/electronic systems on a wide 2 Ek = 1 − 1 − V ′2 c 2 ′ = scale with an intense electromagnetic V ⎛ ⎞ pulse. ⎜ ⎟ Based on the theory of the GCC it ⎟ pc ≅ ⎛ V ′ ⎞ pc 2 ≅ ⎜1 − 1 ⎜ ⎟ is also possible to build a Gravitational ⎜ V ′2 ⎟ V′ ⎝ 2c ⎠ ⎜ 1 + 2 + ... ⎟ Press of ultra-high pressure as shown in ⎝ 2c ⎠ Fig.15.Therefore we conclude that E k >> E k . ′ The chamber 1 and 2 are GCCsConsequently, when the gravity control with air at 1×10-4torr, 300Ksystem of the spacecraft is turned off, ( 6 −1 −8 ) σ (air ) ≈ 10 S .m ; ρ (air ) = 5 × 10 kg .m −3 .occurs an abrupt decrease in the kinetic Thus, for f = 10 Hz and B = 0.107T weenergy of the spacecraft, ΔE k , given by have ⎧ ⎡ σ (air) B 4 ⎤⎫ ′ ΔEk = Ek − Ek ≈ pc ≈ M g c 2 ≈ 1017 J ⎪ ⎢ ⎪ χ air = ⎨1− 2 1+ −1⎥⎬ ≅ −118 ⎪ ⎢ 4πfμ0 ρ(air) c 2 2 ⎥⎪ ⎩ ⎣ ⎦⎭By comparing the energy ΔE k with theinertial energy of the spacecraft, The gravity acceleration above the E i = M i c 2 , we conclude that air of the chamber 1 is r ΔE k ≈ Mg Ei ≈ 10 − 4 M i c 2 g1 = χ stellχ air gˆ ≅ +1.15×103ˆ μ μ (64) Mi Since, in this case, χ steel ≅ 1 ; μ is anˆThe energy ΔE k (several megatons) unitary vector in the opposite direction of rmust be released in very short time g.interval. It is approximately the same Above the air of the chamber 2 theamount of energy that would be released gravity acceleration becomesin the case of collision of the spacecraft ‡ . rHowever, the situation is very different of g2 = (χ stell ) (χair ) gˆ ≅ −1.4 × 105ˆ μ μ (65) 2 2a collision ( M g just becomes suddenlyequal to M i ), and possibly the energy r Therefore the resultant force R acting onΔE k is converted into a High Power m2 , m1 and m isElectromagnetic Pulse.‡ In this case, the collision of the spacecraft wouldrelease ≈1017J (several megatons) and it would besimilar to a powerful kinetic weapon.

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