Gravitational Micro-Thrusters

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https://hal.archives-ouvertes.fr/hal-02178918v2

Preprint submitted on 12 Jul 2019

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Gravitational Micro-Thrusters

Fran de Aquino

To cite this version:

Fran de Aquino. Gravitational Micro-Thrusters. 2019. �hal-02178918v2�

Gravitational Micro-Thrusters

Fran De Aquino

Professor Emeritus of Physics, Maranhao State University, UEMA.

Titular Researcher (R) of National Institute for Space Research, INPE Copyright © 2019 by Fran De Aquino. All Rights Reserved.

Here we show how to produce thrusts of the order of 100kN or more, starting from sets of micro-tubes (diameter<< 1cm) filled with air at low pressure, subjected to gravity g, and a strong magnetic fieldH. Under these conditions, these micro-tubes work as micro-thrusters, where the thrust is produced starting from the local potential gravitational energy.

Key words: Gravitational Mass, Gravitational Interaction, Gravitational Thruster.

INTRODUCTION

In this paper we will show that micro-tubes filled with air at low pressure, subjected to gravity , and a strong magnetic field

( ^{diameter << 1 cm} )

gr

H, can works as micro-thrusters, where the thrust is produced starting from the local potential gravitational energy. In this context, it is also shown that sets of these micro- thrusters can produce thrust of the order of 100kN or more.

THEORY

In a previous paper [1] we shown that there is a correlation between the gravitational mass, , and the rest inertial mass , which is given by

mg m_i0

( )

1 1

1 2 1

1 1

2 1

1 1

2 1

2 2

2

2 0

2

0 0

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ ⎟⎟⎠ −

⎜⎜ ⎞

⎝ +⎛

−

=

⎪⎭=

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟ −

⎠

⎜⎜ ⎞

⎝ +⎛

−

=

⎪⎭=

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟ −

⎠

⎜⎜ ⎞

⎝ +⎛ Δ

−

=

=

c Wn

c m

Un c m

p m

m

r i

r i i

g

ρ χ

where is the variation in the particle’s kinetic momentum; is the electromagnetic energy absorbed or emitted by the particle; is the index of refraction of the particle;

W

is the density of energy on the particle ;

Δp U

n

r

( ^{J / kg} ) ^ρ

^is

the matter density

( ^{kg m}

³

)

^and

c

is the speed of light.

The instantaneous values of the density of electromagnetic energy in an electromagnetic field can be deduced from Maxwell’s equations and has the following expression

( )

2

2 2 2 1 2

1 E H

W= ε + μ

where

E = E

_m

sin ω t

and

H = H sin ω t

are the instantaneous values of the electric field and the magnetic field respectively.

It is known thatB=

μ

H,

E B = ω k

_r [2] and

( ) ( )

³

1 2 1

2 ⎟⎠⎞

⎜⎝

⎛ + +

=

=

=

ωε μ σ

κ ε ω

r r r

c dt

v dz

where is the real part of the propagation vector

k

r

k

r

(also called phase constant);

i

r

ik

k k

k =

r

= +

; ε , μ and σ, are the electromagnetic characteristics of the medium in which the incident (or emitted) radiation is propagating (

ε = ε

_r

ε

₀;

ε

₀=8.854×10⁻¹²F/m;

μ

0

μ

μ =

_r where

μ

₀ =4

π

×10⁻⁷H/m;

σ

is the electrical conductivity in S/m). From Eq. (3), we see that the index of refraction

n

_r

= c v

is given by

( ) 1 ( ) 4

2 1

2

⎟ ⎠ ⎞

⎜ ⎝

⎛ + +

=

= ε

_r

μ

_r

σ ωε

r

v

n c

Equation (3) shows that

ω κ

_r

= v

. Thus,

v

k B

E = ω

_r

=

, i.e.,

( ) ⁵

H v vB E = = μ

Then, Eq. (2) can be rewritten as follows

( )

( ) 6

2

2 2 2 1 2 2 1

2 2 2 1 2 2 2 1

H

H v

H

H H

v W

μ

μ μ ε μ

μ μ

ε

=

= +

=

= +

=

For

σ >> ωε

, Eq. (3) gives

( )

7 2

2 2

2

2 c

v n_r c

ω

=

μσ

=

Substitution of Eqs. (6) and (5) into Eq. (1) gives

2

( ) ⁸

4 1 1 2

1

_{2 2} ⁴

3

⎪⎭

⎪ ⎬

⎫

⎪⎩

⎪ ⎨

⎧

⎥ ⎥

⎦

⎤

⎢ ⎢

⎣

⎡ ⎟⎟ ⎠ −

⎜⎜ ⎞

⎝ + ⎛

−

= H

c f ρ π

σ χ μ

Note that if

H = H

_m

sin ω t

.Then, the average value for

H

² is equal to ¹₂H_m² because

H varies sinusoidaly ( is the maximum value for

H

m

H). On the other hand, we have

H

_rms

= H

_m

2

. Consequently, we can change

H

4 by , and the Eq. (8) can be rewritten as follows

4

Hrms

( ) ⁹

4 1 1 2

1

_{2 2} ⁴

3

⎪⎭

⎪ ⎬

⎫

⎪⎩

⎪ ⎨

⎧

⎥ ⎥

⎦

⎤

⎢ ⎢

⎣

⎡ ⎟⎟ −

⎠

⎜⎜ ⎞

⎝ + ⎛

−

= H

_rms

c f ρ π

σ χ μ

Now consider a metallic cylindrical tube (

φ

internal diameter, height and 0.2mm thick),

hφ

filled with air at low pressure and subjected to gravity, , and an oscillating magnetic field

gr

H

rrms

with frequency . The metallic tube is inside a dielectric, and is electrically charged as shown in Fig.1.

f

If cm

<<1

φ

then, the distances among these electric charges and the atoms of air inside the tube will be very smalls. Consequently, the electrical forces that will act on these atoms will be very strong, and will be sufficient to ionize the oxygen and nitrogen atoms of the air, increasing the electrical conductivity of the air inside the tube.

Fig. – 1 Air Cylindrical tube (φ diameter; h_φ height).

φ

h

φ

S

φ

Upper surface Air at

Low pressure

H r

_rms

μ ; σ

gr

Down surface

- - - - - - - - - - - - - - - -

- - - - - - - - - - - - - -

mm 2 . 0

mm 2 . 0

DIELECTRIC

D I E L E C T R I C

It is known that the electrical conductivity is proportional to both the concentration and the mobility of the ions and the free electrons, and is expressed by

( ) ¹⁰

i i e

e

μ ρ μ

ρ

σ = +

where ρ

_e

and ρ

_i

express respectively the concentrations ( ^{C m}

³

) of electrons and ions;

μ

e

and μ

_i

are respectively the mobilities of the electrons and the ions.

In order to calculate the electrical conductivity of the air inside the metallic tube, we first need to calculate the concentrations ρ

_e

and ρ

_i

.

Since the number of atoms per , , is given by

m3 na

( ) ¹¹

0 s

s

a

A

n N ρ

=

where N

₀

= 6 . 02214129 × 10

²⁶

atoms / kmole , is the Avogadro’s number;

ρ_s

is the matter density (in kg/m

³

) and

A_s

is the molar mass (

kg.kmole⁻¹

) . Then, for ρ

_s=

ρ

_air=1×10⁻⁴kg.m⁻³

( ^{6 . 62 × 10}

⁻²

^Torr )

, Eq. (11) gives

( )( )

( ) 12 /

10 15 . 2

0134 . 28

10 1 10 02214129 .

6

3 21

4 26

0

m atoms A

n N

air air a

×

=

× =

≅ ×

= ρ

⁻

Using techniques of the Statistical Mechanics we can calculate the most probable number of ions, N

_i

, in the volume

π

φ h

φ

V =

₄ ²

of air cylindrical tube, by means of the following expression [3].

( ) ¹³

i

i

a

S N = N

where is the total number of atoms; is the area total of the box

1m2

n N = _a×

S ( ) ^1m

²

and a

_i

is the area of the cell ( )

hφ

^φ . Therefore, Eq. (13) can be rewritten as follows

( ) ¹ ( ) ¹

²

^{( ) ( ) 14}

2

h

_φ

φ

m m N

_i

= n

^a

For φ

=1.6mm

and h

_φ

= 12 cm

, Eq.(14) yields

( ) 15 10

1 .

4

¹⁷

ions N

_i

= ×

Obviously, the number of free electrons will be equal the number of ions, thus we can write that

( ) 16 10

1 .

4

¹⁷

ions N

N

_e

=

_i

= ×

Now, we can calculate the concentrations ρ

_e

and ρ

_i

( ^{C m}

³

) of electrons and ions by means of the following expression

( )

4 ²

^{2 . 7 10}

⁵

^/

³

^{( ) 17}

1

C m

h eN V

eN

_i

i

e

= = = = ×

π

φ

φ

ρ ρ

This corresponds to a strong concentration level in the case of conducting materials.

For these materials, at temperature of 300K, the mobilities μ

_e

and μ

_i

vary from

10

up to

100 m²V⁻¹s⁻¹

[

4]. Assuming that

(Geometric mean of mobility level for conducting materials), the electrical conductivity of the air inside the tube is given by

1 1

30 2 ^{− −}

≅

= _i mV s

e

μ

μ

( ) ^{( )}

( ) ¹⁸

. 10 2

30 10 7 . 2 2

1 7

5

×

−

≅

=

×

= +

=

m S

i i e e

air

ρ μ ρ μ

σ

The pressure,Pr

, exerted on the area _φ, by of the air confined inside the tube, according to Eq.(1), is given by

S

( )

19

0 0

g h h

S g h m S

g m S

g

Pr m^gr ⁱ r ⁱ r _air r

φ φ

φ φ φ

φ

χ χρ

χ

= =

=

=

Substitution of Eq.(9) into Eq. (19) gives

( ) ²⁰

2 4

₂ ⁴

3

2

H h g

P

r _{air air} ^air

fc

^air _{rms air} r

ρ

φ

π σ ρ μ

ρ ⎪⎭

⎪ ⎬

⎫

⎪⎩

⎪ ⎨

⎧

⎥ ⎥

⎦

⎤

⎢ ⎢

⎣

⎡ ⎟⎟ −

⎠

⎜⎜ ⎞

⎝ + ⎛

−

=

For,f=0.2Hz,

σ

_air≅2×10⁷S/m, Eq.(20) gives

[ ]

{ ²

²

^{1 . 75 10}

²⁸

^H

⁴

} ^{h g ( ) 21}

P r

_{air air rms air}

r ρ

φ

ρ

ρ − + × −

=

⁻

For

ρ

_air =1×10⁻⁴kg.m⁻³

( ^{6 . 62 × 10}

⁻²

^Torr )

^{, and}

m A

H_rms=7.96×10⁵ /

( )

^1T , Eq. (21) gives

( ) 22 0164

.

0 h g

P

r r

−

φ

=

Note the sign (-) in Eq. (13). It indicates that, in this case, the pressure Pr

acts on the contrary direction of the gravity . This means that the pressure will be exerted on the upper surface of the cylindrical tube (See Fig.1). Thus,

the cylindrical tube works as a gravitational micro-thruster, producing a thrust , given by

gr

F

( ) 23 0129

. 0 0164

.

0 h S g

²

h g

S P

F

r r r r

φ φ

φ

φ

= − = − φ

=

If

h

_φ

= 12 cm

.;

φ

=1.6mm and , then the intensity of the force

.

2

81 .

9

⁻

= m s

g

Fr

is given by

( ) 24 10

89 .

3

⁸

N

F = ×

⁻

In the case of a plate with gravitational micro-thrusters, the Eq. (24) can be rewritten as follows

Nφ

( ) 25 10

89 .

3

⁸

N

_φ

F

_N

= ×

⁻

For example, if N_φ =

(

560×560

)

=313600, then Eq. (25) gives

( ) ²⁶

2 . 1 10

2 . 1 0122 .

0 N

³

kgf gf

F

_N

= = ×

⁻

=

Assuming that, the 313600 metallic cylindrical tubes (external diameter=

φ

_ex

= φ + 0 . 4 mm = 2 . 0 mm

and height = ) are distributed into a square dielectric plate with sides ,

hφ

l according to the pattern shown

in Fig.2, then we can write that

5 2

.

2

φ

N_φ

l= _ex (See Fig. 2). This means that the dielectric plate will have

( ) ²⁷

10 .

1 m

l =

l1

dielectric plate

l1

(2.2øex)² ... 5 S ... N_ø

( ^2.2 )

^{2 2}

5 l

S = N

^φ

φ

_ex

=

2 5 .

2

φ

^Nφ l = _ex

⇒

Fig. 2 – Distribution of metallic cylindrical tubes into a dielectric plate.

0.55øex 0.55øex 0.55øex 0.55øex

4 Figure 3 shows an experimental set-up

in order to check the total thrust produced by the dielectric plate, with gravitational micro-thrusters, subjected to an oscillating magnetic field, given by

N

φ

( ) ( )

rms

rms rms

turns rms

i

i i

y N H

104

3

01 . 0 300

×

=

=

=

=

Total mass of the system (without coils) = M_total = 30.86 kg

H

_T

= h

_φ

+ 5 mm

Fig. 3 –

Schematic Diagram of the dielectric plate, with

N

φ

Gravitational Micro-Thrusters, on a balance, and inside a magnetic field produced by two external coils.

Vacuum pump

N_ø Gravitational Micro-Thrusters (in white)

(Filled with air at 66.2 m Torr)

(Diameterφ^=1.6mm; heighth_T =12cm;

Air densityρ≅1×10⁻⁴kg.m⁻³) Coil: 300 turns (in X , in order to reduce the effect of distributed capacitance ); # 14 AWG

h_ø

Balance l

H_T

Bottom cover (1mmthick)

y = 1cm

The mass of one dielectric plate (High- density polyethylene (HDPE),

970 kg . m

⁻³), 12.2cm thickness

(

^12.2cm=hφ ^+2mm

)

(See Fig 3), without the micro-thrusters is given by.

m_Mg =l²

(

12.2cm

)( )

970 =143.19kg The mass of one dielectric plate, 12.2cm thickness, with N_φ cylindrical tubes, is given by ^mMg_φ =^mMg −^N_φ

(

970^π₄

φ

ex^2h_φ

)

=28.51^kg The mass of one dielectric plate, 2mm thickness (Bottom cover, see Fig.3), is

m_Mg(2mm) =l²

(

2mm

)( )

970 =2.35kg. Thus, the total mass of the system is given by

( ) kg

m m

M_total= _Mgφ + _Mg2mm =28.51+2.35=30.86 Therefore, the balance (see Fig.3) can have the following characteristics:

g kg capacity

Maximum

1 . 0 accuracy Measuring

35

If the magnetic field

H

_rmsis increased to m

A/ 10 57 .

5 × ⁸

( ^{700 T} )

^{* ( ),}

then Eq. (21) gives

3

4

.

10

1 ×

^{− −}

= kg m

ρ

( ) 28 /

10 66 .

9

³

N m

²

P = − ×

Therefore, the thrust produced by one micro- thruster (diameter

Fφ

mm 6 .

=1

φ

and

heighth_φ =12cm), is given by

( ) 29 0194

.

2

0

N P

r PS

F

_φ

=

_φ

= π

_φ

=

Consequently, the total thrust produced by the system (one plate) with N_φ =313600 micro- thrusters will be given by

( )

30 9

. 6090 N F

N F_Nφ = _{φ φ} =

In practice, we can overlap several similar plates in order to increasing the total thrust. In this case, if the number of plates isN_plates, the result is

( )

31

φ N plates

total N F

F =

For example, if number of overlapping plates (with N_φ =313600 micro-thrusters) is

=27

plates

N , then the total thrust produced

by the stack of plates (~3.4cm total height) will be given by

( )

³²

4 . 164 kN F_total ≅

This thrust is of the order of the thrust of a fifth-generation jet fighter F-22 Raptor, which reaches 160,000N.

* Recently, a magnetic field of 1200 T was generated by the electromagnetic flux-compression (EMFC) technique with a newly developed megagauss generator system [5].

It is important to note that the vertical )

direction ( ^{− g}

^r

of the thrust produced by the plates can be turned of 90°, in order to produce horizontal displacements (See Fig.4 and Fig.5).

Fig. 4 - The vertical direction

( )

−gr of the thrust produced by a plate, with Nφ gravitational micro-thrusters, can be turned of 90°, in order to produce horizontal displacements.

g Ftotal

F

x

Gravitational Micro-Thrusters CG

Air at low pressure

Fx

Fx Fx

Plate, with N_φ Gravitational Micro-Thrusters

Fig. 5 – Horizontal displacement of Fx in several directions, simply rotating the plate with N_φ Gravitational Micro-Thrusters

.

Thus, gravitational micro-thrusters can be used to produce vertical or horizontal thrusts (in respect to ). It is easy to see that, due to the magnitude of the thrust produced by these systems and their versatility, they can be can be used to move several types of vehicles.

gr

References

[1]

of the Relativistic Theory of Quantum Gravity, De Aquino, F. (2010) Mathematical Foundations Pacific Journal of Science and Technology,11 (1), pp. 173-232.

Available at https://hal.archives-ouvertes.fr/hal-

01128520

[2] Halliday, D. and Resnick, R. (1968) Physics, J. Willey & Sons, Portuguese Version, Ed. USP, p.1118.

[3] Beiser, A. (1969) Concepts of Modern Physics, Portuguese version, Ed. Polígono and Ed.

Universidade de S. Paulo., p. 272.

[4] Hayt, W. H. (1974),Engineering Electromagnetics, McGraw-Hill. Portuguese version (1978)Ed.Livros Técnicos e Científicos Editora S.A, RJ, Brasil. P.146. [5] Nakamura, et al., (2018) Record indoor magnetic field of 1200 T generated by electromagnetic flux- compression, Review of Scientific Instruments 89, 095106

https://aip.scitation.org/doi/10.1063/1.5044557 [Accessed 4 July 2019]

6

APPENDIX – Gravitational Thruster

System (using plates with

Nφ

Gravitational Micro-Thrusters) to be used in aerospatial vehicles: the sub-systems S5 and S6 provide the vertical thrust (lift force). The sub-systems S1, S2, S3, S4 provide the horizontal thrusts in any direction. Note that the opposite thrusts (S1,S3), (S2,S4) can stop quickly the vehicle (see top view).

S1 Fx

S2 CG Fx Fx

S4

Fx

S3

Top view

S5 S6 Ceiling

Fv CG Fv

S4 S3 S2

Column 1 Column 2

Floor Side view