Controlling the Gravitational Mass of a Metallic Lamina, and the Gravity Acceleration above it

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Lamina, and the Gravity Acceleration above it

Fran de Aquino

To cite this version:

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Controlling the Gravitational Mass of a Metallic Lamina,

and the Gravity Acceleration above it.

Fran De Aquino

Professor Emeritus of Physics, Maranhao State University, UEMA. Titular Researcher (R) of National Institute for Space Research, INPE

It is proposed a very simple device for controlling the gravitational mass of a metallic lamina, and the gravity acceleration above it. These effects are obtained when a specific extra-low frequency current passes through a specially designed metallic lamina.

Key words:Gravitational Interaction, Gravitational Mass, Gravitational Force Control.

1. Introduction

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 g m 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 ;

p Δ U r

n

(

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 1 2 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

(

)

1

( )

3 1 2 2 _⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ ₊ = = = ωε σ μ ε κ ω r r r c dt dz v

where is the real part of the propagation

vector

r

k

r

(also called phase constant);

i

r

ik

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; 0 8.854 1012F /m; − × =

ε

0

μ

=

r where ). From Eq.

(3), we see that the index of refraction

m / H 7 0 4 10 − × =

π

μ

v

c

n

_r

=

is given by

(

)

1 ( )

4

1

2

_⎟

⎠

⎞

⎜

⎝

⎛

₊

=

ε

r

μ

r

σ

ωε

r

v

c

n

Equation (3) shows that

ω

κ

_r

=

v

. Thus,

v

k

B

E

=

ω

_r

=

, i.e.,

H

v

vB

E

=

μ

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

( )

5 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 1 2 2 1 2 2 1 2 2 1 E c n E c v c E v E v E v E v E v E E W r ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = μ μ μ μ μ μ ε μ μ ε

For

σ

>>

ωε

, Eq. (3) gives

( )

6 2 2 2 2 2 c v c n_r

ω

μσ

= =

Substitution of Eq. (6) into Eq. (5) gives

(

2

)

E2

( )

7

W =

σ

ω

Substitution of Eq. (7) into Eq. (1), yields

( )

8 1 10 032 . 7 1 2 1 1 64 1 2 1 1 4 1 2 1 0 4 3 2 3 27 0 4 3 2 3 2 3 0 0 2 4 3 2 i r i r i g m E f m E f c m E f c m ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × + − = = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − = − ρ σ μ ρ σ μ π μ ρ π σ μ

Note that if

E

=

E

m

sin

ω

t

.Then, the

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E varies sinusoidaly ( is the maximum value for

m

E

E). On the other hand, we have

E

_rms

=

E

_m

2

. Consequently, we can change

E

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

4 rms E

( )

9 1 10 032 . 7 1 2 1 0 4 3 2 3 27 i rms r g E m f m ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × + − = −

ρ

σ

μ

The Ohm's vectorial Law tells us

that

j

_rms

=

σ

E

_rms . Thus, we can write Eq. (9) in the following form:

( )

10

1

10

032 .

7

1

2

1

₀ 3 2 4 27 i rms r g

m

f

j

m

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

⎥

⎦

⎤

⎢

⎣

⎡

−

×

+

−

=

−

σρ

μ

where

j

_rms

=

j

2

[3]. Since

(

)

⎟

( )

11 ⎠

⎞

⎜

⎝

⎛

=

L

V

S

L

V

RS

V

S

R

V

S

i

j

σ

Then, we can write that

( )

12 2 ⎟⎠ ⎞ ⎜ ⎝ ⎛ = L V j_rms

σ

By substitution of Eq. (12) into Eq.(10), we get

( )

_{( )}

13 1 10 758 . 1 1 2 1 ₃ 4 2 3 27 0 ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × + − = = − f L V m m _r i g

ρ

σ

μ

χ

Also, it was shown in the above mentioned paper [1] that, if the weight of a particle in a side of a lamina is

P

m

g

(g

r

=

r perpendicular to the lamina) then the weight of the same particle, in the other side of the lamina is

P

r

′

=

χ

m

_g

g

r

, where

χ

=

m

l_g

m

_il₀ ( and are respectively, the gravitational mass and the rest inertial mass of the lamina). Only when

l g

m

l i m0 1 =

χ

, is that the weight is equal in both sides of the lamina. Thus, the lamina can control the gravity acceleration above it, and in this way, it can work as a Gravity Controller Device.

Since the gravitational mass of a body

above the lamina is , then we can

conclude that 0 i g m m =

( )

g

m

P

′

=

i0

χ

. Therefore, this

means that the gravity acceleration above the lamina isg′=

χ

g.

Here, we describe a very simple device, which works as the mentioned lamina. This device is easy to build, and can be used in order to test the correlation between gravitational mass and inertial mass previously mentioned (Eq. (1)),

and also the modification of the gravity acceleration above the lamina

( )

χ

g

.

2. The Device

Consider the device shown in Fig.1 (a). It is basically a thin Aluminum strip attached to an electrical insulating plate. This strip has been designed over an Aluminum lamina, in order to an electrical current

(

j

rms

;

f

)

to pass through it,

producing the decreasing of its gravitational mass

m

g₍strip), according to Eq. (10). The

Aluminum of this strip has the following characteristics: 99.9% Aluminum;

μ

_r

=

1

; ρ = 2700 kg/m-3;σ = 3.5x107S/m. The Aluminum strip has the following dimensions; Length,

L=3528 mm; Width: l = 5mm; Thickness: Δx

=3μm. The Resistance of the Aluminum strip is:

R

=

L

σ

S

=

L

σ

( )

Δ

xl

=

6 .

7 Ω

and the maximum current density, according to Eq. (12), is:

( )

max 6 2 2 max 3.5 10 / 3.5 / 2 V L A m A mm j rms = = × =

σ

; the maximum current isimax_{( )}rms=jmax_{( )}rmsS=0.05A *

; the Maximum Dissipated Power is: .

( )

watts

Ri

P

_max

=

_max2 _rms

=

0 .

017

By substuting the values of

μ

_r,

σ

,

ρ

and

Linto Eq. (13), we obtain.

( ) ( )

( )

14

1

10

67 .

6

1

2

1

₃ 4 14 0

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

−

⎥

⎦

⎤

⎢

⎣

⎡

×

+

−

=

−

f

V

m

strip i strip g

χ

Consequently, the gravity acceleration above the

Aluminum strip is given by

( )

15

1

10

67 .

6

1

2

1

3 4 14

g

f

V

g

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

−

⎥

⎦

⎤

⎢

⎣

⎡

×

+

−

=

′

_χ

−

The calculated results starting from Eqs.(14) and (15) for f =5

μ

Hz; f =10

μ

Hz;

Hz

f =15

μ

, in the voltage range

0 .

1 V

−

0 .

5 V

are plotted in Table 1.

Figure 1(b) shows an experimental set up in order to control the decreasing of the

Gravitational Mass of the Aluminum strip.

Figure 1(c) shows an experimental set up in order to control the decreasing of the gravity

acceleration above the Aluminum strip.

*

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(a) Device to produce the decreasing in the Gravitational Mass of the Aluminum strip.

(b) Experimental set up, using the device shown in (a), in order to control the decreasing of the Gravitational Mass of the Aluminum strip.

(c) Experimental set up, using the device shown in (a), in order to control the decreasing of the Local Gravity ( g )

above the Aluminum strip (Gravity Controller Device).

Fig. 1 – Experimental set ups for controlling the Gravitational Mass of the Aluminum strip, and the Gravity acceleration above it.

Ø20 AWG 1mm 1mm 11mm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 10mm 10mm 146 mm 144 mm 5mm 250 mm 200 mm 5mm Aluminum strip 99.9% Aluminum ρ = 2700 kg/m-3 σ = 3.5x107 S/m Length: L=3528 mm Width: l =5mm Thickness:Δx =3μm Gravitational Mass (strip) g

m

Rest inertial mass m_i₀₍_strip₎ Ø20 AWG Proof mass ( )

( )

g m P= _g _p χ 1 < χ Dynamometer Ø20 AWG Precision balance Max. 320g 3 digits resolution:1mg Precision balance Max. 320g 3 digits resolution:1mg Function Generator

Frequency range > 1μHz(Sine) Maximum output current > 0.075 App

l

Function Generator

Frequency range > 1μHz(Sine) Maximum output current > 0.075 App

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0.1 0.1428 0.1352 0.9473 9.2930 0.2 0.1428 0.0395 02769 2.6771 0.3 0.1428 -0.2304 -1.6139 -15.8323 0.4 0.1428 -0.6651 -4.6577 -45.6922 0.5 0.1428 -1.2454 -8.7217 -85.5598 0.1 0.1428 0.1418 0.9933 9.7442 0.2 0.1428 0.1279 0.8959 8.7887 0.3 0.1428 0.0794 0.5178 5.0796 0.4 0.1428 -0.0415 -0.2909 -2.8537 0.5 0.1428 -0.2209 -1.5469 -15.1750 0.1 0.1428 0.1425 0.9980 9.7903 0.2 0.1428 0.1383 0.9686 9.5019 0.3 0.1428 0.1207 0.8458 8.2972 0.4 0.1428 0.0779 0.5456 5.3523 0.5 0.1428 0.0014 0.0098 0.0961

Tab. 1 – Calculated results for the Gravitational Mass of the Aluminum stripmg₍strip₎, and for the Gravity acceleration above the Aluminum strip

( )

χ

g .

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Next, we will show that by reducing the thickness of the Aluminum strip to

3 nm

† it is possible to design a similar device for working with frequency up to f =2mHz. In this case the period T of the wave is

T

=

500 s

≅

8 .

3 min

‡ . Let us then consider an Aluminum strip with thickness, width and

length as shown in Fig.2. If the maximum

applied voltage is then,

according to Eq. (13), we have

nm

3

5 mm

3,528mm

volts

V

max

=

22 .

2

( ) ( )

(

)

( )

16

1

10

6 .

1

2

1

10

758 .

1

2

1

3 8 3 4 max 2 3 27 0

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

⎥

⎦

⎤

⎢

⎣

⎡

−

×

+

−

=

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

⎥

⎦

⎤

⎢

⎣

⎡

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

×

+

−

=

− −

f

L

V

m

r strip i strip g

ρ

σ

μ

χ

For f =2mHz, Eq. (16) gives

( ) ( )

( )

17

46 .

0

−

=

strip i strip g

m

χ

Through the Aluminum strip, the maximum intensity of the electrical current is given by

(

)

(

)

(

)

[

(

)(

)

mA m S S L V S j i 3 . 3 10 5 10 3 528 . 3 2 . 22 / 10 5 . 3 7 9 3 max max max = = × × × =

]

= = = − −

σ

The electrical resistance of the Aluminum strip is

†

Ultra Thin Aluminum Nanofoils (foils with nanometers thicknesses) are manufactured, for example by American Elements - The Advanced Materials Manufacturer (See the available nanofoils at:

https://www.americanelements.com/ultra-thin-aluminum-nanofoil-7429-90-5). ‡

In the case of the first device the frequency it were

Hz

f=10

μ

and the period T=105s≅27.77hours.

(

×

)(

[

×

)(

×

)

]

=

Ω

=

− −

6720

10

5

10

3

10

5 .

3

528 .

3

3 9 7

S

L

R

strip

_σ

Therefore, the maximum dissipated power by the strip has now the following value

mW

i

R

P

_stripmax

=

_strip _max2

=

73 .

2

Note that this power is almost the double of the power in the first device (37.7mW).

Let us now verify if the area of the surface of the Aluminum strip (area of the surface of thermal transfer ; 5mm x 3,528mm) is sufficient to transfer to the surrounding air all the heat produced by the strip (in order to avoid the fusion of the strip).

The coefficient of heat transfer, , can be expressed by the following equation [

h

4, 5,6]

(

)

_{( )}

18 T A t Q h nΔ Δ Δ =

where

Δ

Q

Δ

t

(in W) is the dissipated power ; (in m

n

A

2) is the necessary area of the surface of thermal transfer and the difference of temperature between the area of the solid surface and the surrounding fluid (K).

T Δ

When the surrounding fluid is the air , the heat transfer coefficient, , varies from 10

h

up to

K

m

W

.

°

100

2 [7]. Assuming

h

=

10 W

m

2

.

°

K

,

and ΔT =1°K, then for , Eq.

(18) gives

mW

P

_stripmax

=

73 .

2 (

)

_{( )}

19 10 32 . 7 3 2 m T h t Q A_n = × − Δ Δ Δ =

Since the area of the surface of the Aluminum strip is

n

strip

mm

m

A

=

5 ×

3 ,

528 =

17 .

6 ×

10

−3 2

>>

Then, we can conclude that the area of the

Aluminum strip is sufficient to transfer to the surrounding air all the heat produced by it. The

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Fig. 2 – Device using Aluminum strip with 3nm thickness.

Ø20 AWG 1mm 1mm 11mm 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 10mm 10mm 146 mm 144 mm 5mm 250 mm 200 mm 5mm Aluminum strip 99.9% Aluminum ρ = 2700 kg/m-3 σ = 3.5x107 S/m Length: L=3528 mm Width: l =5mm Thickness:Δx =3nm Gravitational Mass (strip) g

m

Rest inertial mass m_i₀₍_strip₎ l

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References

[1] De Aquino, F. (2010) Mathematical Foundations

of the Relativistic Theory of Quantum Gravity, 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] Halliday, D. and Resnick, R. (1968) Physics, J. Willey & Sons, Portuguese Version, Ed. USP, p.1410.

[4] Halliday, D., Resnick, R., Walker, J., (1996) Fundamentos de Física 2 - São Paulo: Livros Técnicos e Científicos Editora, 4a Edição, 1996. [5] Sears, F. W. E Zemansky, M. W. – Física - vol. 2, cap. 15, Ed. Universidade de Brasília, Rio de Janeiro – 1973.

[6] Heat transfer coefficient

.

Available at: https://en.wikipedia.org/wiki/Heat_transfer_coefficient

[7] Overall heat transfer coefficient. Available at: