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Revision as of 22:56, 25 December 2017
Sanchez Solution
In his uglily typeset but brilliantly innovative
http://vixra.org/pdf/1609.0217v3.pdf Calculation of the gravitational constant G using electromagnetic parameters 2016-09-14 ©©-by jesus.sanchez.bilbao@gmail.com peer reviewed, printed, and republished in
http://file.scirp.org/pdf/JHEPGC_2016122915423655.pdf Journal of High Energy Physics, Gravitation and Cosmology, 2017, 3, 87-95 independent researcher Jesús Sánchez from Bilbao derived the equation
α
g
α
2
2
π
=
G
m
e
α
2
2
π
c
h
=
G
m
e
ϵ
0
α
π
q
e
2
=
G
m
e
2
c
h
ϵ
0
2
π
q
e
4
=
G
h
8
π
3
c
3
r
e
2
=
ℓ
P
2
r
e
2
2
π
=
{\displaystyle {\frac {\alpha _{g}}{\alpha ^{2}2\pi }}={\frac {Gm_{e}}{\alpha ^{2}2\pi ch}}={\frac {Gm_{e}\epsilon _{0}}{\alpha \pi q_{e}^{2}}}={\frac {Gm_{e}2ch\epsilon _{0}^{2}}{\pi q_{e}^{4}}}={\frac {Gh}{8\pi ^{3}c^{3}r_{e}^{2}}}={\frac {\ell _{P}^{2}}{r_{e}^{2}2\pi }}=}
=
exp
(
α
π
2
2
−
1
α
2
)
=
e
96.891
=
2
−
139.784
=
10
−
42.079
=
8.33E-43
{\displaystyle =\exp({{\frac {\alpha \pi }{2{\sqrt {2}}}}-{\frac {1}{\alpha {\sqrt {2}}}}})=e^{96.891}=2^{-139.784}=10^{-42.079}={\texttt {8.33E-43}}}
because of
G
h
8
π
3
c
3
r
e
2
=
exp
(
ln
(
G
h
8
π
3
c
3
r
e
2
)
)
=
exp
(
α
π
2
2
−
1
α
2
)
{\displaystyle {\frac {Gh}{8\pi ^{3}c^{3}r_{e}^{2}}}=\exp(\ln({\frac {Gh}{8\pi ^{3}c^{3}r_{e}^{2}}}))=\exp({{\frac {\alpha \pi }{2{\sqrt {2}}}}-{\frac {1}{\alpha {\sqrt {2}}}}})}
because of
ln
(
G
h
8
π
3
c
3
r
e
2
)
=
α
π
2
2
−
1
α
2
{\displaystyle \ln({\frac {Gh}{8\pi ^{3}c^{3}r_{e}^{2}}})={{\frac {\alpha \pi }{2{\sqrt {2}}}}-{\frac {1}{\alpha {\sqrt {2}}}}}}
because of
0
=
α
π
2
2
−
ln
(
G
h
8
π
3
c
3
r
e
2
)
−
1
α
2
{\displaystyle 0={{\frac {\alpha \pi }{2{\sqrt {2}}}}-\ln({\frac {Gh}{8\pi ^{3}c^{3}r_{e}^{2}}})-{\frac {1}{\alpha {\sqrt {2}}}}}}
because of
0
=
∫
r
e
r
e
+
∂
r
(
r
0
π
2
4
r
f
−
r
−
1
G
h
8
π
3
c
3
r
e
2
+
r
−
2
r
f
2
)
∂
r
{\displaystyle 0=\int _{r_{e}}^{r_{e}+\partial r}\left(r^{0}{\frac {{\pi }{\sqrt {2}}}{4r_{f}}}-r^{-1}{\frac {Gh}{8\pi ^{3}c^{3}r_{e}^{2}}}+r^{-2}{\frac {r_{f}}{\sqrt {2}}}\right)\partial r}
because of
0
=
∫
r
e
r
e
+
∂
r
(
∫
0
1
1
−
c
2
∂
c
r
f
/
2
−
G
m
e
2
π
c
2
r
h
2
π
c
m
e
4
π
r
e
2
+
4
π
r
f
/
2
4
π
r
2
)
∂
r
{\displaystyle 0=\int _{r_{e}}^{r_{e}+\partial r}\left({\frac {\int _{0}^{1}{\sqrt {1-c^{2}}}\partial c}{r_{f}/{\sqrt {2}}}}-{\frac {{\frac {Gm_{e}}{{\sqrt {2}}\pi c^{2}r}}{\frac {h}{{\sqrt {2}}\pi cm_{e}}}}{4\pi r_{e}^{2}}}+{\frac {4\pi r_{f}/{\sqrt {2}}}{4\pi r^{2}}}\right)\partial r}
and i m about to explain the rest when rested:
IM Serious (discuss • contribs ) 22:55, 25 December 2017 (UTC) 
