Actually,
\(tanh(\pi)\ne1\)
but,
\(tanh(\pi)=0.996272\)
and
\(\cfrac{1}{tanh(\pi)}=1.033741\)
If we apply this correction to \(G_o\), the gravitational constant calculated before,
\(G_o=\cfrac{299792458^2*ln(cosh(\pi))}{2\pi*9.029022e26 }*\sqrt{3}*{tanh(\pi)}\)
\(G_o=6.69855e-11\)
and the quoted value of \(G_o\) is \(6.67259e(-11)\).
Without the correction,
\(G_o=\cfrac{299792458^2*ln(cosh(\pi))}{2\pi*9.029022e26 }*\sqrt{3}\)
\(G_o=6.72361e-11\)
Which suggests that, the particle is below the limit set by \(tanh(\cfrac{G}{\sqrt{2mc^2}}a_{\psi\,\pi})=1\) and it is likely that when many particles coalesce into one,
\(\left\lceil n \right\rceil =\left\lceil 77.486 \right\rceil =77\)
instead of \(\left\lfloor n \right\rfloor =78\).
From the post "Sticky Particles Too...Many" dated 24 Jun 2016,
\(n\left(\cfrac{a_{\psi\,c}}{a_{\psi\,\pi}}\right)^3=n\left(\cfrac{0.7369}{x}\right)^3=1\)
where \(\pi\) has been replaced by an unknown \(x\) and \(n=77\).
\(77\left(\cfrac{0.7369}{x}\right)^3=1\)
\(x=3.135009\)
So,
\(G_o=\cfrac{299792458^2*ln(cosh(3.135009))}{2\pi*9.029022e26 }*\sqrt{3}\)
\(G_o=6.70562e-11\)
with the correction factor, \(tanh(3.135009)\),
\(G_o=\cfrac{299792458^2*ln(cosh(3.135009))}{2\pi*9.029022e26 }*\sqrt{3}*tanh(3.135009)\)
\(G_o=6.68029e-11\)
And before we get obsessed with decimal number, lunch time!
