One Too Many...

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where $\pi$ has been replaced by an unknown $x$ and $n=77$.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!