Newton Has No $\pi$

If the Newton's Gravitational Constant, $G$ has already incorporated the term $\pi$ that was introduced as a result of $F_{gB}$ in the expression,

$F_{gB}=\pi G\cfrac{m_sm_p}{r^2_{or}}$

then in gravity calculations, there will be significant error from the factor $\pi$.

Seriously.  Because for masses not in spin and orbital motion, the attractive force between them is just,

$F_{g}=G\cfrac{m_sm_p}{r^2}=G_o\cfrac{m_sm_p}{4\pi r^2}$

No $\pi$.

Seriously.  All calculations for masses of heavenly bodies will have to be changed.