Monday, November 17, 2014

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.