If,
\(\cfrac { m_{ e }v^{ 2 } }{ r_{ e } } =Av^{ 2 }+F_{ Le }-F_{ Lp }-F_T\)
\(F_T=\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o }r^{ 2 }_{ e } } \)
\( F_{ Le }=\pi \sum\limits _{ i=1 }^{ n-1 }{ \cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ ei } } }=\sum\limits _{ i=1 }^{ n-1 }{ \cfrac { q^{ 2 } }{ 4\varepsilon _{ o }r^{ 2 }_{ ei } } } \)
\( F_{ Lp }=\pi \cfrac { nq^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ e } }=\cfrac { nq^{ 2 } }{ 4 \varepsilon _{ o }r^{ 2 }_{ e } } \)
the quadratic equation or \(r_e\) becomes,
\( (Av^{ 2 }+\sum \limits_{ i=1 }^{ n-1 }{ \cfrac { q^{ 2 } }{ 4\varepsilon _{ o }r^{ 2 }_{ ei } } } )r^{ 2 }_{ e }-m_{ e }v^{ 2 }r_{ e }-\cfrac { nq^{ 2 } }{ 4 \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ 4\pi \tau _{ o } } =0\)
\( (Av^{ 2 }+\sum \limits_{ i=1 }^{ n-1 }{ \cfrac { q^{ 2 } }{ 4\varepsilon _{ o }r^{ 2 }_{ ei } } } )r^{ 2 }_{ e }-m_{ e }v^{ 2 }r_{ e }-\cfrac { 1 }{ 4 } \left\{ \cfrac { nq^{ 2 } }{ \varepsilon _{ o } } +\cfrac { T_{ e }T_{ n } }{ \pi \tau _{ o } } \right\} =0\)
The y-intercept is negative always.
If we rearrange the terms again,
\(Av^{ 2 }r^{ 2 }_{ e }-m_{ e }v^{ 2 }r_{ e }+\cfrac { 1 }{ 4 } \left\{ \sum\limits _{ i=1 }^{ n-1 }{ \cfrac { q^{ 2 } }{ \varepsilon _{ o }r^{ 2 }_{ ei } } } r^{ 2 }_{ e }-\cfrac { nq^{ 2 } }{ \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ \pi \tau _{ o } } \right\} =0\)
We may get a zero intercept and even a double root, but the term,
\(\sum \limits_{ i=1 }^{ n-1 }{ \cfrac { q^{ 2 } }{ \varepsilon _{ o }r^{ 2 }_{ ei } } } r^{ 2 }_{e}\)
that appears as part of the y-intercept is really awkward and wrong.
There is only one thing to do, summon the spirit of Newton,
\(\cfrac { m_{ e }v^{ 2 } }{ r_{ e } } =Av^{ 2 }+F_{ Le }+F_G-F_{ Lp }-F_T\)
\(F_G=G\cfrac{n.m_pm_e}{r^2_e}\)
\( (Av^{ 2 }+\sum \limits_{ i=1 }^{ n-1 }{ \cfrac { q^{ 2 } }{ 4\varepsilon _{ o }r^{ 2 }_{ ei } } } )r^{ 2 }_{ e }-m_{ e }v^{ 2 }r_{ e }+\cfrac { 1 }{ 4 } \left\{{4nGm_pm_e}- \cfrac { nq^{ 2 } }{ \varepsilon _{ o } } -\cfrac { T_{ e }T_{ n } }{ \pi \tau _{ o } } \right\} =0\)
then the y-intercept can be zero, positive to allow for two roots of \(r_e\) and even positive enough to result in a double root.
If we put in the numbers, however, it seems that we need a positive attractive force desperately.
