Rolling Inside

If it is possible that the electron be inside the proton orbit closer to the center of the orbit, the attraction from the weak field acts against the attraction from the proton and reduces the centrifugal force,

$\cfrac { m_{ e }v^{ 2 } }{ r_{ e } } =\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o }r^{ 2 }_{ e } } -qE^{'}_{or}$

$E_{or}=\cfrac{\tau_o}{2}\cfrac{Tv_{\small{T}}}{\{r^2_{\small{T}}+(r_{or}-r_{e})^2\}^{3/2}}.({r_{or}-r_{e}})$

If we again assume that $r_{e}\lt\lt r_{or}$,

$qE_{or}=A.T$ and,

$A=\cfrac{\tau_o}{2}\cfrac{qv_{\small{T}}}{\{r^2_{\small{T}}+r_{or}^2\}^{3/2}}.{r_{or}}$

then

$m_{ e }v^{ 2 }.r_{ e }=\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } -AT.r^{ 2 }_{ e }$

$AT.r^{ 2 }_{ e }+m_{ e }v^{ 2 }.r_{ e }-\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } } =0$

$r_{ e }=\cfrac { -m_{ e }v^{ 2 }\pm \sqrt { (m_{ e }v^{ 2 })^{ 2 }+4AT\cfrac { q^{ 2 } }{ 4\pi \varepsilon _{ o } }  }  }{ 2AT }$

Consider only $r_{ e }\gt0$,

$r_{ e }=\cfrac { m_{ e }v^{ 2 } }{ 2AT } \left( \sqrt { 1+AT\cfrac { q^{ 2 } }{ \pi \varepsilon _{ o }(m_{ e }v^{ 2 })^{ 2 } }  } -1 \right)$

A illustrative plot shows that $r_e$ decreases monotonously with $T$

In this case the pull of the weak field due to a spinning positive temperature particle is stronger.  The electron rolls inside the proton orbit.

Does this actually happen?