When Drag Is Good For You, Collapsing Universe Otherwise

From the previous post "Drag and A Sense of Lightness"

$F_D=Av^2$

If the helical path of the electron at terminal velocity is due only to the drag force.

$\cfrac{m_ev^2}{r}=Av^2=(1-\cfrac{r_{ec}}{r_e})\cfrac{m_e}{r_{e}}v^2$

$\cfrac { 1 }{ r } =(1-\cfrac { r_{ ec } }{ r_{ e } } )\cfrac { 1 }{ r_{ e } }$

$r=r_{ e }\cfrac{1}{(1-\cfrac { r_{ ec } }{ r_{ e } } )}$

$r$ = 2.5e-11/(1-5.636e-5) = 2.50014e-11 m

Since,

$(1-\cfrac { r_{ ec } }{ r_{ e } } )<1$

$r>r_e$

That means, under normal circumstances, the electron will never be in a helical radius smaller than the atomic radius.  It will not crash into the positive nucleus.  The presence of a positively charged nucleus adds to the centripetal force and the radius of circular motion is smaller.

Have a nice day.