Earth Shield Frequency

From the post "A Shield" dated 27 May 2016,

$f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{gradient|_{\pi}}{m}}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2sech^2(\pi)}{m}}=\cfrac{sech(\pi)}{2\pi}\cfrac{G}{\sqrt{m}}$

$G=\cfrac{\pi \sqrt { 2{ mc^{ 2 } } }  }{a_e}$

$a_E=6371e3m$

$m_E=5.972e24 kg$

$G=$pi*sqrt(2*5.972*(299792458)^2*10^(24))/(6371e3)

$G=5.109e14$

If instead we use half the gradient at $\theta_{\psi\,\pi}=3.135009$

$gradient|_{\pi}=sech^2(3.135009)$

without the constant $G$.

Approximating the tangent with a line with gradient $tan(\beta)$,

$tan(\beta)=sech^2(3.135009)$

$\cfrac{1}{2}\beta=\cfrac{1}{2}tan^{-1}(sech^2(3.135009))$

$tan(\cfrac{1}{2}\beta)=0.00377$

$f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.00377}{m}}$

$f_{res}=0.009772\cfrac{G}{\sqrt{m}}$

$f_{res}=2.0430 Hz$

which is low....

If we instead do not half the gradient at $\theta_{\psi\,\pi}=3.135009$,

$tan(\beta)=sech^2(3.135009)=0.00754$

$f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.00754}{m}}$

$f_{res}=0.01382\cfrac{G}{\sqrt{m}}$

$f_{res}=2.8893 Hz$

If we take the gradient, without $G$ at the origin instead,

$sech^2(0)=1$

$\cfrac{1}{2}\beta=0.39270$

$tan(\cfrac{1}{2}\beta)=0.41421$

$f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.41421}{m}}$

$f_{res}=0.10243\cfrac{G}{\sqrt{m}}$

$f_{res}=21.41 Hz$

What happen when $\psi$ is pushed along its profile?  Maybe $a_{\psi}$ increases...chicken!  Big chicken!

Or,

$tan(\beta)=sech^2(0)=1$

$\cfrac{1}{2}tan(\beta)=0.5$

$f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{G^2*0.5}{m}}$

$f_{res}=0.11254\cfrac{G}{\sqrt{m}}$

$f_{res}=23.53 Hz$

We are comparing $\psi$ to the energy curve of a SHM system, $U=\cfrac{1}{2}kx^2$ the gradient of which $U^{'}=kx$.