So close to a perfect circle

We can write the total energy of the system as a function of the distance from the center:

$U(x) = P/x + (L^2)/2mx^2$

So we can get the frequency for small oscillations:

$dU/dx = -P/x^2 - (L^2)/mx^3 = U'(x)$

$U'(r) = 0$ implies that r is the equilibrium distance

$r = -(L^2)/mP$

Now we apply this "r" to U''(x) = dU'/dx

$U''(r) = 2P/r^3 + 3(L^2)/mr^4$

$U''(r) = 2(P^4)(m^3)/L^6 + 3(P^4)(m^3)/L^6 = 5(P^4)(m^3)/L^6$

We can use that the frequency for small oscillations can be written as

$w = (U''(r)/m)^{1/2}$

$w = (P^2)m(5^{1/2})/L^3$

Note that this frequency defines the change on the maximums and minimums of the trajectory, so we can use that:

$f(L) ~ L^{-3}$

So

$f_1/f_2 = (L_1/L_2)^{-3}$

$f_2 = 0.001Hz$

For that kind of potential, the equilibrium in the radial direction corresponds to the radius of the circular motion, and the small fluctuation will give you a "slight" ellipse, which means the time period of the oscillation is almost equal to the time period of the circular orbit. Consider at the circular orbit:

$mv^2/r \sim 1/r^2 \Rightarrow v^2r = \mbox{const}$

Using that $L \sim vr$ and $f \sim v/r$ , we get $L^3f = \mbox{const}$ , so the frequency is proportional to inverse angular momentum scale cube!

We can write it as:

$f(L)=f(L_1) (L_1/L)^3$

So, we get $f(L_2)=0.001 ~\mbox{Hz}$

At $L_3=0$ , the bare frequency goes to infinity. Indeed, it shows us that sometimes the bare value of a quantity in theoretical physics might not have any physical meaning!