\psi - \log \psi - 1 = 4 \left ( \Xi^{-1} - \log {\Xi^{-1}} - 1 \right )

When \Xi^{-1} >> 1 we have

-\log \psi = 4\Xi^{-1} or \psi = e^{-4/\Xi}

When \Xi^{-1} << 1 we have

\psi = -4 \log \Xi^{-1} = 4 \log \Xi

When \Xi \sim 1 we have

\psi - (\psi - 1) + (\psi-1)^2/2 - 1 = 4(\Xi^{-1} - (\Xi^{-1}-1)+ (\Xi^{-1}-1)^2/2 - 1)

(\psi-1)^2/2 = 2 (\Xi^{-1}-1)^2

\psi= 1 + 2 (\Xi^{-1}-1)

Vis-visa equation

v = \sqrt{\mu} \sqrt{\frac{2}{r}-\frac{1}{a}}

T = 2 \pi \sqrt{\frac{a^3}{\mu}}

v^2=\mu(\frac{2}{r}-\frac{1}{a})

now since \mu = v_k^2 r we have

v^2=v_k^2 (2 - r/a) which can be solved for a

And the angular momentum is

h = r \times v

which can be used to solve for e

(1-e^2)=\frac{h^2}{r^2v_k^2} \frac{r}{a}

We can then calculate the phase using

l=\frac{b^2}{a}=\left ( 1-e^2 \right )a

r=\frac{l}{1+e \cos \theta}

\frac{r}{a} = \frac{(1-e^2)}{1+e \cos \theta}

\cos{\theta}=\frac{(1-e^2)a-r}{re}

General case

FIrst calculate the semi-major axis

c_s^2+v_k^2+2 c_s v_k \sin \phi = v_k^2 (2-r/a)

r/a = 1 - x^2-2x \sin \phi

a = \frac{r}{1-x^2-2x \sin \phi}

Note that this sets a limit on x for material remaining elliptical…

x^2+2x-1=0

x = \sqrt{2}-1 = .414...

Setting a=r gives us the location that launches an elliptic orbit with the same period

x^2+1+2x \sin \phi=2-r/a

x+2 \sin \phi = 0

\phi = \sin^{-1} \left ( -x / 2 \right )

We can then calculate the ellipticity using

h=r v_k (1+x\sin\phi)

e = \sqrt{1 - \frac{h^2}{a r v_k^2}} = \sqrt{1 - (1+x \sin \phi)^2\frac{r}{a}}

e = \sqrt{1 - (1+x \sin \phi)^2 (1 - x^2-2x \sin \phi)}

And the phase using

\cos{\theta}=\frac{(1-e^2)a-r}{re}

\cos{\theta}=\frac{(1+x \sin \phi)^2-1}{ \sqrt{1 - (1+x \sin \phi)^2(1 - x^2-2x \sin \phi)}}

parallel kick

(c_s \pm v_k)^2 = v_k^2 (2-r/a)

(x \pm 1)^2=2-r/a

1 \pm 2 x + x^2 = (2-r/a)

\pm 2x + x^2 = 1 - r/a

a = \frac{r}{1 \mp 2x - x^2}

(1 \mp e)=\frac{r}{a}={1 \mp 2x - x^2}

e = 2x \pm x^2

orthogonal kick

c_s^2 + v_k^2=v_k^2 (2-r/a)

x^2=(1 - r/a)

a = \frac{r}{1-x^2}

h = r \times v = r v_k

(1-e^2)=\frac{h^2}{a\mu} = \frac{r}{a}

e = \sqrt{1 - \frac{r}{a}} = \sqrt{1-x^2}

r=\frac{l}{1+e \cos \theta}

l=\frac{b^2}{a}=\left ( 1-e^2 \right )a

\frac{r}{a} = \frac{(1-e^2)}{1+e \cos \theta}

\cos{\theta}=\frac{(1-e^2)a-r}{re}

1-x^2=\frac{x^2}{1+\sqrt{1-x^2}\cos \theta}

\cos{\theta} = \frac{\frac{x^2}{1-x^2} - 1}{\sqrt{1-x^2}}

\cos{\theta} = \frac{2 x^2 - 1}{\left ( 1-x^2 \right)^{3/2}}

For outer edge, we are headed toward aphelion so perihelion must have \theta must be between 0 and \pi

For inner edge we are headed toward perihelion, so \theta must be between \pi and 2 \pi

Equi-period direction

c_s^2+v_k^2+2 c_s v_k \sin \phi = v_k^2 (2-r/a)

x^2+1+2x \sin \phi=2-r/a

x+2 \sin \phi = 0

\phi = \sin^{-1} \left ( -x / 2 \right )

Arbitrary direction

c_s^2+v_k^2+2 c_s v_k \sin \phi = v_k^2 (2-r/a)

r/a = 1 - x^2-2x \sin \phi

h=r v_k (1+x\sin\phi)

e = \sqrt{1 - \frac{h^2}{a r v_k^2}} = \sqrt{1 - (1+x \sin \phi)^2\frac{r}{a}}

e = \sqrt{1 - (1+x \sin \phi)^2 (1 - x^2-2x \sin \phi)}

\sin{\phi} = 1 \rightarrow e = \sqrt{1 - (1 + x)^2(1-x^2-2x)} = \sqrt{1 - ( (1 + 2x + x^2)(1 - x^2 - 2x))} = \sqrt{1 - (1 - x^2 - 2x + 2x - 2x^3 - 4x^2 + x^2 - x^4 - 2x^3)} = \sqrt{1-(1 - 4x^3 - 4x^2 - x^4 )} = \sqrt{4x^2+4x^3+x^4} = 2x+x^2