Changes between Version 1 and Version 2 of u/johannjc/scratchpad13

Timestamp:

03/11/16 13:47:50 (9 years ago)

Author:

Jonathan

Comment:

—

u/johannjc/scratchpad13

@@ -17 +17 @@
 
 $\psi= 1 + 2 (\Xi^{-1}-1)$
+
+
+== 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} $
+
+$\cos{\theta}=\frac{(1+x \sin \phi)^2-1}{ \sqrt{1 - (1+x \sin \phi)^2(1 - x^2-2x \sin \phi)}}$
+
+
+
+$\sin{\phi} = 1 \rightarrow e = \sqrt{1 - (1 + x)^2(1-x^2-2x)} = \sqrt{1 - ( (1 + 2x + x^2)(1 - x^2 - 2x))} = \sqrt{1 - (1 - x^2 - 2x + 2x - 2x^3 - 4x^2 + x^2 - x^4 - 2x^3)} = \sqrt{1-(1  - 4x^3 - 4x^2 - x^4 )} = \sqrt{4x^2+4x^3+x^4} = 2x+x^2$
+
+
+
+
+
+== 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 ==
+
+$c_s^2+v_k^2+2 c_s v_k \sin \phi = v_k^2 (2-r/a)$
+
+$r/a = 1 - x^2-2x \sin \phi$
+
+$h=r v_k (1+x\sin\phi)$
+
+$e = \sqrt{1 - \frac{h^2}{a r v_k^2}} = \sqrt{1 - (1+x \sin \phi)^2\frac{r}{a}}$
+
+$e = \sqrt{1 - (1+x \sin \phi)^2 (1 - x^2-2x \sin \phi)}$
+
+
+
+$\sin{\phi} = 1 \rightarrow e = \sqrt{1 - (1 + x)^2(1-x^2-2x)} = \sqrt{1 - ( (1 + 2x + x^2)(1 - x^2 - 2x))} = \sqrt{1 - (1 - x^2 - 2x + 2x - 2x^3 - 4x^2 + x^2 - x^4 - 2x^3)} = \sqrt{1-(1  - 4x^3 - 4x^2 - x^4 )} = \sqrt{4x^2+4x^3+x^4} = 2x+x^2$
+
+