Spider Diagram

Draw a spider diagram with countours of \(V_{r}-V_{sys}=0.2,0.4,0.6,0.8\) times \(V_{max}\sin{i}\); show that the last of these forms a closed loop.

First, from Eq.(5.5), \[ V_{r}(R,i;\phi)=V_{sys}+V(R)\sin{i}\cos{\phi} \] which we can re-write it into \[ \frac{V_{r}-V_{sys}}{V_{max}\sin{i}}=\alpha=\frac{V(R)}{V_{max}}\cos{\phi} \] . Where \(V^2(R)=GM\frac{R^2}{(R^2+a^2)^{\frac{3}{2}}}\), and \(V_{max}^2=\frac{2GM}{3\sqrt{3}a}\), hence, \[ \alpha^2=\frac{3\sqrt{3}}{2}\frac{\xi^2\cos{\phi}^2}{(\xi^2+1)^{\frac{3}{2}}} \] where \(\xi=\frac{R}{a}\). By transformation \((\xi,\phi)\rightarrow(x,y)\), \[ x=\xi\cos{\phi} \] \[ y=\xi\sin{\phi} \] , we can re-write \[ y^2=(\frac{27}{4})^{\frac{1}{3}}x^{\frac{4}{3}}\alpha^{\frac{-4}{3}}-x^2-1 \] . This gives us the plot as follows.