The following two pictures show the Kepler orbit of two masses $m_1, m_2$, where $m_1/m_2=7/3$, as I calculated them using $$ r(\varphi)=\frac{a(1-\varepsilon^2)}{1+\varepsilon\cos\varphi},\quad\boldsymbol{x}_1=\frac{m_2}{M} \boldsymbol{r},\quad\boldsymbol{x}_2=-\frac{m_1}{M} \boldsymbol{r} $$ with eccentricity $\varepsilon=0.825,1.25$ and $M=m_1+m_2$. In the pictures, $\times$ denotes the focus, $+$ the origin, that is the barycenter. The brown orbit belongs to $\boldsymbol{r}=\boldsymbol{x}_1-\boldsymbol{x}_2$, while pink belongs to $\boldsymbol{x}_1,\boldsymbol{x}_2$, respectively.

Now, the elliptic case is clear to me and looks as expected. But what about the hyperbolic case? How to explain the $>>$-shaped arcs ($<<$ respectively)? Before moving from one $>$-arc to the other $>$-arc, the object goes to infinity, so is that really the actual orbit or do I have an error in my simulation? Elliptic Kepler orbit Hyperbolic Kepler orbit enter image description here

  • $\begingroup$ You know, thinking about it, you should change the sign in the $r$ equation for the hyperbolic case. The nominator is coming from $ep$ and should stay positive. I wrote my answer assuming that your equation is correct but now I will change it accordingly (the idea stays the same) $\endgroup$ – OON Sep 25 '18 at 8:33
  • $\begingroup$ Ok, I remember now, we take $a<0$ for hyperbolic trajectories. $\endgroup$ – OON Sep 25 '18 at 9:05

The elliptic orbits correspond to the $e<1$ (I'll use $e$ for eccentricity as $\varepsilon$ is usually used for the specific energy). Then for all values of $\phi$ the denominator never vanishes and stays $1+e\cos\phi>0$.

However for $e>1$ when $\cos\phi\mapsto-\frac{1}{e}$ the denominator vanishes and therefore $r$ goes to infinity. This separates two cases, for $-\pi+\arccos{\frac{1}{e}}<\phi<\pi-\arccos{\frac{1}{e}}$ the denominator is positive, whereas otherwise it is negative. The second case gives us this undesirable second hyperbola that corresponds to the bodies repelling each other instead of attracting. It perfectly satisfies the equations of motion too.

For hyperbolic trajectories $a=-\frac{\mu}{2E}<0$ and $e>1$ so the numerator $a(1-e^2)$ is still postitive. That means that unless $-\pi+\arccos{\frac{1}{e}}<\phi<\pi-\arccos{\frac{1}{e}}$ the whole $r$ is negative. That's why this extra hyperbola is in the wrong semiplane considering angles it corresponds to.

However $r$ being the vector $\vec{r}$ norm must be positive therefore these values of $\phi$ represent the unphysical region.

This forbidden region of course can't be reached from the correct hyperbola as it takes infinite time to reach the infinity and "go over it".

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  • $\begingroup$ Great, thanks! That solved it for me. I made another plot, now with just the correct hyperbola; before, I also had the wrong sign in the numerator in the hyperbolic case that did move the origin to the right focus, which is wrong. So now the third picture is correct. $\endgroup$ – Don Fuchs Sep 25 '18 at 14:15

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