We solve for the different types of orbits of the Kepler Problem (two body central force problems).

Introduction

In the previous post we reduced our two body problem to a single effective potential problem:

$$U_{eff}(r) = U(r) + U_{cf}(r) = U(r) + \frac{l^2}{2 \mu r^2}$$

We can now determine the allowed types of orbits. To do this we are going to do some variable substitution, and determine $r(\phi)$ which will allow us to see the orbits more clearly. During this process we will define a new variable called eccentricity that determines the orbit.

Solving for $r(\phi)$

We ended up last time with:

$$\mu \ddot{r} = F(r) + \frac{l^2}{\mu r^3}$$

For reasons I don’t quite understand yet, the problem becomes simpler if we replace $r$ with $1/u$ (something about quasilinear).

We also need to figure out how to write $\ddot{r}$ in terms of $r(\phi)$.

$$\ddot{r} = \frac{d}{dt} \frac{d}{dt} r = \frac{d \phi}{dt} \frac{d}{d\phi} \frac{d \phi}{dt} \frac{d}{d\phi} r = \dot{\phi}^2 \frac{d^2}{d\phi^2} \frac{1}{u} = (\frac{l u^2}{\mu }) \frac{d}{d \phi} (\frac{l u^2}{\mu}) (- \frac{1}{u^2}) \frac{d u}{d \phi} = - \frac{l^2 u^2}{\mu^2} \frac{d^2 u}{d \phi^2}$$

In step four we used the chain rule to partially evaluate $\frac{d}{d \phi} \frac{1}{u} = - \frac{1}{u^2} \frac{d u}{d \phi}$

So now we can substitute our result back into the original equation of motion:

$$u"(\phi) = -u(\phi) - \frac{\mu}{l^2 u(\phi)^2}F$$

Next we plug in our force equation: $F = -G m_1 m_2 / r^2 = -\gamma u^2$:

$$u"(\phi) = -u(\phi) - \frac{\mu}{l^2 u(\phi)^2} (- \gamma u^2) = -u(\phi) - \frac{\gamma \mu}{l^2}$$

This equation is surprisingly simple due to the fact that our force equation is an inverse squared force, the $u(\phi)$ terms cancel out. (Coincidentally I’m in the process of solving a related problem in a physics problem book, page 8 problem 8). I believe this implies that only inverse-square forces have stable orbits, but I haven’t proved it yet.

Since the last term is constant, the solution is just the constant plus a sinusoid:

$$u(\phi) = \frac{1}{r} = \frac{\gamma \mu}{l^2} + A \cos\phi = \frac{\gamma \mu}{l^2} (1 + \epsilon \cos\phi) = \frac{1}{c}(1 + \epsilon \cos \phi)$$

In the last stage we defined a new variable we will call the eccentricity (which will become obvious why soon), and $c = \frac{l^2}{\gamma \mu}$ for bookkeeping.

Solving for r:

$$r(\phi) = \frac{c}{1 + \epsilon \cos \phi}$$

Types of orbits

Bounded

Looking at our previous equation, if we assume $\epsilon < 1$, then r will oscillate between $r_{min} = \frac{c}{1-\epsilon}$ and $r_{max} = \frac{c}{1+\epsilon}$), which looks suspiciously like an ellipse (which is why we call $\epsilon$ the eccentricity). (You can prove it by converting to cartesian coordinates and using the identity $\cos(\tan^{-1}(y/x)) == \frac{1}{\sqrt{1 + y^2/x^2}}$, you will get the equation for an ellipse $x^2/a^2 + y^2/b^2 = 1$).

If $\epsilon = 0$ we will have a constant r, meaning a circle.

Unbounded

We have two possible unbounded states, $\epsilon = 1$ and $\epsilon > 1$.

If $\epsilon = 1$, then r goes to infinity at $\pi$ and $-\pi$. We can show it’s a parabola by converting to cartesian coordinates:

$$r(\phi) = \frac{c}{1 + \cos\phi} = \sqrt{x^2 + y^2} = \frac{c}{1 + \frac{1}{\sqrt{1 + y^2/x^2}}}$$

Plugging into mathematica we get:

$$y^2 = c^2-2cx$$

Which is the equation of a (horizontal) parabola.

If $\epsilon > 1$ we will have a maximum $\phi_{max}$ which defines the range for some sort of hyperbola.

Eccentricity Plot

There’s a fun plot here showing the trajectories of different eccentricities:

https://en.wikipedia.org/wiki/Orbital_eccentricity

Simulation

Not a huge update, but now the simulation shows the eccentricity of the orbit.

https://blog.c0nrad.io/sims/twobody/

Conclusion

This is the last post on two body central force problems. I’m not sure what I’ll do next, but hopefully something just as fun and insightful.