# Tying knots via gravity-assisted spaceship trajectories

Q. Can every knot be realized as the trajectory of a spaceship weaving among a finite number of fixed planets, subject to gravity alone? To make this more specific, let $S$ be a large sphere, containing a finite number of planet-points $P=\{p_1, \ldots, p_n\}$. The planet-points have (in general) different masses, and are fixed in $\mathbb{R}^3$. A spaceship $x$ approaches $S$ from $\infty$, interacts via gravity with the point masses, and eventually exits $S$ to $\infty$. Define the knot $K$ realized by the ship's trajectory as the path of $x$ plus a connection between the two ends at $\infty$. (Assume those two $\infty$-ends are distinct.)

Q'. For any given knot $K$, can one arrange point masses in $P \subset S$ and a line and speed of approach to $S$ so that $x$'s path weaves $K$ by interacting with the planets via gravity alone, i.e., without the use of rocket fuel?

One approach might be to design a "gadget" that mimics a vertex $v$ of a stick knot and $v$'s two incident segments. But preventing the vertex gadgets from interfering with one another might not be straightforward. Cassini gravity-assist trajectory. Image from NASA/JPL.

• You can also ask if every knot type occurs as a periodic orbit in the system. – Thomas Rot Jun 1 '18 at 11:33
• @ThomasRot: Yes. I suspect that is considerably more difficult to achieve. – Joseph O'Rourke Jun 1 '18 at 11:59
• You can make a hyperbolic orbit around a single planet, which on large scale is just a turn. Now tie the knot with a broken line and place masses close to the turns. Note that you can use an arbitrarily small mass to make any turn in a small external field if you place it strategically near the empty space trajectory (this requires proof, of course, but intuitively it is clear). Thus, adding new turns won't spoil the existing ones and you should be able to happily do the induction. I'll try to make some rigorous sense of it later. – fedja Jun 2 '18 at 0:20
• I would expect fedja’s method to be extendable to produce exactly periodic orbits, assuming you close off the induction with some sort of fixpoint theorem. – Geoffrey Irving Jun 2 '18 at 23:18
• @JosephO'Rourke Thank you for the clarification, I meant that my statement about realizing differentiable paths using a Lagrangian needs the paths to be twice differentiable at least for it to be true. I think fedja's intuition is more precise, and correct -- to make the observation slightly more explicit, we can note that the contribution to the spaceship Lagrangian for each planet is of the form $G\frac{m_i}{r_i},$ so as suggested we should be able to place sufficiently small masses sufficiently close to the unperturbed path to reproduce any loop. – Alec Rhea Jun 3 '18 at 6:41