# Orbit through L4 and L5

I was reading the Wikipedia article on Lagrangian points and doing the requisite wiki walk through the various quasi-satellites of Earth when a question occurred to me:

Could there be a stable or Lissajous orbit around the minor body perpendicular to the ecliptic that passes through or near L4 and L5?

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Hmmm...hopefully one of the refugee's from Astro will see this as they include some folks who are fairly conversant with what is known about orbits, but I'm guessing "no" as such an orbit would be in a resonance with the heavy satellite. – dmckee May 14 '12 at 20:55
A solution involving orbiting both would also be an interesting answer. The important part would be passing through the Lagrangian points. – Ignacio Vazquez-Abrams May 14 '12 at 21:00
@IgnacioVazquez-Abrams: See the latest update to my answer. – Rody Oldenhuis May 31 '13 at 13:48

The answer is NO, such an orbit is not possible.

There's horseshoe orbits. These can be made arbitrarily thin, but have to have some thickness to be horseshoe. Therefore, passing exactly through $L_4$ and $L_5$ is not an option, but "close to" is certainly possible.

Horseshoe orbits are co-planar with the minor body. You asked for orbits perpendicular to the ecliptic. At first glance, this would indeed seem possible; when ignoring third-body effects, I can certainly imagine a point mass orbiting the main body in an elliptic orbit, which has the same orbital period as the minor body, where the close approach would make it pass through $L_5$ and $L_4$ (in that order) and the far pass would consume the rest of the orbital period.

The time required for $L_4$ to move directly opposite $L_5$ at time $t$ is simply $\ T/6$ (since 60$^\circ = 360^\circ/6$, with $T$ the orbital period), meaning that the time between the object's pericentre passage and true anomaly 90$^\circ$ must equal $\ T/12$.

These two requirements (a fixed $T$ and fixed time between pericentre passage and true anomaly 90$^\circ$) already fully constrain the shape and size of the orbit. There is however a third requirement, namely, that the distance between the point mass and the main body equals the orbital radius of the minor body when it passes the ecliptic (otherwise, it wouldn't pass through $L_{4,5}$). This means the orbit is over-constrained, and therefore, probably impossible.

Let's get our hands dirty. Given a two-body system in circular orbit,

• $M_0$ the central mass. Define $GM_0 = \mu_0$.
• $M_1$ the minor mass, with $M_1 \ll M_0$.
• $T_S$ the orbital period of the system
• $a_S$ the orbital radius

The orbital period $T$ of the orbit through $L_4$ and $L_5$ perpendicular to the orbital plane of the 2-body system must be equal to $T_S$, so

$$T = T_S\\ 2\pi \sqrt{\frac{a^3}{\mu_0}} = 2\pi \sqrt{\frac{a_S^3}{\mu_0}} \\ a = a_S$$

so we know that the semi-major axis must be equal to the orbital radius of the 2-body system. For any orbit in a 2-body context, the following polar equation applies:

$$r = \frac{a\left(1-e^2\right)}{1 + e\cos(\theta)}$$

with $r$ the distance, $a$ the semi-major axis, $e$ the eccentricity, and $\theta$ the true anomaly. For our out-of-plane orbit, we know that it should cross the orbit of $M_1$ at $\theta = \pi/2$. Substituting this into this equation, and using the fact that and the fact that $a = a_S$ results in

$$r_{\theta=\pi/2} = a_S\left(1-e^2\right)$$

but at that point, the distance must equal the orbital radius, $a_S$,

$$a_S = a_S\left(1-e^2\right)$$

which is only possible if $e=0$ (circular orbit). But since our out-of-plane orbit is not circular (as per the requirement $\theta = \pi/2$ when $t= T/12$ and not $t=T/4$, which is true for circular orbits), we have reached an impossibility.

(I continued a bit to determine what $e$ should be. You can only find it numerically; I found $e \approx 0.5533$).

So in summary: I think those horseshoe orbits are the closest you'll be able to get to what you asked for.

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