A lower bound of the distance between two elliptic orbits

Abstract

We obtain a lower bound of the distance function (MOID) between two noncoplanar bounded Keplerian orbits (either circular or elliptic) with a common focus. This lower bound is positive and vanishes if and only if the orbits intersect. It is expressed explicitly, using only elementary functions of orbital elements, and allows us to significantly increase the speed of processing for large asteroid catalogs. Benchmarks confirm high practical benefits of the lower bound constructed.

Appendix: The distance between two V-sets

The proof of the following lemma is very simple and therefore is omitted.

Lemma

Let $\mathcal {S}_1, \mathcal {S}_2\subset \mathbb {R}^3$ be two arbitrary sets satisfying the following three conditions:

• (i) $\rho (\mathcal {S}_1, \mathcal {S}_2)>0$.

• (ii) There are two points $Q_1\in \mathcal {S}_1$ and $Q_2\in \mathcal {S}_2$ such that $\rho (\mathcal {S}_1, \mathcal {S}_2)=Q_1Q_2$.

• (iii) The pair $(Q_1, Q_2)$ is the only element of a set $\mathcal {S}_1\times \mathcal {S}_2$ that gives $\rho (\mathcal {S}_1, \mathcal {S}_2)=Q_1Q_2$.

Then, for any two subsets $\mathcal {S'}_1\subset \mathcal {S}_1$ and $\mathcal {S'}_2\subset \mathcal {S}_2$ such that $\mathcal {S'}_1\ni Q_1$ and $\mathcal {S'}_2\ni Q_2$ one always has $\rho (\mathcal {S'}_1, \mathcal {S'}_2)=\rho (\mathcal {S}_1, \mathcal {S}_2)$.

For example, if two points M and N lie on skew lines m and n, respectively, and satisfy $\rho (m, n)=MN$, then for any two rays $m'$ and $n'$ (open or closed, no matter) such that $m'\subset m$, $n'\subset n$, $m'\ni M$, $n'\ni N$ we have $\rho (m', n')=\rho (m, n)$.

Consider again two two-dimensional closed half-planes $\alpha \subset \{z\geqslant 0\}$ and $\beta =\{y\geqslant 0; z=0\}$ that form a dihedral angle in $\mathbb {R}^3$ with the plane angle J satisfying $0<J\leqslant \pi /2$ (see Fig. 10). In the positive side of the axis Ox draw points A and B such that $OA<OB$. Given any positive acute angles $\psi$ and $\varphi$, define in the face $\alpha$ a V-set $\mathcal {V}_1$ with a vertex A and an exterior angle $\psi$, and in the face $\beta$ construct a V-set $\mathcal {V}_2$ with a vertex B and an exterior angle $\varphi$ (see Fig. 10). Decompose boundaries of $\mathcal {V}_1$ and $\mathcal {V}_2$ into four closed rays $a', a'', b', b''$, where $a', a''\subset \mathcal {V}_1$, $b', b''\subset \mathcal {V}_2$, in such a way that $a'', b'$ both intersect the plane $\{x=0\}$. Further, draw two straight lines a, b such that $a\supset a'$ and $b\supset b'$. The line a defines in the plane of $\mathcal {V}_1$ a closed half-plane $\mathcal {P}_1$ that (completely) contains $\mathcal {V}_1$. Similarly, define a closed half-plane $\mathcal {P}_2$ with the edge b such that $\mathcal {P}_2\supset \mathcal {V}_2$ (see Fig. 10). Our aim is to prove that

$$\begin{aligned} \rho (\mathcal {V}_1, \mathcal {V}_2)=\rho (a, b). \end{aligned}$$

(23)

First of all, draw two points $H\in a$ and $G\in b$ that give $\rho (a, b)=HG$. It is easy to check that under the conditions

$$\begin{aligned} 0<\varphi ,\psi<\pi /2,\quad 0<J\leqslant \pi /2 \end{aligned}$$

we always have $H_z, G_y>0$. This yields $H\in a'$, $G\in b'$, and hence, we can write

$$\begin{aligned} \mathcal {V}_1\ni H, \mathcal {V}_2\ni G, a\ni H, b\ni G. \end{aligned}$$

(24)

Fig. 10

The distance between half-planes $\mathcal {P}_1$ and $\mathcal {P}_2$ is equal to the distance between their edges a and b. This fact results from elementary similarity and continuity considerations ($AB\rightarrow 0$ implies $\rho (\mathcal {P}_1, \mathcal {P}_2)\rightarrow 0$). See text for the strict proof

Further, show that $\mathcal {P}_1$ and $\mathcal {P}_2$ satisfy all conditions of lemma. For this, we prove that

$$\begin{aligned} \rho (\mathcal {P}_1, \mathcal {P}_2)=HG \end{aligned}$$

(25)

and verify that a pair (H, G) is the only element of a set $\mathcal {P}_1\times \mathcal {P}_2$ satisfying (25). Fix any pair $(Q_1, Q_2)\in \mathcal {P}_1\times \mathcal {P}_2$ distinct from the pair $(H, G)\in \mathcal {P}_1\times \mathcal {P}_2$. It suffices to prove that $Q_1Q_2>HG$. There are three possibilities.

• (A) $Q_1\in a$, $Q_2\in b$.

• (B) One of the points $Q_1, Q_2$ is interior for the half-plain containing it, while another is boundary one.

• (C) $Q_1\notin a$, $Q_2\notin b$.

Consider case (A). Since straight lines a and b are skew, one concludes $Q_1Q_2>HG$.

Consider case (B). Let, for example, $Q_1\notin a$, $Q_2\in b$ (see Fig. 10). Through the point $Q_1$ draw a straight line $c\parallel a$ and denote by C a point where c (dashed in Fig. 10) meets the axis Ox. We have

$$\begin{aligned} Q_1Q_2\geqslant \rho (c,b)=K(\psi , \varphi , J)CB>K(\psi , \varphi , J)AB=\rho (a,b)=HG, \end{aligned}$$

and therefore $Q_1Q_2>HG$.

Case (C) differs from case (B) only in that we have to draw auxiliary straight lines in both half-planes $\mathcal {P}_1$ and $\mathcal {P}_2$.

We see that $\mathcal {P}_1$ and $\mathcal {P}_2$ satisfy all conditions of lemma. In view of (24) by lemma we conclude that

$$\begin{aligned} \rho (\mathcal {V}_1, \mathcal {V}_2)=\rho (\mathcal {P}_1, \mathcal {P}_2),\quad \rho (a, b)=\rho (\mathcal {P}_1, \mathcal {P}_2), \end{aligned}$$

which finally implies (23).