Near distance approximation in astrodynamical applications of Lambert’s theorem

Abstract

The smallness parameter of the approximation method is defined in terms of the non-dimensional initial distance between target and chaser satellite. In the case of a circular target orbit, compact analytical expressions are obtained for the interception travel time up to third order. For eccentric target orbits, an explicit result is worked out to first order, and the tools are prepared for numerical evaluation of higher order contributions. The possible transfer orbits are examined within Lambert’s theorem. For an eventual rendezvous it is assumed that the directions of the angular momenta of the two orbits enclose an acute angle. This assumption, together with the property that the travel time should vanish with vanishing initial distance, leads to a condition on the admissible initial positions of the chaser satellite. The condition is worked out explicitly in the general case of an eccentric target orbit and a non-coplanar transfer orbit. The condition is local. However, since during a rendezvous maneuver, the chaser eventually passes through the local space, the condition propagates to non-local initial distances. As to quantitative accuracy, the third order approximation reproduces the elements of Mars, in the historical problem treated by Gauss, to seven decimals accuracy, and in the case of the International Space Station, the method predicts an encounter error of about 12 m for an initial distance of 70 km.

Appendices

Appendix A: Quotient estimates for circular target orbit

We refer to the definitions (70) and make use of the explicit expressions for the scaled time parameters $\tau _i$, as given in (36)–(38).

$$\begin{aligned} q_2=\frac{\tau _2 \delta \rho }{\tau _1}=\frac{\cos (\Phi )}{1-z}\,\,\frac{W-\sin (\Phi )}{W}\,\delta \rho ,\quad W=\sqrt{1-z+\sin ^2(\Phi )}. \end{aligned}$$

(75)

Since $0\le (W-\sin (\Phi ))/W \le 2$, if $0\le z \le 1$, we obtain the following estimate:

$$\begin{aligned} |q_2|\le \frac{2\delta \rho |\cos (\Phi )|}{1-z},\quad 0\le z <1. \end{aligned}$$

(76)

The third order quotient can be written in the form

$$\begin{aligned} q_3&= \delta \rho ^2 (W-\sin (\Phi )) N/D,\nonumber \\ N&= -25+21 z+4 z^2+4(-9+4 z+3 z^2)\cos (2\Phi )+(5+3 z)\cos (4\Phi )\\&-16 W(1-z) \sin (\Phi )+16 W\sin (3\Phi ),\nonumber \\ D&= 48 W^2 (z-1)\,\left[ \delta \rho \cos (\Phi )\left( \sin (\Phi )- W \right) +W (z-1)\right] .\nonumber \end{aligned}$$

(77)

One can show that ($|N|$ has a maximum at $z=0$ and $\Phi =3 \pi /2$ with $N=16+32 \sqrt{2}$)

$$\begin{aligned} |N|<62; \quad D=48\,W^3(1+q_2)(1-z)^2; \quad 0\le \frac{W-\sin (\Phi }{W} \le 2;\quad \frac{1}{W^2}\le \frac{1}{1-z}.\qquad \end{aligned}$$

(78)

Therefore, we write

$$\begin{aligned} |q_3| \le \frac{31}{12}\delta \rho ^2\frac{1}{(1+q_2)(1-z)^3};\quad |q_2|<1,\quad 0\le z<1. \end{aligned}$$

(79)

We set

$$\begin{aligned} \delta \rho =\epsilon \,\,\delta \rho _0 \quad \mathrm{with}\quad \delta \rho _0=\frac{1}{3}\,(1-z)^{3/2},\,\, 0<\epsilon <1,\quad 0\le z<1, \end{aligned}$$

(80)

and use

$$\begin{aligned} q_2\ge -(2/3)\,\epsilon \sqrt{1-z}\ge -2/3 \end{aligned}$$

(81)

to verify that (76) and (79) are consistent with the inequalities (71).

Appendix B: Eccentric target orbit

The $\delta \rho$ method is extended to elliptic target orbits. The target orbit is described in the principal axes system in terms of the eccentric anomaly $u$ and the Cartesian coordinates $X,Y$ with origin in the gravitational center:

$$\begin{aligned} X=\bar{a}\left( \cos (u)-\bar{e}\right) ,\quad Y=\bar{a}\sqrt{1-\bar{e}^2}\,\sin (u). \end{aligned}$$

(82)

At time $t=0$, the target is assumed at position ${\mathbf {r}}_2$ specified by the angle $u_0\in (0,\pi )$, whereas the chaser is close by at position ${\mathbf {r}}_1$, which we restrict to the orbit plane of the target,

$$\begin{aligned} {\mathbf {r}}_2=\left\{ \bar{a}\left( \cos (u_0)-\bar{e}\right) ,\,\bar{a}\sqrt{1-\bar{e}^2}\sin (u_0)\right\} ;\quad {\mathbf {R}}_1={\mathbf {r}}_2+\bar{a}\delta \rho \left\{ \cos (\Phi ),\, \sin (\Phi )\right\} .\quad \end{aligned}$$

(83)

After the travel time $\Delta t$, the target is at position

$$\begin{aligned} {\mathbf {R}}_2=\left\{ \bar{a}\left( \cos (u_0+\Delta u)-\bar{e}\right) ,\,\bar{a}\sqrt{1-\bar{e}^2}\,\sin (u_0+\Delta u)\right\} . \end{aligned}$$

(84)

Within the Lambert scheme, the positions ${\mathbf {R}}_{1}$, ${\mathbf {R}}_{2}$ together with semimajor axis $a$ specify the chaser orbit, which is constrained by the interception condition (14) with $\Delta t_{tar}=\Delta t_{cha}$.

1.1 Travel time of target

By Kepler’s equation, the travel time can be expressed in terms of the eccentric anomalies as follows:

$$\begin{aligned} \Delta t=\frac{\bar{a}^{3/2}}{2\pi }\left[ \Delta u-\bar{e}\left( \sin (u_0+\Delta u)-\sin (u_0)\right) \right] . \end{aligned}$$

(85)

With the ansatz

$$\begin{aligned} \Delta u=u_1\delta \rho +u_2\delta \rho ^2+u_3\delta \rho ^3, \end{aligned}$$

(86)

one determines the coefficients $u_i$ in terms of $\tau _i\equiv 2\pi \bar{a}^{-3/2}t_i$, $i=1,2,3$, which enter the left-hand side of (85) according to (15). Up to third order in $\delta \rho$, the result is

$$\begin{aligned} u_1&= \frac{1}{U_0} \,\tau _1, \quad U_0=1-\bar{e} \cos (u_0), \\ u_2&= \frac{1}{2U_0^3}\left[ 2 \tau _2 U_0^2-\bar{e} \tau _1^2\sin (u_0)\right] ,\nonumber \\ u_3&= \frac{1}{6U_0^5}\left\{ \bar{e}\tau _1^3\left[ \bar{e}\left( 2-\cos (2 u_0)\right) -\cos (u_0)\right] -6 \bar{e} \tau _1\tau _2\sin (u_0)U_0^2 +6 \tau _3 U_0^4 \right\} . \nonumber \end{aligned}$$

(87)

1.2 Travel time of chaser

To this end, we need the side length $R_1,R_2$, and $c=|{\mathbf {R}}_2-{\mathbf {R}}_1|$ of the Lambert triangle. We write up to third order in $\delta \rho$,

$$\begin{aligned} R_i&= \rho ^{(i)}_{0}+ \rho ^{(i)}_{1}\delta \rho +\rho ^{(i)}_{2}\delta \rho ^2+\rho ^{(i)}_{3}\delta \rho ^3 \quad i=1,2; \nonumber \\ c&= \rho ^{(c)}_{1}\delta \rho +\rho ^{(c)}_{2}\delta \rho ^2+\rho ^{(c)}_{3}\delta \rho ^3, \end{aligned}$$

(88)

and deduce the coefficients from (83), (84), and (87) in the following form:

$$\begin{aligned} \rho ^{(1)}_0&= \bar{a}\,U_0, \nonumber \\ \rho ^{(1)}_1&= \frac{\bar{a}}{U_0}\left[ \sqrt{1-\bar{e}^2}\sin (u_0)\sin (\Phi )+\left( \cos (u_0)-\bar{e}\right) \cos (\Phi )\right] , \nonumber \\ \rho ^{(1)}_2&= \frac{\bar{a}}{4U_0^3} \,F_1 \quad \mathrm{with}\\ F_1&= 1-\cos (2 u_0)\cos (2 \Phi )+\bar{e}^2\left[ \cos ^2(u_0)-\cos (2\Phi )\left( 1+\sin ^2(u_0)\right) \right] \nonumber \\&\quad -4 \bar{e} \cos (u_0)\sin ^2(\Phi )+ 2\sqrt{1-\bar{e}^2}\sin (u_0)\sin (2\Phi )\,(\bar{e}-\cos (u_0)), \nonumber \\ \rho ^{(1)}_3&= \frac{(-\rho ^{(1)}_1)}{4U_0^4}\,F_1.\nonumber \end{aligned}$$

(89)

$$\begin{aligned} \rho ^{(2)}_0&= \bar{a}\,U_0, \nonumber \\ \rho ^{(2)}_1&= \bar{a} \bar{e} u_1\sin (u_0), \\ \rho ^{(2)}_2&= \frac{\bar{a} \bar{e}}{2}\left\{ u_1^2\cos (u_0)+2 u_2 \sin (u_0)\right\} ,\nonumber \\ \rho ^{(2)}_3&= \frac{\bar{a} \bar{e}}{6}\left\{ 6u_1u_2\cos (u_0)-\left( u_1^3-6u_3\right) \sin (u_0)\right\} .\nonumber \end{aligned}$$

(90)

$$\begin{aligned} \rho ^{(c)}_1&= \bar{a}\left\{ \left( \cos (\Phi )+u_1\sin (u_0)\right) ^2+\left( \sqrt{1-\bar{e}^2}u_1 \cos (u_0)-\sin (\Phi )\right) ^2\right\} ^{1/2}, \\ \rho ^{(c)}_2&= \frac{\bar{a}^2}{2\rho ^{(c)}_1}\left\{ \left[ (u_1 \sin (u_0)+\cos (\Phi )\right] \left[ u_1^2 \cos (u_0)+2 u_2 \sin (u_0)\right] \right. \nonumber \\&\quad + \left. \sqrt{1-\bar{e}^2}\left[ u_1^2\sin (u_0)-2 u_2\cos (u_0)\right] \left[ \sin (\Phi )-\sqrt{1-\bar{e}^2} u_1\cos (u_0)\right] \right\} , \nonumber \\ \rho ^{(c)}_3&= \frac{\bar{a}^4}{24(\rho ^{(c)}_1)^3} \{F_2(\rho ^{(c)}_1)^2/\bar{a}^2-3\,F_3^2\}, \nonumber \\ F_2&= 4\sin (\Phi )\sqrt{1-\bar{e}^2}\left[ \left( u_1^3-6 u_3\right) \cos (u_0)+6u_1u_2\sin (u_0)\right] \nonumber \\&\quad +\cos (\Phi )\left[ 24u_1u_2 \cos (u_0)-4\left( u_1^3-6u_3\right) \sin (u_0)\right] \nonumber \\&\quad +\frac{1}{2}\left[ (2-\bar{e}^2)(12 u_2^2+24 u_1u_3-u_1^4)+\bar{e}^2 \cos (2 u_0)\,(7 u_1^4-12u_2^2-24 u_1 u_3)\right. \nonumber \\&\quad +\left. 36 \bar{e}^2 u_1^2u_2\sin (2 u_0)\right] , \nonumber \\ F_3&= \cos (\Phi )\left( u_1^2\cos (u_0)+2 u_2 \sin (u_0)\right) +\sqrt{1-\bar{e}^2}\sin (\Phi )\left( u_1^2 \sin (u_0)-2 u_2 \cos (u_0)\right) \nonumber \\&\quad +u_1\left[ u_2(2-\bar{e}^2)-\bar{e}^2 u_2\cos (2 u_0)+(\bar{e}^2/2)u_1^2 \sin (2 u_0)\right] .\nonumber \end{aligned}$$

(91)

The coefficients $u_1,u_2,u_3$ in (90) and (91) are defined through (87).

The Lambert parameters $\mu$ and $\nu$ are defined in (4) in terms of the distances $R_1,R_2, c$, which to third order in $\delta \rho$, are specified by (4), and (88)–(91). The coefficients $\mu _i$ and $\nu _i$, defined in (26), are easily expressed in the following form:

$$\begin{aligned} \mu _0&= \frac{1}{2}\sqrt{(\rho ^{(1)}_0+\rho ^{(2)}_0)/a}, \nonumber \\ \mu _1&= \left( \rho ^{(1)}_1+\rho ^{(2)}_1+\rho ^{(c)}_1\right) /(8 a \mu _0), \nonumber \\ \mu _2&= \frac{1}{2}\,\mu _0 \left[ \frac{\rho ^{(1)}_2+\rho ^{(2)}_2+\rho ^{(c)}_2}{\rho ^{(1)}_0+\rho ^{(2)}_0}- \left( \frac{\rho ^{(1)}_1+\rho ^{(2)}_1+\rho ^{(c)}_1}{2(\rho ^{(1)}_0+\rho ^{(2)}_0)}\right) ^2\right] , \nonumber \\ \mu _3&= \frac{(-2)}{16 a \mu _0}\,\left\{ \frac{\mu _2}{\mu _0} \left( \rho ^{(1)}_1+\rho ^{(2)}_1+\rho ^{(c)}_1\right) -\left( \rho ^{(1)}_3+\rho ^{(2)}_3+\rho ^{(c)}_3\right) \right\} . \end{aligned}$$

(92)

$$\begin{aligned} \nu _0&= \mu _0, \quad \nu _i=\mu _i\left( \rho ^{(c)}_k)\rightarrow -\rho ^{(c)}_k\right) ,\quad i,k=1,2,3. \end{aligned}$$

(93)

We proceed as in Sect. 5. The travel time of the chaser, $\Delta t^{cha}$, is written in analogy to (15)

$$\begin{aligned} \Delta t^{cha} =t_1^{cha} \delta \rho +t_2^{cha} \delta \rho ^2+t_3^{cha} \delta \rho ^3. \end{aligned}$$

(94)

To first order, we have to evaluate the following expression for the coefficient $t_1^{cha}$, see (33):

$$\begin{aligned} t_1^{cha}=\frac{2 a^{3/2} \mu _0^2(\mu _1-\nu _1)}{\pi \sqrt{1-\mu _0^2}}, \end{aligned}$$

(95)

which, by the interception condition, must be equal to the corresponding travel time coefficient $t_1$ of the target

$$\begin{aligned} t_1^{cha}=t_1\equiv \bar{a}^{3/2}/(2\pi )\,\tau _1. \end{aligned}$$

(96)

On the right hand side of (95), we apply the substitutions according to (92)–(93), then one applies (89)–(91), and eventually (87). We take the square, in order to obtain from the condition (96) a quadratic equation for $\tau _1$ with the solution

$$\begin{aligned} \tau _1=\frac{1}{2}\left[ -w_1\pm \sqrt{w_1^2-4w_0}\right] , \end{aligned}$$

(97)

where

$$\begin{aligned} w_0&= \frac{-1}{1-z}, \quad z=\bar{a}/a,\quad 0\le z<1, \\ w_1&= 2 \frac{\sqrt{1-\bar{e}^2}\cos (u_0)\sin (\Phi )-\sin (u_0)\cos (\Phi )}{(1-z)U_0}.\nonumber \end{aligned}$$

(98)

The radicand of the square root is always positive, since $-4 w_0=4/(1-z)>0$ for $0\le z <1$. Moreover, in order that the travel time is positive to lowest order, we must take the positive sign of the square root:

$$\begin{aligned} \tau _1=\frac{1}{2}\left[ -w_1+\sqrt{w_1^2+4/(1-z)}\right] , \quad 0\le z<1. \end{aligned}$$

(99)

In the limit of a circle orbit of the target, where $\bar{e}=0$, we recover the result (36) with shifted angle $\Phi$:

$$\begin{aligned} \tau _1=\frac{1}{1-z}\left[ \sqrt{1-z+\sin ^2(\Phi -u_0)}-\sin (\Phi -u_0)\right] . \end{aligned}$$

(100)

Continuing to higher orders, we express $\tau _2$ and $\tau _3$ by the corresponding right-hand sides of (33), but use the extended substitutions (89)–(93) for the parameters $\mu _i$ and $\nu _i$. As in the case of circular target orbits, the equations for $\tau _{2}$, $\tau _{3}$ are linear. Explicit expressions, however, are hardly instructive due to their complexity, and in praxis one relies on specific numerical applications.

1.3 AIP of chaser

In the following we derive the generalization of condition (42) for an eccentric target orbit. To first order in $\delta \rho$, we infer from (86) and (87) the phase shift of the eccentric anomaly,

$$\begin{aligned} \Delta u=\delta \rho \tau _1/\left( 1-\bar{e} \cos (u_0)\right) , \end{aligned}$$

(101)

where $\tau _1$ is given by (99). The angles $u_0$ and $u_0+\Delta u$ are mapped to the true anomalies $\varphi _0$ and $\varphi _0+\Delta \varphi$ with the aid of the relation

$$\begin{aligned} \tan \left( \varphi /2\right) =\kappa \tan (u/2),\quad \kappa =\sqrt{(1+\bar{e})/(1-\bar{e})}. \end{aligned}$$

(102)

The result is

$$\begin{aligned} \Delta \varphi =\frac{\sqrt{1-\bar{e}^2}\,\Delta u}{1-\bar{e}\cos (u_0)}+\mathcal{O}(\Delta u^2). \end{aligned}$$

(103)

This gives rise to the coefficient $\phi _1$, see (18),

$$\begin{aligned} \phi _1=F\,\tau _1, \quad F=\sqrt{1-\bar{e}^2}/(1-\bar{e}\cos (u_0))^2, \end{aligned}$$

(104)

and to condition (21) in the form (we have a plane problem with $\Theta =\pi /2$)

$$\begin{aligned} \frac{\sqrt{1-\bar{e}^2}}{2(1-\bar{e}\cos (u_0))^2} \left[ -w_1+\sqrt{w_1^2+\frac{4}{1-z}}\right] >\sin \left( \Phi -\varphi _0(u_0)\right) . \end{aligned}$$

(105)

In the limit of vanishing eccentricity $\bar{e}\rightarrow 0$ of the target orbit with $u_0 \rightarrow \varphi _0$, one recovers (42) with shifted angle $\Phi \rightarrow \Phi '$ and $\bar{a}\rightarrow r_2$,

$$\begin{aligned} \sin (\Phi ') < 1/\sqrt{3-z}, \quad z=r_2/a,\quad \Phi '=\Phi -u_0. \end{aligned}$$

(106)

1.4 Position error

In Table 4, the position error $\Delta r_{err}$ is derived from the time error $\Delta t_{err}$ as follows: The starting position of the target is specified by the eccentric anomaly $u_0$. After the approximate interception time $\Delta t_{\delta \rho }$, the position of the target is given by $u_1$. Now, if the target position at $u_1$ is considered as the end point $P_2$ within the Lambert scheme, then we find with the aid of (4), (5), and (7) , without series expansion, the transfer time $\Delta t_{cha}$. At this time, the target is at position $u_1+\delta u$, where the anomaly shift corresponds to the time difference $\Delta t_{err}=\Delta t_{cha}-\Delta t_{\delta \rho }$. To first order in $\Delta u$ and $\delta u$, one obtains from the Kepler equation (85)

$$\begin{aligned} 2 \pi \frac{\Delta t_{err}}{T_0}=\left( \frac{\bar{a}}{A_0}\right) ^{3/2}\delta u\left[ 1-\bar{e}\cos (u_0) +\mathcal{O}(\Delta u)+ \mathcal{O}(\delta u)\right] . \end{aligned}$$

(107)

From the parameter representation (82) of the target orbit, one finds by the same approximation and after setting $A_0=\bar{a}$

$$\begin{aligned} \Delta r_{err}&= \bar{a}\delta u \sqrt{1-\bar{e}^2 \cos ^2(u_0)}\left[ 1 +\mathcal{O}(\Delta u)+ \mathcal{O}(\delta u)\right] , \end{aligned}$$

(108)

which immediately leads to (69).