A generalization of the equinoctial orbital elements

Abstract

We introduce six quantities that generalize the equinoctial orbital elements when some or all the perturbing forces that act on the propagated body are derived from a potential. Three of the elements define a non-osculating ellipse on the orbital plane, other two fix the orientation of the equinoctial reference frame, and the last allows us to determine the true longitude of the body. The Jacobian matrices of the transformations between the new elements and the position and velocity are explicitly given. As a possible application, we investigate their use in the propagation of Earth’s artificial satellites, showing a remarkable improvement compared to the equinoctial orbital elements.

Appendices

Alternative formulation

An alternative set of generalized equinoctial orbital elements is obtained by replacing $q_1$, $q_2$ with the Euler parameters $e_1$, $e_2$, $e_3$, $e_4$ that define the orientation of the equinoctial reference frame $\varSigma _{\textsf {eq}}$ with respect to $\varSigma$ (see eq 2):

$$\begin{aligned} e_1 := \cos \frac{i}{2}\cos \varOmega ,\quad e_2 := \sin \frac{i}{2},\quad e_3 := \cos \frac{i}{2}\sin \varOmega ,\quad e_4 := 0. \end{aligned}$$

The Euler parameters allow us to partially control the error accumulation during the propagation by monitoring the quantity $e_1^2+e_2^2+e_3^2+e_4^2$. On the other hand, they make the set of GEqOE redundant, increasing the dimension of the state vector from 6 to 7.

If the Euler parameters $e_1$, $e_2$, $e_3$ are used in place of $q_1$, $q_2$, then $\mathbf{e} _X$, $\mathbf{e} _Y$ are computed from the formulae

$$\begin{aligned} \mathbf{e} _X&= \left( e_1^2+e_2^2-e_3^2,\,\, -2e_1e_3,\,\,2e_2e_3\right) ^T,\\ \mathbf{e} _Y&= \left( 2e_1e_3,\,\,e_1^2-e_2^2-e_3^2,\,\, -2e_1e_2\right) ^T. \end{aligned}$$

Moreover, we have

$$\begin{aligned} {\dot{e}}_1&=\frac{1}{2}(w_he_2+w_Xe_3),\\ {\dot{e}}_2&=-\frac{1}{2}(w_he_1-w_Ye_3),\\ {\dot{e}}_3&=-\frac{1}{2}(w_Xe_1+w_Ye_2), \end{aligned}$$

where $w_X$, $w_Y$ are obtained as in (52) and

$$\begin{aligned} w_h = \frac{e_2}{e_1^2+e_3^2}(e_3w_X-e_1w_Y). \end{aligned}$$

While the quantities $e_1$, $e_2$, $e_3$ are defined for any inclination, the time derivatives of $e_1$, $e_2$ become singular for $i=\pi$.

Examples of conversion between Cartesian position and velocity coordinates and the GEqOE

The results reported in this section are obtained using Matlab R2020b.

Consider the four test cases described in Sect. 7 and the corresponding four sets of initial conditions in Table 1. For all of them, the function ${\mathscr {U}}$ is given in (56). The corresponding values of the GEqOE are shown in Table 3. They are obtained by first computing the Cartesian position and velocity coordinates (see Table 2) and then applying the procedure described in Sect. 3. The constants $\mu$, $J_2$, $r_e$ used for the transformations are:

$$\begin{aligned} \begin{aligned} \mu&= 398600.4354360959\,{{\mathrm{km}}}^3/{\mathrm{s}}^2\\ J_2&= 1.08262617385222\times 10^{-3},\\ r_e&= 6378.1366\,{{\mathrm{km}}}. \end{aligned} \end{aligned}$$

(59)

Table 2 Position (km) and velocity (km/s) coordinates with respect to the J2000 frame computed from the four sets of initial conditions shown in Table 1. Some numbers are rounded off

Table 3 GEqOE computed from the four sets of Cartesian position and velocity coordinates, which correspond to the orbital elements in Table 1. We assumed that $GM=\mu$ in the expression of ${\mathscr {U}}$ in (56). The values of the quantities $\nu$ and ${\mathcal {L}}$ ($={\mathcal {L}}_0$) are in rad/s and rad, respectively. Some numbers are rounded off

Next, for a particular example, we show a step-by-step guide to convert to and from GEqOE and Cartesian position and velocity coordinates. The numbers reported below are taken from the output displayed by Matlab R2020b when the long fixed-decimal format is selected. We used the values given by $r_e$ and $(r_e^3/\mu )^{1/2}$ as units of length and time. With this choice, all the equations in the paper can be applied by replacing $\mu$ with 1 and dimensional quantities with the corresponding non-dimensional ones.

Pick the six values of the row labeled by a) in Table 2:

$$\begin{aligned} x&= 7178.1366\,{\mathrm{km}},&{\dot{x}}&= 0\,{\mathrm{km}}/{\mathrm{s}},\\ y&= 0\,{\mathrm{km}},&{\dot{y}}&= 5.269240572916780\,{\mathrm{km}}/{\mathrm{s}},\\ z&= 0\,{\mathrm{km}},&{\dot{z}}&= 5.269240572916780\,{\mathrm{km}}/{\mathrm{s}}. \end{aligned}$$

Following Sect. 3, we get in sequence:

$$\begin{aligned}&\mathbf{e} _r = (1,\,0,\,0)^T,\quad \mathbf{e} _f = (0,\,0.707106781186548,\,0.707106781186548)^T,\\&\mathbf{e} _h = (0,\,-0.707106781186548,\,0.707106781186548)^T,\\&r = 7178.1366\,{\mathrm{km}},\quad {\dot{r}} = 0\,{\mathrm{km}}/{\mathrm{s}},\quad h = 53490.26429528814\,{\mathrm{km}}^2/{\mathrm{s}},\\&{\mathscr {U}} = -0.023732242207310\,{\mathrm{km}}^2/{\mathrm{s}}^2,\quad {\mathscr {E}} = -27.788628457479671\,{\mathrm{km}}^2/{\mathrm{s}}^2,\\&\nu = 0.001039460266303\,\mathrm{rad}/{\mathrm{s}},\\&q_1=0,\quad q_2=0.414213562373095,\\&\mathbf{e} _X=(1,\,0,\,0)^T,\quad \mathbf{e} _Y=(0,\,0.707106781186547,\,0.707106781186548)^T,\\&c = 53467.39881650716\,{\mathrm{km}}^2/{\mathrm{s}},\\&{\varvec{\upsilon }} = (0,\,5.266988131092099,\,5.266988131092099)^T\,{\mathrm{km}}/{\mathrm{s}},\\&\mathbf{g} = (-8.547571013161059\times 10^{-4},\,0,\,0)^T,\\&p_1 = 0,\quad p_2 = -8.547571013161059\times 10^{-4},\\&X = 7178.1366\,{\mathrm{km}},\quad Y = 0\,{\mathrm{km}},\\&{\textit{a}} = 7172.006276704303\,{\mathrm{km}},\quad \alpha = 0.500000091326246,\\&\sin {\mathcal {K}} = 0,\quad \cos {\mathcal {K}} = 1,\\&{\mathcal {L}}=0\,\mathrm{rad}. \end{aligned}$$

Then, we convert the values of $\nu$, $p_1$, $p_2$, ${\mathcal {L}}$, $q_1$, $q_2$ obtained above to position and velocity. Following Sect. 4, we get in sequence:

$$\begin{aligned}&{\mathcal {K}} = 0\,\mathrm{rad},\\&\mathbf{e} _X = (1,\,0,\,0)^T,\quad \mathbf{e} _Y = (0,\,0.707106781186547,\,0.707106781186548)^T,\\&{\textit{a}} = 7172.006276704303\,{\mathrm{km}},\quad \alpha = 0.500000091326246,\\&X = 7178.136600000001\,{\mathrm{km}},\quad Y = 0\,{\mathrm{km}},\\&\mathbf{r} = (7178.136600000001,\,0,\,0)^T\,{\mathrm{km}},\\&r = 7178.136600000001\,{\mathrm{km}},\quad {\dot{r}} = 0\,{\mathrm{km}}/{\mathrm{s}},\\&{\mathscr {U}} = -0.023732242207310\,{\mathrm{km}}^2/{\mathrm{s}}^2,\\&c = 53467.39881650716\,{\mathrm{km}}^2/{\mathrm{s}},\quad h = 53490.26429528814\,{\mathrm{km}}^2/{\mathrm{s}},\\&{\dot{\mathbf{r }}} = (0,\,5.269240572916779,\,5.269240572916779)^T\,{\mathrm{km}}/{\mathrm{s}}. \end{aligned}$$

Further details about Figs. 3, 5, and 7

We report below the true positions and velocities at the final epoch of propagation used to obtain the performance plots of Figs. 3, 5, and 7. These vectors are expressed in the J2000 reference frame.

The $J_2$ value (see eq. 59) of the Earth’s gravitational expansion was taken from the EGM2008 model. The ephemerides of the Sun and the Moon and the values of the parameters $\mu$, $r_e$ (see eq 59) were retrieved from the planetary and lunar ephemerides DE430. Moreover, the precession, nutation, and polar motion of the Earth are completely neglected so that the Earth’s spin axis is fixed and equal to that of the reference epoch J2000.

We denote by $t_0$ and $t_f$ the initial and final epochs of propagation.

• (a) Fig. 3, $t_0=$ 2020 Jan 01 00:00:00 UTC, $t_f=$ 2020 Jan 13 00:00:00 UTC:

  $$\begin{aligned} x_f&= -5398.929377366906\,{\mathrm{km}},&{\dot{x}}_f&= 2.214482567493\,{\mathrm{km}}/{\mathrm{s}},\\ y_f&= -390.257240638229\,{\mathrm{km}},&{\dot{y}}_f&= -6.845637008953\,{\mathrm{km}}/{\mathrm{s}},\\ z_f&= -4693.719111636971\,{\mathrm{km}},&{\dot{z}}_f&= -1.977748618717\,{\mathrm{km}}/{\mathrm{s}}. \end{aligned}$$

• (b) Fig. 5 (top panel), $t_0=$ 2020 Jan 01 00:00:00 UTC, $t_f=$ 2020 Jan 13 00:00:00 UTC:

  $$\begin{aligned} x_f&= -274.761002943290\,{\mathrm{km}},&{\dot{x}}_f&= 7.465328216770\,{\mathrm{km}}/{\mathrm{s}},\\ y_f&= -7154.555995859508\,{\mathrm{km}},&{\dot{y}}_f&= -0.288082051862\,{\mathrm{km}}/{\mathrm{s}},\\ z_f&= -0.095489199987\,{\mathrm{km}},&{\dot{z}}_f&= -0.000288808942\,{\mathrm{km}}/{\mathrm{s}}. \end{aligned}$$

• (c) Fig. 5 (middle panel), $t_0\!=\! 2020$ Jan 01 00:00:00 UTC, $t_f\!=\!2020$ Jan 13 00:00:00 UTC:

  $$\begin{aligned} x_f&= -6127.562058484711\,{\mathrm{km}},&{\dot{x}}_f&= -3.876493609204\,{\mathrm{km}}/{\mathrm{s}},\\ y_f&= 0.290815939820\,{\mathrm{km}},&{\dot{y}}_f&= 0.000242489963\,{\mathrm{km}}/{\mathrm{s}},\\ z_f&= 3725.501491458693\,{\mathrm{km}},&{\dot{z}}_f&= -6.369562182446\,{\mathrm{km}}/{\mathrm{s}}. \end{aligned}$$

• (d) Fig. 5 (bottom panel) and Fig. 7, $t_0=$ 2020 Jan 01 00:00:00 UTC, $t_f=$ 2020 Mar 26 14:27:30 UTC:

  $$\begin{aligned} x_f&= 10732.86105177698\,{\mathrm{km}},&{\dot{x}}_f&= 3.96389903452\,{\mathrm{km}}/{\mathrm{s}},\\ y_f&= 2632.59989195335\,{\mathrm{km}},&{\dot{y}}_f&= 3.86270637636\,{\mathrm{km}}/{\mathrm{s}},\\ z_f&= -1133.57673525621\,{\mathrm{km}},&{\dot{z}}_f&= 5.12156998778\,{\mathrm{km}}/{\mathrm{s}}. \end{aligned}$$

Derivation of eqs (38), (39), and (42)

We set the right-hand sides of (31), (32) equal to the right-hand sides of (33), (34), respectively. By solving the resulting equations for $\cos L$, $\sin L$ and taking into account (19), (20), (22), (40), we find

$$\begin{aligned} \begin{aligned} \cos L&= \frac{{\textit{a}}}{r}\left[ \alpha p_1p_2\sin {\mathcal {K}}+ (1-\alpha p_1^2)\cos {\mathcal {K}}-p_2\right] ,\\ \sin L&= \frac{{\textit{a}}}{r}\left[ \alpha p_1p_2\cos {\mathcal {K}}+ (1-\alpha p_2^2)\sin {\mathcal {K}}-p_1\right] . \end{aligned} \end{aligned}$$

(60)

Noting that

$$\begin{aligned} X = r\cos L,\qquad Y = r\sin L, \end{aligned}$$

(61)

we obtain (42).

System (60) can be solved for $\cos {\mathcal {K}}$, $\sin {\mathcal {K}}$ to yield relations (39). By inserting the expressions given in (39) in the generalized Kepler’s equation (30), we have

$$\begin{aligned} {\mathcal {L}} = {\mathcal {K}}+\frac{1}{{\textit{a}}\beta }(Xp_1-Yp_2), \end{aligned}$$

which corresponds to (38).

Time derivatives of $p_1$, $p_2$

From equations (17), (18), we have

$$\begin{aligned} {\dot{p}}_1&={\dot{g}}\sin \varPsi +g{\dot{\varPsi }}\cos \varPsi , \end{aligned}$$

(62)

$$\begin{aligned} {\dot{p}}_2&={\dot{g}}\cos \varPsi -g{\dot{\varPsi }}\sin \varPsi . \end{aligned}$$

(63)

For the time derivative of g, we use relation (9) and so we need the expressions of $\dot{{\mathscr {E}}}$, ${\dot{c}}$. The former is given in (46), the latter, which is derived from the definition of c provided in (6), results:

$$\begin{aligned} {\dot{c}} = \frac{1}{c}[r^2\dot{{\mathscr {E}}}+r{\dot{r}}(2{\mathscr {U}}-rF_r)]. \end{aligned}$$

(64)

Then, we find

$$\begin{aligned} {\dot{g}} = \frac{1}{\mu ^2g}\left[ (c^2+2{\mathscr {E}}r^2) \dot{{\mathscr {E}}}+2{\mathscr {E}}r{\dot{r}}(2{\mathscr {U}}-rF_r)\right] . \end{aligned}$$

After writing $2{\mathscr {E}}$ as a function of c and g by means of (9), and using

$$\begin{aligned} r = \frac{c^2}{\mu (1+g\cos \theta )},\qquad {\dot{r}} = \frac{\mu }{c}g\sin {\theta }, \end{aligned}$$

which directly follows from (11), (12), we obtain

$$\begin{aligned} {\dot{g}} = \frac{r}{\mu }({\tilde{\varsigma }}\cos \theta +\varsigma g)\dot{{\mathscr {E}}}+ \frac{g^2-1}{c}\varsigma \sin \theta (2{\mathscr {U}}-rF_r), \end{aligned}$$

(65)

where $\varsigma$, ${\tilde{\varsigma }}$ are given by

$$\begin{aligned} \varsigma = \frac{\mu r}{c^2},\qquad {\tilde{\varsigma }} = 1+\varsigma . \end{aligned}$$

From the definition of $\varPsi$, we write

$$\begin{aligned} g{\dot{\varPsi }} = g{\dot{L}} - g{\dot{\theta }}. \end{aligned}$$

(66)

The expression of ${\dot{L}}$ can be derived, for example, from Battin (1999, eqs. 10.78, 10.81, pp. 500–501):

$$\begin{aligned} {\dot{L}} = \frac{h}{r^2}+\frac{r}{h}F_h\tan \frac{i}{2} \sin (\omega +f). \end{aligned}$$

(67)

The time derivative of $\theta$ is obtained by differentiation of both sides of equation

$$\begin{aligned} \tan \theta = \frac{r{\dot{r}}c}{c^2-\mu r}, \end{aligned}$$

which is a consequence of (11), (12). We first use

$$\begin{aligned} \ddot{r}=-\frac{\mu }{r^2}+\frac{c^2}{r^3}-\frac{2{\mathscr {U}}}{r}+F_r, \end{aligned}$$

(68)

to get

$$\begin{aligned} {\dot{\theta }} = \frac{c}{r^2}-\frac{c}{\mu rg}[(2{\mathscr {U}}-rF_r)\cos \theta + {\tilde{\varsigma }}{\dot{c}}\sin \theta ]. \end{aligned}$$

Then, we replace ${\dot{c}}$ with the expression in (64) and find

$$\begin{aligned} {\dot{\theta }} = \frac{c}{r^2}-\frac{1}{\mu g}\Bigl [({\tilde{\varsigma }}r\sin \theta )\dot{{\mathscr {E}}}+ \Bigl (\frac{c}{r}\cos \theta +{\tilde{\varsigma }}{\dot{r}}\sin \theta \Bigr )(2{\mathscr {U}}-rF_r)\Bigr ]. \end{aligned}$$

(69)

From equations (66), (67), (69), we have

$$\begin{aligned} g{\dot{\varPsi }} = \frac{g}{r^2}(h-c)-gw_h+ \frac{{\tilde{\varsigma }}r}{\mu }\dot{{\mathscr {E}}}\sin \theta +\frac{1}{c} ({\tilde{\varsigma }}g+\varsigma \cos \theta )(2{\mathscr {U}}-rF_r), \end{aligned}$$

(70)

where we used the definition of $w_h$ in (14) and made the substitution

$$\begin{aligned} \frac{c}{r}\cos \theta +{\tilde{\varsigma }}{\dot{r}}\sin \theta = \frac{\mu }{c}({\tilde{\varsigma }}g+ \varsigma \cos \theta ). \end{aligned}$$

The expressions of ${\dot{g}}$, $g{\dot{\varPsi }}$, given in (65), (70), are plugged in (62), (63), and by considering the definitions of $p_1$, $p_2$ (see 17, 18) and the relation $L=\varPsi +\theta$, we get

$$\begin{aligned} {\dot{p}}_1&= p_2\biggl (\frac{h-c}{r^2}\!-\!w_h\biggr ) +\frac{1}{c}\Bigl (\frac{r{\dot{r}}}{c}p_1+{\tilde{\varsigma }}p_2+\varsigma \cos L \Bigr )(2{\mathscr {U}}\!-\!rF_r)+\frac{r}{\mu }(\varsigma \,p_1+{\tilde{\varsigma }}\sin L) \dot{{\mathscr {E}}},\\ {\dot{p}}_2&= p_1\biggl (w_h\!-\!\frac{h-c}{r^2}\biggr ) +\frac{1}{c}\Bigl (\frac{r{\dot{r}}}{c}p_2\!-\!{\tilde{\varsigma }}p_1-\varsigma \sin L \Bigr )(2{\mathscr {U}}\!-\!rF_r)+\frac{r}{\mu }(\varsigma \,p_2+{\tilde{\varsigma }}\cos L) \dot{{\mathscr {E}}}. \end{aligned}$$

Finally, in the two equations above we note that

$$\begin{aligned} \frac{r{\dot{r}}}{c}p_1+{\tilde{\varsigma }}p_2+\varsigma \cos L&= \frac{r}{{\textit{a}}}\cos L+2p_2,\\ \frac{r{\dot{r}}}{c}p_2-{\tilde{\varsigma }}p_1-\varsigma \sin L&= -\frac{r}{{\textit{a}}}\sin L-2p_1, \end{aligned}$$

where we used (33), (34).

Time derivative of ${\mathcal {L}}$

From equations (30) and (31), we have

$$\begin{aligned} \dot{{\mathcal {L}}} = \dot{{\mathcal {K}}}\frac{r}{{\textit{a}}}+{\dot{p}}_1\cos {\mathcal {K}}- {\dot{p}}_2\sin {\mathcal {K}}, \end{aligned}$$

(71)

Let us first deal with the term ${\dot{p}}_1\cos {\mathcal {K}}-{\dot{p}}_2\sin {\mathcal {K}}$. Using relations (42) for X, Y in the expressions of ${\dot{p}}_1$, ${\dot{p}}_2$ given in (47), (48), and then considering (31), (32), we get

$$\begin{aligned} \begin{aligned} {\dot{p}}_1\cos {\mathcal {K}}-{\dot{p}}_2\sin {\mathcal {K}}&= \biggl (\frac{h-c}{r^2}-w_h\biggr ) \Bigl (1-\frac{r}{{\textit{a}}}\Bigr )+\frac{1}{c}\biggl [2-\frac{r}{{\textit{a}}}- \alpha \frac{(r{\dot{r}})^2}{\mu {\textit{a}}}\biggr ](2{\mathscr {U}}-rF_r)\\&\quad \, +\frac{r{\dot{r}}\alpha }{\mu c}\biggl [\varrho +r\biggl (\frac{1}{\alpha }- \frac{r}{{\textit{a}}}\biggr )\biggr ]\dot{{\mathscr {E}}}, \end{aligned} \end{aligned}$$

(72)

where $\alpha$, $\beta$ are defined in (40), and

$$\begin{aligned} {\textit{w}}=\sqrt{\frac{\mu }{{\textit{a}}}}. \end{aligned}$$

We compute now the time derivative of ${\mathcal {K}}$. From (15), (29), we have

$$\begin{aligned} \dot{{\mathcal {K}}} = {\dot{L}}+{\dot{G}}-{\dot{\theta }}. \end{aligned}$$

(73)

The expressions of ${\dot{L}}$, ${\dot{\theta }}$ are written in (67), (69). By differentiation of both sides of equation

$$\begin{aligned} \tan G = \frac{r{\dot{r}}}{{\textit{w}}({\textit{a}}-r)}, \end{aligned}$$

which follows from (24), (25), and using (68), we find

$$\begin{aligned} {\dot{G}} = \frac{{\textit{w}}}{r}-\frac{1}{g{\textit{w}}{\textit{a}}}\Bigl [\frac{\mu r\sin \theta }{2c{\textit{a}}^2}(r+{\textit{a}})\dot{{\textit{a}}}+(\cos G)(2{\mathscr {U}}-rF_r)\Bigr ]. \end{aligned}$$

Then, considering that

$$\begin{aligned} \cos G = \frac{\cos \theta + g}{1 + g\cos \theta },\qquad \dot{{\textit{a}}} = \frac{2{\textit{a}}^2}{\mu }\dot{{\mathscr {E}}}, \end{aligned}$$

we can write

$$\begin{aligned} {\dot{G}} = \frac{{\textit{w}}}{r}-\frac{r}{cg{\textit{w}}{\textit{a}}} \Bigl [\sin \theta (r+{\textit{a}})\dot{{\mathscr {E}}}+\frac{\mu }{c}(\cos \theta +g)(2{\mathscr {U}}-rF_r)\Bigr ]. \end{aligned}$$

(74)

From equations (73) and (67), (69), (74), we obtain

$$\begin{aligned} \dot{{\mathcal {K}}} = \frac{{\textit{w}}}{r}+\frac{h-c}{r^2}-w_h +\frac{1}{c}\Bigl [1+\alpha \Bigl (1-\frac{r}{{\textit{a}}}\Bigr )\Bigr ](2{\mathscr {U}}-F_rr) -\frac{r{\dot{r}}\alpha }{\mu {\textit{w}}}\Bigl (1-\frac{r}{{\textit{a}}\beta }\Bigr )\dot{{\mathscr {E}}}. \end{aligned}$$

(75)

Finally, by making the substitutions

$$\begin{aligned} {\dot{r}}^2=-\frac{\mu }{{\textit{a}}}+\frac{2\mu }{r}-\frac{c^2}{r^2},\qquad 1-\frac{r}{{\textit{a}}\beta } = 1-\beta \varsigma , \end{aligned}$$

in equations (72), (75), respectively, and taking into account the relation $\alpha \beta =1-\alpha$, we find from equation (71) the expression of $\dot{{\mathcal {L}}}$ reported in (49).