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Mutual potential between two rigid bodies with arbitrary shapes and mass distributions

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Abstract

Formulae to compute the mutual potential, force, and torque between two rigid bodies are given. These formulae are expressed in Cartesian coordinates using inertia integrals. They are valid for rigid bodies with arbitrary shapes and mass distributions. By using recursive relations, these formulae can be easily implemented on computers. Comparisons with previous studies show their superiority in computation speed. Using the algorithm as a tool, the planar problem of two ellipsoids is studied. Generally, potential truncated at the second order is good enough for a qualitative description of the mutual dynamics. However, for ellipsoids with very large non-spherical terms, higher order terms of the potential should be considered, at the cost of a higher computational cost. Explicit formulae of the potential truncated to the fourth order are given.

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Notes

  1. Potential truncated at the \(4\hbox {th}\) order is given in Appendix 1, along with the force and the torque. Note that these formulae are only valid for the planar case.

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Acknowledgments

This work was finished during the first author’s visit to the Colorado Center for Astrodynamics Research (CCAR). X.Y.H. thanks the support from national Natural Science Foundation of China (11322330) and national Basic Research Program of China (2013CB834100). The authors thank Gwenaël Boué and the other anonymous reviewer for their many useful comments.

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Authors and Affiliations

  1. School of Astronomy and Space Science, Nanjing University, Nanjing, 210093, China

    Xiyun Hou & Xiaosheng Xin

  2. Department of Aerospace Engineering Sciences, University of Colorado Boulder, Boulder, CO, 80309, USA

    Daniel J. Scheeres

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  1. Xiyun Hou
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  2. Daniel J. Scheeres
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  3. Xiaosheng Xin
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Correspondence to Xiyun Hou.

Appendices

Appendix 1: explicit expressions of the \(4\hbox {th}\) order potential for the planar two ellipsoids problem

We provide formulae for the \(4\hbox {th}\) order potential here. In these formulae, we define

$$\begin{aligned} \delta =\theta -\phi \end{aligned}$$

where \(\theta \) is the quadrant angle of the position vector \({\mathbf {R}}\) in the body-fixed frame of A, and \(\phi \) is defined in Sect. 5. The relative geometry of the system is defined in Fig. 15.

Fig. 15
figure 15

A picture illustrating the relative geometry of the system and the variables used

Full size image

In the following, the mutual potential is defined by Eq. (19) where

$$\begin{aligned} \tilde{U}_0= & {} T^{000}_AT^{000}_B \\ \tilde{U}_2= & {} A_1+A_2\cos 2\theta +A_3\cos 2\delta \\ \tilde{U}_4= & {} B_1+B_2\cos 2\theta +B_3\cos 4\theta +B_4\cos 2\delta +B_5\cos 4\delta +B_6\cos 2\phi +B_7\cos (2\theta +2\delta ) \end{aligned}$$

where

$$\begin{aligned} A_1= & {} \frac{1}{4}\left( T^{200}_B+T^{020}_B-2T^{002}_B\right) T^{000}_A+\frac{1}{4}\left( T^{200}_A+T^{020}_A-2T^{002}_A\right) T^{000}_B\\ A_2= & {} \frac{3}{4}\left( T^{200}_A-T^{020}_A\right) T^{000}_B,\\ A_3= & {} \frac{3}{4}\left( T^{200}_B-T^{020}_B\right) T^{000}_A\\ B_1= & {} \frac{9}{64}T^{000}_B\left( T^{400}_A+T^{040}_A+\frac{8}{3}T^{004}_A+2T^{220}_A-8T^{022}_A-8T^{202}_A\right) \\&+\,\frac{9}{64}T^{000}_A\left( T^{400}_B+T^{040}_B+\frac{8}{3}T^{004}_B+2T^{220}_B-8T^{022}_B-8T^{202}_B\right) \\&-\,\frac{9}{8}T^{002}_B\left( T^{200}_A+T^{020}_A-T^{002}_A\right) -\frac{9}{8}T^{002}_A\left( T^{200}_B+T^{020}_B-T^{002}_B\right) \\&+\,\frac{9}{16}\left( T^{200}_A+T^{020}_A\right) \left( T^{200}_B+T^{020}_B\right) \\ B_2= & {} \frac{5}{16}T^{000}_B\left( T^{400}_A-T^{040}_A+6T^{022}_A-6T^{202}_A\right) \\&+\,\frac{15}{16}\left( T^{200}_A-T^{020}_A\right) \left( T^{200}_B+T^{020}_B-2T^{002}_B\right) \\ B_3= & {} \frac{35}{16}T^{00}\left( T^{400}_A+T^{040}_A-6T^{220}_A\right) \\ B_4= & {} \frac{5}{16}T^{000}_A\left( T^{400}_B-T^{040}_B+6T^{022}_B-6T^{202}_B\right) \\&+\,\frac{15}{16}\left( T^{200}_B-T^{020}_B\right) \left( T^{200}_A+T^{020}_A-2T^{002}_A\right) \\ B_5= & {} \frac{35}{64}T^{000}_A\left( T^{400}_B+T^{040}_B-6T^{220}_B\right) \\ B_6= & {} \frac{9}{32}\left( T^{200}_A-T^{020}_A\right) \left( T^{200}_B-T^{020}_B\right) \\ B_7= & {} \frac{105}{32}\left( T^{200}_A-T^{020}_A\right) \left( T^{200}_B-T^{020}_B\right) \end{aligned}$$

For the \(2\hbox {nd}\) order potential, it’s obvious that \(U_{\min }({\mathbf {R}})\) corresponds to the maximum value of \(\tilde{U}_2\). This means \(\delta =0\) (i.e., \(\theta =\phi \)). As a result, we have

$$\begin{aligned} U_{\min }({\mathbf {R}})=- G\left( \frac{\tilde{U}_0}{R}+\frac{\tilde{U}^{\max }_2}{R^3}\right) \end{aligned}$$

where

$$\begin{aligned} \tilde{U}^{max}_2=A_1+A_2\cos 2\theta +A_3 \end{aligned}$$

For the potential truncated at the 4th order, \(U_{\min }({\mathbf {R}})\) usually doesn’t exactly happen at \(\delta =0\), especially for small values of R. The exact value of \(\phi \) for \(U_{\min }({\mathbf {R}})\) can be obtained by the following condition

$$\begin{aligned} \frac{\partial {U\left( {\mathbf {R}},\phi \right) }}{\partial {\phi }}=0 \end{aligned}$$

Usually there are two solutions to this equation. The minimum value happens for \(\delta \) close to \(\theta \).

Equations of motion are already given by Eq. (30). The forces are given by

$$\begin{aligned} \frac{\partial {U}}{\partial {X}}= & {} G\left[ X\left( \frac{\tilde{U}_0}{R^3}+\frac{3\tilde{U}_2}{R^5}+\frac{5\tilde{U_4}}{R^7}\right) - \left( \frac{1}{R^3}\frac{\partial {\tilde{U}}_2}{\partial {\theta }}+\frac{1}{R^5}\frac{\partial {\tilde{U}}_4}{\partial {\theta }}\right) \frac{\partial {\theta }}{\partial {X}}\right] \\ \frac{\partial {U}}{\partial {Y}}= & {} G\left[ Y\left( \frac{\tilde{U}_0}{R^3}+\frac{3\tilde{U}_2}{R^5}+\frac{5\tilde{U_4}}{R^7}\right) - \left( \frac{1}{R^3}\frac{\partial {\tilde{U}}_2}{\partial {\theta }}+\frac{1}{R^5}\frac{\partial {\tilde{U}}_4}{\partial {\theta }}\right) \frac{\partial {\theta }}{\partial {Y}}\right] \end{aligned}$$

where

$$\begin{aligned} \frac{\partial {\theta }}{\partial {X}}=-\frac{Y}{R^2},\quad \frac{\partial {\theta }}{\partial {Y}}=\frac{X}{R^2} \end{aligned}$$

The torque \(M_z^B\) is given by

$$\begin{aligned} M_z^B=-\frac{\partial {U}}{\partial {\theta _B}}= -\frac{\partial {U}}{\partial {\phi }}\frac{\partial {\phi }}{\partial {\theta }_B}= -\frac{\partial {U}}{\partial {\phi }}=G\left[ \frac{1}{R^3}\frac{\partial {\tilde{U}_2}}{\partial {\phi }} +\frac{1}{R^5}\frac{\partial {\tilde{U}_4}}{\partial {\phi }}\right] \end{aligned}$$

and the torque \(M_z^A\) is given by

$$\begin{aligned} M_z^A=-\frac{\partial {U}}{\partial {\theta _A}}=-\frac{\partial {U}}{\partial {\theta }}\frac{\partial {\theta }}{\partial {\theta }_A} -\frac{\partial {U}}{\partial {\phi }}\frac{\partial {\phi }}{\partial {\theta }_A}=\frac{\partial {U}}{\partial {\theta }} +\frac{\partial {U}}{\partial {\phi }}={\mathbf {R}}\times \frac{\partial {U}}{\partial {{\mathbf {R}}}}-M^B_z \\ =-G\left[ \frac{1}{R^3}\frac{\partial {\tilde{U}_2}}{\partial {\theta }}+\frac{1}{R^5}\frac{\partial {\tilde{U}_4}}{\partial {\theta }}\right] -G\left[ \frac{1}{R^3}\frac{\partial {\tilde{U}_2}}{\partial {\phi }}+\frac{1}{R^5}\frac{\partial {\tilde{U}_4}}{\partial {\phi }}\right] \end{aligned}$$

We have

$$\begin{aligned} \frac{\partial {\tilde{U}_2}}{\partial {\theta }}= & {} -2A_2\sin 2\theta -2A_3\sin 2\delta \\ \frac{\partial {\tilde{U}_2}}{\partial {\phi }}= & {} 2A_3\sin 2\delta \frac{\partial {\tilde{U}_4}}{\partial {\theta }}=-2B_2\sin 2\theta \\&-\,4B_3\sin 4\theta -2B_4\sin 2\delta -4B_5\sin 4\delta -4B_7\sin \left( 2\theta +2\delta \right) \\ \frac{\partial {\tilde{U}_4}}{\partial {\phi }}= & {} 2B_4\sin 2\delta +4B_5\sin 4\delta -2B_6\sin 2\phi +2B_7\sin (2\theta +2\delta ) \end{aligned}$$

The accuracy of these formulae is checked by a handmade program. Integrated orbits using these formulae are compared with the ones integrated by using Eq. (21) (or. Eq. (27)) which needs the intermediate variables \(T^{\prime lmn}_B\). Details of these comparisons are omitted. These formulae should be useful for future studies. Compared with the \(2\hbox {nd}\) order potential, increase of the computation cost is acceptable.

One remark is made here. Utilizing the fact that

$$\begin{aligned} \theta =-\phi _1,\quad \delta =-\phi _2 \end{aligned}$$

The second order potential is identical to that given in (Scheeres 2009) which used a different set of variables. For researchers who are interested in this set of variables, the 4th order potential can be easily obtained from the formulae given here.

Appendix 2: the matrix \(\tilde{A}^{\prime }\)

From Eq. (32), neglecting the obvious zero terms for the contact equilibrium, we have the following variational equations

$$\begin{aligned} \left\{ \begin{array}{ll} \delta \ddot{R}=\left( \dot{\theta }+\dot{\theta }_A\right) ^2\delta R+2R\left( \dot{\theta }+\dot{\theta }_A\right) \delta \dot{\theta }+2R\left( \dot{\theta }+\dot{\theta }_A\right) \delta \dot{\theta }_A -\frac{1}{m}U_{RR}\delta R \\ \delta \ddot{\theta }=-\left[ \left( \frac{1}{I^A_z}+\frac{1}{mR^2}\right) U_{\theta \theta }+\frac{1}{I^A_z}U_{\theta \phi }\right] \delta \theta -\left[ \left( \frac{1}{I^A_z}+\frac{1}{mR^2}\right) U_{\theta \phi }+\frac{1}{I^A_z}U_{\phi \phi }\right] \delta \phi -\frac{2\left( \dot{\theta }+\dot{\theta }_A\right) }{R}\delta \dot{R} \\ \delta \ddot{\phi }=-\left[ \left( \frac{1}{I^A_z}+\frac{1}{I^B_z}\right) U_{\theta \phi }+\frac{1}{I^A_z}U_{\theta \theta }\right] \delta \theta -\left[ \left( \frac{1}{I^A_z}+\frac{1}{I^B_z}\right) U_{\phi \phi }+\frac{1}{I^A_z}U_{\theta \phi }\right] \delta \phi \end{array} \right. \end{aligned}$$

Utilizing the fact that the total angular momentum of the system conserves, we have

$$\begin{aligned} \delta \dot{\theta }_A=-\frac{2mR \left( \dot{\theta }_A+\dot{\theta }\right) }{I_z}\delta R- \frac{mR^2}{I_z}\delta \dot{\theta }-\frac{I^B_z}{I_z}\delta \dot{\phi } \end{aligned}$$

Substituting the relation to the above equation, we have

$$\begin{aligned} \left\{ \begin{array}{ll} \delta \ddot{R}=a_{41}\delta R+a_{45}\delta \dot{\theta }+a_{46}\delta \dot{\phi } \\ \delta \ddot{\theta }=a_{52}\delta \theta +a_{53}\delta \phi +a_{54}\delta \dot{R} \\ \delta \ddot{\phi }=a_{62}\delta \theta +a_{63}\delta \phi \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} a_{41}= & {} \dot{\theta }^2_A\left( 1-\frac{4mR^2}{I_z}\right) -\frac{1}{m}U_{RR}, \\ a_{45}= & {} 2R\dot{\theta }_A\left( 1-\frac{mR^2}{I_z}\right) , \\ a_{46}= & {} -\frac{2R\dot{\theta }_AI^B_z}{I_z}, \\ a_{52}= & {} -\left[ \left( \frac{1}{I^A_z}+\frac{1}{mR^2}\right) U_{\theta \theta }+\frac{1}{I^A_z}U_{\theta \phi }\right] , \\ a_{53}= & {} -\left[ \left( \frac{1}{I^A_z}+\frac{1}{mR^2}\right) U_{\theta \phi }+\frac{1}{I^A_z}U_{\phi \phi }\right] , \\ a_{54}= & {} -\frac{2\dot{\theta }_A}{R}, \\ a_{62}= & {} -\left[ \left( \frac{1}{I^A_z}+\frac{1}{I^B_z}\right) U_{\theta \phi }+\frac{1}{I^A_z}U_{\theta \theta }\right] , \\ a_{63}= & {} -\left[ \left( \frac{1}{I^A_z}+\frac{1}{I^B_z}\right) U_{\phi \phi }+\frac{1}{I^A_z}U_{\theta \phi }\right] \end{aligned}$$

The matrix \(\tilde{A}^{\prime }\) is a plain fact from the above equation. For the equilibrium case studied in this paper (i.e., two ellipsoids aligning with each other along their long axes), we have

$$\begin{aligned} \dot{\theta }_A^2= & {} \frac{G}{m}\left[ \frac{\tilde{U}_0}{R^3}+\frac{3\left( A_1+A_2+A_3\right) }{R^5}+\frac{5\left( B_1+B_2+B_3+B_4+B_5+B_6+B_7\right) }{R^7}\right] \\ U_{RR}= & {} -G\left[ \frac{2\tilde{U}_0}{R^3}+\frac{12\left( A_1+A_2+A_3\right) }{R^5}+\frac{30\left( B_1+B_2+B_3+B_4+B_5+B_6+B_7\right) }{R^7}\right] \\ U_{\theta \theta }= & {} 4G\left[ \frac{\left( A_2+A_3\right) }{R^3}+\frac{\left( B_2+4B_3+B_4+4B_5+B_6+4B_7\right) }{R^5}\right] \\ U_{\theta \phi }= & {} -4G\left[ \frac{A_3}{R^3}+\frac{\left( B_4+4B_5+2B_7\right) }{R^5}\right] \\ U_{\phi \phi }= & {} 4G\left[ \frac{A_3}{R^3}+\frac{\left( B_4+4B_5+B_6+B_7\right) }{R^5}\right] \end{aligned}$$

The eigenvalues of the system satisfies the following equation

$$\begin{aligned}&\lambda ^6-\left( a_{41}+a_{52}+a_{63}+a_{45}a_{54}\right) \lambda ^4 +\left( a_{41}a_{63}+a_{52}a_{63}+a_{41}a_{52}-a_{53}a_{62}+a_{45}a_{54}a_{63}\right. \\&\left. \quad -\,a_{46}a_{62}a_{54}\right) \lambda ^2 + \left( a_{53}a_{62}a_{41}-a_{41}a_{52}a_{63}\right) =0 \end{aligned}$$

From this equation, we compute the eigenvalues and generate the left frame of Fig. 6.

For the equilibrium case studied in this paper, the system energy is

$$\begin{aligned} E=\frac{H^2}{2I_p}+U=\frac{1}{2}(mR^2+I^A_z+I^B_z)\frac{G}{m} \left[ \frac{\tilde{U}_0}{R^3}+\frac{3D_1}{R^5}+\frac{5D_2}{R^7}\right] -G\left[ \frac{\tilde{U}_0}{R}+\frac{D_1}{R^3}+\frac{D_2}{R^5}\right] \end{aligned}$$

where

$$\begin{aligned} D_1= & {} A_1+A_2+A_3 \\ D_2= & {} B_1+B_2+B_3+B_4+B_5+B_6+B_7 \end{aligned}$$

The condition \(E=0\) gives out the following equation

$$\begin{aligned}&5D_2\left( I^A_z+I^B_z\right) \left( \frac{1}{R^2}\right) ^3+\left[ 3D_1\left( I^A_z+I^B_z\right) +3mD_2\right] \left( \frac{1}{R^2}\right) ^2\\&\quad +\,\left[ (I^A_z+I^B_z)\tilde{U}_0+mD_1\right] \left( \frac{1}{R^2}\right) -m\tilde{U}_0=0 \end{aligned}$$

From this equation, we compute the \(R_H \ge 1\) in the right frame of Fig. 6.

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Hou, X., Scheeres, D.J. & Xin, X. Mutual potential between two rigid bodies with arbitrary shapes and mass distributions. Celest Mech Dyn Astr 127, 369–395 (2017). https://doi.org/10.1007/s10569-016-9731-y

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