Controlled Spacecraft Trajectories in the Context of a Mission to a Binary Asteroid System

Abstract

The NASA DART and ESA Hera missions to binary asteroid 65803 Didymos have generated a lot of interest in the study of spacecraft dynamics in the vicinity of binary asteroids. The combination of the effect of the irregular shape and the rotation of the primary bodies makes them not only non-linear, but also non-autonomous systems. This study uses a technique to design reference trajectories in the context of a mission to a binary asteroid system using a fourth-order gravitational potential model for the two primary bodies. The rotation of the primary bodies, their elliptical mutual motion and the solar radiation pressure are also considered in the model, which makes this study unique. It then compares the control effort required when these reference trajectories are used with that required when reference trajectories are built with simpler models. The goal is to examine how the choice of the model used to compute reference trajectories influences the control effort required to keep the spacecraft close to them.

Appendix: Calculation of the inertia integrals based on a polyhedron shape model

The expressions for the inertia integrals can be calculated by integration for simple geometrical shapes, such as a triaxial ellipsoid. They can also be calculated based on the polyhedron shape of the primary bodies for arbitrary shape. A method to calculate the second order inertia integrals is developed by [12] and summarized here. The fourth order inertia integrals can also be calculated using a similar method.

A polyhedron shape model is composed of tetrahedrons having their triangular faces composing the external shape of the body. Each face is described by the vector of coordinates of the vertices composing it, named here D, E and F as in Fig. 24. The edges G and H are defined as follow:

$$ \begin{array}{ll} &\boldsymbol{G} = \boldsymbol{E}-\boldsymbol{D}\\ &\boldsymbol{H} = \boldsymbol{F}-\boldsymbol{D} \end{array} $$

(24)

Fig. 24

Facet with vertices and edges

All of the coordinates of the model are defined with respect to the origin O which does not necessarily correspond to the center of mass of the arbitrarily shaped body.

The first things to define are the area of a facet, the volume of a tetrahedron and the center of mass of the body modeled using a polyhedron shape.

The area of a facet, ΔS is the norm of the cross product of the edges G and H divided by 2:

$$ \begin{array}{ll} &\boldsymbol{N} = \boldsymbol{G}\times\boldsymbol{H}\\ &{\Delta} S = N/2 \end{array} $$

(25)

The total surface of the body is then the sum of all the ΔSs.

The volume of a tetrahedron is 1/3 its height multiplied by the surface of its facet:

$$ {\Delta} V = \boldsymbol{D}/3 \cdot \left( \boldsymbol{G} \times \boldsymbol{H}\right)/2 $$

(26)

Here again, the total volume of the body is the sum of the ΔV s.

Considering that the body has uniform density ρ, each tetrahedron can be considered having a mass of ΔM = ρΔV. The centroid of a tetrahedron is calculated by:

$$ {\Delta} \boldsymbol{R} = \left( \boldsymbol{D}+\boldsymbol{E}+\boldsymbol{F}+\boldsymbol{O}\right)/4 $$

(27)

The center of mass R of the body as a whole is calculated as follow:

$$ \boldsymbol{R} = \sum {\Delta} M {\Delta} \boldsymbol{R}/M = \sum {\Delta} V {\Delta} \boldsymbol{R}/V $$

(28)

Second Order Inertia Integrals and Moments of Inertia

The calculation of the second moments of inertia is well described by [12] and will be summarized here. The second order moments of inertia can be represented in the inertia matrix of the body:

$$ \boldsymbol{I}=\begin{bmatrix} I_{xx}& -I_{xy}& -I_{xz}\\-I_{xy}&I_{yy}&-I_{yz}\\-I_{xz}&-I_{yz}&I_{zz} \end{bmatrix} $$

(29)

since I_jk = I_kj, there are only six values to calculate to get the full inertia matrix. Each term of the inertia matrix can be calculated are follow:

$$ \begin{array}{ll} &I_{xx}= J_{yy}+J_{zz}\\ &I_{yy}= J_{xx}+J_{zz}\\ &I_{zz}= J_{yy}+J_{xx}\\ &I_{xy}= J_{xy}\\ &I_{yz}= J_{yz}\\ &I_{zx}= J_{zx} \end{array} $$

(30)

where:

$$ J_{jk}=\rho\int\int\int jk dV $$

(31)

where the triple integral in x, y, and z is done over the volume of the body with a uniform density ρ. The results are the second order inertia integrals of the body.

Dobrovolskis [12], simplified the volume integral using Gauss’ theorem which states that the integral of a value q = ∇⋅Q over a closed volume V is equal to the integral of Q over its boundary S:

$$ \int\int\int qdV = \int\int\boldsymbol{Q}\cdot d\boldsymbol{S} $$

(32)

In the present case, dS is a vector in the direction normal to the surface S with a magnitude equal to the element of area dS. Taking the vector Q parallel to the radius vector r going from the origin to an arbitrary point on the external facet assures that only the external surface contribute to the integral. The simplest choice of Q is then jkr/5. The products of inertia then become:

$$ J_{jk}=\frac{\rho}{5}\int\int jk\boldsymbol{r}\cdot d\boldsymbol{S} $$

(33)

Considering that:

$$ \boldsymbol{r}\cdot d\boldsymbol{S} = \left( \boldsymbol{r}\cdot\boldsymbol{N}/N\right) dS = \left( 6{\Delta} V/N\right) dS $$

(34)

the integral to evaluate is then:

$$ {\Delta} J_{jk} = \rho \frac{6}{5}\frac{\Delta V}{N}\int\int jk dS $$

(35)

To simplify the integral, let us define dimensionless coordinates $\left (g,h\right )$ in the plane of the facet D, E, F such that:

$$ \boldsymbol{r} = \boldsymbol{D}+g\boldsymbol{G}+h\boldsymbol{H} $$

(36)

We then have:

$$ \begin{array}{ll} &x = D_{x}+gG_{x}+hH_{x}\\ &y = D_{y}+gG_{y}+hH_{y}\\ &z = D_{z}+gG_{z}+hH_{z} \end{array} $$

(37)

The element of area dS then becomes Ndgdh. Inserting the last results into Eq. 35 gives:

$$ \begin{array}{ll} {\Delta} J_{jk}&=\displaystyle\frac{6}{5}\rho{\Delta} V\int\int jk dgdh\\ &=\displaystyle\frac{6}{5}\rho {\Delta} V \int\int\left[D_{j} D_{k}+g^{2}G_{j} G_{k} +h^{2} H_{j} H_{k}+g\left( D_{j} G_{k}+D_{k} G_{j}\right)\right.\\ &\left. +h\left( D_{j} H_{k}+D_{k} H_{j}\right)+gh\left( G_{j} H_{k}+G_{k} H_{j}\right)\right] dgdh \end{array} $$

(38)

With the limit of integration being either 0 < g < 1 − h, 0 < h < 1 or 0 < h < 1 − g, 0 < g < 1, it can be shown that:

$$ \begin{array}{ll} &\displaystyle \int\int dgdh = \frac{1}{2}\\ &\displaystyle \int\int g dgdh = \int\int h dgdh = \frac{1}{6}\\ &\displaystyle \int\int g^{2} dgdh = \int\int h^{2} dgdh = \frac{1}{12}\\ &\displaystyle \int\int gh dgdh = \frac{1}{24} \end{array} $$

(39)

Substituting in Eq. 38 and applying G = E −D and H = F −D, gives the final result:

$$ {\Delta} J_{jk} = \frac{\rho{\Delta} V}{20}\left[2D_{j} D_{k} +2E_{j} E_{k}+2F_{j} F_{k}+D_{j} E_{k}+D_{k} E_{j}+D_{j} F_{k}+D_{k} F_{j}+E_{j} F_{k}+E_{k} F_{j}\right] $$

(40)

As previously, J_jk is the sum of the ΔJ_jk.

At this point, the inertia matrix is populated, but the origin of the reference frame used to define the position of the vertices is not necessarily the center of mass of the body and the reference frame is not necessarily aligned with the principal axes. First, let us use the parallel axis theorem to calculate the equivalent moments and product of inertia for a reference frame having its origin at the center of mass of the body. By using the previously calculated center of mass of the body with coordinates X, Y and Z:

$$ \boldsymbol{I}^{\prime} = \boldsymbol{I}-M_{a} \begin{bmatrix} Y^{2}+Z^{2}&-XY&-XZ\\-XY&X^{2}+Z^{2}&-YZ\\-XZ&-YZ&X^{2}+Y^{2} \end{bmatrix} $$

(41)

Because the inertia matrix is composed of real numbers and is symmetric, its eigenvalues are also real numbers and its eigenvectors are orthogonal. The direction cosine matrix (DCM) describing the orientation of the actual reference frame relative to reference frame aligned with the principal axes is then the matrix composed of the eigenvectors of the inertia matrix with the reference frame aligned with the center of mass of the body. Transposing it gives the DCM describing the reference frame having origin at the principal axes of the body relative to the original reference frame. Knowing that, it is then possible to calculate the coordinates of each vertices into the reference frame aligned with the principal axes of the body (body i fixed reference frame)and having its origin at the center of mass of the body. Applying Eq. 40 to the modified coordinates show that the products of inertia have now negligible values compared to the principal moments of inertia. These modified coordinates will be used to calculate the fourth order inertia integrals of the body.

Fourth Order Inertia Integrals

Using the modified coordinates for the calculation of the fourth order inertia integrals permits to only calculate the following six values, the other ones being null:J_xxxx, J_yyyy, J_zzzz, J_xxyy, J_yyzz, J_zzxx. The same method than the one described in [12] is used here. In the fourth order case, Q is (jk)²r/7. The integrals that need to be evaluated are then:

$$ {\Delta} J_{jjkk}= \rho \frac{6}{7}\frac{\Delta V}{N}\int\int jjkk dgdh $$

(42)

The required extra integrals for the calculation of the fourth order inertia integrals are then:

$$ \begin{array}{ll} &\displaystyle \int\int g^{3} dgdh = \int\int h^{3} dgdh = \frac{1}{20}\\ &\displaystyle \int\int g^{4} dgdh = \int\int h^{4} dgdh = \frac{1}{30}\\ &\displaystyle \int\int gh^{2} dgdh = \int\int g^{2}h dgdh = \frac{1}{60}\\ &\displaystyle \int\int gh^{3} dgdh = \int\int g^{3}h dgdh = \frac{1}{120}\\ &\displaystyle \int\int g^{2}h^{2} dgdh = \frac{1}{180} \end{array} $$

(43)

Substituting these results into Eq. 42 and after some algebra, the resulting calculation of the fourth order inertia integrals is:

$$ \begin{array}{ll} &{\kern-.4pc}\displaystyle{\Delta} J_{jjkk} = \frac{6}{7}\rho{\Delta} V \left[\frac{1}{2}{D^{2}_{j}} {D^{2}_{k}} +\frac{1}{12}\left( {D^{2}_{j}}\left( {G^{2}_{k}}+{H^{2}_{k}}\right) +4D_{j} D_{k}\left( G_{j} G_{k} + H_{j} H_{k}\right) + {D^{2}_{k}} \left( {G^{2}_{j}} +{H^{2}_{j}}\right) \right)\right.\\ &{\kern2.4pc}\displaystyle+\frac{1}{3}\left( {D^{2}_{j}} D_{k} \left( G_{k} +H_{k}\right) + D_{j} {D^{2}_{k}} \left( G_{j} + H_{j}\right) \right)\\ &{\kern2.4pc}\displaystyle+\frac{1}{10}\left( D_{k} \left( {G^{2}_{j}} G_{k} + {H^{2}_{j}} H_{k}\right) + D_{j} \left( G_{j} {G^{2}_{k}} + H_{j} {H^{2}_{k}}\right)\right)\\ &{\kern2.4pc}\displaystyle+\frac{1}{30}\left( {G^{2}_{j}} {G^{2}_{k}}+{H^{2}_{j}} {H^{2}_{k}}\right)+\frac{1}{12}\left( {D^{2}_{j}} G_{k} H_{k} +2D_{j} G_{j} D_{k} H_{k} +2D_{j} H_{j} D_{k} G_{k} + G_{j} H_{j} {D^{2}_{k}}\right)\\ &{\kern2.4pc}\displaystyle+\frac{1}{30}\left( D_{k}\left( {G^{2}_{j}} H_{k}+{H^{2}_{j}} G_{k}+2G_{j} H_{j} G_{k} + 2G_{j} H_{j} H_{k}\right)\right.\\ &{\kern2.4pc}\left.\displaystyle+D_{j} \left( H_{j} {G^{2}_{k}} + G_{j} {H^{2}_{k}}+2G_{j} G_{k} H_{k} +2 H_{j} G_{k} H_{k}\right)\right)\\ &{\kern2.4pc}\left.\displaystyle+\frac{1}{60}\left( H_{j} G_{j}\left( {G^{2}_{k}} {}+{}{H^{2}_{k}}\right)+H_{k} G_{k}\left( {G^{2}_{j}} {}+{} {H^{2}_{j}}\right)\right){}+{}\frac{1}{180}\left( {G^{2}_{j}} {H^{2}_{k}} + {H^{2}_{j}} {G^{2}_{k}} +4G_{j} H_{j} G_{k} H_{k}\right)\right] \end{array} $$

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The special case of J_jjjj is given by:

$$ \begin{array}{ll} &{\kern-.5pc}\displaystyle{\Delta} J_{jjjj} = \frac{6}{7}\rho{\Delta} V \left[\frac{1}{2}{D^{4}_{j}}+\frac{1}{2}{D^{2}_{j}} \left( {G^{2}_{j}}+{H^{2}_{j}}+G_{j} H_{j}\right)+\frac{2}{3}{D^{3}_{j}}\left( G_{j}+H_{j}\right)\right.\\ &{\kern2pc}\left. +\displaystyle\frac{1}{5}D_{j}\left( {G^{3}_{j}} +{H^{3}_{j}}+{G^{2}_{j}} H_{j} +G_{j} {H^{2}_{j}}\right)+\frac{1}{30}\left( {G^{4}_{j}}+{H^{4}_{j}} +{G^{2}_{j}} {H^{2}_{j}}+{G^{3}_{j}} H_{j}+{H^{3}_{j}} G_{j}\right)\right] \end{array} $$

(45)