A Closed-Form Solution to the Minimum V 2

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1 Celestial Mechanics and Dynamical Astronomy manuscrit No. (will be inserted by the editor) Martín Avendaño Daniele Mortari A Closed-Form Solution to the Minimum V tot Lambert s Problem Received: Month Day, Year / Acceted: date Abstract A closed form solution to the minimum Vtot Lambert roblem between two assigned ositions in two distinct orbits is resented. Motivation comes from the need of comuting otimal orbit transfer matrices to solve re-configuration roblems of satellite constellations and the comlexity associated in facing these roblems with the minimization of V tot. Extensive numerical tests show that the difference in fuel consumtion between the solutions obtained by minimizing Vtot and V tot does not exceed 17%. The Vtot solution can be adoted as starting oint to find the minimum V tot. The solving equation for minimum Vtot Lambert roblem is a quartic olynomial in term of the angular momentum modulus of the otimal transfer orbit. The root selection is discussed and the singular case, occurring when the initial and final radii are arallel, is analytically solved. A numerical examle for the general case (orbit transfer ork-cho between two non-colanar ellitical orbits) and two examles for the singular case (Hohmann and GTO transfers) are rovided. 1 Introduction Imulsive orbit transfer [1 3] is an imortant research area in celestial mechanics with an existing rich literature. The variety of studies erformed in Martín Avendaño Texas A&M University, Deartment of Mathematics 03 Milner - College Station, TX Tel.: +1 (979) , Fax: +1 (979) avendano@math.tamu.edu Daniele Mortari Texas A&M University, Deartment of Aerosace Engineering 611C H.R. Bright Bldg - College Station, TX Tel.: +1 (979) , Fax: +1 (979) mortari@aero.tamu.edu

2 this area is wide, covering different roblems, such as the existence of an overall otimal solution for the n-imulse maneuver roblem [4], a different roof of the Hohmann transfer [5], as well as a technique based on re-comuted tables for the colanar orbit transfer case [6]. Lambert s roblem is a articular instance of the -imulse orbit transfer roblem, a well-known roblem in celestial mechanics. The roblem consists of finding the orbital elements of a transfer orbit connecting a dearture osition (radius R 1 ) with an arrival osition (radius R ) under the assumtion that the orbit transfer must be erformed on an assigned time interval T (time of flight). In a more comact form, the roblem can be stated as it follows: Given initial and final radius vectors, find the conic section with rescribed time-of-flight between initial and final ositions. This roblem is extensively solved to find the otimal dearture and time-of-flight for a generic two-imulse orbit transfer, where the otimality usually imlies the minimization of the fuel consumtion. A tyical examle is the building of the so called ork-cho lots [14] allowing the reliminary analysis to identify dearture and arrival times of rendezvous roblems. Because of its imortance, since its initial formulation, several mathematical aroaches have been develoed to solve this roblem in the most efficient and robust way. Among all the imortant contributions given to this roblem, we can list [7], roviding a universal solution using second order Halley s iteration, [15], resenting a uniform method (indeendent from the orbit), and [16, 17] showing an iterative technique to comute the solution with 6 digits of accuracy. Alternatively, references [8 1] have faced the otimal two-imulse orbit transfer roblem in a slightly different way, known as the minimum V tot Lambert roblem, where the constraint of erforming the orbit transfer in a rescribed time interval is substituted by finding the orbit transfer associated with minimum fuel consumtion. The existing roosed techniques to solve this roblem can be classified in two distinct categories. The first one rooses iterative techniques characterized by different convergence seed and comutational load, while the aroaches belonging to the second category all arrive at a comlicate equation whose solution requires numerical technique (e.g., sixth degree olynomial root solver). So far no closed form solution has been found to the minimum V tot Lambert roblem, and even in the most recent attemts to solve it (e.g., Ref. [13]), the solution is obtained through numerical techniques whose convergence (or divergence) is a function of the initial oint selection. On the other hand, the convergence seed is also a function of the articular roblem to be solved. The geometry of this minimum V tot Lambert roblem, is rovided in Fig. 1. The otimal orbit transfer roblem is a roblem that must be extensively solved in satellite re-configuration roblems, where the transfer orbit cost matrices must be evaluated to solve the otimal combinatory roblem that is finding the destination orbit to each initial orbit. Motivated by this secific reason, this aer rooses to solve the Vtot Lambert roblem, a roblem we rove that can be solved in a closed form. The urose is to minimize the following cost function: V tot = V 1 + V, (1)

3 3 where V 1 and V reresent the initial and final orbit transfer imulses, resectively. Minimization of V tot means fuel minimization while Vtot minimization does not. However, the question of how much more fuel the Vtot minimization requires is a question that is here addressed. Assume that the minimum V tot and Vtot require two different two-imulse maneuvers, and q, resectively. It is clear that V tot () V tot (q) and that Vtot(q) Vtot(). Using the elementary inequalities, a + b (a + b ) and a + b a + b, we can derive the following estimation V tot () V tot (q) Vtot(q) Vtot() V tot (). () From a strictly mathematical oint of view, Eq. () means that the manoeuver minimizing Vtot requires u to 41.5% more fuel that the manoeuver minimizing V tot. This uer bound estimation is, indeed, substantially far from what exerienced in ractical cases. Extensive numerical tests have shown that Eq. () overestimates the real difference by obtaining an uer bound value of less than 17%. In addition, this uer bound value aears to be associated with those cases where the orbital lane change is dramatic. Most of the numerical tests show minimal differences and in many cases, as for the Hohmann transfer case, the minimizations of V tot and Vtot give the same solution. Problem definition Let us consider the minimum-fuel two-imulse orbit transfer roblem between orbit 1 and orbit, whose geometry in rovided in Fig. 1. Initially, the sacecraft is on orbit 1 with radius R 1 = R 1 r 1 and velocity V 1, where it alies the first imulse V 1 = W 1 V 1 and moves into the transfer orbit with velocity W 1. Finally, it reaches the target orbit at radius R = R r with velocity W, where it alies the second imulse, V = V W. The transfer orbit lane is assigned as the lane where the dearture and arrival vectors lie. In this lane we have two transfer trajectories allowing the sacecraft to go from R 1 to R. In terms of true anomaly variations, these two transfer trajectories comlement each other since the variation can be greater than π or less than π, deending on the selected direction of the angular momentum. In other words, the comutation of the true anomaly variation must be consistent with the definition of the angular momentum direction. We have two distinct directions for the angular momentum h ± = ± R 1 R R 1 R (3) The direction of the angular momentum also establishes the transfer direction and, consequently, the dearture and arrival true anomalies, ϕ 1 and ϕ,

4 4 Fig. 1 Radii and velocities definitions resectively. In order to be consistent with the angular momentum direction, the difference between the two true anomalies, ϕ = ϕ ϕ 1, must satisfy sin ϕ ± = r (h ± r 1 ) and cos ϕ = r 1 r (4) where ϕ + + ϕ = π. Initial and final velocities, V 1 and V, can be slit in radial, tangential, and normal comonents, as follows { V1 = (V 1 r 1 )r 1 + (V 1 s 1 )s 1 + (V 1 h)h = V 1r r 1 + V 1s s 1 + V 1n h V = (V r )r + (V s )s + (V h)h = V r r + V s s + V n h (5) where s 1 = h r 1 and s = h r. This minimum Vtot Lambert roblem imlies the minimization of the following quantity Vtot = (W 1 V 1 ) (W 1 V 1 ) + (V W ) (V W ) = = V1 + V (V 1 W 1 + V W ) + W1 + W From the energy equation alied to the transfer orbit we obtain W 1 = µ R 1 µ a and W = µ R µ a where a is the unknown semi-major axis of the transfer orbit. Substituting Eq. (7) in Eq. (6), the transfer cost function to minimize becomes (6) (7) V tot = C (V 1 W 1 + V W ) µ a (8)

5 5 where C = V1 + V + µ + µ includes all known constants. This means R 1 R that our roblem becomes to maximize the cost function G = V 1 W 1 + V W + µ (1 e ) where e and are the eccentricity and the semi-arameter of the transfer orbit. The olar reresentation of the orbital radius allows us to write and R = R 1 = (9) 1 + e cos ϕ 1 e cos ϕ 1 = R 1 1 (10) 1 + e cos(ϕ 1 + ϕ) e cos(ϕ 1 + ϕ) = R 1 (11) By exanding cos(ϕ 1 + ϕ) = cos ϕ 1 cos ϕ sin ϕ 1 sin ϕ, we can combine Eq. (10) and Eq. (11) into [( ) ( )] 1 e sin ϕ 1 = 1 cos ϕ 1 (1) sin ϕ R 1 R Equation (10) can be rewritten as e cos ϕ 1 = e cos[(ϕ 1 + ϕ) ϕ]; therefore we can write e [cos(ϕ 1 + ϕ) cos ϕ + sin(ϕ 1 + ϕ) sin ϕ] = R 1 1 from which we obtain e sin(ϕ 1 + ϕ) = 1 sin ϕ [( ) ( ) ] 1 1 cos ϕ R 1 R (13) The two scalar roducts aearing in Eq. (9) can be evaluated in any reference frame. For instance, they can be evaluated in the orbital reference frame of the transfer orbit. This yields T V 1r cos ϕ 1 V 1s sin ϕ 1 µ sin ϕ 1 V 1 W 1 = V 1r sin ϕ 1 + V 1s cos ϕ 1 e + cos ϕ 1 0 = V 1n µ [ V 1r e sin ϕ 1 + V 1s (1 + e cos ϕ 1 ) ] Substituting Eq. (10) and Eq. (1) in Eq. (14) we obtain = (14) V 1 W 1 = K 1 + L 1 (15) where K 1 = µ sin ϕ L 1 = (1 cos ϕ) µ sin ϕ ( cos ϕ 1 ) µ V 1r + V 1s R 1 R R 1 (16) V 1r

6 6 Analogously, for the V W we obtain T V r cos ϕ V s sin ϕ µ sin ϕ V W = V r sin ϕ + V s cos ϕ e + cos ϕ 0 = V n = µ {V r e sin(ϕ 1 + ϕ) + [1 + e cos(ϕ 1 + ϕ)] V s } (17) Substituting Eq. (11) and Eq. (13) in Eq. (17) we obtain where V W = K + L (18) ( ) µ 1 cos ϕ µ K = V r + V s sin ϕ R 1 R R (19) µ (cos ϕ 1) L = V r sin ϕ The eccentricity can also be exressed in terms of the semi-arameter of the transfer orbit. Using Eq. (10) and Eq. (1) we can evaluate the quantity where 1 e = 1 (e sin ϕ 1 ) (e cos ϕ 1 ) = α + β + γ (0) α = 1 [( 1 cos ϕ R1 sin 1 )] ϕ R 1 R β = ( (1 cos ϕ) cos ϕ R 1 sin 1 ) ϕ R 1 R (1 cos ϕ) γ = sin ϕ (1) Using Eq. (15), Eq. (18), and Eq. (0), we can now re-write Eq. (9) in terms of only the semi-arameter, G = (K 1 + K ) + L ( 1 + L + µ α + β + γ ). () The semi-arameter is related to the modulus of the angular momentum as h = µ. Therefore, we can re-write Eq. () as follows G = K 1 + K µ (L1 + L ) h + + α h + µ β + µ γ µ h h (3) Stationary conditions imly G (h) = 0. Multilying G (h) by h 3 /(α) we get a quartic with missing quadratic term F (h) = h 4 + c 3 h 3 + c 1 h + c 0 = 0 (4) where c 3 = K 1 + K α µ µ (L1 + L ) c 1 =, and c 0 = µ γ α α. (5)

7 7 This quartic equation is solved using Ferrari s method as described in the aendix. A simle insection of Eq. (1) shows that both α and γ are negative. Therefore the indeendent coefficient c 0 of the quartic is also negative, by Eq. (5). This imlies that Eq. (4) has at least one ositive and one negative root, because F (0) = c 0 < 0 and lim h F (h) = +. For the same reason, one of the ositive solutions satisfies F (h) > 0, which means that G (h) = αf (h)/h 3 < 0; i.e., it is a local maximum of G as we needed. We can discard the negative and comlex roots, since they carry no hysical meaning (h reresents the magnitude of the angular momentum vector). As we initially said, we have to consider two ossible orientations h ± for the angular momentum. Consequently, we have to follow this rocedure twice and eventually solve two quadratic equations. However, a simle analysis of the whole algorithm reveals that these two quartic equations are identical excet the coefficients c 1 and c 3 that aear with a different sign. More recisely, if F (h) = 0 is the quartic associated with h +, then F ( h) = 0 is the quartic associated with h. From this observation we conclude that we only need to solve one quartic equation, and kee all the real solutions. The ositive roots corresond to the magnitude of the angular momentum (which is ointing in the direction h + ) and the negative roots corresond (if we change them into ositive) to the magnitude of the angular momentum (ointing as h ). When the otimal solution h ot has been found, the semi-arameter can be recovered as = h ot/µ, and the eccentricity can be comuted by e = 1 α β γ (6) using the values of α, β, and γ rovided by Eq. (1). The value for ϕ 1 is comuted using Eq. (10) and Eq. (1). Finally ϕ = ϕ 1 + ϕ. Secial care is needed here when h ot is negative: do not forget to use ϕ instead of ϕ..1 Pork-cho examle Using the theory so far exlained (and the solution rovided for the singular case that is resented in the next section) it is ossible to roduce orkchos to identify the best dearture and arrival orbital ositions for twoimulse orbit transfer roblem between two generic ellitic orbits. To make an examle with real satellites we have selected the orbits associated with the two satellites ALSAT 1 and ARIANE 44L, resectively, whose Two-Line Elements 1 (TLE) are given as it follows ALSAT U 0054A ARIANE 44L U 91075N NORAD Two-Line Element Sets are available for most of the existing satellites in the non-archival ublications htt:// and exlained in Section.4. of Ref. [].

8 8 V 1 =.156 km/s Time (UTC) 15-Se-008 at 1:38:40 R x = R y = R z = V x = V y = V z = V x = V y = V z = V r = V t = V h = V = km/s Time (UTC) 15-Se-008 at 14:05:00 R x = R y = R z = V x = V y = V z = V x =.798 V y =.408 V z =.631 V r = V t = V h = Table 1 Imulses v (km/s) km/s Figure shows the three dimensional (left) and contour (right) lots for the minimum Vtot transfer cost for the two selected orbits. The V tot (km/s) is rovided as a function of the dearture and arrival mean anomalies, resectively. Fig. 3-D and contour ma orkcho of v 1 + v The otimal orbit transfer is identified in Fig. by a black vertical bar in the 3-D lot and by a white X in the contour lot. The details of the otimal orbit transfer are given in Table 1. These are: 1) the imulse magnitude and time (UTC), ) the Cartesian osition [R x, R y, R z ], 3) the Cartesian velocity [V x, V y, V z ], 4) the Cartesian comonents of the imulse [ V x, V y, V z ], and 5) the radial, tangential, and normal comonent of the imulse [ V r, V t, V h ].

9 9 Finally, Fig. 3 shows the three-dimensional geometry of the selected orbits and the otimal orbit transfer trajectory. Fig. 3 3-D view of the otimal orbit transfer for the orkcho examle 3 Singularity The method reviously described becomes singular when dearture and arrival radii are arallel. In this case the vector roduct in Eq. (3) vanishes and the direction of the angular momentum becomes undefined. This haens, for instance, in the Hohmann transfer case. It is clear that if ϕ = 0 and R 1 R then it is imossible to go from R 1 to R by means of a simle twoimulse manoeuver. The remaining singular case, when ϕ = π, is solved in this section. If ϕ = π then ϕ = ϕ 1 + π, and the semi-arameter of any orbit assing through R 1 and R is determined R 1 = from which we obtain and R = (7) 1 + e cos ϕ 1 1 e cos ϕ 1 = R 1 R R 1 + R and e cos ϕ 1 = R R 1 R + R 1. (8) The scalar roduct between radius and velocity vector satisfies and from which we derive R 1 W 1 = h e sin ϕ e cos ϕ 1 = (R + R 1 ) h e sin ϕ 1 R (9) R W = (R + R 1 ) h e sin ϕ 1 R 1 (30) r 1 W 1 + r W = 0. (31) Equation (31) is useful to solve the singularity case because it tells us that the comonents of the transfer velocities along the direction r 1 (or r ) are identical. Equation (31) imlies the geometrical situation deicted in Fig. 4,

10 10 Fig. 4 Singularity case: geometric roerty rovided by Eq. (31) showing the transfer orbit velocities comonents in the (unknown) transfer orbit lane. In other words we can write ξ = W 1r = W r (3) for the radial comonents while the normal comonents can be derived from the angular momentum h = R 1 W 1 = R W (33) Let ϑ be the angle between the dearture orbit lane and the orbit transfer lane. We can introduce the [ r 1, s 1, h 1 ] reference frame, where h 1 = R 1 V 1 and s 1 = h 1 r 1. The transformation matrix moving R 1 V 1 from this reference frame to the inertial reference frame is C = [ r 1. s 1. h 1 ] (34) In the [ r 1, s 1, h 1 ] reference frame the initial and final velocities, V 1 and V, have comonents V 1r V 1 r 1 V r V r 1 V (1) 1 = V 1s 0 = V 1 s 1 and V(1) = V s 0 V = V s 1 h V h (35) 1 while the transfer velocity vectors can be exressed as ξ W (1) 1 = R 1 (ϑ) W 1 and W(1) = R 1 (ϑ) 0 ξ W 0 (36) where R 1 (ϑ) = cos ϑ sin ϑ is the matrix erforming rigid rotation 0 sin ϑ cos ϑ about the first axis (direction r 1 ) by the angle ϑ.

11 11 The cost function to minimize is rovided by Eq. (6). Using Eq. (33) and Eq. (36), the term (W 1 + W ) can be written as W1 + W = ξ + h R1 + h R (37) while the V 1 W 1 and V W scalar roducts can be evaluated in the [ r 1, s 1, h 1 ] reference frame as V (1) 1 W (1) h 1 = +V 1r ξ + V 1s cos ϑ R 1 V (1) W (1) = V r ξ V s h R cos ϑ V h h R sin ϑ (38) Substituting Eq. (37) and Eq. (38) in Eq. (6) we obtain a cost function in term of the two unknowns, ξ and ϑ, only. The derivatives with resect to the variables are Vtot = 0 ϑ Vtot = 0 ξ = V 1r + V r ξ ( V1s V ) s sin ϑ = V h cos ϑ (39) R 1 R R (40) The knowledge of the angle ϑ and the velocity ξ, allow us to obtain the solution for the singular case W 1 = C W (1) 1 and W = C W (1) (41) where W (1) 1 and W (1) are given by Eq. (36) and C by Eq. (34). 3.1 Hohmann transfer examle An imortant examle that can be solved using the theory resented for the singular case is the Hohmann transfer. For this case we have = R 1 R and h = µ R1 R µ =. R 1 + R R 1 + R The velocities satisfy V 1r = V r = V h = 0, V 1s = µ/r 1, and V s = µ/r Using these values, Eq. (40) gives ξ = 0 while Eq. (39) rovides ϑ = 0. Finally, Eq. (33) and Eq. (36) rovide W 1 = h µ R 0 = and W 1 = W 1 (4) R 1 R 1 (R 1 + R ) 0 which is the correct solution of the Hohmann transfer.

12 1 3. GTO examle The roblem of orbit transfer from an inclined circular arking orbit (inclination i, Radius R 1 ) to the Geo-Synchronous equatorial circular orbit (Radius R ), known as the GTO transfer roblem, is here considered as another examle of alication of the theory develoed for the singular case. For this examle, the dearture and arrival velocities vectors as exressed in the [ r 1, s 1, h 1 ] reference frame are which imlies V (1) µ 1 = R ξ = 0 and tan ϑ = and V (1) µ = R sin i (R /R 1 ) 3/ + cos i 0 cos i sin i (43) (44) The angle ϑ reresents the orbit lane change contribution during the first imulse (the lane change at second imulse will be i+ϑ). For a 8 inclined (Cae Kennedy) 500 km arking orbit, Eq. (44) rovides the correct value of ϑ = Fig. 5 GTO transfer. Difference between min( v 1 + v ) and min( v 1 + v ). The red line indicates the min( v 1 + v ) solution. Figure 5 shows, for the GTO transfer roblem, the otimal solution obtained by minimizing v 1 + v (right lot) and the one obtained by minimizing v 1 + v (left lot). The red vertical line indicates the solution obtained by minimizing v 1 + v. 4 Conclusions This aer resents a closed-form solution for the minimum V tot Lambert roblem. Motivation comes from the need of building transfer cost matrices to solve the combinatory roblems of reconfiguring satellite constellations. The resulting rocedure resented constitutes an easy tool to imlement that can be useful when extensive minimum V tot roblems must be solved, as in the re-configuration roblems of satellite constellations.

13 13 The solution rovided by minimizing Vtot does not always minimize the fuel consumtion, but it reresents a good aroximation, and it can be used as the starting oint for finding (e.g., using gradient descent aroach) the true otimal two-imulse manoeuver with minimum V tot. The difference between Vtot and V tot minimizations roblems is numerically found to be bounded with a maximum value lower than 17%. Since the method does not require any iterations, the number of stes for the whole rocedure (comlexity) is constant. This is a crucial roerty when the algorithm has to be extensively used within other otimization rograms. An examle is the building of fast ork-cho lots for twoimulse transfer between assigned orbits. Two examles for the singular case (occurring when initial and final radii are arallel) are rovided for Hohmann and GTO transfers. Aendix: Solution of Algebraic Quartic Equation Consider again Eq. (4), here reeated h 4 + c 3 h 3 + c 1 h + c 0 = 0 This equation can be solved using Ferrari s method, which is summarized in this aendix for the secific case of a quartic with missing quadratic coefficient, c = 0. By setting a = 3 c 3 8, b = c c 1, and c = 3 c c 1 c 3 + c 0 (45) 4 then, if b = 0, the four solutions are rovided by h = c 3 4 ± a ± a 4 c (if b = 0). (46) Otherwise, after evaluating the quantities P = a a3 c, Q = a c 3 b 8, and R = Q Q ± 4 + P 3 7, (47) we can comute U = 3 R and y = Finally, the four solutions are { = 5 a/6 3 Q (if U = 0) = 5 a/6 + U P/(3 U) (if U 0) h 1, = c 3 a + y ± (3 a + y + b/ a + y) 4 + h 3,4 = c 3 a + y ± (3 a + y b/ a + y) 4 (48) (49)

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