Tangent-Impulse Interception for a Hyperbolic Target

The two-body interception problem with an upper-bounded tangent impulse for the interceptor on an elliptic parking orbit to collide with a nonmaneuvering target on a hyperbolic orbit is studied. Firstly, four special initial true anomalies whose velocity vectors are parallel to either of the lines of asymptotes for the target hyperbolic orbit are obtained by using Newton-Raphson method. For different impulse points, the solution-existence ranges of the target true anomaly for any conic transfer are discussed in detail. Then, the time-of-flight equation is solved by the secant method for a single-variable piecewise function about the target true anomaly. Considering the sphere of influence of the Earth and the upper bound on the fuel, all feasible solutions are obtained for different impulse points. Finally, a numerical example is provided to apply the proposed technique for all feasible solutions and the global minimum-time solution with initial coasting time.


Introduction
The earth-crossing comets, asteroids, space debris, and other spacecraft are potentially hazardous objects (PHOs) since they may closely encounter with the Earth.Usually, an active spacecraft is required to intercept, impact, or deflect the PHO [1][2][3][4].The impact and deflection mission is mainly used when the PHO is beyond the sphere of influence of the Earth.If this mission is failed, the PHO will be on a hyperbolic trajectory during the Earth flyby; thus the interception problem from a parking orbit for a hyperbolic target needs to be solved.
Assume the nonmaneuvering target is subject only to a Newtonian gravitational attraction and the Lambert approach is the most fundamental and the simplest method to accomplish the two-body interception problem with impulse maneuvers [5][6][7].If the impulse point and the interception point are both assigned, the time of flight (TOF) for the target can be obtained by solving Kepler's equation of hyperbolic orbit, and then the interception problem is equivalent to Lambert's problem [8][9][10][11].The required velocity vector of the transfer trajectory at the impulse point can be obtained with the Lambert approach.It should be notified that the transfer trajectory can be elliptic, parabolic, and hyperbolic.
When the initial coasting time is considered, the impulse point may be not at the initial time, then interception optimization problems to save time and fuel are studied by many researchers [12][13][14].
However, no constraints are imposed on the impulse direction for the Lambert approach when both impulse and interception points are assigned.If two coplanar orbits are tangent and the velocity vectors at their common point are in the same direction, then the task of nulling the relative velocity will be simple [8].The cotangent transfer indicates that the transfer orbit is tangent to both the initial orbit and the final orbit.The classic Hohmann transfer [15] is a twoimpulse cotangent transfer between two coplanar circular orbits, and it is the minimum-energy one among all the twoimpulse transfers.This result is also satisfied for two coplanar elliptic orbits sharing the same apsidal line.For two coplanar elliptic orbits with different apsidal lines, the numerical solution [16] and the closed-form solution [17,18] have been obtained for the cotangent transfer problem.In addition, Zhang et al. solved the two-impulse cotangent rendezvous problem between two coplanar elliptic orbits [19] and between elliptic and hyperbolic orbits [20], respectively.Different from two impulses are required for the rendezvous problem, a single impulse can fulfill the interception mission.Thus, the tangent-impulse interception problem requires the same TOF for two spacecraft and a transfer orbit tangent to the initial orbit.Then, for a given impulse point, the interception point is not arbitrary and needs to be solved.
This paper studies the coplanar tangent-impulse interception problem from an interceptor on elliptic parking orbits to a target on hyperbolic orbits.Based on four special initial true anomalies, the solution-existence ranges of the target true anomaly are obtained for different impulse points.Then the one-tangent-impulse interception trajectories are obtained by solving a single-variable piecewise function.

Problem Statement
Assume that the interceptor moves on the initial elliptic parking orbit and the target moves on the target hyperbolic orbit, and the initial and target orbits are coplanar with different apsidal lines.At the initial time, the interceptor is at  10 and the target is at  20 (see Figure 1).Let the motion direction of both orbits be counterclockwise.The coasting arc for the interceptor is from  10 to  1 , and the initial coasting time before the tangent impulse is  1 .The coasting arc for the target is from  20 to  2 , and the coasting time of the target is  2 .The tangent impulse indicates that the transfer orbit is tangent to the initial elliptic orbit at  1 , and the transfer time is  3 .For the orbital interception problem, the flight time of the interceptor and that of the target are the same.For convenience, let where   ( = 1, 2, 3) denotes the coasting time with zero revolution and   ( = 1, 3) denotes the orbital period, where the subscripts "1", "2, " and "3" denote the initial, target, and transfer orbits, respectively.For the upper-bounded impulse interception, the minimum-time solution is interesting; then the revolution number for the transfer orbit is set to be 0. Then the orbital interception problem with a tangent impulse is to solve the impulse point  1 and the interception point  2 such that the following expression is satisfied: in which  is a positive integer indicates that the interceptor moves multiple revolutions before the first tangent impulse on point  1 .
For the two-impulse cotangent rendezvous problem, if the final true anomaly  2 is given, there is a single closedform solution for the initial true anomaly  1 ; then  in (1) is a function only of  2 for zero revolution.However, for the one-tangent-impulse interception problem, from the above analysis, it is known that  is a function of two variables, the initial true anomaly  1 of the impulse point,and the target true anomaly  2 of the interception point.Usually, there is an upper-bound constraint on the fuel consumption ΔV; then the interception problem is to solve all values of  1 and  2 such that (2) is satisfied and the fuel consumption is less than a given value.In this paper, firstly the interception problem will be solved at a given  1 of the impulse point; then all feasible solutions will be obtained at different  1 within one initial orbital period.

Time-of-Flight Equations
where  is the gravitational parameter,  is the radius of ,  is the true anomaly,  = (1− 2 ) is the semilatus rectum,  is the semimajor axis,  is the eccentricity, and  is the hyperbolic anomaly, which is related to the true anomaly by from which it is known that Then the derivative of  2 with respect to  2 is Furthermore, the derivative of  2 with respect to  2 is (7)

Transfer Time between the Initial and Target
Orbits.The initial flight-path angle at  1 can be written as [8]  = arctan ( ) .
Let the transfer angle be , and [9] where V 1 denotes the magnitude of the velocity vector at point  1 on the transfer orbit.For different values of , there are three cases for the transfer time [9,19].
(1) If  ∈ (0, 2), the transfer trajectory is an elliptic orbit and the transfer time is (2) If  > 2, the transfer trajectory is a hyperbolic orbit and the transfer time is (3) If  = 2, the transfer trajectory is a parabolic orbit and the transfer time is It should be notified that (( 11)-( 13)) are functions independent of V 1 .The derivative of the transfer time  3 with respect to the true anomaly  2 is where the partial derivatives  3 / For a given impulse point, the initial coasting time  1 for the interceptor is independent of  2 .Then the derivative of  with respect to  2 is

Conditions of Solution Existence
The above section gives the transfer time for any conic orbit.A necessary condition is  > 0. Since the initial flight-path angle  ∈ (−/2, /2), the initial flight-direction angle is For the coplanar orbits, the transfer angle is where  is the argument of perigee.Let where  is the semilatus rectum.There are three cases for  > 0 [18]: where the four-quadrant inverse tangent function  = atan2( 1 ,  2 ) is defined as the angle satisfying both sin  =  1 /√ Then there are also three cases to be discussed [18]: ) . ( In addition, the range of  1 for existing elliptic transfers needs to be solved.If there exists some range of  2 , for 0 <  < 2, the following inequality must be satisfied: Substituting ( 20) into (22) yields Note that the flight-direction angle  ∈ (0, /2); then cos(2) < 1 is satisfied for any  1 ; thus (23) can be rewritten as From (( 16), ( 17)) it is known that  and  are functions only of  1 , then  is a function only of  1 .The values of  1 for the equality  = 0 can be obtained by numerical methods; finally the range for inequality  > 0 in (24) is obtained.

Special Initial True Anomalies and the Solution-Existence Range
There are four special points on the initial orbit, whose directions of velocity vectors are parallel to either of the lines of asymptotes for the target hyperbolic orbit.These initial true anomalies Note that the flight-direction angle is a function only of the true anomaly; then the above equation can be rewritten as which is a function only of  1,1 .Thus, the classic Newton-Raphson iterative algorithm can be used to obtain the numerical solution for (26).
From the geometry of Figure 2(b), it is known that which can be rewritten in a function only of  1,2 as Similarly, from the geometries of Figures 2(c) and 2(d), it is known that the function only of  1,3 is and the function only of  1,4 is All the values of  1, , ( = 1, 2, 3, 4) are obtained by Newton-Raphson iterative algorithm.
Once the special initial true anomalies are obtained, the solution-existence range of target true anomaly  2 can be discussed and obtained as follows.
(1) If  1 ∈ [ 1,1 ,  1,2 ], there is only one intersection point for the velocity-vector line and the target hyperbolic orbit.Assume that the true anomaly of the intersection point on the target hyperbolic orbit is  2 , which can be obtained by (19); then the target true anomaly is in the range (− 2 max ,  2 ).
(2) If  1 ∈ ( 1,2 ,  1,3 ), there are two intersection points for the velocity-vector line and the target Mathematical Problems in Engineering hyperbolic orbit.Assume that the true anomalies of the intersection points on the target hyperbolic orbit are  2 and  2 ; then the solution-existence range for any conic transfer orbit is in the range ( 2 ,  2 ).The solution-existence range for elliptic transfer orbits is obtained as ( 2 ,  2 ) ⊂ ( 2 ,  2 ) by ( 21).Then the solution-existence range for any conic transfer orbit is ( 2 ,  2 ). ( , there is only one intersection point for the velocity-vector line and the target hyperbolic orbit.Assume that the true anomaly of the intersection point on the target hyperbolic orbit is  2 , which is obtained by (19); then the target true anomaly is in the range ( 2 ,  2 max ).The solutionexistence range for any conic transfer orbit is obtained as follows.(a) If there is no range for elliptic transfer orbits, then the solution-existence range for any conic transfer orbit is ( 2 ,  2 max ).(b) If there exists a range ( 2 ,  2 ) ⊂ ( 2 ,  2 max ), for elliptic transfer orbits, then the range for any conic transfer orbit can be ( 2 ,  2 ) or ( 2 ,  2 max ).When  3 by (( 11)-( 13)) is positive for an arbitrarily selected  2 ∈ ( 2 ,  2 max ), for example,  2 = ( 2 +  2 max )/2, the solutionexistence range is ( 2 ,  2 max ); otherwise, the range is ), there are two intersection points for the velocity-vector line and the target hyperbolic orbit.Assume that the true anomalies of the intersection points on the target hyperbolic orbit are  2 and  2 ; then the target true anomaly must be in the range ( 2 ,  2 max ).There is no range for elliptic transfer orbits, and the solution-existence range for any conic (hyperbolic) transfer orbit is ( 2 ,  2 max ).
From the above discussions it is known that there may be no solution-existence range for any conic transfer orbit only when  1 ∈ [ 1,1 ,  1,2 ], and there is certainly no solution-existence range for elliptic transfer orbits when Let the solution-existence range for any conic transfer by the above method be ( 2 ,  2 ).The target true anomaly  2 at the impulse time is obtained by the hyperbolic orbit propagation.Then there are three cases.(1) If  2 <  2 , then the final solution should be obtained in the range then the final solution should be obtained in the range ( 2 ,  2 ).(3) If  2 >  2 , then there is no solution.

Solving the Problem
For a given impulse point, the solution-existence range for any conic transfer and that for elliptic transfers are obtained in Section 5. Assume that the final solution-existence range for any conic transfer considering the target true anomaly at the impulse time is ( 2,1 ,  2,2 ).The interception problem can be solved by numerical methods.
The value of   can be calculated by (15); then the solutions of  2 for   ( 2 ) = 0 can be obtained by a numerical iterative algorithm, for example, the golden section search.If there is no solution for   ( 2 ) = 0, then  decreases or increases monotonically.Then for the range ( 2,1 ,  2,2 ), if ⌊( 1 + )⌋ = ⌊( 2 − )⌋ or max(⌊( 1 + )⌋, ⌊( 2 − )⌋) < 0, where the function ⌊⌋ denotes the nearest integer less than or equal to , and  is a small positive number, then there is no solution for (2).Otherwise, there are two cases.
In the patched-conic approximation method, the radius of the Earth's sphere of influence is  SOI = 925, 000 km; then the true anomaly range of the solution is  2 ∈ (− SOI ,  SOI ), where  SOI = arccos([ 2 / SOI − 1]/ 2 ).Thus, the solutions beyond this range should be removed.
The above section gives a method to obtain all solutions for a given initial impulse point  1 .For each solution, the corresponding fuel consumption is . (32) The solution whose cost is greater than the given maximal value should be removed.Thus, the feasible solution needs to satisfy two conditions, including the radius of the Earth's sphere of influence and the given maximal value of fuel.When considering the initial coasting time before  the impulse, the impulse point is free, then for each initial impulse point, all feasible solutions can be obtained and the minimum-time one is with the minimum value of  2 .
Reference [20] dealt with the cotangent rendezvous between two coplanar elliptic and hyperbolic orbits (Problem 1), whereas this paper is for the tangent-impulse interception problem from an interceptor on elliptic parking orbits to a target on hyperbolic orbits (Problem 2).Three main differences between these two problems are listed as follows.
(1) For Problem 1, there is a single closed-form solution for an arbitrary point on the target hyperbolic orbit; that is,  2 ∈ (− 2 max ,  2 max ), where  2 max = arccos(−1/ 2 ); thus there is only one degree of freedom in (2) for zero revolution; however, in a specified (e.g., counterclockwise) motion direction, the solution-existence range is not the whole range of target true anomaly.For Problem 2, there is no certain relationship between the tangent-impulse point  1 and the interception point  2 .In other words, there are two degrees of freedom in (2) with zero revolutions.
(2) For Problem 1, with nonintersecting case (counterclockwise), the solution-existence range for cotangent rendezvous is  2 ∈ ( 2 ,  2 max ), where ( 2 ,  2 ) is obtained by the solution-existence range for elliptic transfers [20].But for Problem 2 with nonintersecting case (counterclockwise), the solution-existence range of  2 can be obtained by the proposed method for different  1 , which is dependent on four special initial true anomalies whose velocity vectors are parallel to the lines of asymptotes for the target hyperbolic orbit.Moreover, there may be no solution-existence range for any conic transfer orbit when  1 ∈ [ 1,1 ,  1,2 ].
(3) For Problem 1, with a given revolution number  3 of the transfer orbit, usually there is only one orbit solution (see the examples in [20]).However, for For different impulse points, all feasible solutions satisfying  2 ∈ (−126.9425∘ , 126.9425 ∘ ) and the fuel constraint and the corresponding value of  2 =  for (2) can be obtained by (31) and are listed in Table 1.The total interception time is  2 =  1 +  3 ; the corresponding costs are also listed in Table 1.It should be notified that there is no feasible solution when  1 = 30 ∘ +   ⋅ 20 ∘ ,   = 1, 2, 3, 4, 5.By using the numerical optimization algorithm, the minimum-time impulse point is at  1 = 170.5235∘ , the solution is  2 = −110.8795∘ , the cost is 4.9999 km/s, and the total interception time is 19810.6s.The minimum-time interception trajectory is hyperbolic and it is near to that in Figure 5.

Conclusions
This paper studies the interception problem for a hyperbolic target with an upper-bounded tangent impulse.Based on four special initial true anomalies, the solution-existence range of the target true anomaly for any conic transfer is obtained.Then the problem is solved by the secant method for a time-of-flight equation.For a given impulse point, there may be zero, one, many, or infinite solutions; however, when considering the sphere of influence of the Earth, there are always finite solutions.For different impulse points, all feasible solutions and the global minimum-time solution are obtained.This proposed method is valid for an assigned impulse point and the time optimization with the initial coasting time.The interception with a tangent-impulse technique can be used in orbital interception task for potentially hazardous objects during the Earth flyby with simple control.

Figure 1 :
Figure 1: Geometrical interpretation of tangent-impulse interception for a hyperbolic target.

Figure 2 :
Figure 2: Geometry interpretation for special initial true anomalies.

Figure 3 :
Figure 3: The curves of  versus  2 for different  1 .

Table 1 :
Feasible solutions of  2 for different  1 and the corresponding interception time  2 and ΔV. , the corresponding costs is 2.9698 km/s, the corresponding transfer time is 20927.9s, and the total interception time is 62895.4s.The value of  is 2.7280; thus the trajectory is hyperbolic.