Research Article Tangent-Impulse Interception for a Hyperbolic Target

Mathematical Problems in Engineering Volume 14, Article ID 837849, 1 pages http://dx.doi.org/1.1155/14/837849 Research Article Tangent-Impulse Interception for a Hyperbolic Target Dongzhe Wang, Gang Zhang, and Xibin Cao Research Center of Satellite Technology, Harbin Institute of Technology, Harbin 158, China Correspondence should be addressed to Gang Zhang; zhanggang311@163.com Received 3 November 13; Accepted 3 March 14; Published 8 April 14 Academic Editor: Antonio F. Bertachini A. Prado Copyright 14 Dongzhe Wang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 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. 1. Introduction The earth-crossing comets, asteroids, space debris, and other spacecraft are potentially hazardous objects (PHOs) since theymaycloselyencounterwiththeearth.usually,anactive spacecraft is required to intercept, impact, or deflect the PHO [1 4]. The impact and deflection mission is mainly used when thephoisbeyondthesphereofinfluenceoftheearth.ifthis 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 7]. If the impulse point and the interception pointarebothassigned,thetimeofflight(tof)forthetarget can be obtained by solving Kepler s equation of hyperbolic orbit, and then the interception problem is equivalent to Lambert s problem [8 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 [1 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 thefinalorbit.theclassichohmanntransfer[15] isatwoimpulse 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]havebeen 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 [], respectively. Different from two impulses are required for the rendezvous

Mathematical Problems in Engineering 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 atargetonhyperbolicorbits.basedonfourspecialinitialtrue 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. Transfer orbit Target hyperbolic orbit Initial elliptic orbit P 1 θ r r 1 P P 1. 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 P 1 and the target is at P (see Figure 1). Let the motion direction of both orbits be counterclockwise. The coasting arc for the interceptor is from P 1 to P 1, and the initial coasting time before the tangent impulse is t 1.Thecoastingarcforthe target is from P to P, and the coasting time of the target is t. The tangent impulse indicates that the transfer orbit is tangent to the initial elliptic orbit at P 1,andthetransfer time is t 3. For the orbital interception problem, the flight time of the interceptor and that of the target are the same. For convenience, let η 1 T 1 (t t 1 t 3 ), (1) where t j (j = 1,, 3) denotes the coasting time with zero revolution and T j (j = 1, 3) denotes the orbital period, where thesubscripts 1,, and 3 denotetheinitial,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. Then the orbital interception problem with a tangent impulse is to solve the impulse point P 1 and the interception point P such that the following expression is satisfied: η N = {, 1,,...} () in which η is a positive integer indicates that the interceptor moves multiple revolutions before the first tangent impulse on point P 1. For the two-impulse cotangent rendezvous problem, if thefinaltrueanomalyf is given, there is a single closedform solution for the initial true anomaly f 1 ;thenη in (1) is afunctiononlyoff for zero revolution. However, for the one-tangent-impulse interception problem, from the above analysis, it is known that η is a function of two variables, the initial true anomaly f 1 of the impulse point,and the target true anomaly f oftheinterceptionpoint.usually,thereisan upper-bound constraint on the fuel consumption ΔV; then the interception problem is to solve all values of f 1 and f such that () is satisfied and the fuel consumption is less than a givenvalue.inthispaper,firstlytheinterceptionproblemwill γ 1 1 Figure 1: Geometrical interpretation of tangent-impulse interception for a hyperbolic target. be solved at a given f 1 of the impulse point; then all feasible solutions will be obtained at different f 1 within one initial orbital period. P 3. Time-of-Flight Equations 3.1. Coasting Time for the Initial/Target Orbit. For the initial elliptic orbit, the coasting time from P 1 to P 1 can be obtained by Kepler s equation. For the target hyperbolic orbit, the coasting time from P to P is [16] t = r r sin (f f ) μp + a3 μ [sinh (H H ) (H H )] =r p sin (f f ) μ 1+e cos f + a3 μ [sinh (H H ) (H H )], where μ is the gravitational parameter, r is the radius of P, f is thetrueanomaly,p=a(1 e ) is the semilatus rectum, a is the semimajor axis, e is the eccentricity, and H is the hyperbolic anomaly,whichisrelatedtothetrueanomalyby sinh H = e 1sin f 1+e cos f, cosh H = e + cos f 1+e cos f (3) (4)

Mathematical Problems in Engineering 3 from which it is known that H = log (sinh H + cosh H ) e = log ( 1sin f +e + cos f ). 1+e cos f Then the derivative of H with respect to f is dh e = 1. (6) df 1+e cos f Furthermore, the derivative of t with respect to f is dt =r df p cos (f f )+e cos f μ (1 + e cos f ) + a3 μ [cosh (H H ) 1] e 1 1+e cos f. 3.. Transfer Time between the Initial and Target Orbits. The initial flight-path angle at P 1 canbewrittenas[8] γ=arctan ( e 1 sin f 1 ). (8) 1+e 1 cos f 1 Let the transfer angle be θ,and[9] λ= (5) (7) r (1 cos θ) r 1 cos γ r cos (θ + γ) cos γ, (9) V 1t = μλ, (1) r 1 where V 1t denotes the magnitude of the velocity vector at point P 1 on the transfer orbit. For different values of λ,there are three cases for the transfer time [9, 19]. (1) If λ (, ), the transfer trajectory is an elliptic orbit and the transfer time is t 3 = μ ( r 1 )3/ tan 1 /λ 1 ( λ cos γcot (θ/) sin γ ) +r r 1 μλ tan γ (1 cos θ) + (1 λ) sin θ. ( λ) cos γ (11) () If λ>, the transfer trajectory is a hyperbolic orbit and the transfer time is t 3 =r r 1 μλ tan γ (1 cos θ) + (1 λ) sin θ ( λ) cos γ 1 μ ( r 1 λ )3/ ln sin γ cos γcot (θ/) (1 /λ)1/ sin γ cos γcot (θ/) + (1 /λ) 1/. (1) (3) If λ=, the transfer trajectory is a parabolic orbit and the transfer time is t 3 = 3 r3 1 μλ [ 3 cos γcot (θ/) [cos γcot (θ/) sin γ] + 1 [cos γcot (θ/) sin γ] 3 ]. (13) It should be notified that ((11) (13)) are functions independent of V 1t. The derivative of the transfer time t 3 with respect to the true anomaly f is dt 3 =( t 3 + t 3 λ ) dr +( t 3 df r λ r df θ + t 3 λ λ θ ) dθ, (14) df where the partial derivatives t 3 / r, t 3 / λ, t 3 / θ, λ/ r, and λ/ θ and the full derivative dr /df are obtained in [19, ]anddθ/df =1. The partial derivatives t 3 / λ, t 3 / θ are different for elliptic transfers and hyperbolic transfers, but the partial derivatives t 3 / r, λ/ r,and λ/ θ and the full derivative dr /df are the same for elliptic orbits and hyperbolic orbits. Note that the target true anomalies f for parabolic orbits are not continuous; thus the expression dt 3 /df is not derived for λ=. For a given impulse point, the initial coasting time t 1 for the interceptor is independent of f. Then the derivative of η with respect to f is η dη df = 1 T 1 ( dt df dt 3 df ). (15) 4. Conditions of Solution Existence The above section gives the transfer time for any conic orbit. A necessary condition is λ>. Since the initial flight-path angle γ ( π/, π/), the initial flight-direction angle is γ= π γ. (16) For the coplanar orbits, the transfer angle is θ=f +χ, where χ ω ω 1 f 1, (17) where ω is the argument of perigee. Let r c 1 = sin (χ γ) + e 1 sin γ, p c = cos (χ γ), c 3 = r 1 p sin γ, (18) where p is the semilatus rectum. There are three cases for λ > [18]: (1) if c 3 / c1 +c < 1, f [, π); () if

4 Mathematical Problems in Engineering c 3 / c1 +c > 1,thereisnosolutionforf ; (3) if 1 c 3 / c1 +c 1, then the range of f for λ>is arcsin ( c 3 ) atan(c 1,c ) c1 +c <f <π arcsin ( c 3 c 1 +c ) atan(c 1,c ), (19) where the four-quadrant inverse tangent function α = atan(b 1,b ) isdefinedastheanglesatisfyingbothsinα = b 1 / b1 +b and cos α=b / b1 +b. For any elliptical transfer orbit, <λ<is required. Let c 4 = p r 1 cos (γ) + e (1 cos (γ)) cos χ, c 5 = p r 1 sin (γ) + e (1 cos (γ)) sin χ, c 6 = p r 1 + cos (γ) 1. () Thentherearealsothreecasestobediscussed[18]: (1) if c 6 / c4 +c 5 < 1, f [, π); () if c 6 / c4 +c 5 >1,thereis no solution for f ; (3) if 1 c 6 / c4 +c 5 1, then the range of f for <λ<can be expressed as χ atan(c 4,c 5 )+arcsin ( c 6 c 4 +c 5 ) <f <π χ atan(c 4,c 5 ) arcsin ( c 6 c 4 +c 5 ). (1) In addition, the range of f 1 for existing elliptic transfers needs to be solved. If there exists some range of f,for< λ<, the following inequality must be satisfied: Substituting ()into()yields [1 cos (γ)] c 4 +c 5 c 6. () Note that the flight-direction angle γ (, π/); then cos(γ) < 1 is satisfied for any f 1 ;thus(3)canberewritten as F (e 1)[1 cos (γ)] + p p 1 (1 + e 1 cos f 1 )[1+e cos (γ χ)]. (4) From ((16), (17)) it is known that γ and χ are functions only of f 1,thenF is a function only of f 1.Thevaluesoff 1 for the equality F=can be obtained by numerical methods; finally the range for inequality F>in (4)isobtained. 5. 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 f 1,sj (j=1,,3,4)aresolvedinthefollowing paragraphs. Let u sj = f 1,sj +ω 1, (j = 1,,3,4), u max = f max +ω = arccos( 1/e )+ω,andu min = arccos(1/e )+ ω. From the geometry of Figure (a) it is known that u s1 + γ s1 =u max f 1,s1 +ω 1 + γ s1 = arccos ( 1 (5) )+ω e. Note that the flight-direction angle is a function only of the true anomaly; then the above equation can be rewritten as f 1,s1 +ω 1 + π arctan ( e 1 sin f 1,s1 ) 1+e 1 cos f 1,s1 = arccos ( 1 e )+ω (6) which is a function only of f 1,s1.Thus,theclassicNewton- Raphson iterative algorithm can be used to obtain the numerical solution for (6). From the geometry of Figure (b),itisknownthat f 1,s +ω 1 + γ s [arccos ( 1 e )+ω ]=π (7) which can be rewritten in a function only of f 1,s as f 1,s +ω 1 π arctan ( e 1 sin f 1,s ) 1+e 1 cos f 1,s = arccos ( 1 e )+ω. (8) Similarly,fromthe geometriesoffigures(c) and (d),it is known that the function only of f 1,s3 is {(e 1)[1 cos (γ)] + p p 1 (1 + e 1 cos f 1 ) [1+e cos (γ χ)]}. (3) f 1,s3 +ω 1 π arctan ( e 1 sin f 1,s3 ) 1+e 1 cos f 1,s3 (9) = arccos ( 1 )+ω e

Mathematical Problems in Engineering 5 Parallel line of asymptotes Target hyperbolic orbit Parallel line of asymptotes Target hyperbolic orbit γ s1 P s1 r s1 u s1 u max γ s P s u min r s u s Initial elliptic orbit Initial elliptic orbit (a) f 1,s1 (b) f 1,s Target hyperbolic orbit Target hyperbolic orbit Initial elliptic orbit Initial elliptic orbit u s3 u s4 u max r s3 F r 1 s4 u min P s3 P s4 γ s4 γ s3 Parallel line of asymptotes Parallel line of asymptotes (c) f 1,s3 (d) f 1,s4 Figure : Geometry interpretation for special initial true anomalies. and the function only of f 1,s4 is f 1,s4 +ω 1 3π arctan ( e 1 sin f 1,s4 ) 1+e 1 cos f 1,s4 = arccos ( 1 e )+ω. (3) All the values of f 1,sj, (j=1,,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 f can be discussed and obtained as follows. (1) If f 1 [f 1,s1,f 1,s ], 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 f q, which can be obtained by (19); then the target true anomaly is in the range ( f max,f q ). The solution-existence range for any conic transfer orbit is obtained as follows. (a) If there is no range for elliptic transfer orbits, there is also no range for any conic transfer orbit. (b) If there exists a range (f a,f b ) ( f max,f q ) for elliptic transfer orbits, which is obtained by (1), then the range for any conic transfer orbit can be (f a,f b ) or ( f max,f b ).When t 3 by (11) (13) is positive for an arbitrarily selected f ( f max,f a ), for example, f = ( f max + f a )/, the solution-existence range is ( f max,f b ); otherwise, the range is (f a,f b ). () If f 1 (f 1,s,f 1,s3 ), there are two intersection points for the velocity-vector line and the target

6 Mathematical Problems in Engineering hyperbolic orbit. Assume that the true anomalies of the intersection points on the target hyperbolic orbit are f p and f q ; then the solution-existence range for any conic transfer orbit is in the range (f p,f q ).The solution-existence range for elliptic transfer orbits is obtained as (f a,f b ) (f p,f q ) by (1). Then the solution-existence range for any conic transfer orbit is (f p,f b ). (3) If f 1 [f 1,s3,f 1,s4 ], 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 f p,whichisobtainedby(19); then the target true anomaly is in the range (f p,f max ).Thesolutionexistence 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 (f p,f max ). (b) If there exists a range (f a,f b ) (f p,f max ), for elliptic transfer orbits, then the range for any conic transfer orbit can be (f p,f b ) or (f p,f max ).Whent 3 by ((11) (13)) is positive for an arbitrarily selected f (f b,f max ), for example, f = (f b +f max )/, thesolutionexistence range is (f p,f max );otherwise,therangeis (f p,f b ). (4) If f 1 (f 1,s4,f 1,s1 ), 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 f p and f q ; then the target true anomaly must be in the range (f q,f max ).Thereisnorangeforelliptic transfer orbits, and the solution-existence range for any conic (hyperbolic) transfer orbit is (f q,f max ). From the above discussions it is known that there may be no solution-existence range for any conic transfer orbit only when f 1 [f 1,s1,f 1,s ], and there is certainly no solution-existence range for elliptic transfer orbits when f 1 [f 1,s4,f 1,s1 ]. Let the solution-existence range for any conic transfer by the above method be (f m,f n ).Thetargettrueanomaly f i at the impulse time is obtained by the hyperbolic orbit propagation. Then there are three cases. (1) If f i <f m,then thefinalsolutionshouldbeobtainedintherange(f m,f n ). () If f i [f m,f n ], then the final solution should be obtained in the range (f i,f n ). (3) If f i >f n, then there is no solution. 6. 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 (f,g1,f,g ). The interception problem canbesolvedbynumericalmethods. The value of η can be calculated by (15); then the solutions of f for η (f )=canbeobtainedbyanumericaliterative algorithm, for example, the golden section search. If there is no solution for η (f )=,thenη decreases or increases monotonically. Then for the range (f,g1,f,g ),if η(f g1 + δ) = η(f g δ) or max( η(f g1 + δ), η(f g δ) ) <, where the function y denotes the nearest integer less than or equal to y, andδ is a small positive number, then there is no solution for (). Otherwise, there are two cases. (1) If min( η(f g1 + δ), η(f g δ) ), thereare η(f g δ) η(f g1 + δ) solutions, which can be obtained by the secant method as f,n+ =f,n+1 (f,n+1 f,n ) η(f,n+1 ) N 1 η(f,n+1 ) η(f,n ), n N, (31) where the initial guesses are f, =f g1 +δ, f,1 =f g δ, N 1 = η(f g1 +δ) +1, η(f g1 +δ) +,..., η(f g δ) if η(f g1 +δ) < η(f g δ) and N 1 = η(f g δ) +1, η(f g δ) +,..., η(f g1 +δ) if η(f g1 + δ) > η(f g δ). () If min( η(f g1 + δ), η(f g δ) ) <, thereare [max( η(f g1 + δ), η(f g δ) ) + 1] solutions, which can be obtained by (31) withn 1 =,1,..., max( η(f g1 + δ), η(f g δ) ). Notethatη(f g1 ) instead of η(f g1 +δ)should be used for the range [f g1,f g ). If there exist k n 1solutions for η (f )=and these k n extreme points for η(f ) are obtained as f,k1 <f,k < < f,kn, then there are k n +1piecewise ranges, that is, (f,g1,f,k1 ), [f,k1,f,k ),...,[f,kn,f,g ).Foreachpiecewise range, the function of η decreases or increases monotonically. Then the solution for each piecewise range can be obtained by (31) with different initial guesses. If lim (f f,g )η=+ (see Figure 3(d)), there are infinite solutions in the range [f,kn,f,g ), and the minimum-time solution (or with the minimum f )willbeobtained. In the patched-conic approximation method, the radius of the Earth s sphere of influence is R SOI = 95, km; then thetrueanomalyrangeofthesolutionisf ( f SOI,f SOI ), where f SOI = arccos([p /R SOI 1]/e ).Thus,thesolutions beyondthisrangeshouldberemoved. The above section gives a method to obtain all solutions for a given initial impulse point f 1. For each solution, the corresponding fuel consumption is ΔV = V 1t V 1 = μ λ(1+e p 1 cos f 1 ) μ (1 + e1 1 p +e 1 cos f 1 ). 1 (3) 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

Mathematical Problems in Engineering 7..5 η..4 η.6.5.8 η 1 1 1 8 6 4 4 6 1 1 3 4 True anomaly f ( ) (a) f 1 = 16 5 4 4 6 8 1 1 14 True anomaly f ( ) (c) f 1 = 4 η 1 15 1 5 5 1 True anomaly f ( ) 1 9 8 7 6 5 4 3 1 (b) f 1 = 17 6 7 8 9 1 11 1 13 True anomaly f ( ) (d) f 1 = 33 Figure 3: The curves of η versus f for different f 1. theimpulse,theimpulsepointisfree,thenforeachinitial impulse point, all feasible solutions can be obtained and the minimum-time one is with the minimum value of f. Reference [] 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 ). 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, f ( f max,f max ), where f max = arccos( 1/e ); thus there is only one degree of freedom in () for zero revolution; however, in a specified (e.g., counterclockwise) motion direction, thesolution-existencerangeisnotthewholerange of target true anomaly. For Problem, there is no certain relationship between the tangent-impulse point f 1 and the interception point f.inotherwords, there are two degrees of freedom in () withzero revolutions. () For Problem 1, with nonintersecting case (counterclockwise), the solution-existence range for cotangent rendezvous is f (f a,f max ),where(f a,f b ) is obtained by the solution-existence range for elliptic transfers [].ButforProblemwithnonintersecting case (counterclockwise), the solution-existence range of f can be obtained by the proposed method for different f 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 f 1 [f 1,s1,f 1,s ]. (3) For Problem 1, with a given revolution number N 3 of the transfer orbit, usually there is only one orbit solution (see the examples in []). However, for

8 Mathematical Problems in Engineering Problem, with an upper-bounded impulse, the minimum-time solution is interesting; then N 3 is set to be. For a given impulse point f 1, there may exist multiple solutions of f (see Figures 3(a) and 3(b)). With the initial coasting time, all feasible solutions for different impulse points are obtained and the minimum-time one is with the minimum value of f. 7. Numerical Examples Assume that the interceptor s orbit elements are a 1 =R E + 4, km, e 1 =.6, ω 1 =1,andf 1 =6 and the target s orbit elements are a = R E 1, km, e = 1.6, ω =, and f = 1,whereR E = 6378.13 km is the Earth s radius.thegravitationalparameterusedfortheexampleis μ = 398,6.4415 km 3 /s.theupperboundofthetangent impulse is set to be 5 km/s. Four special initial true anomalies are obtained as f 1,s1 = 45.4187, f 1,s = 158.13, f 1,s3 = 191.9457,andf 1,s4 = 84.5334 via (6), (8), (9), and (3). To obtain elliptic transfers, the range of the initial true anomaly of the impulse point is f 1 (79.944, 49.968 ) via (4). In addition, considering the radius of the Earth s sphere of influence, that is, 95, km, the range of true anomaly is f ( 16.945, 16.945 ). If the impulse is imposed at the initial time, that is, f 1 = f 1 =6,thenf 1 [f 1,s1,f 1,s ] is satisfied, and there is no range for elliptic transfers; thus no solution-existence range exists for f. If the impulse is imposed at f 1 = 16,thetarget true anomaly at the impulse time is f = 118.3547 by solving Kepler s equation for hyperbolic orbits, and the plots of η versus f canbeobtainedinfigure 3(a). The final solution-existence range for any conic transfer considering the target true anomaly at the impulse time is (f,g1,f,g ) = ( 118.3547, 7.48 ). There are two solutions, that is, 9.8589 and.796 ;thecorresponding revolution numbers are both and the corresponding costs are 1.318 km/s and 1.11 km/s, respectively. The minimumtime solution is 9.8589, the corresponding transfer time is 197.4s,andthetotalinterceptiontimet is 831.4 s via (3). The value of λ is obtained as 1.1178 by (9); thus the trajectory is elliptic and it is plotted in Figure 4. If the impulse is imposed at f 1 = 17,thetargettrue anomaly at the impulse time is f = 117.788,theplotsof η versus f canbeobtainedinfigure 3(b).Thefinalsolutionexistence range is (f,g1,f,g ) = ( 116.347, 88.311 ). There are three solutions, 115.8579, 19.93, and 53.43, all the corresponding revolution numbers are, and the corresponding costs are 1.1998 km/s, 3.988 km/s, and1.177km/s,respectively.theenergycostof 115.8579 is muchgreaterthan5km/s;thusitcannotbeused.thefeasible minimum-time solution is 19.93,thecorresponding transfer time is 13188.3 s, and the total interception time is 158.8 s. The value of λ is.7636; thus the trajectory is hyperbolic and it is plotted in Figure 5. If the impulse is imposed at f 1 = 4,thetargettrue anomaly at the impulse time is f = 11.5985 ;theplotsof η versus f can be obtained in Figure 3(c). Thereisasingle Y (km) 4 6 8 1 1 14 16 1 4 15 1 5 5 X (km) Initial Target Transfer P P 1 P P 1 1 4 Figure 4: Tangent-impulse interception with elliptic transfer trajectory. Y (km) 4 4 6 8 1 1 1 4 P 1 14 P 16 1.5 1.5.5 1 X (km) 1 5 Initial Target Transfer Figure 5: Tangent-impulse interception with hyperbolic transfer trajectory. solution 115.336, the corresponding revolution number is, the cost is 1.7743 km/s, the transfer time is 36318. s, and the total interception time is 548.1 s. The value of λ is 1.9316; thus the trajectory is elliptic. If the impulse is imposed at f 1 = 33,thetargettrue anomaly at the impulse time is f = 11.89,theplotsofη versus f canbeobtainedinfigure 3(d),andthereareinfinite solutions. However, when considering the sphere of influence of the Earth, there are three solutions. The minimum-time solution is 118.6911, the corresponding revolution number P P 1

Mathematical Problems in Engineering 9 Table 1: Feasible solutions of f for different f 1 and the corresponding interception time t and ΔV. f 1 ( ) Solutions of (f ( ), N 1 ) t (s) ΔV (km/s) 15 ( 56.6897, ) 318.4.935 ( 39.9134, ) 3844.3.8897 17 ( 19.93, ) 158.8 3.988 (53.43, 1) 35834.5 1.177 19 (89.163, ) 398.1 1.4596 1 (14.3983, ) 43665.1 1.6333 3 (11.3591, ) 4993.6 1.7338 5 (117.713, ) 59847. 1.8148 7 (11.9959, ) 849.5 1.98 9 (15.9555, ) 1641.3.3 31 (14.4383, 1) 111489.9.976 33 (118.6911, 1) 6895.4.9698 (15.751, ) 15667.5.687 (16.966, 3) 36873.8.6464 35 (114.511, 1) 5541. 4.469 (14.88, ) 1853. 3.547 (16.145, 3) 16386. 3.419 (16.754, 4) 17544.3 3.3698 1 (16.1114, 4) 168589.6 4.9766 (16.669, 5) 8476.8 4.8911 is 1, the corresponding costs is.9698 km/s, the corresponding transfer time is 97.9 s, and the total interception time is 6895.4 s. The value of λ is.78; thus the trajectory is hyperbolic. For different impulse points, all feasible solutions satisfying f ( 16.945, 16.945 ) and the fuel constraint and the corresponding value of N =ηfor () canbeobtained by (31)andarelistedinTable 1. The total interception time is t =t 1 +t 3 ; the corresponding costs are also listed in Table 1. It should be notified that there is no feasible solution when f 1 =3 +k j, k j = 1,,3,4,5.Byusingthenumerical optimization algorithm, the minimum-time impulse point is at f 1 = 17.535,thesolutionisf = 11.8795,thecostis 4.9999 km/s, and the total interception time is 1981.6 s. The minimum-time interception trajectory is hyperbolic and it is near to that in Figure 5. 8. 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, theremaybezero,one,many,orinfinitesolutions;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. Conflict of Interests The authors declare that there is no conflict of interests regarding the publication of this paper. Acknowledgments ThisworkissupportedinpartbytheChinaPostdoctoralScience Foundation Funded Project (Grant no. 1M5753), the Heilongjiang Postdoctoral Science Foundation (Grant no. LBH-Z115), and the National Natural Scientific Foundation of China (Grant no. 617396). References [1] J. C. Solem, Interception of comets and asteroids on collision course with Earth, Spacecraft and Rockets,vol.3,no., pp. 8, 1993. [] I.M.Ross,S.-Y.Park,andS.D.V.Porter, Gravitationaleffectsof earth in optimizing ΔV for deflecting earth-crossing asteroids, Spacecraft and Rockets, vol.38,no.5,pp.759 764, 1. [3] B. Dachwald and B. Wie, Solar sail trajectory optimization for intercepting, impacting, and deflecting near-earth asteroids, in Proceedings of the AIAA Guidance, Navigation, and Control Conference and Exhibit, pp. 3331 3348, San Francisco, Calif, USA, August 5. [4] D. Izzo, Optimization of interplanetary trajectories for impulsive and continuous asteroid deflection, Guidance, Control, and Dynamics,vol.3,no.,pp.41 48,7. [5] J. Park, R. G. Rhinehart, and P. T. Kabamba, Miss analysis in Lambert interceptions with application to a new guidance law, Mathematical Problems in Engineering: Theory, Methods and Applications,vol.6,no.-3,pp.5 65,. [6] Y. Ulybyshev, Direct high-speed interception: analytic solutions, qualitative analysis, and applications, Spacecraft and Rockets, vol. 38, no. 3, pp. 351 359, 1. [7] S. P. Burns and J. J. Scherock, Lambert guidance routine designed to match position and velocity of ballistic target, Guidance, Control, and Dynamics, vol.7,no.6,pp. 989 996, 4. [8] R.H.Battin,An Introduction to the Mathematics and Methods of Astrodynamics, American Institute of Aeronautics and Astronautics (AIAA), Reston, Va, USA, Revised edition, 1999. [9] S. L. Nelson and P. Zarchan, Alternative approach to the solution of Lambert s problem, Guidance, Control, and Dynamics,vol.15,no.4,pp.13 19,199. [1] P. V. Arlulkar and S. D. Naik, Solution based on dynamical approach for multiple-revolution lambert problem, Guidance, Control, and Dynamics, vol.34,no.3,pp.9 93, 11. [11] G. Zhang, D. Mortari, and D. Zhou, Constrained multiplerevolution Lambert s problem, Guidance, Control, and Dynamics,vol.33,no.6,pp.1779 1786,1.

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