Spacecraft Dynamics and Control

Transcription

1 Spacecraft Dynamics and Control Matthew M. Peet Arizona State University Lecture 10: Rendezvous and Targeting - Lambert s Problem

2 Introduction In this Lecture, you will learn: Introduction to Lambert s Problem The Rendezvous Problem The Targeting Problem Fixed-Time interception Solution to Lambert s Problem Focus as a function of semi-major axis, a Time-of-Flight as a function of semi-major axis, a Fixed-Time interception Calculating v. Numerical Problem: Suppose we are in an equatorial parking orbit or radius r. Given a target with position r and velocity v, calculate the v required to intercept the target before it reaches the surface of the earth. M. Peet Lecture 10: Spacecraft Dynamics / 30

3 Problems we Have Solved How to change our orbit to a desired one. Raise orbit Inclination change Rendezvous? OK, when mission is not time-sensitive. Must use phasing... We need θ = π T hohman T outer, where θ = n inner n outer. M. Peet Lecture 10: Spacecraft Dynamics 3 / 30

4 The Problem with phasing Problem: We have to wait. Remember what happened to the Death star? M. Peet Lecture 10: Spacecraft Dynamics 4 / 30

5 Asteroid Interception Suppose that: Our time to intercept is limited. The target trajectory is known. Problem: Design an orbit starting from r 0 which intersects the orbit of the asteroid at the same time as the asteroid. Before the asteroid intersects the earth (when r(t) = 6378) M. Peet Lecture 10: Spacecraft Dynamics 5 / 30

6 Missile Defense Problem: ICBM s have re-entry speeds in excess of 8km/s (Mach 6). Patriot missiles can achieve max of Mach 5. Objective: Intercept ballistic trajectory before missile re-entry Before the missile intersects the atmosphere When r(t) = 6378km+ = 00km Complications: Plane changes may be required. The required time-to-intercept may be small. Hohman transfer is not possible M. Peet Lecture 10: Spacecraft Dynamics 6 / 30

7 The Targeting Problem Step 1: Determine the orbit of the target Step 1 can be accomplished one of two ways: Method 1: 1. Given r(t 0 ) and v(t 0 ), find a, e, i, ω p, Ω and f(t 0 ) we have covered this approach in Lecture 6.. Unfortunately, it is difficult to measure v Method : 1. Given two observations r(t 1 ) and r(t ), find a, e, i, ω p, Ω and f(t 0 ). Alternatively, find v(t1) and v(t ). This is referred to as Lambert s problem (the topic of this lecture) Note: This is a boundary-value problem: We know some states at two points. In contrast to the initial value problem, where we know all states at the initial time. Unlike initial-value problems, boundary-value problems cannot always be solved. M. Peet Lecture 10: Spacecraft Dynamics 7 / 30

8 Carl Friedrich Gauss ( ) The Problem of orbit determination was originally solved by C. F. Gauss Mathematician First Astronomer Second Boring/Conservative. Considered by many as one of the greatest mathematicians Professor of Astronomy in Göttingen He was OK. Discovered Gaussian Distributions Gauss Law M. Peet Lecture 10: Spacecraft Dynamics 8 / 30

9 Discovery and Rediscovery of Ceres The pseudo-planet Ceres was discovered by G. Piazzi Observed 1 times between Jan. 1 and Feb. 11, 1801 Planet was then lost. Complication: Observation was only declination and right-ascension. Observations were only spread over 1% of the orbit. No ranging info. For this case, three observations are needed. C. F. Gauss solved the orbit determination problem and correctly predicted the location. Planet was re-found on Dec 31, 1801 in the correct location. M. Peet Lecture 10: Spacecraft Dynamics 9 / 30

10 The Targeting Problem Step : Determine the desired position of the target Once we have found the orbit of the target, we can determine where the target will be at the desired time of impact, t f. Procedure: The difference t f t 0 is the Time of Flight (TOF) Calculate M(t f ) = M(t 0 ) + n(t f t 0 ) Use M(t f ) to find E(t f ). Use E(t f ) to find f(t f ). Use f(t f ) to find r(t f ). M. Peet Lecture 10: Spacecraft Dynamics 10 / 30

11 The Targeting Problem Step 3: Find the Intercept Trajectory For a given Initial Position, r 1 Final Position, r Time of Flight, T OF the transfer orbit is uniquely determined. Challenge: Find that orbit!!! Difficulties: Where is the second focus? May require initial plane-change. May use LOTS of fuel. Figure: For given P 1 and P and TOF, the transfer ellipse is uniquely determined. On the Plus Side: We know the change in true anomaly, f... For this geometry, TOF only depends on a. M. Peet Lecture 10: Spacecraft Dynamics 11 / 30

12 The Targeting Problem Step 4: Calculate the v Once we have found the transfer orbit, Calculate v tr (t 0 ) of the transfer orbit. Calculate our current velocity, v(t 0 ) If a ground-launch, use rotation of the earth. If in orbit, use orbital elements. Calculate v = v tr (t 0 ) v(t 0 ) M. Peet Lecture 10: Spacecraft Dynamics 1 / 30

13 Finding the transfer orbit (The Hard Part) Semi-major axis, a and Focus The difficult part is to determine a. TOF only depends on a The value of a dictates the location of the focus. Focal Circle 1: For given r i and a, The set of potential locations for the second focus is a circle. Let r1 = r be the distance to the known focus. Then rf1 = a r 1 is the distance to the unknown focus. The unknown focus lies on a circle of radius r f1 around our current position. Figure: Potential Locations of Second Focus M. Peet Lecture 10: Spacecraft Dynamics 13 / 30

14 Finding Semi-major axis, a and Focus Focal Circle : The same geometry holds for the final position vector, r f. The set of potential locations for the second focus is a circle. Let r = r f be the distance to the known focus. Then rf = a r is the distance to the unknown focus. The unknown focus lies on a circle of radius r f around the target position. Figure: Potential Locations of Second Focus using Final Destination M. Peet Lecture 10: Spacecraft Dynamics 14 / 30

15 Finding Semi-major axis, a and Focus Geometry: By intersecting the two circles about r i and r f, We find two potential locations for the focus We can discard the more distant focus. Figure: Potential Locations of Second Focus for a given a M. Peet Lecture 10: Spacecraft Dynamics 15 / 30

16 Finding Semi-major axis, a and Focus Note: As we vary a, the set of foci form a hyperbola. Conclusions There is a minimum-energy transfer corresponding to minimum a. amin = r 0+r+c 4 Focus lies on the line between r1 and r. But we don t want a minimum energy transfer! Still, its nice to have a lower bound. Question: How to use TOF to find a? Figure: Potential Locations of Second Focus for a given a M. Peet Lecture 10: Spacecraft Dynamics 16 / 30

17 Finding Semi-major axis, a and Focus We know θ = f(t f ) f(t 0 ) - change in true anomaly. t = T OF How to find a? Kepler s Equation: µ a 3 t = E(t f ) E(t 0 ) e(sin E(t f ) sin E(t 0 )) Problem: We know f, but not E! E depends on e as well as f and a We don t know e Figure: Geometry of the Problem M. Peet Lecture 10: Spacecraft Dynamics 17 / 30

18 Finding Semi-major axis, a and Focus Solution First: Calculate some lengths c = r 1 r is the chord. s = c+r1+r is the semi-perimeter. NOT semiparameter. Then we get Lambert s Equation: a t = 3 (α β (sin α sin β)) µ where [ α ] sin = s a, sin [ ] β s c = a Conclusion: We can express TOF, solely as a function of a. albeit through a complicated function. Figure: Geometry of the Problem M. Peet Lecture 10: Spacecraft Dynamics 18 / 30

19 Solving Lambert s Equation Bisection There are several ways to solve Lambert s Equation Newton Iteration More Complicated than Kepler s Equation Series Expansion Probably the easiest... Bisection Relatively Slow, but easy to understand Figure: Geometry of the Problem M. Peet Lecture 10: Spacecraft Dynamics 19 / 30

20 Solving Lambert s Equation via Bisection a Define g(a) = 3 µ (α β (sin α sin β)). Root-Finding Problem: Find a : such that g(a) = t Bisection Algorithm: 1 Choose a min = s = r0+r+c 4 Choose a max >> a min 3 Set a = amax+amin 4 If g(a) > t, set a min = a 5 If g(a) < t, set a max = a 6 Goto 3 This is guaranteed to converge to the unique solution (if it exists). We assume Elliptic solutions. F(x) F(a 1 ) F(a ) F(a 3 ) a 1 F(b ) b 1 x F(b 1 ) M. Peet Lecture 10: Spacecraft Dynamics 0 / 30

21 Bisection Some Implementation Notes Make Sure a Solution Exists!! First calculate the minimum TOF. This corresponds to a parabolic trajectory ( t p = s µ ( s c s ) 3 ) Can get there even faster by using a hyperbolic approach (Not Covered). One might also calculate the Maximum TOF a 3 min t max = µ (α max β max (sin α max sin β max )) where sin [ αmax ] s =, a min sin [ βmax ] s c = a min One can exceed this by going the long way around (Not Covered Here) M. Peet Lecture 10: Spacecraft Dynamics 1 / 30

22 Calculating v(t 0 ) and v(t f ) Once we have a, calculating v is not difficult. where A = v(t 0 ) = (B + A) u c + (B A) u 1, v(t f ) = (B + A) u c (B A) u µ ( α ) ( ) µ β 4a cot, B = 4a cot and the unit vectors u 1 and u point to positions 1 and. u 1 = r(t 0) r 1, u = r(t f ) r u c points from position 1 to. u c = r(t f ) r(t 0 ) c M. Peet Lecture 10: Spacecraft Dynamics / 30

23 Calculating v Once we have found the transfer orbit, Calculate v tr (t 0 ) of the transfer orbit. Calculate our current velocity, v(t 0 ) If a ground-launch, use rotation of the earth. If in orbit, use orbital elements. Calculate v = v tr (t 0 ) v(t 0 ) M. Peet Lecture 10: Spacecraft Dynamics 3 / 30

24 Numerical Example of Missile Targeting Problem: Suppose that Brasil launches an ICBM at Bangkok, Thailand. We have an interceptor in the air with position and velocity r = [ ] km v = [ ] km/s. We have tracked the missile at r t = [ ] km heading v = [ ] km/s. Question: Determine the v required to intercept the missile before re-entry, which occurs in 30 minutes M. Peet Lecture 10: Spacecraft Dynamics 4 / 30

25 Numerical Example of Missile Targeting The first step is to determine the position of the ICBM in t + 30min. Recall: To propagate an orbit in time: 1. Use r t and v t to find the orbital elements, including M(t 0 ).. Propagate Mean anomaly M(t f ) = M(t 0 ) + n t where t = 1800s. 3. Use M(t f ) to find true anomaly, f(t f ). Requires iteration to solve Kepler s Equation. 4. Use the orbital elements, including f(t f ) to find r(t f ) M. Peet Lecture 10: Spacecraft Dynamics 5 / 30

26 Numerical Example of Missile Targeting The next step is to determine whether an intercept orbit is feasible using TOF=30min. Geometry of the Problem: r 1 = r = 6, 980km, r = r t (t f ) = 1, 8km, c = r r t (t f ) = 7, 080km, s = c + r 1 + r = 13, 171km Minimum Flight Time: Using the formula, the minimum (parabolic) flight time is ( t min = t p = ( ) s 3 3 ) s c 1 = 18.min 3 µ s Thus we have more than enough time. Maximum Flight Time: Geometry yields a minimum semi-major axis of a min = s = 6, 586km Plugging this into Lambert s equation yields a maximum flight time of t max = 37.3min. M. Peet Lecture 10: Spacecraft Dynamics 6 / 30

27 Numerical Example of Missile Targeting What remains is to solve Lambert s equation: a t = 3 (α β (sin α sin β)) µ where [ α ] s sin = a, sin [ ] β s c = a Initialize our search parameters using a [a l, a h ] = [a min, s]. 1. a 1 = a l+a h = 8, 3 - T OF = 1.14min - too low, decrease a 1.1 Set a h = a 1. a = a l+a h = 7, T OF = 4min - too low, decrease a.1 Set a h = a 3. a 3 = a l+a h = 6, T OF = 6.76min - too low, decrease a 3.1 Set a h = a 3 4. K. a k = a l+a h = 6, T OF = 9.99 K.1 Close Enough! M. Peet Lecture 10: Spacecraft Dynamics 7 / 30

28 Numerical Example of Missile Targeting Now we need to calculate v. where v(t 0 ) = (B + A) u c + (B A) u 1, v(t f ) = (B + A) u c (B A) u A = µ ( α ) ( ) µ β 4a cot =.597, B = 4a cot = and the unit vectors u 1 =.5, u =.8414, u c = which yields Calculating v v t (t 0 ) = [ ] km/s v = v t (t 0 ) v = [ ] km/s For a total impulse of 6km/s. M. Peet Lecture 10: Spacecraft Dynamics 8 / 30

29 Numerical Example of Missile Targeting Figure: Intercept Trajectory M. Peet Lecture 10: Spacecraft Dynamics 9 / 30

30 Summary This Lecture you have learned: Introduction to Lambert s Problem The Rendezvous Problem The Targeting Problem Fixed-Time interception Solution to Lambert s Problem Focus as a function of semi-major axis, a Time-of-Flight as a function of semi-major axis, a Fixed-Time interception Calculating v. Next Lecture: Rocketry. M. Peet Lecture 10: Spacecraft Dynamics 30 / 30