Spacecraft Dynamics and Control
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1 Spacecraft Dynamics and Control Matthew M. Peet Arizona State University Lecture 5: Hyperbolic Orbits
2 Introduction In this Lecture, you will learn: Hyperbolic orbits Hyperbolic Anomaly Kepler s Equation, Again. How to find r and v A Mission Design Example Introduction to the Orbital Plane M. Peet Lecture 5: Spacecraft Dynamics 2 / 31
3 Given t, find r and v For elliptic orbits: 1. Given time, t, solve for Mean Anomaly M(t) = nt 2. Given Mean Anomaly, solve for Eccentric Anomaly M(t) = E e sin E 3. Given eccentric anomaly, solve for true anomaly 4. Given true anomaly, solve for r tan f 2 = 1 + e 1 e tan E 2 r(t) = a(1 e2 ) 1 + e cos f(t) Does this work for Hyperbolic Orbits? Lets recall the angles. M. Peet Lecture 5: Spacecraft Dynamics 3 / 31
4 What are these Angles? True Anomaly, f (θ) The angle the position vector, r makes with the eccentricity vector, e, measured COUNTERCLOCKWISE. The angle the position vector makes with periapse. M. Peet Lecture 5: Spacecraft Dynamics 4 / 31
5 What are these Angles? Eccentric Anomaly, E Measured from center of ellipse to a auxiliary reference circle. M. Peet Lecture 5: Spacecraft Dynamics 5 / 31
6 What are these Angles? Mean Anomaly M(t) = 2π t T = 2π A P F V A Ellipse The fraction of area of the ellipse which has been swept out, in radians. M. Peet Lecture 5: Spacecraft Dynamics 6 / 31
7 Relationships between M, E, and f M vs. E M. Peet Lecture 5: Spacecraft Dynamics 7 / 31
8 Relationships between M, E, and f M vs. f M. Peet Lecture 5: Spacecraft Dynamics 8 / 31
9 Problems with Hyperbolic Orbits The orbit does not repeat (no period, T ) We can t use a T = 2π 3 µ What is mean motion, n? No reference circle Eccentric Anomaly is Undefined Note: In our treatment of hyperbolae, we do NOT use the Universal Variable approach of Prussing/Conway and others. M. Peet Lecture 5: Spacecraft Dynamics 9 / 31
10 Solutions for Hyperbolic Orbits Eccentricity Eccentricity is still and e = e. e = 1 µ ( r h µ r ) r M. Peet Lecture 5: Spacecraft Dynamics 10 / 31
11 Solutions for Hyperbolic Orbits Semimajor axis Semimajor axis can still be defined by energy as E = µ 2a = 1 2 v2 excess The periapse is still r p = a(1 e) M. Peet Lecture 5: Spacecraft Dynamics 11 / 31
12 Solutions for Hyperbolic Orbits The Polar Equation Hyperbolic Orbits still satisfy the polar equation r = a(1 e2 ) 1 + e cos f M. Peet Lecture 5: Spacecraft Dynamics 12 / 31
13 Solutions for Hyperbolic Orbits Velocity Velocity can still be calculated from the vis - viva equation ( 2 v = µ r 1 ) a M. Peet Lecture 5: Spacecraft Dynamics 13 / 31
14 Solutions for Hyperbolic Orbits Reference Hyperbola Hyperbolic Anomaly is defined by the projection onto a reference hyperbola. defined using the reference hyperbola, tangent at perigee x 2 y 2 = a 2 Hyperbolic anomaly is the hyperbolic angle using the area enclosed by the center of the hyperbola, the point of perifocus and the point on the reference hyperbola directly above the position vector. M. Peet Lecture 5: Spacecraft Dynamics 14 / 31
15 Recall your Hyperbolic Trig. Cosh and Sinh Consider x 2 y 2 = 1 cosh and sinh relate area swept out by the reference hyperbola to lengths. Yet another branch of mathematics developed for solving orbits (Lambert). M. Peet Lecture 5: Spacecraft Dynamics 15 / 31
16 Hyperbolic Anomaly Hyperbolic Trig (which I won t get into) gives a relationship to true anomaly, which is ( ) ( ) H e 1 f tanh = 2 e + 1 tan 2 Alternatively, tan ( ) ( ) f e + 1 H = 2 e 1 tanh 2 M. Peet Lecture 5: Spacecraft Dynamics 16 / 31
17 Hyperbolic Anomaly Using hyperbolic anomaly, we can give a simpler form of the polar equation. r = a(1 e cosh H) M. Peet Lecture 5: Spacecraft Dynamics 17 / 31
18 Hyperbolic Kepler s Equation To solve for position, we redefine mean motion, n, and mean anomaly, M, to get µ n = a 3 Definition 1 (Hyperbolic Kepler s Equation). nt = µ a 3 t = M = e sinh(h) H If we want to solve this, we get a different Newton iteration. Newton Iteration for Hyperbolic Anomaly: with H 1 = M. H k+1 = H k + M e sinh(h k) + H k e cosh(h k ) 1 M. Peet Lecture 5: Spacecraft Dynamics 18 / 31
19 Relationship between M and f for Hyperbolic Orbits M. Peet Lecture 5: Spacecraft Dynamics 19 / 31
20 Example: Jupiter Flyby Problem: Suppose we want to make a flyby of Jupiter. The relative velocity at approach is v = 10km/s. To achieve the proper turning angle, we need an eccentricity of e = Radiation limits our time within radius r = 100, 000km to 1 hour (radius of Jupiter is 71, 000km). Will the spacecraft survive the flyby? M. Peet Lecture 5: Spacecraft Dynamics 20 / 31
21 Example: Jupiter Flyby M. Peet Lecture 5: Spacecraft Dynamics 21 / 31
22 Example Continued Solution: First solve for a and p. µ = 1.267E8. The total energy of the orbit is given by E tot = 1 2 v2 The total energy is expressed as which yields E = µ 2a = 1 2 v2 a = µ v 2 = 1.267E6 The parameter is p = a(1 e 2 ) = E5 M. Peet Lecture 5: Spacecraft Dynamics 22 / 31
23 Example Continued We need to find the time between r 1 = 100, 000km and r 2 = 100, 000km. Find f at each of these points. Start with the conic equation: r(t) = p 1 + e cos f(t) Solving for f, ( 1 f 1,2 = cos 1 e r ) = ±64.8 deg ep M. Peet Lecture 5: Spacecraft Dynamics 23 / 31
24 Example Continued Given the true anomalies, f 1,2, we want to find the associated times, t 1,2. Only solve for t 2, get t 1 by symmetry. First find Hyperbolic Anomaly, H 2 = tanh 1 ( e 1 e + 1 tan Now use Hyperbolic anomaly to find mean anomaly ( ) ) f2 = M 2 = e sinh(h 2 ) H 2 =.0085 This is the easy direction. No Newton iteration required. t 2 is now easy to find a t 2 = M 3 2 µ = Finally, we conclude t = 2 t 2 = 2153s = 35min. So the spacecraft survives. M. Peet Lecture 5: Spacecraft Dynamics 24 / 31
25 The Orbital Elements So far, all orbits are parameterized by 3 parameters semimajor axis, a eccentricity, e true anomaly, f a and e define the geometry of the orbit. f describes the position within the orbit (a proxy for time). M. Peet Lecture 5: Spacecraft Dynamics 25 / 31
26 The Orbital Elements Note: We have shown how to use a, e and f to find the scalars r and v. Question: How do we find the vectors r and v? Answer: We have to determine how the orbit is oriented in space. Orientation is determined by vectors e and h. We need 3 new orbital elements Orientation can be determined by 3 rotations. M. Peet Lecture 5: Spacecraft Dynamics 26 / 31
27 Inclination, i Angle the orbital plane makes with the reference plane. M. Peet Lecture 5: Spacecraft Dynamics 27 / 31
28 Right Ascension of Ascending Node, Ω Angle measured from reference direction in the reference plane to intersection with orbital plane. M. Peet Lecture 5: Spacecraft Dynamics 28 / 31
29 Argument of Periapse, ω Angle measured from reference plane to point of periapse. M. Peet Lecture 5: Spacecraft Dynamics 29 / 31
30 Summary This Lecture you have learned: Hyperbolic orbits Hyperbolic Anomaly Kepler s Equation, Again. How to find r and v A Mission Design Example Introduction to the Orbital Plane M. Peet Lecture 5: Spacecraft Dynamics 30 / 31
31 Summary M. Peet Lecture 5: Spacecraft Dynamics 31 / 31
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