Spacecraft and Aircraft Dynamics - Lecture 5 - Illinois Institute of ...

Spacecraft and Aircraft DynamicsMatthew M. PeetIllinois Institute of TechnologyLecture 5: Hyperbolic Orbits

IntroductionIn this Lecture, you will learn:Hyperbolic orbits• Hyperbolic Anomaly• Kepler’s Equation, Again.• How to find r and vA Mission Design ExampleIntroduction to the Orbital PlaneM. Peet Lecture 5: Spacecraft Dynamics 2 / 26

Given t, find r and vFor elliptic orbits:1. Given time, t, solve for Mean AnomalyM(t) = nt2. Given Mean Anomaly, solve for Eccentric AnomalyM(t) = E −esinE3. Given eccentric anomaly, solve for true anomaly4. Given true anomaly, solve for rtan f 2 = √1+e1−e tan E 2r(t) = a(1−e2 )1+ecosf(t)Does this work for Hyperbolic Orbits? Lets recall the angles.M. Peet Lecture 5: Spacecraft Dynamics 3 / 26

What are these Angles?Eccentric Anomaly, E• Measured from center of ellipse to a auxiliary reference circle.M. Peet Lecture 5: Spacecraft Dynamics 5 / 26

What are these Angles?Mean AnomalyM(t) = 2π t T = 2π A PFVA Ellipse• The fraction of area of the ellipse which has been swept out, in radians.M. Peet Lecture 5: Spacecraft Dynamics 6 / 26

Relationships between M, E, and fM vs. EM. Peet Lecture 5: Spacecraft Dynamics 7 / 26

Relationships between M, E, and fM vs. fM. Peet Lecture 5: Spacecraft Dynamics 8 / 26

Problems with Hyperbolic Orbits• The orbit does not repeat (noperiod, T)◮ We can’t use√µT = 2πa 3◮ What is mean motion, n?• No reference circle◮ Eccentric Anomaly is UndefinedNote: In our treatment of hyperbolae, we do NOT use the Universal Variableapproach of Prussing/Conway and others.M. Peet Lecture 5: Spacecraft Dynamics 9 / 26

Solutions for Hyperbolic OrbitsReference HyperbolaWe will not get into details!• defined using the reference hyperbola, tangent at perigeex 2 −y 2 = 1M. Peet Lecture 5: Spacecraft Dynamics 10 / 26

Recall your Hyperbolic Trig.Cosh and SinhRelate area of reference hyperbola to lengths.• Yet another branch of mathematics developed for solving orbits (Lambert).M. Peet Lecture 5: Spacecraft Dynamics 11 / 26

Hyperbolic Anomaly• Hyperbolic Anomaly, H is a measure of Area.• Hyperbolic Trig gives a relationship to true anomaly, which is( ) √ ( )H e−1 ftanh =2 e+1 tan 2• Alternatively,( √ ( )f e+1 Htan =2)e−1 tanh 2M. Peet Lecture 5: Spacecraft Dynamics 12 / 26

Hyperbolic Kepler’s EquationTo solve for position, we redefine mean motion, n, and mean anomaly, M, to getDefinition 1 (Hyperbolic Kepler’s Equation).M =√ µ−a3t = esinh(H)−HNewton Iteration for Hyperbolic Anomaly:with H 1 = M.H k+1 = H k + M −esinh(H k)+H kecosh(H k )−1M. Peet Lecture 5: Spacecraft Dynamics 13 / 26

Relationship between M and f for Hyperbolic OrbitsM. Peet Lecture 5: Spacecraft Dynamics 14 / 26

Example: Jupiter FlybyProblem: Suppose we want to make a flyby of Jupiter. The relative velocity atapproach is v ∞ = 10km/s. To achieve the proper turning angle, we need aneccentricity of e = 1.07. Radiation limits our time within radius r = 100,000kmto 1 hour (radius of Jupiter is 71,000km). Will the spacecraft survive the flyby?M. Peet Lecture 5: Spacecraft Dynamics 15 / 26

Example: Jupiter FlybyM. Peet Lecture 5: Spacecraft Dynamics 16 / 26

Example ContinuedSolution: First solve for a and p. µ = 1.267E8.• The total energy of the orbit isgiven byE tot = 1 2 v2 ∞• The total energy is expressed aswhich yieldsE = − µ 2a = 1 2 v2 ∞a = − µv 2 ∞= −1.267E6• The parameter isp = a(1−e 2 ) = 1.8359E5M. Peet Lecture 5: Spacecraft Dynamics 17 / 26

Example ContinuedWe need to find the time between r 1 = 100,000km and r 2 = 100,000km. Findf at each of these points.• Start with the conic equation:r(t) =p1+ecosf(t)• Solving for f,( 1f 1,2 = cos −1 e − r )= ±64.8degepM. Peet Lecture 5: Spacecraft Dynamics 18 / 26

Example ContinuedGiven 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,(√ ( e−1H 2 = tanh −1 e+1 tan f22• Now use Hyperbolic anomaly to find mean anomalyM 2 = esinh(H 2 )−H 2 = .0085) ) = .1173◮ This is the “easy” direction.◮ No Newton iteration required.• t 2 is now easy to findt 2 = M 2√−a 3µ = 1076.6Finally, we conclude ∆t = 2∗t 2 = 2153s = 35min.So the spacecraft survives.M. Peet Lecture 5: Spacecraft Dynamics 19 / 26

The Orbital ElementsSo 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 20 / 26

The Orbital ElementsNote: 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 21 / 26

Inclination, iAngle the orbital plane makes with the reference plane.M. Peet Lecture 5: Spacecraft Dynamics 22 / 26

Right Ascension of Ascending Node, ΩAngle measured from reference direction in the reference plane to intersectionwith orbital plane.M. Peet Lecture 5: Spacecraft Dynamics 23 / 26

Argument of Periapse, ωAngle measured from reference plane to point of periapse.M. Peet Lecture 5: Spacecraft Dynamics 24 / 26

SummaryThis Lecture you have learned:Hyperbolic orbits• Hyperbolic Anomaly• Kepler’s Equation, Again.• How to find r and vA Mission Design ExampleIntroduction to the Orbital PlaneM. Peet Lecture 5: Spacecraft Dynamics 25 / 26

SummaryM. Peet Lecture 5: Spacecraft Dynamics 26 / 26