ARO309 - Astronautics and Spacecraft Design Winter 2014 Try Lam CalPoly Pomona Aerospace Engineering

Lecture 03: Numerical Integrations Chapter TBD

Mathematical Models (Equations) Introductions In this lecture we will look at how dynamical system problems are solved numerically Real Life Problem In Astodynamics Problems we usually deal with Initial Value Problem (IVP) Mathematical Models (Equations) Given: Find: Numerical Algorithm

2nd Order Runge Kutta Given an 1st order ODE Applying Taylor Series expansion to 2nd order Since: and

2nd Order Runge Kutta Now we have

2nd Order Runge Kutta Or The Coefficient in Butcher Tableaux form. Note:

2nd Order Runge Kutta Graphically slope=k2 yn+1 (Approximate Soln) slope=k1=f yo Exact Solution to t1

Gernal Runge Kutta (explicit) The explicit RK method is given by

Runge Kutta (explicit) Note that RK must satisfy: Butcher Table

Runge Kutta (explicit) RK4 where

RK4 Example Goal: Integrate the 2-Body Problem using RK4 for a single step with step size = 60 sec Equations of Motion (EOM): State Equation (around Earth):

RK4 Example From the problem:

RK4 Example

RK4 Example RK4 fails for large time steps (range plot example)

Lecture 04: Two-Body Dynamics: Orbit Position as a Function of Time Chapter 3

Introductions Chapter 2 (Lection 1 and 2) relates position as a function of θ (true anomaly) but not time Time was only introduced when referring to orbit period Here we attempt to find the relations between position of the S/C and time  Kepler’s Equation

Time versus True Anomaly Recall from Chapter 2 Since Then Integrating from 0 (assuming tp = 0) to t and from 0 to θ

Time versus True Anomaly

Time versus True Anomaly Simple Case: Circular Orbits (e=0) If e = 0, then therefore Since for a circular orbit we have then FOR CIRCULAR ORBIT OR

Time versus True Anomaly Elliptical Orbits (0<e<1) Then a = 1 and b = e, therefore we have b < a Me = Mean anomaly for the ellipse

Time versus True Anomaly Elliptical Orbits (0<e<1) Therefore we have From the orbit period of an ellipse we know (or can derive) that Therefore we can solve for me as function orbit period as OR where n = mean motion = 2π/Te

Time versus True Anomaly Elliptical Orbits (0<e<1) We need to fine out Me still ? Let’s introduce another variable E = eccentric anomaly

Time versus True Anomaly Elliptical Orbits (0<e<1) OR This relates E and θ, but it leaves the quadrant of the solution unknown and you get two values of E for the equation. To eliminate this ambiguity we use the following identity Therefore or

Time versus True Anomaly Elliptical Orbits (0<e<1) We need to fine out Me still E This is Kepler’s Equation

Time versus True Anomaly Elliptical Orbits (0<e<1) To find t given Δθ Given orbital parameters, find e and h (assume θ = 0 deg) Find E: Find T (orbit period):

Time versus True Anomaly Elliptical Orbits (0<e<1) To find t given Δθ Fine Me: Find t: Question: What if you are going from a θ = θa to θ = θb? Answer: Find the time from θ = 0 to θ = θa and the time from θ = 0 to θ = θb. Then subtract the differences.

Time versus True Anomaly Elliptical Orbits (0<e<1) To find θ given Δt Given orbital parameters, find e and h (assume θ = 0 deg Find T (orbit period): Find Me: Find E using Newton’s method (or a transcendental solver)

Time versus True Anomaly Elliptical Orbits (0<e<1) To find θ given Δt Using Newton’s Method: Initialize E = Eo: Find f(E): Find f’(E): If abs( f(E) / f’(E) ) > TOL, then repeat with Else Econverged = En For Me < 180 deg For Me > 180 deg

Time versus True Anomaly Elliptical Orbits (0<e<1) To find θ given Δt After finding the converged E, then find θ

Time versus True Anomaly Parabolic Orbits (e = 1) Then a = 1 and b = e, therefore we have b = a MP = Parabolic Mean Anomaly

Time versus True Anomaly Parabolic Orbits (e = 1) STEPS: Find h Find MP Find θ Thus given t or Δt we can find MP To fine θ we can find the root of the below equation Which has one real root

Time versus True Anomaly Hyperbolic Orbits (e > 1) Then a = 1 and b = e, therefore we have b > a

Time versus True Anomaly Hyperbolic Orbits (e > 1) Where the Hyperbolic mean anomaly is Thus we have Similar with Ellipse we will intro a new variable, F, the hyperbolic eccentric anomaly to help solve for the Mean Hyperbolic anomaly, Mh.

Time versus True Anomaly Hyperbolic Orbits (e > 1) Hyperbolic eccentric anomaly for the Hyperbola Since:

Time versus True Anomaly Hyperbolic Orbits (e > 1) We now have Solving for F and since we now have Using the following trig identities for sine and cosine

Time versus True Anomaly Hyperbolic Orbits (e > 1) We now have Therefore we now have: This is Kepler’s Equation for Hyperbola Similar to Elliptical orbits we can solve for F as a function of θ, which is found to be. Thus given θ we can find F, and Mh, and finally t.

Time versus True Anomaly Hyperbolic Orbits (e > 1) If time, t, was given and θ is to be found then we have to solve for Kepler’s equation for hyperbola iteratively using Newton’s method STEPS TO FIND θ (given t) Set initial F0 = Mh where Find f and f’ If abs( f / f’ ) > TOL, repeat steps with updated F Else, Fconverged = Fi. Now find θ

Universal Variables What happens if you don’t know what type of orbit you are in? Why use 3 set of equations? Kepler’s equation can be written in terms of a universal variable or universal anomaly, Χ, and Kepler’s equation becomes the universal Kepler’s equation. If α < 0, then orbit is hyperbolic If α = 0, then orbit is parabolic If α > 0, then orbit is elliptical Where

Universal Variables Stumpff functions or for z = αΧ2 ,

Universal Variables To use Newton’s method we need to define the following function and it’s derivative Iterate with the following algorithm with

Universal Variables Relation ship between X and the orbits For t0 = 0 at periapsis

Universal Variables Example 3.6 (Textbook: Curtis’s) Find h and e Since , then

Universal Variables Example 3.6 (Textbook: Curtis’s) Therefore So X0 is the initial X to use for the Newton’s method to find the converged X

Universal Variables Example 3.6 (Textbook: Curtis’s)

Universal Variables Example 3.6 (Textbook: Curtis’s) Thus we accept the X value of X = 128.5 where

Lagrange Coefficients II Recall Lagrange Coefficients in terms of f and g coefficients From the universal anomaly X we can find the f and g coefficients

Lagrange Coefficients II Where and Steps finding state at a future Δθ using Lagrange Coefficients Find r0 and v0 from the given position and velocity vector Find vr0 and α Find X Find f and g Find r, where r = f r0 + g v0 Find fdot and gdot Find v