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

2. Lecture 03: Numerical Integrations
Chapter TBD

3. 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

4. 2nd Order Runge Kutta Given an 1st order ODE
Applying Taylor Series expansion to 2nd order
Since:
and

5. 2nd Order Runge Kutta
Now we have

6. 2nd Order Runge Kutta Or
The Coefficient in Butcher Tableaux form. Note:

7. 2nd Order Runge Kutta Graphically slope=k2 yn+1 (Approximate Soln)
slope=k1=f yo
Exact Solution to t1

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

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

10. Runge Kutta (explicit)
RK4
where

11. 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):

12. RK4 Example
From the problem:

13. RK4 Example

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

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

16. 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

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

18. Time versus True Anomaly

19. 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

20. 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

21. 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

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

23. 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

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

25. 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):

26. 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.

27. 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)

28. 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

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

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

31. 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

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

33. 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.

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

35. 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

36. 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.

37. 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

38. 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

39. Universal Variables
Stumpff functions
or for z = αΧ2 ,

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

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

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

43. 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

44. Universal Variables
Example 3.6 (Textbook: Curtis’s)

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

46. 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

47. 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