1. ASEN 5050 SPACEFLIGHT DYNAMICS Intro to STK, More 2-Body
Prof. Jeffrey S. Parker
University of Colorado – Boulder
Lecture 6: The Two Body Problem

2. Announcements Friday: no class at 9:00. STK Lab 1
ITLL 1B10, 2B10, or Visions Lab – or on your own computer.
Groups of 1-3.
Intro to STK today
Alan will be in ITLL 2B10 to help answer questions from 1-3.
Due Friday 9/26 at 9:00 am
Homework #2 is due Friday 9/12 at 9:00 am
D2L or under my door (ECNT 418). I’ll have Alan pick them up around 9:05 or 9:10 so don’t be late!
Homework #3 is due Friday 9/19 at 9:00 am
Concept Quiz #5 will be available at 10:00 am, due Monday morning at 8:00 am.
Reading: Chapters 1 and 2
Lecture 6: The Two Body Problem

3. Quiz 4
Lecture 6: The Two Body Problem

4. Quiz 4
Lecture 6: The Two Body Problem

5. Quiz 4
Lecture 6: The Two Body Problem

6. Quiz 4
Lecture 6: The Two Body Problem

7. Challenge #3 We examined Pluto’s and Neptune’s orbits last time.
Question: since Pluto sometimes travels interior to Neptune’s orbit, could they ever collide?
If so, what sort of order of duration do we need to wait until it may statistically happen? Years? Millennia? Eons?
Lecture 6: The Two Body Problem

8. Derivation of Kepler’s Equation
If t is given:
If n is given:
M = n(t - tp)
Another Useful Relation:
Lecture 6: The Two Body Problem

9. Solving Kepler’s Equation
Given M, solve for E
What would a plot of this look like?
Circular orbit?
Elliptical?
Lecture 6: The Two Body Problem

10. Solving Kepler’s Equation
Given M, solve for E
To see what we have to do, here are a few plots of this relationship:
~ parabolic
Near-circular
Lecture 6: The Two Body Problem

11. Solving Kepler’s Equation
Let’s say e = 0.1 and M = 100 deg (1.745 rad)
What is E that produces M=100?
Lecture 6: The Two Body Problem

12. Solving Kepler’s Equation
Let’s say e = 0.1 and M = 100 deg (1.745 rad)
Subtract M from all values
Find zero-crossing
What is E that produces Value=0?
Lecture 6: The Two Body Problem

13. Solving Kepler’s Equation
Let’s say e = 0.1 and M = 100 deg (1.745 rad)
Step 1: Initial guess (E0 = M)
Step 1: Value = deg
Lecture 6: The Two Body Problem

14. Solving Kepler’s Equation
Let’s say e = 0.1 and M = 100 deg (1.745 rad)
Step 2: Modify guess and find the zero
Lecture 6: The Two Body Problem

15. Solving Kepler’s Equation
Let’s say e = 0.1 and M = 100 deg (1.745 rad)
Newton-Raphson
Derivative line
Lecture 6: The Two Body Problem

16. Solving Kepler’s Equation
Let’s say e = 0.1 and M = 100 deg (1.745 rad)
Newton-Raphson
Derivative line
Guess #2
Lecture 6: The Two Body Problem

17. Solving Kepler’s Equation
Newton Raphson Method applied to Kepler’s Equation
Lecture 6: The Two Body Problem

18. Solving Kepler’s Equation
Let’s say e = 0.1 and M = 100 deg (1.745 rad)
Guess 1: E0 = 100 deg
Error 1: deg
Guess 2: E1 = deg
Error 2: deg
Guess 3: E2 = deg
Error 3: deg
Guess 4: E3 = deg
Error 4: deg
Lecture 6: The Two Body Problem

19. Solving Kepler’s Equation
Let’s say e = 0.1
Any tricks or challenges? No.
Lecture 6: The Two Body Problem

20. Solving Kepler’s Equation
Let’s say e = 0.95
Any tricks or challenges? Not really.
Lecture 6: The Two Body Problem

21. Solving Kepler’s Equation
Let’s say e = 0.95, M = 180 deg
Iteration History:
E0 = deg
Error = deg
Lecture 6: The Two Body Problem

22. Solving Kepler’s Equation
Let’s say e = 0.95, M = 300 deg
Iteration History:
E0 = deg
Value = deg
E1 = deg
Error = deg
E2 = deg
Error = deg
E3 = deg
Error = deg
E4 = deg
Error = deg
E5 = deg
Error = deg 2 1 3
Lecture 6: The Two Body Problem

23. Solving Kepler’s Equation
Iteration History:
E0 = deg
Value = deg
E1 = deg
Error = deg
E2 = deg
Error = deg
E3 = deg
Error = deg
E4 = deg
Error = deg
E5 = deg
Error = deg
E6 = deg
Error = deg
Let’s say e = 0.95, M = 350 deg
Lecture 6: The Two Body Problem

24. Solving Kepler’s Equation
How many iterations does it take?
e = 0
Lecture 6: The Two Body Problem

25. Solving Kepler’s Equation
How many iterations does it take?
e = 0.5
Lecture 6: The Two Body Problem

26. Solving Kepler’s Equation
How many iterations does it take?
e = 0.9
Lecture 6: The Two Body Problem

27. Solving Kepler’s Equation
How many iterations does it take?
e = 0.95
Lecture 6: The Two Body Problem

28. Solving Kepler’s Equation
How many iterations does it take?
e = 0.99
Not good!
Lecture 6: The Two Body Problem

29. Kepler’s Equation Algorithm 2 in Vallado:
Lecture 6: The Two Body Problem

30. Orbital Elements
(Vallado, 1997)
Lecture 6: The Two Body Problem

31. Orbital Elements Now, let’s define our other orbital elements.
The inclination, i, refers to the tilt of the orbit plane. It is the angle between and , and varies from 0-180°.
Lecture 6: The Two Body Problem

32. Orbital Elements
The right ascension of the ascending node, W, is the angle in the equatorial plane from to the ascending node. The ascending node is the point on the equator where the satellite passes from South to North (opposite for the descending node).
The line of nodes connects the ascending and descending nodes. The node vector, , points towards the ascending node and is denoted:
The node lies between 0° and 360°.
Lecture 6: The Two Body Problem

33. Orbital Elements
The argument of periapse, w, measured from the ascending node, locates the closest point of the orbit (periapse) and is the angle between and .
Lecture 6: The Two Body Problem

34. Orbital Elements
The true anomaly, n, is the angle between periapse and the satellite position; thus:
( is positive going away from periapse, negative coming towards periapse.)
Lecture 6: The Two Body Problem

35. Special Cases
Elliptical Equatorial Orbits – W is undefined, so we use true longitude of periapse, ,
This is equivalent to astronomers’ longitude of periapse, , where
Lecture 6: The Two Body Problem

36. Special Cases
Circular Orbits – w is undefined, use argument of latitude, u, where:
Lecture 6: The Two Body Problem

37. Special Cases Circular Equatorial – w and W undefined
Lecture 6: The Two Body Problem

38. Two Line Element Sets
(Vallado, 1997)
Available on class web page. (Can be read by many programs including STK.)
, are “Kozai” means. B* is a drag parameter.
Lecture 6: The Two Body Problem

39. Two Line Element Sets Example
U 86017A
Epoch: Dec 18, h 50min sec UTC
Errors can be as large as a km or more.
Lecture 6: The Two Body Problem

40. Orbital Elements from and (and t)
Algorithm 9 in the book
First compute the following vectors
Compute the energy: p
Lecture 6: The Two Body Problem

41. Orbital Elements from and (and t)
Test using Example 2-5 in book
Also,
Lecture 6: The Two Body Problem

42. STK
STK! Systems Tool Kit version 10.0
Lecture 6: The Two Body Problem

43. Welcome Splash
Lecture 6: The Two Body Problem

44. STK Boot-up
Lecture 6: The Two Body Problem

45. Save Save frequently Save each scenario in its own directory
Please note: each scenario includes many files, for each object. If two scenarios are in the same directory and share objects, they may not behave properly.
Lecture 6: The Two Body Problem

46. Lecture 6: The Two Body Problem

47. A satellite
Lecture 6: The Two Body Problem

48. View
Lecture 6: The Two Body Problem

49. ASEN 5050 SPACEFLIGHT DYNAMICS Coordinate and Time Systems
Prof. Jeffrey S. Parker
University of Colorado - Boulder
Lecture 6: The Two Body Problem

50. Coordinate Systems
Given a full state, with position and velocity known.
Or, given the full set of coordinate elements.
What coordinate system is this state represented in?
Could be any non-rotating coordinate system!
Earth J2000 or ecliptic J2000 or Mars, etc.
Lecture 6: The Two Body Problem

51. Coordinate Systems Celestial Sphere
Celestial poles intersect Earth’s rotation axis.
Celestial equator extends Earth equator.
Direction of objects measured with right ascension (a) and declination (d).
Lecture 6: The Two Body Problem

52. Coordinate Systems
The Vernal Equinox defines the reference direction. A.k.a. The Line of Aries
The ecliptic is defined as the mean plane of the Earth’s orbit about the Sun.
The angle between the Earth’s mean equator and the ecliptic is called the obliquity of the ecliptic, e~23.5.
Lecture 6: The Two Body Problem

53. Coordinate Frames Inertial: fixed orientation in space Rotating
Inertial coordinate frames are typically tied to hundreds of observations of quasars and other very distant near-fixed objects in the sky.
Rotating
Constant angular velocity: mean spin motion of a planet
Osculating angular velocity: accurate spin motion of a planet
Lecture 6: The Two Body Problem

54. Coordinate Systems Coordinate Systems = Frame + Origin
Inertial coordinate systems require that the system be non-accelerating.
Inertial frame + non-accelerating origin
“Inertial” coordinate systems are usually just non-rotating coordinate systems.
Is the Earth-centered J2000 coordinate system inertial?
Lecture 6: The Two Body Problem

55. Useful Coordinate Systems
ICRF
International Celestial Reference Frame, a realization of the ICR System.
Defined by IAU (International Astronomical Union)
Tied to the observations of a selection of 212 well-known quasars and other distant bright radio objects.
Each is known to within 0.5 milliarcsec
Fixed as well as possible to the observable universe.
Motion of quasars is averaged out.
Coordinate axes known to within 0.02 milliarcsec
Quasi-inertial reference frame (rotates a little)
Center: Barycenter of the Solar System
Lecture 6: The Two Body Problem

56. Useful Coordinate Systems
ICRF2
Second International Celestial Reference Frame, consistent with the first but with better observational data.
Defined by IAU in 2009.
Tied to the observations of a selection of 295 well-known quasars and other distant bright radio objects (97 of which are in ICRF1).
Each is known to within 0.1 milliarcsec
Fixed as well as possible to the observable universe.
Motion of quasars is averaged out.
Coordinate axes known to within 0.01 milliarcsec
Quasi-inertial reference frame (rotates a little)
Center: Barycenter of the Solar System
Lecture 6: The Two Body Problem

57. Useful Coordinate Systems
EME2000 / J2000 / ECI
Earth-centered Mean Equator and Equinox of J2000
Center = Earth
Frame = Inertial (very similar to ICRF)
X = Vernal Equinox at 1/1/ :00:00 TT (Terrestrial Time)
Z = Spin axis of Earth at same time
Y = Completes right-handed coordinate frame
Lecture 6: The Two Body Problem

58. Useful Coordinate Systems
EMO2000
Earth-centered Mean Orbit and Equinox of J2000
Center = Earth
Frame = Inertial
X = Vernal Equinox at 1/1/ :00:00 TT (Terrestrial Time)
Z = Orbit normal vector at same time
Y = Completes right-handed coordinate frame
This differs from EME2000 by ~ degrees.
Lecture 6: The Two Body Problem

59. Useful Coordinate Systems
Note that J2000 is very similar to ICRF and ICRF2
The pole of the J2000 frame differs from the ICRF pole by ~18 milliarcsec
The right ascension of the J2000 x-axis differs from the ICRF by 78 milliarcsec
JPL’s DE405 / DE421 ephemerides are defined to be consistent with the ICRF, but are usually referred to as “EME2000.” They are very similar, but not actually the same.
Lecture 6: The Two Body Problem

60. Useful Coordinate Systems
ECF / ECEF / Earth Fixed / International Terrestrial Reference Frame (ITRF)
Earth-centered Earth Fixed
Center = Earth
Frame = Rotating and osculating (including precession, nutation, etc)
X = Osculating vector from center of Earth toward the equator along the Prime Meridian
Z = Osculating spin-axis vector
Y = Completes right-handed coordinate frame
Lecture 6: The Two Body Problem

61. Useful Coordinate Systems
Earth Rotation
The angular velocity vector ωE is not constant in direction or magnitude
Direction: polar motion
Chandler period: 430 days
Solar period: 365 days
Magnitude: related to length of day (LOD)
Components of ωE depend on observations; difficult to predict over long periods
Lecture 6: The Two Body Problem

62. Useful Coordinate Systems
Principal Axis Frames
Planet-centered Rotating System
Center = Planet
Frame:
X = Points in the direction of the minimum moment of inertia, i.e., the prime meridian principal axis.
Z = Points in the direction of maximum moment of inertia (for Earth and Moon, this is the North Pole principal axis).
Y = Completes right-handed coordinate frame
Lecture 6: The Two Body Problem

63. Useful Coordinate Systems
IAU Systems
Center: Planet
Frame: Either inertial or fixed
Z = Points in the direction of the spin axis of the body.
Note: by convention, all z-axes point in the solar system North direction (same hemisphere as Earth’s North).
Low-degree polynomial approximations are used to compute the pole vector for most planets wrt ICRF.
Longitude defined relative to a fixed surface feature for rigid bodies.
Lecture 6: The Two Body Problem

64. Useful Coordinate Systems
Example:
Lat and Lon of Greenwich, England, shown in EME2000.
Greenwich defined in IAU Earth frame to be at a constant lat and lon at the J2000 epoch.
Lecture 6: The Two Body Problem

65. Useful Coordinate Systems
Synodic Coordinate Systems
Earth-Moon, Sun-Earth/Moon, Jupiter-Europa, etc
Center = Barycenter of two masses
Frame:
X = Points from larger mass to the smaller mass.
Z = Points in the direction of angular momentum.
Y = Completes right-handed coordinate frame
Lecture 6: The Two Body Problem

66. Announcements Friday: no class at 9:00. STK Lab 1
ITLL 1B10, 2B10, or Visions Lab – or on your own computer.
Groups of 1-3.
Intro to STK today
Alan will be in ITLL 2B10 to help answer questions from 1-3.
Due Friday 9/26 at 9:00 am
Homework #2 is due Friday 9/12 at 9:00 am
D2L or under my door (ECNT 418). I’ll have Alan pick them up around 9:05 or 9:10 so don’t be late!
Homework #3 is due Friday 9/19 at 9:00 am
Concept Quiz #5 will be available at 10:00 am, due Monday morning at 8:00 am.
Reading: Chapters 1 and 2
Lecture 6: The Two Body Problem