1. Basic Orbital Mechanics Dr. Andrew Ketsdever MAE 5595

2. Conic Sections EccentricityConic = 0Circle 0 - 1Ellipse = 1Parabola > 1Hyperbola

3. Elliptical Orbit Geometry

4. Conic Sections

5. Classical Orbital Elements Semi-Major Axis, a –Size Eccentricity, e –Shape Kepler’s 3 rd Law

6. Classical Orbital Elements Inclination –Tilt InclinationOrbit = 90ºPolar 0º or 180ºEquatorial 0º - 90ºPrograde 90º - 180ºRetrograde

7. Classical Orbital Elements Right Ascension of the Ascending Node (RAAN)

8. Classical Orbital Elements Argument of Perigee

9. Classical Orbital Elements True Anomaly

10. Computing COEs From a R and V vector –Can compute the 6 COEs –Also works in reverse (given COEs compute R and V) –Example:

11. COEs a = 7965.1 km e = 0.0584 i = 90º  = 270º  = 90º = 0º Mission: Probably remote sensing or a spy satellite because it’s in a low, polar orbit.

12. Ground Tracks Ground Track Slides Courtesy of Major David French

13. COE Determination Semimajor axis Δ longitude ΔNΔN P = ΔN 15º / hr

14. COE Determination Eccentricity

15. COE Determination Inclination i = highest latitude

16. COE Determination Argument of Perigee ω = 90º

17. COE Determination True Anomaly

18. Orbit Examples

19. Molniya

20. Geostationary

21. Geosynchronous

22. e = 0 i = 0  Geosynchronous e = 0.4  = 180  e = 0.6  = 90 

23. Orbit Prediction Kepler’s Problem –If we know where a satellite (or planet) is today, where in its orbit will it be tomorrow? –Kepler devised a series of mathematical expressions to solve this particular problem Eccentric Anomaly Mean Anomaly True Anomaly II. The line joining the planet to the Sun sweeps out equal areas in equal times as the planet travels around the ellipse.

24. Orbit Prediction Kepler defined the Eccentric Anomaly to relate elliptical motion to circular motion He also defined Mean Anomaly to make the circular motion constant Convert unsteady elliptical motion into unsteady circular motion into steady circular motion…

25. Orbit Prediction

26. Orbital Prediction Given a = 7000 km e = 0.05 = 270º Find the time of flight to final = 50º

27. Orbital Prediction n = 0.001078 rad/sec E initial = 272.87º E future = 47.84º M initial = 275.73º M future = 45.72º TOF = 2104.58 sec or 35.08 min