## Orbital Angles (Called Anomalies)

• P = period of Orbit
• t = time for position of interest
• T = time of periapsis passage
• e = eccentricity of the orbit
• True (Angle) Anomaly, φ
the angle of the object past periapsis with center the focus
• Mean (Angle) Anomaly, M = 2π(t - T)/P
Actually the time past periapsis put in angle format
• Eccentric (Angle) Anomaly, E
The angle of the object position projected on the reference circle.

### Relationships

Find Time given Position
1. Determine Eccentric Anomaly
cos E = (e + cos φ)/(1 + e cos φ)
2. Kepler's Equation
M = E - e sin E
t = (M/2π)P + T

Example 1:For e = 0.3, φ = 90° find the time past periapsis
cos E = (0.3 - 0)/1 + 0.3) = 0.2308: E = 76.66° = 1.338 rad
M = 1.338 - 0.3 sin(76.7°) = 1.046 rad
t - T = (M/π)P = 0.166 P
For a cirular orbit this would be 0.25 P

Find Position given Time: M = 2π(t-T)/P
1. Kepler's Equation
M = E - e sin E
Solve for E given M - use interative procedure.
2. cos(φ) = (cos E - e)/(1 - e cos E)

Example 2: For e = 0.3, (t - T)/P = 0.25, find the angle past periapsis
M = 0.25 2π = 1.571
solve 1.571 = E - 0.3 sin E by interative use of the previous solution
E = M + e sin E

• 1st: E = 1.571 + 0.3 sin 90° = 1.871 rad = 107.2°
• 2nd: E = 1.571 + 0.3 sin 107.2° = 1.858 rad = 106.4°
• 3rd: E = 1.571 + 0.3 sin 106.4° = 1.859 rad = 106.5°
• 4th: E = 1.571 + 0.3 sin 106.5° = 1.859 rad
Solution is E = 1.859 rad
cosφ = [cos(106.5°)- 0.3]/[1 - 0.3 cos(106.5°)] = -0.455
φ = 117.1°
For a cicular orbit this would be 90°

Note: All angles are in radians
1 radian = 360/2π = 57.3 degrees

June 10, 2005 - L.Bogan