Transcript Powerpoint-3_Basic Orbital Mechanics

Basic Orbital Mechanics
Dr. Andrew Ketsdever
MAE 5595
Conic Sections
Eccentricity
Conic
=0
Circle
0-1
Ellipse
=1
Parabola
>1
Hyperbola
Elliptical Orbit Geometry
Conic Sections
V



 
2 R
2a
2
Vcircular 

R
Classical Orbital Elements
• Semi-Major Axis, a
– Size
 

2a
• Eccentricity, e
– Shape
Period  2
a3

2c
e
2a
Kepler’s 3rd Law
Classical Orbital Elements
• Inclination
– Tilt
Inclination
Orbit
= 90º
Polar
0º or 180º Equatorial
0º - 90º
Prograde
90º - 180º Retrograde
hZ
cos i 
h
Classical Orbital Elements
• Right Ascension of
the Ascending Node
(RAAN)
 ˆ 
n  K h
nX
cos  
n
If n y  0, then0    180
If n y  0, then180   360
Classical Orbital Elements
• Argument of Perigee
 
n e
cos  
ne
If e Z  0, then0    180
If e Z  0, then180    360
Classical Orbital Elements
• True Anomaly
 
eR
cos 
eR
 
If (R  V)  0, then0    180
 
If (R  V)  0, then180  360
Computing COEs
• From a R and V vector
– Can compute the 6 COEs
– Also works in reverse (given COEs compute
R and V)
– Example:
R  0 Iˆ  0 Jˆ  7500Kˆ
km
V  0 Iˆ  7.5 Jˆ  0 Kˆ
km
sec
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.
Ground Tracks
Ground Track Slides Courtesy of Major David French
COE Determination
P  2
a3

ΔN
Δ longitude
P=
ΔN
15º / hr
Semimajor axis
COE Determination
Eccentricity
COE Determination
i = highest latitude
Inclination
COE Determination
ω = 90º
Argument of Perigee
COE Determination
True Anomaly
Orbit Examples
Molniya
Geostationary
Geosynchronous
Geosynchronous
e=0
e = 0.4
e=0
i = 0
 = 180
e = 0.6
 = 90
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.
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…
Orbit Prediction
 , E, M are always in thesame half - plane
  E  M for  0 or 180
Orbital Prediction
• Given
a = 7000 km
e = 0.05
 = 270º
Find the time of flight to final = 50º
Orbital Prediction
•
•
•
•
•
•
n = 0.001078 rad/sec
Einitial = 272.87º
Efuture = 47.84º
Minitial = 275.73º
Mfuture = 45.72º
TOF = 2104.58 sec or
35.08 min
n

a3
e  cos 
cos E 
1  e cos 
M  E  e sin E
TOF 
M f  M i  2k
n