1. 5.2 Computing Orbiting Elements
Describing Orbits
5.2 Computing Orbiting Elements
In This Section You’ll Learn to…
Determine all six orbital elements, given only the position, R, and velocity, V, of a spacecraft at one particular time.

2. Finding Semimajor Axis, a
(5-3) g=(V^2)/2-u/R
V= magnitude of the spacecraft’s velocity vector (km/s)
u=gravitational parameter (km^3/s^2)
R=magnitude of the spacecraft’s position vector (km)
g=mechanical energy
So if we know the magnitude of R and of V, we can solve for the energy and thus the semimajor axis.
(5-4) a=-u/2g
Recall the semimajor axis, a, tells us the orbit’s size and depends on the orbit’s specific mechanical energy, E.

3. Finding Eccentricity, e
(5-5) e=1/u[(V^2-u/R)R (R*V)V]
e=eccentricity vector (unitless, points at perigee)
u=gravitational parameter (km^3/s^2)=3.986*10^5 km^3/s^2 for Earth.
V=magnitude of V (km/s)
R=magnitude of R (km)
R=position vector (km)
V=velocity vector (km/s)
An eccentricity vector, e, that points from Earth’s center to perigee and whose magnitude equals the eccentricity, e. e relates to position, R, and velocity, V.

4. Finding Inclination, i (5-6) A*B=ABcos0 (5-7) 0=cos^-1(A*B/AB)
Figure Find the angle Between two Vectors. When we take the projection of vector A on vector B (a cos0) and multiply it times the magnitude of B (B), we get the dot product of A*B, and use it to find the value for angle 0.
(5-7) 0=cos^-1(A*B/AB)
Figure Inverse Cosine. An inverse cosine gives two possible answers: 0 and (360-0)

5. (5-8) i=cos^-1(K*h/Kh) K=unit vector through the North Pole
i=inclination (deg or rad)
K=unit vector through the North Pole
h=specific angular momentum vector (km^2/s)
K=magnitude of K=1
h=magnitude of h (km^2/s)
Figure Inclination. Recall inclination, i, is the angle between the K unit vector and the specific angular momentum vector, h.
Figure Inclination. Recall inclination, i, is the angle between the K unit vector and the specific angular momentum vector, h.

6. Finding Right Ascension of the Ascending Node, Omega
(5-9) n=K*h
n=ascending node vector (km^2/s, points at the ascending node)
K=unit vector through the North Pole
h=specific angular momentum vector (km^2/s)
(5-10) e=cos^-1(I*n/I*n)
e =right ascension of the ascending node( deg or rad)
I=unit vector in the principal direction
n=ascending node vector(km^2/s, points at the ascending node)
I=magnitude of I=I
n=magnitude of n (km^2/s)
Figure Finding the Ascending Node. We can find the ascending node vector, n , by using the right-hand rule. Point your index finger at K and your middle finger at h. Your thumb will point in the direction of n.
Figure Quadrant Check for Omega. We can find the quadrant for the right ascension of the ascending node, Omega, by looking at the sign of the J component of n, n. If n is greater than zero, Omega is between 0 and 180 degrees. If n is less than zero, omega is between 180 and 360 degrees.

7. Finding Argument of Perigee, w
(5-11) w=cos^-1(n*e/ne)
w=argument of perigee (deg or rad)
n=ascending node vector (km^2/s, points at the ascending node)
e=eccentricity vector (unitless, points at perigee)
n=magnitude of n (km^2/s)
e=magnitude of e (unitless)
Figure Quadrant Check for the Argument of Perigee, w. We check the quadrant for the argument of perigee, w, by looking at the K component of the eccentricity vector, e. If e is greater than zero, perigee lies above the equator; thus, w is between 0 degrees and 180 degrees. If e is less than zero, perigee lies in the Southern Hemisphere; and, w is between 180 degrees and 360 degrees.

8. Finding True Anomaly, v (5-12) v=cos^-1(e*R/e*R)
v=true anomaly (deg or rad)
e=eccentricity vector (unitless, points at perigee)
R=position vector (km)
e=magnitude of e (unitless)
R=magnitude of R (km)
If (R*V)>0 (o>0) then 0 < v < 180
If (R*V)<0 (o<0) then 180 < v < 360
Figure Finding True Anomaly, v. We find the true anomaly, v as the angle between the eccentricity vector, e, and the spacecraft’s position vector, R.
Figure Quadrant Check for True Anomaly, v. To resolve the quadrant for true anomaly, v, check the sign on the flight-path angle, o. If o is positive, the spacecraft is moving away from perigee, so true anomaly is between 0 and 180 degree. If o is negative, the spacecraft is moving perigee, so true anomaly is between 180 degrees and 360 degrees.

10. Ending of Section 5.2
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