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.

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.

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.

Finding Inclination, i (5-6) A*B=ABcos0 (5-7) 0=cos^-1(A*B/AB) Figure 5-17. 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 5-18. Inverse Cosine. An inverse cosine gives two possible answers: 0 and (360-0)

(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 5-19. Inclination. Recall inclination, i, is the angle between the K unit vector and the specific angular momentum vector, h. Figure 5-19. Inclination. Recall inclination, i, is the angle between the K unit vector and the specific angular momentum vector, h.

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 5-20. 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 5-21. 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.

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 5-23. 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.

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 5-24. 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 5-25. 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.

Ending of Section 5.2 Questions???