Time of Flight
4.5 Orbit Propagation Using Lagrangian Coefficients
Therefore, θ2= 64 97
Radius at the propagate position is determined using the trajectory equation:
r2= p 1 + e cosθ2
= a 1−e2 1 + e cosθ2
= 9, 116 1 km
h = f gh−f gh (4.58) Hence, the scalar coefficients ofh in Eq. (4.57) must satisfy
1 = f g−f g (4.59)
Therefore, we can use Eq. (4.59) and three known Lagrangian coefficients to determine the fourth coefficient.
To begin our analysis, let us express the initial position and velocity vectors in the peri-focal (orPQW) frame [see Eqs. (3.18) and (3.23)]:
r0= r0cosθ0P + r0sinθ0Q (4.60a) v0=−μ
h sinθ0P +μ
h e+ cosθ0 Q (4.60b)
Recall that the unit vectorP is along the periapsis direction, Q is in the orbital plane and 90 from periapsis in the direction of motion, andW is normal to the orbit (along the angular momentum vectorh). We wish to invert Eq. (4.60) and solve for unit vectors P andQ in terms of the initial state (r0,v0). One way to do this is to group the four terms on the right-hand side of Eq. (4.60) in a two-dimensional matrix:
C = r0cosθ0 r0sinθ0
−μ
h sinθ0 μ
h e+ cosθ0
(4.61)
The inverse of matrixC is
C−1= 1 det C
μ
h e+ cosθ0 −r0sinθ0
μ
hsinθ0 r0cosθ0
(4.62)
where the determinant ofC is angular momentum h [the reader should note that the cross productr0×v0 withr0andv0expressed in terms of theirP and Q components leads tor0×v0=h = det C W]. Using Eq. (4.62), we can write the P and Q unit vectors in terms ofr0andv0:
P = μ
h2 e+ cosθ0 r0−r0
hsinθ0v0 (4.63a)
Q = μ
h2sinθ0r0+r0
hcosθ0v0 (4.63b)
Now, write an expression forr(t) and v(t) vectors at an arbitrary future time using peri-focal coordinates
r t = r cosθP + r sinθQ (4.64a)
v t = −μ
h sinθP +μ
h e+ cosθ Q (4.64b)
Equation (4.64) has the same format as Eq. (4.60); however, Eq. (4.64) uses the propa-gatedradius and true anomaly r and θ, respectively. Substituting Eq. (4.63) for P and Q in Eq. (4.64) and collecting terms yields
r t =μr cosθ e + cosθ0 +μr sinθ sinθ0
h2 r0+−r cosθ r0sinθ0 + r sinθ r0cosθ0
h v0
(4.65a) v t = −μ2sinθ e + cosθ0 +μ2 e+ cosθ sinθ0
h3 r0
+μr0sinθ sinθ0 +μr0 e+ cosθ cosθ0
h2 v0
(4.65b)
Comparing Eqs. (4.65) and (4.55), we can identify the four Lagrangian coefficients:
f=μrcosθ e + cosθ0 +μrsinθ sinθ0
h2 (4.66)
g=−r cosθ r0sinθ0 + r sinθ r0cosθ0
h (4.67)
f= −μ2sinθ e + cosθ0 +μ2 e+ cosθ sinθ0
h3 (4.68)
g=μr0sinθ sinθ0 +μr0 e+ cosθ cosθ0
h2 (4.69)
Note that all four Lagrangian coefficients depend on products of sine and cosine ofθ0and θ. Therefore, these combinations may be written as trigonometric functions of differ-ences in true anomaly (i.e., Δθ = θ−θ0). Furthermore, we may substitute h = pμ, e= p/r−1 /cosθ, and e = p/r0−1 /cosθ0 (from the trajectory equation) in order to express the Lagrangian coefficients in terms of p. Using these substitutions, Eqs.
(4.66)–(4.69) are reduced to (somewhat) simpler expressions:
f= 1−r
p 1−cosΔθ (4.70)
g=rr0sinΔθ
pμ (4.71)
f = μ p
1−cosΔθ
p −1
r−1 r0
tanΔθ
2 (4.72)
g= 1−r0
p 1−cosΔθ (4.73)
whereΔθ = θ−θ0is the change in true anomaly between position vectorsr and r0. Note that whenΔθ = 0 (i.e., no propagation ahead in time), Eqs. (4.70)–(4.73) show that f = 1, g= 0, f = 0, and g = 1. Hence, when Δθ = 0 we obtain r t = f r0+ gv0=r0 and v t = f r0+ gv0=v0as expected.
We can now summarize our orbit-propagation method:
1) Given the initial state (r0,v0), compute the parameter p from angular momentum (h = r0×v0) and semimajor axis a from energy (ξ = v20/2−μ/r0=−μ/2a). Determine eccentricity e from p and a (e = 1−p/a).
2) Determine the initial true anomalyθ0from the trajectory equation r0= p/ 1 + e cosθ0 .
3) Select the propagated (future) true anomaly θ and compute the propagated radius using the trajectory equation r = p/ 1 + e cosθ . 4) Compute the four Lagrangian coefficients
using Eqs. (4.70)–(4.73) with knowledge of the four valuesΔθ = θ−θ0, r0, r, and p.
5) Calculate the propagated state (r, v) using the Lagrangian coefficients in Eq. (4.55).
Figure 4.12 shows orbit propagation from (r0,v0) to (r, v) using the Lagrangian coefficients.
It is useful to step back and reexamine the propagation method previously summarized.
Because we know the initial state (r0,v0), we can compute the corresponding six orbital ele-ments (a, e, i,Ω, ω, θ0). When we select the pro-pagated (future) true anomalyθ (step 3), we can simply compute the propagated (r, v) vectors by using the six orbital elements (a, e, i,Ω, ω, θ) and the transformation methods outlined in Chapter 3. So, orbit propagation using the Lagrangian coefficients provides an alternative technique (and perhaps a more compact method) for determining the future state (r, v) from the initial state (r0,v0) for a selected orbit positionθ. The key here is that the orbit is known and we have selected the future angular positionθ. The more fundamental (and more difficult) problem is Lambert’s problem: determine the orbit given two position vectorsr0andr and the corresponding TOF.
As previously mentioned, Lambert’s problem will be addressed in the next section.
One way to formulate Lambert’s problem is to express the Lagrangian coefficients in terms of change in eccentric anomaly ΔE and TOF. To do so, we start by rewriting the initial position vector in perifocal coordinates, Eq. (4.60), in terms of initial eccentric anomaly E0
r0= a cos E0−ae P + a 1−e2sin E0Q (4.74)
where the P coordinate (r0cosθ0) is obtained from Eq. (4.4) and the Q coordinate (r0sinθ0) is the product of Eqs. (4.12) and (4.13). Next, take the time derivative of Eq. (4.74)
v0=r0=−aE0sin E0P + aE0 1−e2cos E0Q (4.75)
Recall that we determined the time rate of eccentric anomaly in the derivation of Kepler’s equation in Section 4.2. Equation (4.14) is repeated below (with the appropriate subscript)
E0= μ a3
1 1−ecosE0
(4.76) r0
v r ∆θ
v0 0
0 0 0
v r v
v r r
g f. . g f
+
= +
=
Figure 4.12 Orbit propagation from state (r0,v0) to (r,v).
Equation (4.12) shows that 1−ecosE0= r0/a. Substituting this result into Eq. (4.76), we obtain a simple expression for E0
E0= 1 r0
μ
a (4.77)
Substituting Eq. (4.77) for E0 in Eq. (4.75) yields v0=− μa
r0 sin E0P + μa
r0 1−e2cos E0Q (4.78) The propagated state (r, v) at a future eccentric anomaly E is computed using Eqs. (4.74) and (4.78)
r = acosE −ae P + a 1−e2sin EQ (4.79a) v = − μa
r sin EP + μa
r 1−e2cos EQ (4.79b)
The subsequent steps follow the same procedure that was outlined at the beginning of this section: (1) invert Eqs. (4.74) and (4.78) and solve for theP and Q unit vectors;
(2) substitute the P and Q vectors into Eq. (4.79); and (3) express the trigonometric products as sine and cosine of the difference in eccentric anomalyΔE = E −E0. We will not present the details of these steps. The resulting Lagrangian coefficients in terms of ΔE are
f = 1−a
r0 1−cosΔE (4.80)
g= t−t0 − a3
μ ΔE −sinΔE (4.81)
f=− μasinΔE
rr0 (4.82)
g= 1−a
r 1−cosΔE (4.83)
whereΔE = E −E0is the change in eccentric anomaly between position vectorsr and r0. Equations (4.80)–(4.83) determine the Lagrangian coefficients in terms of the difference in eccentric anomalyΔE. Note that the g coefficient, Eq. (4.81), depends on the flight time t−t0from positionr0tor. In addition, note that when ΔE = 0 and t −t0= 0 (i.e., no prop-agation), Eqs. (4.80)–(4.83) show that f = 1, g = 0, f = 0, and g = 1 as expected.
In summary, we may compute the four Lagrangian coefficients using the eccentric anomaly difference ΔE [i.e., Eqs. (4.80)–(4.83)], or the true anomaly difference Δθ [i.e., Eqs. (4.70)–(4.73)]. Both sets of equations yield identical results as long as ΔE
corresponds toΔθ and eccentricity e. The following two examples illustrate orbit prop-agation with the Lagrangian coefficients computed usingΔθ and ΔE.
Example 4.7 A satellite has the following initial position and velocity vectors in the Earth-centered inertial (ECI) frame:
r0=
−15,634 4,689 7,407
km, v0=
−4 6954
−2 3777 0 6497
km/s
Propagate the orbit and determine the position and velocity vectors (r, v) corresponding to a change in true anomaly ofΔθ = 80 .
We can compute the Lagrangian coefficients using Eqs. (4.70)–(4.73) and true anomaly differenceΔθ. To do so, we need to calculate the parameter p and radii r0and r. Because the radius is computed from the trajectory equation, we require eccentricity e and the initial and propagated true anomalies, θ0 and θ. Therefore, we begin by computing the basic orbital characteristics (energy and angular momentum) from the initial state (r0,v0):
Initial radius r0= r0 = 17, 924 1 km Initial velocity v0= v0 = 5 3031 km/s Energy ξ = v20
2 −μ
r0=−8 1771 km2/s2 Semimajor axis a = −μ
2ξ = 24, 373 0 km
Angular momentum h = r0×v0= 20,658 −24,621 59,190 Tkm2/s Parameter p =h2
μ = 11, 380 8 km Eccentricity e = 1−p
a= 0 7301
We need to compute the propagated radius (magnitude r) using the trajectory equation and propagated true anomalyθ = θ0+Δθ. Because we know r0, p, and e, we can compute the initial true anomaly from the trajectory equation:
r0= p 1 + e cosθ0
cosθ0=1 e
p
r0−1 = −0 5
Because cosθ0< 0, the initial true anomalyθ0could be in the second or third quadrant.
The position and velocity dot product yieldsr0 v0= r0r0= 67,071 km2/s > 0, and there-foreθ0is in the second quadrant because radial velocity is positive. Therefore, initial true anomaly isθ0= cos−1 −0 5 = 120 and propagated true anomaly is θ = θ0+Δθ = 200 . Propagated radius is r = p/ 1 + e cosθ = 36,253.4 km.
Now we have all four values required for the Lagrangian coefficients. Using Eqs.
(4.70)–(4.73), we can calculate
f = 1−r
p 1−cosΔθ = – 1 632334 g=rr0sinΔθ
pμ = 9, 501 2766 s f= μ
p
1−cosΔθ
p −1
r−1
r0 tanΔθ
2 = – 5 345888 10– 5 s– 1 g= 1−r0
p 1−cosΔθ = – 0 301453
Finally, the propagated state is computed by using the Lagrangian coefficients in Eq. (4.55):
r = f r0+ gv0=
−19,092
−30,245
−5,918 km
v = f r0+ gv0=
2 2512 0 4661
−0 5918 km/s
The reader should note that the Lagrangian coefficients f and g are dimensionless by observing their respective defining equations and the propagation equations shown above. Furthermore, the units of the Lagrangian coefficient f are per second, which is the time derivative of coefficient f. The units of g are seconds, which is the time integral of coefficient g.
As a final note, the interested reader can determine the classical orbital elements by transforming either state (r0,v0) or (r, v) using the methods presented in Chapter 3.
This particular orbit is (approximately) a geostationary transfer orbit (GTO) with a perigee altitude of 200 km and apogee altitude reaching geostationary orbit (note that e= 0.7301). The remaining classical orbital elements are i = 28.5 , Ω = 40 , and ω = 0 (i.e., the perigee direction is along the ascending node). Figure 4.13 shows the GTO for this example. Note the initial position at θ0= 120 and propagated position at θ = 200 .
Example 4.8 Repeat the orbit-propagation problem outlined in Example 4.7 using the Lagrangian coefficients defined by the change in eccentric anomaly,ΔE.
Here the Lagrangian coefficients are determined by Eqs. (4.80)–(4.83) and hence we need a, r0, r,ΔE, and TOF t −t0. We calculated the first three parameters in Example 4.7. Equation (4.33) will yield the eccentric anomalies corresponding to the initial and propagated true anomalies,θ0= 120 andθ = 200 :
Initial position tanE0
2 = 1−e 1 + etanθ0
2 = 0 684102 E0= 1 1200 rad Propagated position tanE
2= 1−e 1 + etanθ
2=– 2 239944 E= 3 9814 rad
Note that the“calculator” inverse-tangent operation for the propagated position would yield E =−2 3018 rad (third quadrant) which is less than E0. BecauseΔE = E −E0> 0, we added 2π to obtain E = 3.9814 rad (third quadrant). Using these values, the difference in eccentric anomaly isΔE = E −E0= 2.7814 rad (or 159.4 ).
Next, we use Kepler’s equation (4.21) to compute the TOF from r0tor t−t0=1
n M−M0
The mean motion is n = μ/a3= 1.6592(10–4) rad/s, and the two mean anomalies are Initial mean anomaly M0= E0−esinE0= 0 5195 rad
Propagated mean anomaly M= E−esinE = 4 5249 rad
and, therefore, the flight time is t−t0= 24,140.5 s (or 6.7 h). Using Eqs. (4.80)–(4.83), we obtain the four Lagrangian coefficients:
f= 1−a r0
1−cosΔE = – 1 632334
g= t−t0 − a3
μ ΔE −sinΔE = 9,501 2766 s f=− μasinΔE
rr0 = – 5 345888 10– 5 s– 1 g= 1−a
r 1−cosΔE = – 0 301453
The above Lagrangian coefficients are identical to the f, g, f , and g coefficients computed in Example 4.7 using the difference in true anomaly. Of course, this result should have been anticipated becauseΔE = 159.4 was computed using the true anomalies corre-sponding toΔθ = 80 in Example 4.7. In other words, both ΔE = 159.4 and Δθ = 80 rep-resent the same change in orbital position fromr0tor. The propagated position and velocity vectors (r, v) are computed using the Lagrangian coefficients in Eq. (4.55) and the results are identical to the solution of Example 4.7.
Examples 4.7 and 4.8 show that orbit propagation is probably easier to perform by using the Lagrangian coefficients defined in terms of difference in true anomalyΔθ. Cal-culating the coefficients in terms ofΔE requires the extra steps of computing eccentric
θ r
θ0 r0 v0
v
Figure 4.13 Geostationary transfer orbit propagation from state (r0,v0) to (r,v) (Example 4.7).
anomalies from true anomalies and the TOF. The real advantage of theΔE formulation is the case where flight time betweenr0andr is known. This scenario is Lambert’s problem, which we describe in the next section.