Non-Keplerian Motion
5.3 General Perturbation Methods
5.3.2 Secular Perturbations due to Oblateness (J 2 )
da dt = 2
na
∂R
∂σ (5.48)
de dt=1−e2
na2e
∂R
∂σ− 1−e2 na2e
∂R
∂ω (5.49)
di
dt= 1
na2 1−e2sini cosi∂R
∂ω−∂R
∂Ω (5.50)
dΩ
dt = 1
na2 1−e2sini
∂R
∂i (5.51)
dω
dt = 1−e2 na2e
∂R
∂e− cosi na2 1−e2sin i
∂R
∂i (5.52)
dσ dt =− 2
na
∂R
∂a−1−e2 na2e
∂R
∂e (5.53)
Equations (5.48)–(5.53) are Lagrange’s variation of parameter equations, and they are in terms of the partial derivatives of the disturbance potential R. Recall that R must be a conservative disturbing function such as a non-spherical gravitational body or third-body gravity. Close inspection of Lagrange’s equations shows that the disturbing poten-tial R must be expressed in terms of the orbital elements so that we can compute the partial derivatives.
At this point, the reader may have lost sight of the fact that the original objective of this section is to develop a general perturbation method that results in analytical expressions for the effects of perturbations. Lagrange’s variation of parameter equations (5.48)–
(5.53) are highly nonlinear first-order differential equations– finding analytical solutions for each orbital element does not appear to be an easy task! The next subsection presents analytical expressions for the“mean” or “averaged” effect of Earth’s oblateness.
periodic changes due to J2, but in all three cases the net changes were zero after an orbital revolution. Figures 5.3d and 5.3e showed that elementsΩ and ω exhibited periodic var-iations about a“mean” or secular change that is linear with time.
It is possible to analytically derive the secular changes in the orbital elements due to Earth’s oblateness (or J2) effect. To begin, we must express the disturbing function R, Eq. (5.54), in terms of the orbital elements. The sine of the geocentric latitude is com-puted from spherical trigonometry
sinϕ = sin ω + θ sini (5.55)
Using this expression for sinϕ in Eq. (5.15), the second-order Legendre polynomial is P2 sinϕ =1
2 3 sin2 ω + θ sin2i−1 (5.56) Substituting Eq. (5.56) into the oblateness disturbing function, Eq. (5.54) becomes
R=− μ 2rJ2 RE
r
2
3sin2 ω + θ sin2i−1 (5.57) Finally, we must substitute the trajectory equation
r= p
1 + e cosθ= a 1−e2 1 + e cosθ into Eq. (5.57) to obtain the disturbing function
R=− μ
2a3 1−e2 3J2R2E 1 + e cosθ 3 3sin2 ω + θ sin2i−1 (5.58) Equation (5.58) presents R as a function of the orbital elements. We could take partial derivatives of R and use them in Lagrange’s variation of parameter equations. However, this step would not result in analytical expressions because the first-order differential equations would remain highly nonlinear. Because we want the secular changes in the elements, we need to“average out” the “fast” variable that causes periodic fluctuations during an orbital revolution. The“fast” variable is the angular position of the satellite.
The“mean” disturbance function is its average value over one orbit, that is,
R= 1 2π
2π
0
RdM (5.59)
where the over-bar indicates the“mean” or “average” value. Here the “averaging” inte-gration is performed using mean anomaly M as the angular variable. Recall from Chapter 4 that the time-rate of mean anomaly is the mean motion, or
dM
dt = n (5.60)
Therefore, the differential in the mean anomaly is
dM= ndt (5.61)
The differential time dt can be found using the angular momentum relationship expressed as the product of radial position r and transverse velocity vθ= rθ
h= r2θ = r2dθ
dt (5.62)
Therefore, dt = r2dθ/h, and Eq. (5.61) becomes dM=nr2
h dθ (5.63)
Substituting Eqs. (5.63) and (5.58) into the mean disturbance function (5.59) yields
R= 1 2π
2π
0
−nr2μ
2ha3 1−e2 3J2R2E 1 + e cosθ 3 3 sin2 ω + θ sin2i−1 dθ (5.64) Finally, we substitute h = μa 1−e2 and the trajectory equation for r in Eq. (5.64) and perform the integration to obtain
R= n2J2R2E
4 1−e2 3/2 2−3sin2i (5.65)
Equation (5.65) is the mean disturbance function where the periodic fluctuations (due to angular positionθ) have been averaged out. If we use the partial derivatives of R in Lagrange’s planetary equations (5.48)–(5.53), we get the mean or averaged rates for the orbital elements over one revolution. Because R is not a function of elementsΩ, ω, or σ, the corresponding partial derivatives ∂R/∂Ω, ∂R/∂ω, and ∂R/∂σ are zero. Con-sequently, Eqs. (5.48)–(5.50) show that the mean rates for semimajor axis, eccentricity, and inclination are zero:
da
dt = 0 (5.66)
de
dt= 0 (5.67)
di
dt= 0 (5.68)
Equations (5.66)–(5.68) confirm the numerical results presented in Example 5.1: the orbital elements a, e, and i exhibit zero change when averaged over a single orbital rev-olution when J2is the only perturbation (again, see Figures 5.3a–c). Equations (5.66)–
(5.68) are results we obtained from our general perturbation analysis: that is, the secular change in elements a, e, and i due to Earth oblateness is zero.
Note that Lagrange’s planetary equations for the longitude of the ascending node and argument of perigee, Eqs. (5.51) and (5.52), contain the partial derivatives ∂R/∂i and
∂R/∂e. Taking the partial derivative of Eq. (5.65) with respect to inclination yields
∂R
∂i = −3n2J2R2E
2 1−e2 3/2sini cosi (5.69)
Substituting Eq. (5.69) into Eq. (5.51), yields the mean rate in the longitude of the ascend-ing node:
dΩ
dt = −3nJ2
2 1−e2 2 RE
a
2
cosi (5.70)
Equation (5.70) is the secular rate of change for longitude of the ascending node. This average or mean rate is the linear“drift” shown by the sloping dashed line in Figure 5.3d.
Equation (5.70) is sometimes called the nodal regression because the oblateness (J2) effect causes the ascending node vector to regress or rotate from east to west for a direct orbit (i.e., inclination i < 90 ). Figure 5.4 shows a direct Earth orbit where the zonal harmonic J2 is causing the ascending noden to rotate westward. Of course, the nodal regression indi-cates that the orbital plane is rotating clockwise about the polar axis as viewed from above the North Pole.
Figure 5.5 shows the secular drift rate dΩ/dt as computed by Eq. (5.70) for circular Earth orbits. Here the dimensionless zonal harmonic coefficient is J2= 0.0010826267 and the equatorial Earth radius is RE = 6,378.14 km. The reader should note that the dimensions of Eq. (5.70) are radians per second because the ratio RE/a is dimensionless and the mean motion n is an angular velocity in radians per second [mean motion n solely depends on semimajor axis a as shown by Eq. (5.23); do not confuse mean motion nwith the magnitude of the ascending node vectorn!]. The nodal regression presented in Figure 5.5 has been converted to units of degrees per day. Figure 5.5 and Eq. (5.70) show that dΩ/dt is negative for direct orbits, zero for polar orbits (i = 90 ), and positive for retrograde orbits (i > 90 ). Furthermore, the nodal drift rate is significant for LEOs.
Ascending noden
Nodal regression
I
J K
h i
Ω
dt dΩ< 0
Figure 5.4 Nodal regression due to zonal harmonic J2(Earth oblateness).
For example, a 325-km altitude circular LEO with inclination i = 28.5 will exhibit a nodal regression of–7.4 deg/day. This motion of the orbital plane must be taken into account when planning orbital transfers or orbital rendezvous maneuvers.
Because Earth oblateness has a net zero effect on a, e, and i, we can consider the nodal regression dΩ/dt as constant for a given Earth orbit. Therefore, Eq. (5.70) can be inte-grated to yield
Ω t = Ω0+dΩ
dtt (5.71)
Equation (5.71) is the time history of the mean (or average) longitude of the ascending node. The initial valueΩ0is the longitude of the ascending node at time t = 0. It should be clear that Eq. (5.71) only accounts for the secular drift inΩ and does not include the periodic variations during each orbital revolution. The following example illustrates nodal regression for an Earth orbit.
Example 5.2 The International Space Station (ISS) has the following orbital elements at time t = 0
Semimajor axis a = 6,790.6 km Eccentricity e = 0.0005
Inclination i = 51.65
Longitude of the ascending nodeΩ0= 295
Compute the longitude of the ascending node for the ISS 7 days after this epoch.
0 30 60 90 120 150 180
Inclination, deg –10
–8 –6 –4 –2 0 2 4 6 8 10
Nodal regression rate, deg/day
325 km 1000 km 1500 km
3000 km
Labels are altitudes of circular Earth orbits
Figure 5.5 Nodal regression dΩ/dt vs. inclination for circular Earth orbits.
The ISS orbit is nearly circular at an altitude of about 412 km. Mean motion n is the critical value needed in the nodal-regression equation (5.70). Using Eq. (5.23), the mean motion is
n= μ
a3= 0 001128 rad/s
Using Eq. (5.70) with J2= 0.0010826267 and RE= 6,378.14 km, we obtain the mean nodal regression
dΩ
dt = −3nJ2
2 1−e2 2 RE
a
2
cosi =−1 0029 10−6 rad/s
Converting the nodal regression to degrees per day yields dΩ/dt = –4.965 deg/day.
Therefore, the longitude of the ascending node at time t = 7 days is Ω t = Ω0+dΩ
dtt= 295 + – 4 965 deg/day 7days = 260 245
The orbital plane of the ISS has rotated nearly 35 westward in 1 week. This simple exam-ple shows that two-body (Keexam-plerian) motion does not hold for a LEO such as the ISS. The orbital plane is not fixed in inertial space; in reality the plane rotates westward because the Earth is not a perfect sphere.
The Earth-oblateness effect may be used advantageously to design a sun-synchronous orbitwhere the orbital plane rotates at the same rate as the Earth’s rotation rate about the sun. Figure 5.6a shows a“top-down” view of the Earth’s orbit about the sun. Earth’s mean motion (or“average” angular velocity) is one revolution every year or (360 )/(365.25 days)
= 0.986 deg/day (of course, the Earth’s heliocentric orbit is elliptical and hence its angular velocity slightly varies with time). Figure 5.6a also shows a top-down view of a nearly polar Earth orbit with a mean nodal rate dΩ/dt that matches Earth’s mean motion about the sun (because the Earth orbit is nearly polar it appears as an“edge” in Figure 5.6a). The Earth orbit shown in Figure 5.6a is a sun-synchronous orbit because its orbital plane rotates at the same rate as Earth’s mean motion about the sun, that is, dΩ/dt = 0.986 deg/day. If the orbital plane is initially orientated so that it is perpendicular to the sun–Earth line (as in Figure 5.6a), then the Earth orbit will maintain this orientation with respect to the sun as the Earth follows its heliocentric path. Hence, a satellite in a sun-synchronous orbit shown in Figure 5.6a will always be in sunlight and will never expe-rience an Earth eclipse. Figure 5.6b shows an Earth satellite in a polar orbit (i = 90 ) where the J2zonal harmonic has a zero net effect on the orientation of the orbital plane (i.e., dΩ/dt = 0). An Earth satellite in a polar orbit may initially experience all-sunlight con-ditions (see time t = 0 in Figure 5.6b), but 3 months later nearly half of its orbit will be in the Earth’s shadow (see time t = 91 days in Figure 5.6b).
Equation (5.70) shows that we have the freedom to select semimajor axis a, eccentricity e, and inclination i in order to establish a sun-synchronous orbit with a mean ascending node rate dΩ/dt = 0.986 deg/day. First, Eq. (5.70) and Figure 5.5 show that a sun-synchronous orbit must be slightly retrograde (i > 90 ) so that the nodal rotation rate is positive (eastward). The following example involves computing the orbital elements for a sun-synchronous orbit.
Example 5.3 Determine the orbital inclination required for a circular sun-synchronous orbit. Plot inclination vs. orbital altitudes (up to 1,500 km) for circular sun-synchronous orbits.
We know that the mean rate of the longitude of the ascending node must match the Earth’s mean motion about the sun (i.e., 360 deg/year = 0.9856 deg/day). Using Eq. (5.70), we have
dΩ
dt = −3nJ2
2 1−e2 2 RE
a
2
cos i = 0 9856 deg/day = 1 991021 10−7 rad/s
Because we are interested in circular orbits, we can set e = 0 and a = r (radius). Sub-stituting n = μ/a3= μ/r3 (for a circular orbit) and solving for inclination, we obtain
sun
Earth at
Earth at
t= 0 Orbital
plane
i= 98° Earth at
t= 46 days 45°
45°
t= 91 days
(a)
Earth’s angular rate about the sun
0.986 deg/day
0.986 deg/day
0.986 deg/day
sun
Earth at t = 0 Orbital
plane i = 90°
Earth at t= 46 days
Earth at t= 91 days
(b)
Earth’s angular rate about the sun
0.986 deg/day
0
≈
≈
Ω 0.986 deg/day
≈
≈
Ω
≈ Ω
= Ω
Figure 5.6 (a) Sun-synchronous orbit and (b) polar orbit with a stationary orbital plane.
i= cos−1 −2Ω 3J2R2E
r7 μ
Using J2= 0.0010826267, RE = 6,378.14 km, andΩ = 1.991021(10−7) rad/s, we can determine the sun-synchronous orbital inclination for a given radius r. Figure 5.7 shows sun-synchronous orbit inclination for circular altitudes ranging from 325 km (LEO) to 1,500 km. We see that inclination must be retrograde for a sun-synchronous orbit (as expected) and that inclination increases with altitude. The sun-synchronous orbit shown in Figure 5.6a has inclination i = 98 , and therefore its circular orbital altitude must be about 654 km (see Figure 5.7).
The previous analysis has identified the secular change in longitude of the ascending node due to oblateness. Earth oblateness also causes a secular drift rate in argument of perigee. Equation (5.52) shows that the time-rate ofω depends on partial derivatives of the disturbance function with respect to eccentricity and inclination. Equation (5.69) provides the partial derivative ∂R/∂i. The partial derivative of Eq. (5.65) with respect to eccentricity is
∂R
∂e = 3n2J2R2Ee
4 1−e2 5/2 2−3sin2i (5.72)
Substituting Eqs. (5.69) and (5.72) into Eq. (5.52) yields the mean rate in the argument of perigee:
dω
dt = 3nJ2 4 1−e2 2
RE a
2
4−5sin2i (5.73)
200 400 600 800 1000 1200 1400 1600
Circular orbit altitude, km 96
97 98 99 100 101 102 103
Inclination for sun-synchronous orbit, deg
Figure 5.7 Inclination required for a circular sun-synchronous orbit (Example 5.3).
Equation (5.73) is the secular rate of change for argument of perigee caused by Earth oblateness. This mean rate is the linear“drift” shown by the dashed line in Figure 5.3e (Example 5.1). Equation (5.73) is the apsidal rotation because the oblateness (J2) effect causes the eccentricity vectore (or apse line) to rotate in the orbital plane. Integrating Eq. (5.73) yields the time history of the mean argument of perigeeω.
ω t = ω0+dω
dtt (5.74)
Figure 5.8 shows a direct Earth orbit (i < 90 ) where the zonal harmonic J2is causing the eccentricity vectore to rotate about the angular momentum vector h in the direction of orbital motion. Equation (5.73) shows that the mean argument of perigee rate is pos-itivewhen the sine of the inclination satisfies the condition sini < 4/5. This condition is satisfied for direct orbits with inclination i < 63.4 and retrograde orbits with inclination i> 116.6 . Because the direct elliptical Earth orbit shown in Figure 5.8 has inclination i< 63.4 , the secular change dω/dt is positive and vector e rotates counter-clockwise in the orbital plane.
Figure 5.9 shows the apsidal rotation rate, Eq. (5.73), as a function of inclination for elliptical Earth orbits. The four elliptical orbits presented in Fig. 5.9 have a common per-igee altitude of 325 km. Figure 5.9 clearly shows that Earth oblateness causes positive apsidal rotation (dω/dt > 0) for inclinations i < 63.4 and i > 116.6 , and negative apsidal rotation when 63.4 < i < 116.6 . Equation (5.73) and Figure 5.9 show that the apsidal rotation is maximized (and positive) for equatorial orbits (i = 0 and i = 180 ). Apsidal rotation is zero at i = 63.4 (direct) and i = 116.6 (retrograde) because both inclinations
e,Perigee direction
Apsidal rotation
> 0
I
J K
h i< 63.4ο
Ascending noden
ω dt
dω
Figure 5.8 Apsidal rotation due to zonal harmonic J2(Earth oblateness).
satisfy the condition sini = 4/5 (or 4−5sin2i= 0). Figure 5.10 shows the Molniya orbit used by Russia for communication satellites (see Section 3.4 in Chapter 3). This highly elliptical orbit has i = 63.4 and argument of perigeeω = –90 . Therefore, the apse line does not rotate within the orbital plane and apogee of a Molniya orbit remains at the highest possible latitude (63.4oN in this case).
0 30 60 90 120 150 180
Inclination, deg –5
0 5 10 15 20
Apsidal rotation rate, deg/day
500 km 1000 km
1500 km 3000 km
Labels are apogee altitudes for elliptical Earth orbits with 325 km perigee altitude
i = 63.4 deg i = 116.6 deg
Figure 5.9 Apsidal rotation dω/dt vs. inclination for elliptical Earth orbits.
Ascending node, n J
Perigee, e I
K
Apogee
h i = 63.4°
Ω
Figure 5.10 Molniya orbit.
Example 5.4 Compute the secular drift rates for the longitude of the ascending node and argument of perigee for the satellite orbit in Example 5.1. Compare the analytical secular change in Ω and ω with the numerical simulation results presented in Figures 5.3d and 5.3e.
Recall that the orbital elements from Example 5.1 are Semimajor axis a = 8,059 km Eccentricity e = 0.15
Inclination i = 20 The mean motion of this orbit is
n= μ
a3= 0 000872664 rad/s
Using this mean motion with J2= 0.0010826267 and RE= 6,378.14 km in Eq. (5.70), the secular change in ascending node is
dΩ
dt = −3nJ2
2 1−e2 2 RE
a
2
cosi =– 8 7297 10−7 rad/s Or, in degrees per day we have dΩ/dt = −4 32 deg/day
Recall that Figure 5.3d (Example 5.1) shows the periodic and secular changes in the lon-gitude of the ascending node resulting from numerical integration (i.e., special perturba-tion methods). The dashed line in Figure 5.3d is the secular drift inΩ. The longitude of the ascending node decreases approximately–1.8 over 10 h (or 0.4167 days) so the approx-imate secular rate is (–1.8 )/(0.4167 days) = –4.32 deg/day. Hence, an approximate linear fit through the numerically simulated responseΩ(t) shows a good match with the analytically determined mean drift rate dΩ/dt.
Equation (5.73) gives us the secular change in argument of perigee dω
dt = 3nJ2
4 1−e2 2 RE
a
2
4−5sin2i = 1 5863 10−6 rad/s Or, in degrees per day dω/dt = 7 85 deg/day
Figure 5.3e shows that argument of perigee increases by about 3.25 over 10 h or 7.8 deg/day. Again, the analytical secular change exhibits a good match with the approx-imate linear fit through the simulation results.