Frozen orbit

In orbital mechanics, a frozen orbit is an orbit for an artificial satellite in which natural drifting due to the central body's shape has been minimized by careful selection of the orbital parameters. Typically, this is an orbit in which, over a long period of time, the satellite's altitude remains constant at the same point in each orbit.^[1] Changes in the inclination, position of the apsis of the orbit, and eccentricity have been minimized by choosing initial values so that their perturbations cancel out.,^[2] which results in a long-term stable orbit that minimizes the use of station-keeping propellant.

Background and motivation

For most spacecraft, changes to orbits are caused by the oblateness of the Earth, gravitational attraction from the sun and moon, solar radiation pressure, and air drag. These are called "perturbing forces". They must be counteracted by maneuvers to keep the spacecraft in the desired orbit. For a geostationary spacecraft, correction maneuvers on the order of 40–50 m/s per year are required to counteract the gravitational forces from the sun and moon which move the orbital plane away from the equatorial plane of the Earth.

For Sun-synchronous spacecraft, intentional shifting of the orbit plane (called "precession") can be used for the benefit of the mission. For these missions, a near-circular orbit with an altitude of 600–900 km is used. An appropriate inclination (97.8-99.0 degrees) is selected so that the precession of the orbital plane is equal to the rate of movement of the Earth around the Sun, about 1 degree per day.

As a result, the spacecraft will pass over points on the Earth that have the same time of day during every orbit. For instance, if the orbit is "square to the Sun", the vehicle will always pass over points at which it is 6 a.m. on the north-bound portion, and 6 p.m. on the south-bound portion (or vice versa). This is called a "Dawn-Dusk" orbit. Alternatively, if the Sun lies in the orbital plane, the vehicle will always pass over places where it is midday on the north-bound leg, and places where it is midnight on the south-bound leg (or vice versa). These are called "Noon-Midnight" orbits. Such orbits are desirable for many Earth observation missions such as weather, imagery, and mapping.

The perturbing force caused by the oblateness of the Earth will in general perturb not only the orbital plane but also the eccentricity vector of the orbit. There exists, however, an almost circular orbit for which there are no secular/long periodic perturbations of the eccentricity vector, only periodic perturbations with period equal to the orbital period. Such an orbit is then perfectly periodic (except for the orbital plane precession) and it is therefore called a "frozen orbit". Such an orbit is often the preferred choice for an Earth observation mission where repeated observations of the same area of the Earth should be made under as constant observation conditions as possible.

The Earth observation satellites ERS-1, ERS-2 and Envisat are operated in Sun-synchronous frozen orbits.

Lunar frozen orbits

Through a study of many lunar orbiting satellites, scientists have discovered that most low lunar orbits (LLO) are unstable.^[3] Four frozen lunar orbits have been identified at 27°, 50°, 76°, and 86° inclination. NASA expounded on this in 2006:

Lunar mascons make most low lunar orbits unstable ... As a satellite passes 50 or 60 miles overhead, the mascons pull it forward, back, left, right, or down, the exact direction and magnitude of the tugging depends on the satellite's trajectory. Absent any periodic boosts from onboard rockets to correct the orbit, most satellites released into low lunar orbits (under about 60 miles or 100 km) will eventually crash into the Moon. ... [There are] a number of 'frozen orbits' where a spacecraft can stay in a low lunar orbit indefinitely. They occur at four inclinations: 27°, 50°, 76°, and 86°"—the last one being nearly over the lunar poles. The orbit of the relatively long-lived Apollo 15 subsatellite PFS-1 had an inclination of 28°, which turned out to be close to the inclination of one of the frozen orbits—but less fortunate PFS-2 had an orbital inclination of only 11°.^[4]

Classical theory

The classical theory of frozen orbits is essentially based on the analytical perturbation analysis for artificial satellites of Dirk Brouwer made under contract with NASA and published in 1959.^[5]

This analysis can be carried out as follows:

In the article orbital perturbation analysis the secular perturbation of the orbital pole ${\displaystyle \Delta {\hat {z}}\,}$  from the ${\displaystyle J_{2}\,}$  term of the geopotential model is shown to be

${\displaystyle \Delta {\hat {z}}\ =\ -2\pi \ {\frac {J_{2}}{\mu \ p^{2}}}\ {\frac {3}{2}}\ \cos i\ \sin i\quad {\hat {g}}}$

(1)

which can be expressed in terms of orbital elements thus:

${\displaystyle \Delta i\ =\ 0}$

(2)

${\displaystyle \Delta \Omega \ =\ -2\pi \ {\frac {J_{2}}{\mu \ p^{2}}}\ {\frac {3}{2}}\ \cos i}$

(3)

Making a similar analysis for the ${\displaystyle J_{3}\,}$  term (corresponding to the fact that the earth is slightly pear shaped), one gets

${\displaystyle \Delta {\hat {z}}\ =\ 2\pi \ {\frac {J_{3}}{\mu \ p^{3}}}\ {\frac {3}{2}}\ \cos i\ \left(\ e_{h}\ \left(1-{\frac {15}{4}}\ \sin ^{2}i\right)\ {\hat {g}}\ -\ e_{g}\ \left(1-{\frac {5}{4}}\ \sin ^{2}i\right)\ {\hat {h}}\right)}$

(4)

which can be expressed in terms of orbital elements as

${\displaystyle \Delta i\ =\ -2\pi \ {\frac {J_{3}}{\mu \ p^{3}}}\ {\frac {3}{2}}\ \cos i\ e_{g}\ \left(1-{\frac {5}{4}}\ \sin ^{2}i\right)}$

(5)

${\displaystyle \Delta \Omega \ =\ 2\pi \ {\frac {J_{3}}{\mu \ p^{3}}}\ {\frac {3}{2}}\ {\frac {\cos i}{\sin i}}\ \ e_{h}\ \left(1-{\frac {15}{4}}\ \sin ^{2}i\right)}$

(6)

In the same article the secular perturbation of the components of the eccentricity vector caused by the ${\displaystyle J_{2}\,}$  is shown to be:

${\displaystyle {\begin{aligned}\left(\Delta e_{g},\Delta e_{h}\right)\ =\ &-2\pi \ {\frac {J_{2}}{\mu \ p^{2}}}\ {\frac {3}{2}}\left({\frac {3}{2}}\ \sin ^{2}i\ -\ 1\right)\ (-e_{h},e_{g})\ +\ 2\pi \ {\frac {J_{2}}{\mu \ p^{2}}}\ {\frac {3}{2}}\ \cos ^{2}i\left(-e_{h},e_{g}\right)\ =\\&-2\pi \ {\frac {J_{2}}{\mu \ p^{2}}}\ 3\left({\frac {5}{4}}\ \sin ^{2}i\ -\ 1\right)\ \left(-e_{h},e_{g}\right)\end{aligned}}}$

(7)

where:

• The first term is the in-plane perturbation of the eccentricity vector caused by the in-plane component of the perturbing force
• The second term is the effect of the new position of the ascending node in the new orbital plane, the orbital plane being perturbed by the out-of-plane force component

Making the analysis for the ${\displaystyle J_{3}\,}$  term one gets for the first term, i.e. for the perturbation of the eccentricity vector from the in-plane force component

${\displaystyle 2\pi \ {\frac {J_{3}}{\mu \ p^{3}}}\ {\frac {3}{2}}\ \sin i\ \left({\frac {5}{4}}\ \sin ^{2}i\ -\ 1\right)\left(\left(1-{e_{g}}^{2}+4\ {e_{h}}^{2}\right)\ {\hat {g}}\ -\ 5\ e_{g}\ e_{h}\ {\hat {h}}\right)}$

(8)

For inclinations in the range 97.8–99.0 deg, the ${\displaystyle \Delta \Omega \,}$  value given by (6) is much smaller than the value given by (3) and can be ignored. Similarly the quadratic terms of the eccentricity vector components in (8) can be ignored for almost circular orbits, i.e. (8) can be approximated with

${\displaystyle 2\pi \ {\frac {J_{3}}{\mu \ p^{3}}}\ {\frac {3}{2}}\ \sin i\ \left({\frac {5}{4}}\ \sin ^{2}i\ -\ 1\right)\ {\hat {g}}}$

(9)

Adding the ${\displaystyle J_{3}\,}$  contribution ${\displaystyle 2\pi \ {\frac {J_{3}}{\mu \ p^{3}}}\ {\frac {3}{2}}\ \sin i\ \left({\frac {5}{4}}\ \sin ^{2}i\ -\ 1\right)\ (1,\ 0)}$ 

to (7) one gets

${\displaystyle \left(\Delta e_{g},\Delta e_{h}\right)\ =\ -2\pi \ {\frac {J_{2}}{\mu \ p^{2}}}\ 3\left({\frac {5}{4}}\ \sin ^{2}i\ -\ 1\right)\ \left(-\left(e_{h}+{\frac {J_{3}\ \sin i}{J_{2}\ 2\ p}}\right),\ e_{g}\right)}$

(10)

Now the difference equation shows that the eccentricity vector will describe a circle centered at the point ${\displaystyle \left(\ 0,\ -{\frac {J_{3}\ \sin i}{J_{2}\ 2\ p}}\ \right)\,}$ ; the polar argument of the eccentricity vector increases with ${\displaystyle -2\pi \ {\frac {J_{2}}{\mu \ p^{2}}}\ 3\left({\frac {5}{4}}\ \sin ^{2}i\ -\ 1\right)\,}$  radians between consecutive orbits.

As

${\displaystyle \mu =398600.440{\text{ km}}^{3}/s^{2}\,}$ 

${\displaystyle J_{2}=1.7555\ 10^{10}{\text{ km}}^{5}/s^{2}\,}$ 

${\displaystyle J_{3}=-2.619\ 10^{11}{\text{ km}}^{6}/s^{2}\,}$ 

one gets for a polar orbit (${\displaystyle i=90^{\circ }\,}$ ) with ${\displaystyle p=7200{\text{ km}}\,}$  that the centre of the circle is at ${\displaystyle (0,\ 0.001036)\,}$  and the change of polar argument is 0.00400 radians per orbit.

The latter figure means that the eccentricity vector will have described a full circle in 1569 orbits. Selecting the initial mean eccentricity vector as ${\displaystyle (0,\ 0.001036)\,}$  the mean eccentricity vector will stay constant for successive orbits, i.e. the orbit is frozen because the secular perturbations of the ${\displaystyle J_{2}\,}$  term given by (7) and of the ${\displaystyle J_{3}\,}$  term given by (9) cancel out.

In terms of classical orbital elements, this means that a frozen orbit should have the following mean elements:

${\displaystyle e=-{\frac {J_{3}\ \sin i}{J_{2}\ 2\ p}}\,}$ 

${\displaystyle \omega =\ 90^{\circ }\,}$ 

Modern theory

The modern theory of frozen orbits is based on the algorithm given in a 1989 article by Mats Rosengren.^[6]

For this the analytical expression (7) is used to iteratively update the initial (mean) eccentricity vector to obtain that the (mean) eccentricity vector several orbits later computed by the precise numerical propagation takes precisely the same value. In this way the secular perturbation of the eccentricity vector caused by the ${\displaystyle J_{2}\,}$  term is used to counteract all secular perturbations, not only those (dominating) caused by the ${\displaystyle J_{3}\,}$  term. One such additional secular perturbation that in this way can be compensated for is the one caused by the solar radiation pressure, this perturbation is discussed in the article "Orbital perturbation analysis (spacecraft)".

Applying this algorithm for the case discussed above, i.e. a polar orbit (${\displaystyle i=90^{\circ }\,}$ ) with ${\displaystyle p=7200{\text{ km}}\,}$  ignoring all perturbing forces other than the ${\displaystyle J_{2}\,}$  and the ${\displaystyle J_{3}\,}$  forces for the numerical propagation one gets exactly the same optimal average eccentricity vector as with the "classical theory", i.e. ${\displaystyle (0,\ 0.001036)\,}$ .

When we also include the forces due to the higher zonal terms the optimal value changes to ${\displaystyle (0,\ 0.001285)\,}$ .

Assuming in addition a reasonable solar pressure (a "cross-sectional-area" of 0.05 m²/kg, the direction to the sun in the direction towards the ascending node) the optimal value for the average eccentricity vector becomes ${\displaystyle (0.000069,\ 0.001285)\,}$  which corresponds to :${\displaystyle \omega =\ 87^{\circ }\,}$ , i.e. the optimal value is not ${\displaystyle \omega =\ 90^{\circ }}$  anymore.

This algorithm is implemented in the orbit control software used for the Earth observation satellites ERS-1, ERS-2 and Envisat

Derivation of the closed form expressions for the J₃ perturbation

The main perturbing force to be counteracted in order to have a frozen orbit is the "${\displaystyle J_{3}\,}$  force", i.e. the gravitational force caused by an imperfect symmetry north–south of the Earth, and the "classical theory" is based on the closed form expression for this "${\displaystyle J_{3}\,}$  perturbation". With the "modern theory" this explicit closed form expression is not directly used but it is certainly still worthwhile to derive it.

The derivation of this expression can be done as follows:

The potential from a zonal term is rotational symmetric around the polar axis of the Earth and corresponding force is entirely in a longitudinal plane with one component ${\displaystyle F_{r}\ {\hat {r}}\,}$  in the radial direction and one component ${\displaystyle F_{\lambda }\ {\hat {\lambda }}\,}$  with the unit vector ${\displaystyle {\hat {\lambda }}\,}$  orthogonal to the radial direction towards north. These directions ${\displaystyle {\hat {r}}\,}$  and ${\displaystyle {\hat {\lambda }}\,}$  are illustrated in Figure 1.

 

Figure 1: The unit vectors ${\displaystyle {\hat {\phi }},\ {\hat {\lambda }},\ {\hat {r}}}$ 

In the article Geopotential model it is shown that these force components caused by the ${\displaystyle J_{3}\,}$  term are

${\displaystyle {\begin{aligned}&F_{r}=J_{3}\ {\frac {1}{r^{5}}}\ 2\ \sin \lambda \ \left(5\sin ^{2}\lambda \ -\ 3\right)\\&F_{\lambda }=-J_{3}\ {\frac {1}{r^{5}}}\ {\frac {3}{2}}\ \cos \lambda \ \left(5\ \sin ^{2}\lambda \ -1\right)\end{aligned}}}$

(11)

To be able to apply relations derived in the article Orbital perturbation analysis (spacecraft) the force component ${\displaystyle F_{\lambda }\ {\hat {\lambda }}\,}$  must be split into two orthogonal components ${\displaystyle F_{t}\ {\hat {t}}}$  and ${\displaystyle F_{z}\ {\hat {z}}}$  as illustrated in figure 2

 

Figure 2: The unit vector ${\displaystyle {\hat {t}}\,}$  orthogonal to ${\displaystyle {\hat {r}}\,}$  in the direction of motion and the orbital pole ${\displaystyle {\hat {z}}\,}$ . The force component ${\displaystyle F_{\lambda }}$  is marked as "F"

Let ${\displaystyle {\hat {a}},\ {\hat {b}},\ {\hat {n}}\,}$  make up a rectangular coordinate system with origin in the center of the Earth (in the center of the Reference ellipsoid) such that ${\displaystyle {\hat {n}}\,}$  points in the direction north and such that ${\displaystyle {\hat {a}},\ {\hat {b}}\,}$  are in the equatorial plane of the Earth with ${\displaystyle {\hat {a}}\,}$  pointing towards the ascending node, i.e. towards the blue point of Figure 2.

The components of the unit vectors

${\displaystyle {\hat {r}},\ {\hat {t}},\ {\hat {z}}\,}$ 

making up the local coordinate system (of which ${\displaystyle {\hat {t}},\ {\hat {z}},}$  are illustrated in figure 2), and expressing their relation with ${\displaystyle {\hat {a}},\ {\hat {b}},\ {\hat {n}}\,}$ , are as follows:

${\displaystyle r_{a}=\cos u\,}$ 

${\displaystyle r_{b}=\cos i\ \sin u\,}$ 

${\displaystyle r_{n}=\sin i\ \sin u\,}$ 

${\displaystyle t_{a}=-\sin u\,}$ 

${\displaystyle t_{b}=\cos i\ \cos u\,}$ 

${\displaystyle t_{n}=\sin i\ \cos u\,}$ 

${\displaystyle z_{a}=0\,}$ 

${\displaystyle z_{b}=-\sin i\,}$ 

${\displaystyle z_{n}=\cos i\,}$ 

where ${\displaystyle u\,}$  is the polar argument of ${\displaystyle {\hat {r}}\,}$  relative the orthogonal unit vectors ${\displaystyle {\hat {g}}={\hat {a}}\,}$  and ${\displaystyle {\hat {h}}=\cos i\ {\hat {b}}\ +\ \sin i\ {\hat {n}}\,}$  in the orbital plane

Firstly

${\displaystyle \sin \lambda =\ r_{n}\ =\ \sin i\ \sin u\,}$ 

where ${\displaystyle \lambda \,}$  is the angle between the equator plane and ${\displaystyle {\hat {r}}\,}$  (between the green points of figure 2) and from equation (12) of the article Geopotential model one therefore obtains

${\displaystyle F_{r}=J_{3}\ {\frac {1}{r^{5}}}\ 2\ \sin i\ \sin u\,\ \left(5\sin ^{2}i\ \sin ^{2}u\ -\ 3\right)}$

(12)

Secondly the projection of direction north, ${\displaystyle {\hat {n}}\,}$ , on the plane spanned by ${\displaystyle {\hat {t}},\ {\hat {z}},}$  is

${\displaystyle \sin i\ \cos u\ {\hat {t}}\ +\ \cos i\ {\hat {z}}\,}$ 

and this projection is

${\displaystyle \cos \lambda \ {\hat {\lambda }}\,}$ 

where ${\displaystyle {\hat {\lambda }}\,}$  is the unit vector ${\displaystyle {\hat {\lambda }}}$  orthogonal to the radial direction towards north illustrated in figure 1.

From equation (11) we see that

${\displaystyle F_{\lambda }\ {\hat {\lambda }}\ =-J_{3}\ {\frac {1}{r^{5}}}\ {\frac {3}{2}}\ \left(5\ \sin ^{2}\lambda \ -1\right)\ \cos \lambda \ {\hat {\lambda }}\ =\ -J_{3}\ {\frac {1}{r^{5}}}\ {\frac {3}{2}}\ \left(5\ \sin ^{2}\lambda \ -1\right)\ (\sin i\ \cos u\ {\hat {t}}\ +\ \cos i\ {\hat {z}})\,}$ 

and therefore:

${\displaystyle F_{t}=\ -J_{3}\ {\frac {1}{r^{5}}}\ {\frac {3}{2}}\ \left(5\ \sin ^{2}i\ \sin ^{2}u\ -1\right)\ \sin i\ \cos u}$

(13)

${\displaystyle F_{z}=\ -J_{3}\ {\frac {1}{r^{5}}}\ {\frac {3}{2}}\ \left(5\ \sin ^{2}i\ \sin ^{2}u\ -1\right)\ \cos i}$

(14)

In the article Orbital perturbation analysis (spacecraft) it is further shown that the secular perturbation of the orbital pole ${\displaystyle {\hat {z}}\,}$  is

${\displaystyle \Delta {\hat {z}}\ =\ {\frac {1}{\mu p}}\left[{\hat {g}}\int \limits _{0}^{2\pi }F_{z}r^{3}\cos u\ du+\ {\hat {h}}\int \limits _{0}^{2\pi }F_{z}r^{3}\sin u\ du\right]\quad \times \ {\hat {z}}}$

(15)

Introducing the expression for ${\displaystyle F_{z}\,}$  of (14) in (15) one gets

${\displaystyle {\begin{aligned}&\Delta {\hat {z}}\ =-{\frac {J_{3}}{\mu \ p^{3}}}\ {\frac {3}{2}}\ \cos i\ \cdot \\&\left[{\hat {g}}\int \limits _{0}^{2\pi }{\left({\frac {p}{r}}\right)}^{2}\left(5\ \sin ^{2}i\ \sin ^{2}u\ -1\right)\cos u\ du\ +{\hat {h}}\int \limits _{0}^{2\pi }{\left({\frac {p}{r}}\right)}^{2}\left(5\ \sin ^{2}i\ \sin ^{2}u\ -1\right)\sin u\ du\right]\quad \times \ {\hat {z}}\end{aligned}}}$

(16)

The fraction ${\displaystyle {\frac {p}{r}}\,}$  is

${\displaystyle {\frac {p}{r}}\ =\ 1+e\cdot \cos \theta \ =\ 1+e_{g}\cdot \cos u+e_{h}\cdot \sin u}$ 

where

${\displaystyle e_{g}=\ e\ \cos \omega }$ 

${\displaystyle e_{h}=\ e\ \sin \omega }$ 

are the components of the eccentricity vector in the ${\displaystyle {\hat {g}},\ {\hat {h}}\,}$  coordinate system.

As all integrals of type

${\displaystyle \int \limits _{0}^{2\pi }\cos ^{m}u\ \sin ^{n}u\ du\,}$ 

are zero if not both ${\displaystyle n\,}$  and ${\displaystyle m\,}$  are even, we see that

${\displaystyle {\begin{aligned}\int \limits _{0}^{2\pi }{\left({\frac {p}{r}}\right)}^{2}\left(5\ \sin ^{2}i\ \sin ^{2}u\ -1\right)\cos u\ du&=\ 2\ e_{g}\ \left(5\ \sin ^{2}i\ \int \limits _{0}^{2\pi }\sin ^{2}u\cos ^{2}u\ du\ -\int \limits _{0}^{2\pi }\cos ^{2}u\ du\right)\\&=\ 2\pi \ e_{g}\ ({\frac {5}{4}}\sin ^{2}i-1)\end{aligned}}}$

(17)

and

${\displaystyle {\begin{aligned}\int \limits _{0}^{2\pi }{\left({\frac {p}{r}}\right)}^{2}\left(5\ \sin ^{2}i\ \sin ^{2}u\ -1\right)\sin u\ du&=\ 2\ e_{h}\ \left(5\ \sin ^{2}i\ \int \limits _{0}^{2\pi }\sin ^{4}u\ du\ -\int \limits _{0}^{2\pi }\sin ^{2}u\ du\right)\\&=\ 2\pi \ e_{h}\ ({\frac {15}{4}}\sin ^{2}i-1)\end{aligned}}}$

(18)

It follows that

${\displaystyle {\begin{aligned}\Delta {\hat {z}}\ &=\ 2\pi \ {\frac {J_{3}}{\mu \ p^{3}}}\ {\frac {3}{2}}\ \cos i\ \left[e_{g}\ (1-{\frac {5}{4}}\sin ^{2}i)\ {\hat {g}}+\ e_{h}\ (1-{\frac {15}{4}}\sin ^{2}i)\ {\hat {h}}\right]\quad \times \ {\hat {z}}\\&=\ 2\pi \ {\frac {J_{3}}{\mu \ p^{3}}}\ {\frac {3}{2}}\ \cos i\ \left[\ e_{h}\ (1-{\frac {15}{4}}\sin ^{2}i)\ {\hat {g}}\ -e_{g}\ (1-{\frac {5}{4}}\sin ^{2}i)\ {\hat {h}}\right]\end{aligned}}}$

(19)

where

${\displaystyle {\hat {g}}\,}$  and ${\displaystyle {\hat {h}}\,}$  are the base vectors of the rectangular coordinate system in the plane of the reference Kepler orbit with ${\displaystyle {\hat {g}}\,}$  in the equatorial plane towards the ascending node and ${\displaystyle u\,}$  is the polar argument relative this equatorial coordinate system

${\displaystyle f_{z}\,}$  is the force component (per unit mass) in the direction of the orbit pole ${\displaystyle {\hat {z}}\,}$ 

In the article Orbital perturbation analysis (spacecraft) it is shown that the secular perturbation of the eccentricity vector is

${\displaystyle \Delta {\bar {e}}\ ={\frac {1}{\mu }}\ \int \limits _{0}^{2\pi }\left(-{\hat {t}}\ f_{r}\ +\ \left(2\ {\hat {r}}-{\frac {V_{r}}{V_{t}}}\ {\hat {t}}\right)\ f_{t}\right)\ r^{2}\ du}$

(20)

where

• ${\displaystyle {\hat {r}},{\hat {t}}\,}$  is the usual local coordinate system with unit vector ${\displaystyle {\hat {r}}\,}$  directed away from the Earth
• ${\displaystyle V_{r}={\sqrt {\frac {\mu }{p}}}\cdot e\cdot \sin \theta }$  - the velocity component in direction ${\displaystyle {\hat {r}}\,}$ 
• ${\displaystyle V_{t}={\sqrt {\frac {\mu }{p}}}\cdot (1+e\cdot \cos \theta )}$  - the velocity component in direction ${\displaystyle {\hat {t}}\,}$ 

Introducing the expression for ${\displaystyle F_{r},\ F_{t}\,}$  of (12) and (13) in (20) one gets

${\displaystyle {\begin{aligned}&\Delta {\bar {e}}\ ={\frac {J_{3}}{\mu \ p^{3}}}\ \sin i\ \cdot \\&\int \limits _{0}^{2\pi }\left(-{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{3}\ 2\ \sin u\,\ \left(5\sin ^{2}i\ \sin ^{2}u\ -\ 3\right)\ -\ \left(2\ {\hat {r}}-{\frac {V_{r}}{V_{t}}}\ {\hat {t}}\right)\ {\left({\frac {p}{r}}\right)}^{3}\ {\frac {3}{2}}\ \left(5\ \sin ^{2}i\ \sin ^{2}u\ -1\right)\ \cos u\right)du\end{aligned}}}$

(21)

Using that

${\displaystyle {\frac {V_{r}}{V_{t}}}={\frac {e_{g}\cdot \sin u\ -\ e_{h}\cdot \cos u}{\frac {p}{r}}}}$ 

the integral above can be split in 8 terms:

${\displaystyle {\begin{aligned}&\int \limits _{0}^{2\pi }\left(-{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{3}\ 2\ \sin u\,\ \left(5\sin ^{2}i\ \sin ^{2}u\ -\ 3\right)\ -\ \left(2\ {\hat {r}}-{\frac {V_{r}}{V_{t}}}\ {\hat {t}}\right)\ {\left({\frac {p}{r}}\right)}^{3}\ {\frac {3}{2}}\ \left(5\ \sin ^{2}i\ \sin ^{2}u\ -1\right)\ \cos u\right)\ du\ =\\&-10\sin ^{2}i\ \int \limits _{0}^{2\pi }{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{3}\ \sin ^{3}u\ du\\&+6\ \int \limits _{0}^{2\pi }{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{3}\ \sin u\ du\\&-15\ \sin ^{2}i\int \limits _{0}^{2\pi }{\hat {r}}\ {\left({\frac {p}{r}}\right)}^{3}\ \sin ^{2}u\ \cos u\ du\\&+3\ \int \limits _{0}^{2\pi }{\hat {r}}\ {\left({\frac {p}{r}}\right)}^{3}\ \cos u\ du\\&+{\frac {15}{2}}\sin ^{2}i\ e_{g}\int \limits _{0}^{2\pi }{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{2}\ \ \ \sin ^{3}u\ \cos u\ du\\&-{\frac {15}{2}}\sin ^{2}i\ e_{h}\ \int \limits _{0}^{2\pi }{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{2}\ \ \ \sin ^{2}u\ \cos ^{2}u\ du\\&-{\frac {3}{2}}\ e_{g}\ \int \limits _{0}^{2\pi }\ {\hat {t}}\ {\left({\frac {p}{r}}\right)}^{2}\ \ \sin u\ \cos u\ du\\&+{\frac {3}{2}}\ e_{h}\ \int \limits _{0}^{2\pi }{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{2}\ \ \cos ^{2}u\ du\end{aligned}}}$

(22)

Given that

${\displaystyle {\hat {r}}=\cos u\ {\hat {g}}\ +\ \sin u\ {\hat {h}}}$ 

${\displaystyle {\hat {t}}=-\sin u\ {\hat {g}}\ +\ \cos u\ {\hat {h}}}$ 

we obtain

${\displaystyle {\frac {p}{r}}\ =\ 1+e\cdot \cos \theta \ =\ 1+e_{g}\cdot \cos u+e_{h}\cdot \sin u}$ 

and that all integrals of type

${\displaystyle \int \limits _{0}^{2\pi }\cos ^{m}u\ \sin ^{n}u\ du\,}$ 

are zero if not both ${\displaystyle n\,}$  and ${\displaystyle m\,}$  are even:

Term 1

${\displaystyle {\begin{aligned}&\int \limits _{0}^{2\pi }{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{3}\ \sin ^{3}u\ du\ =-{\hat {g}}\ \int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{3}\ \sin ^{4}u\ du\ +{\hat {h}}\int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{3}\ \sin ^{3}u\ \cos u\ du\ =\\&-{\hat {g}}\ \left(\int \limits _{0}^{2\pi }\ \sin ^{4}u\ du\ +\ 3\ {e_{g}}^{2}\ \int \limits _{0}^{2\pi }\ \cos ^{2}u\ \sin ^{4}u\ du\ \ +\ 3\ {e_{h}}^{2}\ \int \limits _{0}^{2\pi }\ \sin ^{6}u\ du\ \right)\\&+{\hat {h}}\ 6\ e_{g}\ e_{h}\ \int \limits _{0}^{2\pi }\ \cos ^{2}u\ \sin ^{4}u\ du=\\&-{\hat {g}}\ \left(2\pi \left({\frac {3}{8}}\ +\ {\frac {3}{16}}\ {e_{g}}^{2}\ +\ {\frac {15}{16}}\ {e_{h}}^{2}\right)\right)+{\hat {h}}\ \left(2\pi \left({\frac {3}{8}}\ e_{g}\ e_{h}\right)\right)\end{aligned}}}$

(23)

Term 2

${\displaystyle {\begin{aligned}&\int \limits _{0}^{2\pi }{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{3}\ \sin u\ du\ =-{\hat {g}}\ \int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{3}\ \sin ^{2}u\ du\ +{\hat {h}}\int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{3}\ \sin u\ \cos u\ du\ =\\&-{\hat {g}}\ \left(\int \limits _{0}^{2\pi }\ \sin ^{2}u\ du\ +\ 3\ {e_{g}}^{2}\ \int \limits _{0}^{2\pi }\ \cos ^{2}u\ \sin ^{2}u\ du\ \ +\ 3\ {e_{h}}^{2}\ \int \limits _{0}^{2\pi }\ \sin ^{6}u\ du\ \right)\\&+{\hat {h}}\ 6\ e_{g}\ e_{h}\ \int \limits _{0}^{2\pi }\ \cos ^{2}u\ \sin ^{2}u\ du=\\&-{\hat {g}}\ \left(2\pi \left({\frac {1}{2}}\ +\ {\frac {3}{8}}\ {e_{g}}^{2}\ +\ {\frac {9}{8}}\ {e_{h}}^{2}\right)\right)+{\hat {h}}\ \left(2\pi \left({\frac {3}{4}}\ e_{g}\ e_{h}\right)\right)\end{aligned}}}$

(24)

Term 3

${\displaystyle {\begin{aligned}&\int \limits _{0}^{2\pi }{\hat {r}}\ {\left({\frac {p}{r}}\right)}^{3}\ \sin ^{2}u\ \cos u\ du\ ={\hat {g}}\ \int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{3}\ \sin ^{2}u\cos ^{2}u\ du\ +{\hat {h}}\int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{3}\ \sin ^{3}u\ \cos u\ du\ =\\&{\hat {g}}\ \left(\int \limits _{0}^{2\pi }\ \sin ^{2}u\cos ^{2}u\ du\ +\ 3\ {e_{g}}^{2}\ \int \limits _{0}^{2\pi }\ \cos ^{4}u\ \sin ^{2}u\ du\ \ +\ 3\ {e_{h}}^{2}\ \int \limits _{0}^{2\pi }\ \sin ^{4}u\ \cos ^{2}u\ du\ \right)\\&+{\hat {h}}\ 6\ e_{g}\ e_{h}\ \int \limits _{0}^{2\pi }\ \cos ^{2}u\ \sin ^{4}u\ du=\\&{\hat {g}}\ \left(2\pi \left({\frac {1}{8}}\ +\ {\frac {3}{16}}\ {e_{g}}^{2}\ +\ {\frac {3}{16}}\ {e_{h}}^{2}\right)\right)+{\hat {h}}\ \left(2\pi \left({\frac {3}{8}}\ e_{g}\ e_{h}\right)\right)\end{aligned}}}$

(25)

Term 4

${\displaystyle {\begin{aligned}&\int \limits _{0}^{2\pi }{\hat {r}}\ {\left({\frac {p}{r}}\right)}^{3}\ \cos u\ du\ ={\hat {g}}\ \int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{3}\ \cos ^{2}u\ du\ +{\hat {h}}\int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{3}\ \sin u\ \cos u\ du\ =\\&{\hat {g}}\ \left(\int \limits _{0}^{2\pi }\cos ^{2}u\ du\ +\ 3\ {e_{g}}^{2}\ \int \limits _{0}^{2\pi }\ \cos ^{4}u\ du\ \ +\ 3\ {e_{h}}^{2}\ \int \limits _{0}^{2\pi }\ \sin ^{2}u\ \cos ^{2}u\ du\ \right)\\&+{\hat {h}}\ 6\ e_{g}\ e_{h}\ \int \limits _{0}^{2\pi }\ \cos ^{2}u\ \sin ^{2}u\ du=\\&{\hat {g}}\ \left(2\pi \left({\frac {1}{2}}\ +\ {\frac {9}{8}}\ {e_{g}}^{2}\ +\ {\frac {3}{8}}\ {e_{h}}^{2}\right)\right)+{\hat {h}}\ \left(2\pi \left({\frac {3}{4}}\ e_{g}\ e_{h}\right)\right)\end{aligned}}}$

(26)

Term 5

${\displaystyle {\begin{aligned}&\int \limits _{0}^{2\pi }{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{2}\ \sin ^{3}u\ \cos u\ du\ =-{\hat {g}}\ \int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{2}\ \sin ^{4}u\ \cos u\ du\ +{\hat {h}}\int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{2}\ \sin ^{3}u\ \cos ^{2}u\ du\ =\\&-{\hat {g}}\ 2\ e_{g}\ \int \limits _{0}^{2\pi }\ \sin ^{4}u\ \cos ^{2}u\ du+{\hat {h}}\ 2\ e_{h}\ \int \limits _{0}^{2\pi }\ \sin ^{4}u\ \cos ^{2}u\ du=-{\hat {g}}\ \left(2\pi {\frac {1}{8}}\ e_{g}\right)+{\hat {h}}\ \left(2\pi {\frac {1}{8}}\ e_{h}\right)\end{aligned}}}$

(27)

Term 6

${\displaystyle {\begin{aligned}&\int \limits _{0}^{2\pi }{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{2}\ \sin ^{2}u\ \cos ^{2}u\ du\ =-{\hat {g}}\ \int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{2}\ \sin ^{3}u\ \cos ^{2}u\ du\ +{\hat {h}}\int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{2}\ \sin ^{2}u\ \cos ^{3}u\ du\ =\\&-{\hat {g}}\ 2\ e_{h}\ \int \limits _{0}^{2\pi }\ \sin ^{4}u\ \cos ^{2}u\ du+{\hat {h}}\ 2\ e_{g}\ \int \limits _{0}^{2\pi }\ \sin ^{2}u\ \cos ^{4}u\ du=-{\hat {g}}\ \left(2\pi {\frac {1}{8}}\ e_{h}\right)+{\hat {h}}\ \left(2\pi {\frac {1}{8}}\ e_{g}\right)\end{aligned}}}$

(28)

Term 7

${\displaystyle {\begin{aligned}&\int \limits _{0}^{2\pi }{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{2}\ \sin u\ \cos u\ du\ =-{\hat {g}}\ \int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{2}\ \sin ^{2}u\ \cos u\ du\ +{\hat {h}}\int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{2}\ \sin u\ \cos ^{2}u\ du\ =\\&-{\hat {g}}\ 2\ e_{g}\ \int \limits _{0}^{2\pi }\ \sin ^{2}u\ \cos ^{2}u\ du+{\hat {h}}\ 2\ e_{h}\ \int \limits _{0}^{2\pi }\ \sin ^{2}u\ \cos ^{2}u\ du=-{\hat {g}}\ \left(2\pi {\frac {1}{4}}\ e_{g}\right)+{\hat {h}}\ \left(2\pi {\frac {1}{4}}\ e_{h}\right)\end{aligned}}}$

(29)

Term 8

${\displaystyle {\begin{aligned}&\int \limits _{0}^{2\pi }{\hat {t}}\ {\left({\frac {p}{r}}\right)}^{2}\ \cos ^{2}u\ du\ =-{\hat {g}}\ \int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{2}\ \sin u\ \cos ^{2}u\ du\ +{\hat {h}}\int \limits _{0}^{2\pi }\ {\left({\frac {p}{r}}\right)}^{2}\ \cos ^{3}u\ du\ =\\&-{\hat {g}}\ 2\ e_{h}\ \int \limits _{0}^{2\pi }\ \sin ^{2}u\ \cos ^{2}u\ du+{\hat {h}}\ 2\ e_{g}\ \int \limits _{0}^{2\pi }\ \cos ^{4}u\ du=-{\hat {g}}\ \left(2\pi {\frac {1}{4}}\ e_{h}\right)+{\hat {h}}\ \left(2\pi {\frac {3}{4}}\ e_{g}\right)\end{aligned}}}$

(30)

As

${\displaystyle {\begin{aligned}&-10\sin ^{2}i\ \left(-{\hat {g}}\ \left({\frac {3}{8}}\ +\ {\frac {3}{16}}\ {e_{g}}^{2}\ +\ {\frac {15}{16}}\ {e_{h}}^{2}\right)+{\hat {h}}\ \left({\frac {3}{8}}\ e_{g}\ e_{h}\right)\right)\\&+6\ \left(-{\hat {g}}\ \left({\frac {1}{2}}\ +\ {\frac {3}{8}}\ {e_{g}}^{2}\ +\ {\frac {9}{8}}\ {e_{h}}^{2}\right)+{\hat {h}}\ \left({\frac {3}{4}}\ e_{g}\ e_{h}\right)\right)\\&-15\sin ^{2}i\ \left({\hat {g}}\ \left({\frac {1}{8}}\ +\ {\frac {3}{16}}\ {e_{g}}^{2}\ +\ {\frac {3}{16}}\ {e_{h}}^{2}\right)+{\hat {h}}\ \left({\frac {3}{8}}\ e_{g}\ e_{h}\right)\right)\\&+3\left({\hat {g}}\ \left({\frac {1}{2}}\ +\ {\frac {9}{8}}\ {e_{g}}^{2}\ +\ {\frac {3}{8}}\ {e_{h}}^{2}\right)+{\hat {h}}\ \left({\frac {3}{4}}\ e_{g}\ e_{h}\right)\right)\\&+{\frac {15}{2}}\sin ^{2}i\ e_{g}\ \left(-{\hat {g}}\ \left({\frac {1}{8}}\ e_{g}\right)+{\hat {h}}\ \left({\frac {1}{8}}\ e_{h}\right)\right)\\&-{\frac {15}{2}}\sin ^{2}i\ e_{h}\ \left(-{\hat {g}}\ \left({\frac {1}{8}}\ e_{h}\right)+{\hat {h}}\ \left({\frac {1}{8}}\ e_{g}\right)\right)\\&-{\frac {3}{2}}\ e_{g}\ \left(-{\hat {g}}\ \left({\frac {1}{4}}\ e_{g}\right)+{\hat {h}}\ \left({\frac {1}{4}}\ e_{h}\right)\right)\\&+{\frac {3}{2}}\ e_{h}\ \left(-{\hat {g}}\ \left({\frac {1}{4}}\ e_{h}\right)+{\hat {h}}\ \left({\frac {3}{4}}\ e_{g}\right)\right)=\\&{\frac {3}{2}}\ \left({\frac {5}{4}}\ \sin ^{2}i\ -\ 1\right)\left((1-{e_{g}}^{2}\ +\ 4\ {e_{h}}^{2}){\hat {g}}\ -\ 5\ e_{g}\ e_{h}\ {\hat {h}}\right)\end{aligned}}}$

(31)

It follows that

December 20, 2023 Latest articles January 01, 1970 155th Motor Rifle Division January 01, 1970 1567 in England January 01, 1970 1564 in England January 01, 1970 1568 in England January 01, 1970 156th Infantry Regiment (United States) January 01, 1970 1576 Plot January 01, 1970 1578 in England January 01, 1970 1579 in England January 01, 1970 1581 in architecture January 01, 1970 1581 in music Most Read January 01, 1970 MedWatch January 01, 1970 Meda River January 01, 1970 Medaglia d'Argento al Valore Militare January 01, 1970 Medal for the Liberation of Korea January 01, 1970 Medalist Golf Club frozen, orbit, orbital, mechanics, frozen, orbit, orbit, artificial, satellite, which, natural, drifting, central, body, shape, been, minimized, careful, selection, orbital, parameters, typically, this, orbit, which, over, long, period, time, satellite, altitu. In orbital mechanics a frozen orbit is an orbit for an artificial satellite in which natural drifting due to the central body s shape has been minimized by careful selection of the orbital parameters Typically this is an orbit in which over a long period of time the satellite s altitude remains constant at the same point in each orbit 1 Changes in the inclination position of the apsis of the orbit and eccentricity have been minimized by choosing initial values so that their perturbations cancel out 2 which results in a long term stable orbit that minimizes the use of station keeping propellant Contents 1 Background and motivation 1 1 Lunar frozen orbits 2 Classical theory 3 Modern theory 4 Derivation of the closed form expressions for the J3 perturbation 5 References 6 Further readingBackground and motivation editFor most spacecraft changes to orbits are caused by the oblateness of the Earth gravitational attraction from the sun and moon solar radiation pressure and air drag These are called perturbing forces They must be counteracted by maneuvers to keep the spacecraft in the desired orbit For a geostationary spacecraft correction maneuvers on the order of 40 50 m s per year are required to counteract the gravitational forces from the sun and moon which move the orbital plane away from the equatorial plane of the Earth For Sun synchronous spacecraft intentional shifting of the orbit plane called precession can be used for the benefit of the mission For these missions a near circular orbit with an altitude of 600 900 km is used An appropriate inclination 97 8 99 0 degrees is selected so that the precession of the orbital plane is equal to the rate of movement of the Earth around the Sun about 1 degree per day As a result the spacecraft will pass over points on the Earth that have the same time of day during every orbit For instance if the orbit is square to the Sun the vehicle will always pass over points at which it is 6 a m on the north bound portion and 6 p m on the south bound portion or vice versa This is called a Dawn Dusk orbit Alternatively if the Sun lies in the orbital plane the vehicle will always pass over places where it is midday on the north bound leg and places where it is midnight on the south bound leg or vice versa These are called Noon Midnight orbits Such orbits are desirable for many Earth observation missions such as weather imagery and mapping The perturbing force caused by the oblateness of the Earth will in general perturb not only the orbital plane but also the eccentricity vector of the orbit There exists however an almost circular orbit for which there are no secular long periodic perturbations of the eccentricity vector only periodic perturbations with period equal to the orbital period Such an orbit is then perfectly periodic except for the orbital plane precession and it is therefore called a frozen orbit Such an orbit is often the preferred choice for an Earth observation mission where repeated observations of the same area of the Earth should be made under as constant observation conditions as possible The Earth observation satellites ERS 1 ERS 2 and Envisat are operated in Sun synchronous frozen orbits Lunar frozen orbits edit Through a study of many lunar orbiting satellites scientists have discovered that most low lunar orbits LLO are unstable 3 Four frozen lunar orbits have been identified at 27 50 76 and 86 inclination NASA expounded on this in 2006 Lunar mascons make most low lunar orbits unstable As a satellite passes 50 or 60 miles overhead the mascons pull it forward back left right or down the exact direction and magnitude of the tugging depends on the satellite s trajectory Absent any periodic boosts from onboard rockets to correct the orbit most satellites released into low lunar orbits under about 60 miles or 100 km will eventually crash into the Moon There are a number of frozen orbits where a spacecraft can stay in a low lunar orbit indefinitely They occur at four inclinations 27 50 76 and 86 the last one being nearly over the lunar poles The orbit of the relatively long lived Apollo 15 subsatellite PFS 1 had an inclination of 28 which turned out to be close to the inclination of one of the frozen orbits but less fortunate PFS 2 had an orbital inclination of only 11 4 Classical theory editThis section may require cleanup to meet Wikipedia s quality standards The specific problem is Wikipedia Is Not A Textbook The provided derivation is lacking context for why it would be useful to readers Please help improve this section if you can September 2013 Learn how and when to remove this template message The classical theory of frozen orbits is essentially based on the analytical perturbation analysis for artificial satellites of Dirk Brouwer made under contract with NASA and published in 1959 5 This analysis can be carried out as follows In the article orbital perturbation analysis the secular perturbation of the orbital pole D z displaystyle Delta hat z nbsp from the J 2 displaystyle J 2 nbsp term of the geopotential model is shown to be D z 2 p J 2 m p 2 3 2 cos i sin i g displaystyle Delta hat z 2 pi frac J 2 mu p 2 frac 3 2 cos i sin i quad hat g nbsp 1 dd which can be expressed in terms of orbital elements thus D i 0 displaystyle Delta i 0 nbsp 2 dd D W 2 p J 2 m p 2 3 2 cos i displaystyle Delta Omega 2 pi frac J 2 mu p 2 frac 3 2 cos i nbsp 3 dd Making a similar analysis for the J 3 displaystyle J 3 nbsp term corresponding to the fact that the earth is slightly pear shaped one gets D z 2 p J 3 m p 3 3 2 cos i e h 1 15 4 sin 2 i g e g 1 5 4 sin 2 i h displaystyle Delta hat z 2 pi frac J 3 mu p 3 frac 3 2 cos i left e h left 1 frac 15 4 sin 2 i right hat g e g left 1 frac 5 4 sin 2 i right hat h right nbsp 4 dd which can be expressed in terms of orbital elements as D i 2 p J 3 m p 3 3 2 cos i e g 1 5 4 sin 2 i displaystyle Delta i 2 pi frac J 3 mu p 3 frac 3 2 cos i e g left 1 frac 5 4 sin 2 i right nbsp 5 dd D W 2 p J 3 m p 3 3 2 cos i sin i e h 1 15 4 sin 2 i displaystyle Delta Omega 2 pi frac J 3 mu p 3 frac 3 2 frac cos i sin i e h left 1 frac 15 4 sin 2 i right nbsp 6 dd In the same article the secular perturbation of the components of the eccentricity vector caused by the J 2 displaystyle J 2 nbsp is shown to be D e g D e h 2 p J 2 m p 2 3 2 3 2 sin 2 i 1 e h e g 2 p J 2 m p 2 3 2 cos 2 i e h e g 2 p J 2 m p 2 3 5 4 sin 2 i 1 e h e g displaystyle begin aligned left Delta e g Delta e h right amp 2 pi frac J 2 mu p 2 frac 3 2 left frac 3 2 sin 2 i 1 right e h e g 2 pi frac J 2 mu p 2 frac 3 2 cos 2 i left e h e g right amp 2 pi frac J 2 mu p 2 3 left frac 5 4 sin 2 i 1 right left e h e g right end aligned nbsp 7 dd where The first term is the in plane perturbation of the eccentricity vector caused by the in plane component of the perturbing force The second term is the effect of the new position of the ascending node in the new orbital plane the orbital plane being perturbed by the out of plane force componentMaking the analysis for the J 3 displaystyle J 3 nbsp term one gets for the first term i e for the perturbation of the eccentricity vector from the in plane force component 2 p J 3 m p 3 3 2 sin i 5 4 sin 2 i 1 1 e g 2 4 e h 2 g 5 e g e h h displaystyle 2 pi frac J 3 mu p 3 frac 3 2 sin i left frac 5 4 sin 2 i 1 right left left 1 e g 2 4 e h 2 right hat g 5 e g e h hat h right nbsp 8 dd For inclinations in the range 97 8 99 0 deg the D W displaystyle Delta Omega nbsp value given by 6 is much smaller than the value given by 3 and can be ignored Similarly the quadratic terms of the eccentricity vector components in 8 can be ignored for almost circular orbits i e 8 can be approximated with 2 p J 3 m p 3 3 2 sin i 5 4 sin 2 i 1 g displaystyle 2 pi frac J 3 mu p 3 frac 3 2 sin i left frac 5 4 sin 2 i 1 right hat g nbsp 9 dd Adding the J 3 displaystyle J 3 nbsp contribution 2 p J 3 m p 3 3 2 sin i 5 4 sin 2 i 1 1 0 displaystyle 2 pi frac J 3 mu p 3 frac 3 2 sin i left frac 5 4 sin 2 i 1 right 1 0 nbsp to 7 one gets D e g D e h 2 p J 2 m p 2 3 5 4 sin 2 i 1 e h J 3 sin i J 2 2 p e g displaystyle left Delta e g Delta e h right 2 pi frac J 2 mu p 2 3 left frac 5 4 sin 2 i 1 right left left e h frac J 3 sin i J 2 2 p right e g right nbsp 10 dd Now the difference equation shows that the eccentricity vector will describe a circle centered at the point 0 J 3 sin i J 2 2 p displaystyle left 0 frac J 3 sin i J 2 2 p right nbsp the polar argument of the eccentricity vector increases with 2 p J 2 m p 2 3 5 4 sin 2 i 1 displaystyle 2 pi frac J 2 mu p 2 3 left frac 5 4 sin 2 i 1 right nbsp radians between consecutive orbits As m 398600 440 km 3 s 2 displaystyle mu 398600 440 text km 3 s 2 nbsp J 2 1 7555 10 10 km 5 s 2 displaystyle J 2 1 7555 10 10 text km 5 s 2 nbsp J 3 2 619 10 11 km 6 s 2 displaystyle J 3 2 619 10 11 text km 6 s 2 nbsp one gets for a polar orbit i 90 displaystyle i 90 circ nbsp with p 7200 km displaystyle p 7200 text km nbsp that the centre of the circle is at 0 0 001036 displaystyle 0 0 001036 nbsp and the change of polar argument is 0 00400 radians per orbit The latter figure means that the eccentricity vector will have described a full circle in 1569 orbits Selecting the initial mean eccentricity vector as 0 0 001036 displaystyle 0 0 001036 nbsp the mean eccentricity vector will stay constant for successive orbits i e the orbit is frozen because the secular perturbations of the J 2 displaystyle J 2 nbsp term given by 7 and of the J 3 displaystyle J 3 nbsp term given by 9 cancel out In terms of classical orbital elements this means that a frozen orbit should have the following mean elements e J 3 sin i J 2 2 p displaystyle e frac J 3 sin i J 2 2 p nbsp w 90 displaystyle omega 90 circ nbsp Modern theory editThe modern theory of frozen orbits is based on the algorithm given in a 1989 article by Mats Rosengren 6 For this the analytical expression 7 is used to iteratively update the initial mean eccentricity vector to obtain that the mean eccentricity vector several orbits later computed by the precise numerical propagation takes precisely the same value In this way the secular perturbation of the eccentricity vector caused by the J 2 displaystyle J 2 nbsp term is used to counteract all secular perturbations not only those dominating caused by the J 3 displaystyle J 3 nbsp term One such additional secular perturbation that in this way can be compensated for is the one caused by the solar radiation pressure this perturbation is discussed in the article Orbital perturbation analysis spacecraft Applying this algorithm for the case discussed above i e a polar orbit i 90 displaystyle i 90 circ nbsp with p 7200 km displaystyle p 7200 text km nbsp ignoring all perturbing forces other than the J 2 displaystyle J 2 nbsp and the J 3 displaystyle J 3 nbsp forces for the numerical propagation one gets exactly the same optimal average eccentricity vector as with the classical theory i e 0 0 001036 displaystyle 0 0 001036 nbsp When we also include the forces due to the higher zonal terms the optimal value changes to 0 0 001285 displaystyle 0 0 001285 nbsp Assuming in addition a reasonable solar pressure a cross sectional area of 0 05 m2 kg the direction to the sun in the direction towards the ascending node the optimal value for the average eccentricity vector becomes 0 000069 0 001285 displaystyle 0 000069 0 001285 nbsp which corresponds to w 87 displaystyle omega 87 circ nbsp i e the optimal value is not w 90 displaystyle omega 90 circ nbsp anymore This algorithm is implemented in the orbit control software used for the Earth observation satellites ERS 1 ERS 2 and EnvisatDerivation of the closed form expressions for the J3 perturbation editThis article may be too technical for most readers to understand Please help improve it to make it understandable to non experts without removing the technical details September 2013 Learn how and when to remove this template message The main perturbing force to be counteracted in order to have a frozen orbit is the J 3 displaystyle J 3 nbsp force i e the gravitational force caused by an imperfect symmetry north south of the Earth and the classical theory is based on the closed form expression for this J 3 displaystyle J 3 nbsp perturbation With the modern theory this explicit closed form expression is not directly used but it is certainly still worthwhile to derive it The derivation of this expression can be done as follows The potential from a zonal term is rotational symmetric around the polar axis of the Earth and corresponding force is entirely in a longitudinal plane with one component F r r displaystyle F r hat r nbsp in the radial direction and one component F l l displaystyle F lambda hat lambda nbsp with the unit vector l displaystyle hat lambda nbsp orthogonal to the radial direction towards north These directions r displaystyle hat r nbsp and l displaystyle hat lambda nbsp are illustrated in Figure 1 nbsp Figure 1 The unit vectors ϕ l r displaystyle hat phi hat lambda hat r nbsp In the article Geopotential model it is shown that these force components caused by the J 3 displaystyle J 3 nbsp term are F r J 3 1 r 5 2 sin l 5 sin 2 l 3 F l J 3 1 r 5 3 2 cos l 5 sin 2 l 1 displaystyle begin aligned amp F r J 3 frac 1 r 5 2 sin lambda left 5 sin 2 lambda 3 right amp F lambda J 3 frac 1 r 5 frac 3 2 cos lambda left 5 sin 2 lambda 1 right end aligned nbsp 11 dd To be able to apply relations derived in the article Orbital perturbation analysis spacecraft the force component F l l displaystyle F lambda hat lambda nbsp must be split into two orthogonal components F t t displaystyle F t hat t nbsp and F z z displaystyle F z hat z nbsp as illustrated in figure 2 nbsp Figure 2 The unit vector t displaystyle hat t nbsp orthogonal to r displaystyle hat r nbsp in the direction of motion and the orbital pole z displaystyle hat z nbsp The force component F l displaystyle F lambda nbsp is marked as F Let a b n displaystyle hat a hat b hat n nbsp make up a rectangular coordinate system with origin in the center of the Earth in the center of the Reference ellipsoid such that n displaystyle hat n nbsp points in the direction north and such that a b displaystyle hat a hat b nbsp are in the equatorial plane of the Earth with a displaystyle hat a nbsp pointing towards the ascending node i e towards the blue point of Figure 2 The components of the unit vectors r t z displaystyle hat r hat t hat z nbsp making up the local coordinate system of which t z displaystyle hat t hat z nbsp are illustrated in figure 2 and expressing their relation with a b n displaystyle hat a hat b hat n nbsp are as follows r a cos u displaystyle r a cos u nbsp r b cos i sin u displaystyle r b cos i sin u nbsp r n sin i sin u displaystyle r n sin i sin u nbsp t a sin u displaystyle t a sin u nbsp t b cos i cos u displaystyle t b cos i cos u nbsp t n sin i cos u displaystyle t n sin i cos u nbsp z a 0 displaystyle z a 0 nbsp z b sin i displaystyle z b sin i nbsp z n cos i displaystyle z n cos i nbsp where u displaystyle u nbsp is the polar argument of r displaystyle hat r nbsp relative the orthogonal unit vectors g a displaystyle hat g hat a nbsp and h cos i b sin i n displaystyle hat h cos i hat b sin i hat n nbsp in the orbital planeFirstly sin l r n sin i sin u displaystyle sin lambda r n sin i sin u nbsp where l displaystyle lambda nbsp is the angle between the equator plane and r displaystyle hat r nbsp between the green points of figure 2 and from equation 12 of the article Geopotential model one therefore obtains F r J 3 1 r 5 2 sin i sin u 5 sin 2 i sin 2 u 3 displaystyle F r J 3 frac 1 r 5 2 sin i sin u left 5 sin 2 i sin 2 u 3 right nbsp 12 dd Secondly the projection of direction north n displaystyle hat n nbsp on the plane spanned by t z displaystyle hat t hat z nbsp is sin i cos u t cos i z displaystyle sin i cos u hat t cos i hat z nbsp and this projection is cos l l displaystyle cos lambda hat lambda nbsp where l displaystyle hat lambda nbsp is the unit vector l displaystyle hat lambda nbsp orthogonal to the radial direction towards north illustrated in figure 1 From equation 11 we see that F l l J 3 1 r 5 3 2 5 sin 2 l 1 cos l l J 3 1 r 5 3 2 5 sin 2 l 1 sin i cos u t cos i z displaystyle F lambda hat lambda J 3 frac 1 r 5 frac 3 2 left 5 sin 2 lambda 1 right cos lambda hat lambda J 3 frac 1 r 5 frac 3 2 left 5 sin 2 lambda 1 right sin i cos u hat t cos i hat z nbsp and therefore F t J 3 1 r 5 3 2 5 sin 2 i sin 2 u 1 sin i cos u displaystyle F t J 3 frac 1 r 5 frac 3 2 left 5 sin 2 i sin 2 u 1 right sin i cos u nbsp 13 dd F z J 3 1 r 5 3 2 5 sin 2 i sin 2 u 1 cos i displaystyle F z J 3 frac 1 r 5 frac 3 2 left 5 sin 2 i sin 2 u 1 right cos i nbsp 14 dd In the article Orbital perturbation analysis spacecraft it is further shown that the secular perturbation of the orbital pole z displaystyle hat z nbsp is D z 1 m p g 0 2 p F z r 3 cos u d u h 0 2 p F z r 3 sin u d u z displaystyle Delta hat z frac 1 mu p left hat g int limits 0 2 pi F z r 3 cos u du hat h int limits 0 2 pi F z r 3 sin u du right quad times hat z nbsp 15 dd Introducing the expression for F z displaystyle F z nbsp of 14 in 15 one gets D z J 3 m p 3 3 2 cos i g 0 2 p p r 2 5 sin 2 i sin 2 u 1 cos u d u h 0 2 p p r 2 5 sin 2 i sin 2 u 1 sin u d u z displaystyle begin aligned amp Delta hat z frac J 3 mu p 3 frac 3 2 cos i cdot amp left hat g int limits 0 2 pi left frac p r right 2 left 5 sin 2 i sin 2 u 1 right cos u du hat h int limits 0 2 pi left frac p r right 2 left 5 sin 2 i sin 2 u 1 right sin u du right quad times hat z end aligned nbsp 16 dd The fraction p r displaystyle frac p r nbsp is p r 1 e cos 8 1 e g cos u e h sin u displaystyle frac p r 1 e cdot cos theta 1 e g cdot cos u e h cdot sin u nbsp where e g e cos w displaystyle e g e cos omega nbsp e h e sin w displaystyle e h e sin omega nbsp are the components of the eccentricity vector in the g h displaystyle hat g hat h nbsp coordinate system As all integrals of type 0 2 p cos m u sin n u d u displaystyle int limits 0 2 pi cos m u sin n u du nbsp are zero if not both n displaystyle n nbsp and m displaystyle m nbsp are even we see that 0 2 p p r 2 5 sin 2 i sin 2 u 1 cos u d u 2 e g 5 sin 2 i 0 2 p sin 2 u cos 2 u d u 0 2 p cos 2 u d u 2 p e g 5 4 sin 2 i 1 displaystyle begin aligned int limits 0 2 pi left frac p r right 2 left 5 sin 2 i sin 2 u 1 right cos u du amp 2 e g left 5 sin 2 i int limits 0 2 pi sin 2 u cos 2 u du int limits 0 2 pi cos 2 u du right amp 2 pi e g frac 5 4 sin 2 i 1 end aligned nbsp 17 dd and 0 2 p p r 2 5 sin 2 i sin 2 u 1 sin u d u 2 e h 5 sin 2 i 0 2 p sin 4 u d u 0 2 p sin 2 u d u 2 p e h 15 4 sin 2 i 1 displaystyle begin aligned int limits 0 2 pi left frac p r right 2 left 5 sin 2 i sin 2 u 1 right sin u du amp 2 e h left 5 sin 2 i int limits 0 2 pi sin 4 u du int limits 0 2 pi sin 2 u du right amp 2 pi e h frac 15 4 sin 2 i 1 end aligned nbsp 18 dd It follows that D z 2 p J 3 m p 3 3 2 cos i e g 1 5 4 sin 2 i g e h 1 15 4 sin 2 i h z 2 p J 3 m p 3 3 2 cos i e h 1 15 4 sin 2 i g e g 1 5 4 sin 2 i h displaystyle begin aligned Delta hat z amp 2 pi frac J 3 mu p 3 frac 3 2 cos i left e g 1 frac 5 4 sin 2 i hat g e h 1 frac 15 4 sin 2 i hat h right quad times hat z amp 2 pi frac J 3 mu p 3 frac 3 2 cos i left e h 1 frac 15 4 sin 2 i hat g e g 1 frac 5 4 sin 2 i hat h right end aligned nbsp 19 dd where g displaystyle hat g nbsp and h displaystyle hat h nbsp are the base vectors of the rectangular coordinate system in the plane of the reference Kepler orbit with g displaystyle hat g nbsp in the equatorial plane towards the ascending node and u displaystyle u nbsp is the polar argument relative this equatorial coordinate systemf z displaystyle f z nbsp is the force component per unit mass in the direction of the orbit pole z displaystyle hat z nbsp In the article Orbital perturbation analysis spacecraft it is shown that the secular perturbation of the eccentricity vector is D e 1 m 0 2 p t f r 2 r V r V t t f t r 2 d u displaystyle Delta bar e frac 1 mu int limits 0 2 pi left hat t f r left 2 hat r frac V r V t hat t right f t right r 2 du nbsp 20 dd where r t displaystyle hat r hat t nbsp is the usual local coordinate system with unit vector r displaystyle hat r nbsp directed away from the Earth V r m p e sin 8 displaystyle V r sqrt frac mu p cdot e cdot sin theta nbsp the velocity component in direction r displaystyle hat r nbsp V t m p 1 e cos 8 displaystyle V t sqrt frac mu p cdot 1 e cdot cos theta nbsp the velocity component in direction t displaystyle hat t nbsp Introducing the expression for F r F t displaystyle F r F t nbsp of 12 and 13 in 20 one gets D e J 3 m p 3 sin i 0 2 p t p r 3 2 sin u 5 sin 2 i sin 2 u 3 2 r V r V t t p r 3 3 2 5 sin 2 i sin 2 u 1 cos u d u displaystyle begin aligned amp Delta bar e frac J 3 mu p 3 sin i cdot amp int limits 0 2 pi left hat t left frac p r right 3 2 sin u left 5 sin 2 i sin 2 u 3 right left 2 hat r frac V r V t hat t right left frac p r right 3 frac 3 2 left 5 sin 2 i sin 2 u 1 right cos u right du end aligned nbsp 21 dd Using that V r V t e g sin u e h cos u p r displaystyle frac V r V t frac e g cdot sin u e h cdot cos u frac p r nbsp the integral above can be split in 8 terms 0 2 p t p r 3 2 sin u 5 sin 2 i sin 2 u 3 2 r V r V t t p r 3 3 2 5 sin 2 i sin 2 u 1 cos u d u 10 sin 2 i 0 2 p t p r 3 sin 3 u d u 6 0 2 p t p r 3 sin u d u 15 sin 2 i 0 2 p r p r 3 sin 2 u cos u d u 3 0 2 p r p r 3 cos u d u 15 2 sin 2 i e g 0 2 p t p r 2 sin 3 u cos u d u 15 2 sin 2 i e h 0 2 p t p r 2 sin 2 u cos 2 u d u 3 2 e g 0 2 p t p r 2 sin u cos u d u 3 2 e h 0 2 p t p r 2 cos 2 u d u displaystyle begin aligned amp int limits 0 2 pi left hat t left frac p r right 3 2 sin u left 5 sin 2 i sin 2 u 3 right left 2 hat r frac V r V t hat t right left frac p r right 3 frac 3 2 left 5 sin 2 i sin 2 u 1 right cos u right du amp 10 sin 2 i int limits 0 2 pi hat t left frac p r right 3 sin 3 u du amp 6 int limits 0 2 pi hat t left frac p r right 3 sin u du amp 15 sin 2 i int limits 0 2 pi hat r left frac p r right 3 sin 2 u cos u du amp 3 int limits 0 2 pi hat r left frac p r right 3 cos u du amp frac 15 2 sin 2 i e g int limits 0 2 pi hat t left frac p r right 2 sin 3 u cos u du amp frac 15 2 sin 2 i e h int limits 0 2 pi hat t left frac p r right 2 sin 2 u cos 2 u du amp frac 3 2 e g int limits 0 2 pi hat t left frac p r right 2 sin u cos u du amp frac 3 2 e h int limits 0 2 pi hat t left frac p r right 2 cos 2 u du end aligned nbsp 22 dd Given that r cos u g sin u h displaystyle hat r cos u hat g sin u hat h nbsp t sin u g cos u h displaystyle hat t sin u hat g cos u hat h nbsp we obtain p r 1 e cos 8 1 e g cos u e h sin u displaystyle frac p r 1 e cdot cos theta 1 e g cdot cos u e h cdot sin u nbsp and that all integrals of type 0 2 p cos m u sin n u d u displaystyle int limits 0 2 pi cos m u sin n u du nbsp are zero if not both n displaystyle n nbsp and m displaystyle m nbsp are even Term 1 0 2 p t p r 3 sin 3 u d u g 0 2 p p r 3 sin 4 u d u h 0 2 p p r 3 sin 3 u cos u d u g 0 2 p sin 4 u d u 3 e g 2 0 2 p cos 2 u sin 4 u d u 3 e h 2 0 2 p sin 6 u d u h 6 e g e h 0 2 p cos 2 u sin 4 u d u g 2 p 3 8 3 16 e g 2 15 16 e h 2 h 2 p 3 8 e g e h displaystyle begin aligned amp int limits 0 2 pi hat t left frac p r right 3 sin 3 u du hat g int limits 0 2 pi left frac p r right 3 sin 4 u du hat h int limits 0 2 pi left frac p r right 3 sin 3 u cos u du amp hat g left int limits 0 2 pi sin 4 u du 3 e g 2 int limits 0 2 pi cos 2 u sin 4 u du 3 e h 2 int limits 0 2 pi sin 6 u du right amp hat h 6 e g e h int limits 0 2 pi cos 2 u sin 4 u du amp hat g left 2 pi left frac 3 8 frac 3 16 e g 2 frac 15 16 e h 2 right right hat h left 2 pi left frac 3 8 e g e h right right end aligned nbsp 23 dd Term 2 0 2 p t p r 3 sin u d u g 0 2 p p r 3 sin 2 u d u h 0 2 p p r 3 sin u cos u d u g 0 2 p sin 2 u d u 3 e g 2 0 2 p cos 2 u sin 2 u d u 3 e h 2 0 2 p sin 6 u d u h 6 e g e h 0 2 p cos 2 u sin 2 u d u g 2 p 1 2 3 8 e g 2 9 8 e h 2 h 2 p 3 4 e g e h displaystyle begin aligned amp int limits 0 2 pi hat t left frac p r right 3 sin u du hat g int limits 0 2 pi left frac p r right 3 sin 2 u du hat h int limits 0 2 pi left frac p r right 3 sin u cos u du amp hat g left int limits 0 2 pi sin 2 u du 3 e g 2 int limits 0 2 pi cos 2 u sin 2 u du 3 e h 2 int limits 0 2 pi sin 6 u du right amp hat h 6 e g e h int limits 0 2 pi cos 2 u sin 2 u du amp hat g left 2 pi left frac 1 2 frac 3 8 e g 2 frac 9 8 e h 2 right right hat h left 2 pi left frac 3 4 e g e h right right end aligned nbsp 24 dd Term 3 0 2 p r p r 3 sin 2 u cos u d u g 0 2 p p r 3 sin 2 u cos 2 u d u h 0 2 p p r 3 sin 3 u cos u d u g 0 2 p sin 2 u cos 2 u d u 3 e g 2 0 2 p cos 4 u sin 2 u d u 3 e h 2 0 2 p sin 4 u cos 2 u d u h 6 e g e h 0 2 p cos 2 u sin 4 u d u g 2 p 1 8 3 16 e g 2 3 16 e h 2 h 2 p 3 8 e g e h displaystyle begin aligned amp int limits 0 2 pi hat r left frac p r right 3 sin 2 u cos u du hat g int limits 0 2 pi left frac p r right 3 sin 2 u cos 2 u du hat h int limits 0 2 pi left frac p r right 3 sin 3 u cos u du amp hat g left int limits 0 2 pi sin 2 u cos 2 u du 3 e g 2 int limits 0 2 pi cos 4 u sin 2 u du 3 e h 2 int limits 0 2 pi sin 4 u cos 2 u du right amp hat h 6 e g e h int limits 0 2 pi cos 2 u sin 4 u du amp hat g left 2 pi left frac 1 8 frac 3 16 e g 2 frac 3 16 e h 2 right right hat h left 2 pi left frac 3 8 e g e h right right end aligned nbsp 25 dd Term 4 0 2 p r p r 3 cos u d u g 0 2 p p r 3 cos 2 u d u h 0 2 p p r 3 sin u cos u d u g 0 2 p cos 2 u d u 3 e g 2 0 2 p cos 4 u d u 3 e h 2 0 2 p sin 2 u cos 2 u d u h 6 e g e h 0 2 p cos 2 u sin 2 u d u g 2 p 1 2 9 8 e g 2 3 8 e h 2 h 2 p 3 4 e g e h displaystyle begin aligned amp int limits 0 2 pi hat r left frac p r right 3 cos u du hat g int limits 0 2 pi left frac p r right 3 cos 2 u du hat h int limits 0 2 pi left frac p r right 3 sin u cos u du amp hat g left int limits 0 2 pi cos 2 u du 3 e g 2 int limits 0 2 pi cos 4 u du 3 e h 2 int limits 0 2 pi sin 2 u cos 2 u du right amp hat h 6 e g e h int limits 0 2 pi cos 2 u sin 2 u du amp hat g left 2 pi left frac 1 2 frac 9 8 e g 2 frac 3 8 e h 2 right right hat h left 2 pi left frac 3 4 e g e h right right end aligned nbsp 26 dd Term 5 0 2 p t p r 2 sin 3 u cos u d u g 0 2 p p r 2 sin 4 u cos u d u h 0 2 p p r 2 sin 3 u cos 2 u d u g 2 e g 0 2 p sin 4 u cos 2 u d u h 2 e h 0 2 p sin 4 u cos 2 u d u g 2 p 1 8 e g h 2 p 1 8 e h displaystyle begin aligned amp int limits 0 2 pi hat t left frac p r right 2 sin 3 u cos u du hat g int limits 0 2 pi left frac p r right 2 sin 4 u cos u du hat h int limits 0 2 pi left frac p r right 2 sin 3 u cos 2 u du amp hat g 2 e g int limits 0 2 pi sin 4 u cos 2 u du hat h 2 e h int limits 0 2 pi sin 4 u cos 2 u du hat g left 2 pi frac 1 8 e g right hat h left 2 pi frac 1 8 e h right end aligned nbsp 27 dd Term 6 0 2 p t p r 2 sin 2 u cos 2 u d u g 0 2 p p r 2 sin 3 u cos 2 u d u h 0 2 p p r 2 sin 2 u cos 3 u d u g 2 e h 0 2 p sin 4 u cos 2 u d u h 2 e g 0 2 p sin 2 u cos 4 u d u g 2 p 1 8 e h h 2 p 1 8 e g displaystyle begin aligned amp int limits 0 2 pi hat t left frac p r right 2 sin 2 u cos 2 u du hat g int limits 0 2 pi left frac p r right 2 sin 3 u cos 2 u du hat h int limits 0 2 pi left frac p r right 2 sin 2 u cos 3 u du amp hat g 2 e h int limits 0 2 pi sin 4 u cos 2 u du hat h 2 e g int limits 0 2 pi sin 2 u cos 4 u du hat g left 2 pi frac 1 8 e h right hat h left 2 pi frac 1 8 e g right end aligned nbsp 28 dd Term 7 0 2 p t p r 2 sin u cos u d u g 0 2 p p r 2 sin 2 u cos u d u h 0 2 p p r 2 sin u cos 2 u d u g 2 e g 0 2 p sin 2 u cos 2 u d u h 2 e h 0 2 p sin 2 u cos 2 u d u g 2 p 1 4 e g h 2 p 1 4 e h displaystyle begin aligned amp int limits 0 2 pi hat t left frac p r right 2 sin u cos u du hat g int limits 0 2 pi left frac p r right 2 sin 2 u cos u du hat h int limits 0 2 pi left frac p r right 2 sin u cos 2 u du amp hat g 2 e g int limits 0 2 pi sin 2 u cos 2 u du hat h 2 e h int limits 0 2 pi sin 2 u cos 2 u du hat g left 2 pi frac 1 4 e g right hat h left 2 pi frac 1 4 e h right end aligned nbsp 29 dd Term 8 0 2 p t p r 2 cos 2 u d u g 0 2 p p r 2 sin u cos 2 u d u h 0 2 p p r 2 cos 3 u d u g 2 e h 0 2 p sin 2 u cos 2 u d u h 2 e g 0 2 p cos 4 u d u g 2 p 1 4 e h h 2 p 3 4 e g displaystyle begin aligned amp int limits 0 2 pi hat t left frac p r right 2 cos 2 u du hat g int limits 0 2 pi left frac p r right 2 sin u cos 2 u du hat h int limits 0 2 pi left frac p r right 2 cos 3 u du amp hat g 2 e h int limits 0 2 pi sin 2 u cos 2 u du hat h 2 e g int limits 0 2 pi cos 4 u du hat g left 2 pi frac 1 4 e h right hat h left 2 pi frac 3 4 e g right end aligned nbsp 30 dd As 10 sin 2 i g 3 8 3 16 e g 2 15 16 e h 2 h 3 8 e g e h 6 g 1 2 3 8 e g 2 9 8 e h 2 h 3 4 e g e h 15 sin 2 i g 1 8 3 16 e g 2 3 16 e h 2 h 3 8 e g e h 3 g 1 2 9 8 e g 2 3 8 e h 2 h 3 4 e g e h 15 2 sin 2 i e g g 1 8 e g h 1 8 e h 15 2 sin 2 i e h g 1 8 e h h 1 8 e g 3 2 e g g 1 4 e g h 1 4 e h 3 2 e h g 1 4 e h h 3 4 e g 3 2 5 4 sin 2 i 1 1 e g 2 4 e h 2 g 5 e g e h h displaystyle begin aligned amp 10 sin 2 i left hat g left frac 3 8 frac 3 16 e g 2 frac 15 16 e h 2 right hat h left frac 3 8 e g e h right right amp 6 left hat g left frac 1 2 frac 3 8 e g 2 frac 9 8 e h 2 right hat h left frac 3 4 e g e h right right amp 15 sin 2 i left hat g left frac 1 8 frac 3 16 e g 2 frac 3 16 e h 2 right hat h left frac 3 8 e g e h right right amp 3 left hat g left frac 1 2 frac 9 8 e g 2 frac 3 8 e h 2 right hat h left frac 3 4 e g e h right right amp frac 15 2 sin 2 i e g left hat g left frac 1 8 e g right hat h left frac 1 8 e h right right amp frac 15 2 sin 2 i e h left hat g left frac 1 8 e h right hat h left frac 1 8 e g right right amp frac 3 2 e g left hat g left frac 1 4 e g right hat h left frac 1 4 e h right right amp frac 3 2 e h left hat g left frac 1 4 e h right hat h left frac 3 4 e g right right amp frac 3 2 left frac 5 4 sin 2 i 1 right left 1 e g 2 4 e h 2 hat g 5 e g e h hat h right end aligned nbsp 31 dd It follows that D e J 3 m p 3 sin i mundero, wikipedia, wiki, book, books, library, article , read, download, free, free download, mp3, video, mp4, 3gp, jpg, jpeg, gif, png, picture, music, song, movie, book, game, games. © Copyright 2021, All Rights Reserved  |   www.wiki3.en-us.nina.az Rss Feed SiteMap Ping DMCA Contact Us HomePage Closed Call: