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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 where, over a long period of time, the altitude remains constant at the same point in each orbit[1] — changes in the inclination, position of the lowest point of the orbit, and eccentricity have been minimized by choosing initial values so that their perturbations cancel out.[2] This 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 reading

Background and motivation[edit] 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[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:[4]

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 poor PFS-2 was cursed with an inclination of only 11°.

Classical theory[edit]

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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

Δ

z ^

displaystyle Delta hat z ,

from the

J

2

displaystyle J_ 2 ,

term of the geopotential model is shown to be

Δ

z ^

  =   − 2 π  

J

2

μ  

p

2

 

3 2

  cos ⁡ i   sin ⁡ i

g ^

displaystyle Delta hat z = -2pi frac J_ 2 mu p^ 2 frac 3 2 cos i sin iquad hat g

 

 

 

 

(1)

which can be expressed in terms of orbital elements thusly:

Δ i   =   0

displaystyle Delta i = 0

 

 

 

 

(2)

Δ Ω   =   − 2 π  

J

2

μ  

p

2

 

3 2

  cos ⁡ i

displaystyle Delta Omega = -2pi frac J_ 2 mu p^ 2 frac 3 2 cos i

 

 

 

 

(3)

Making a similar analysis for the

J

3

displaystyle J_ 3 ,

term (corresponding to the fact that the earth is slightly pear shaped), one gets

Δ

z ^

  =   2 π  

J

3

μ  

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 = 2pi frac J_ 3 mu p^ 3 frac 3 2 cos i left( e_ h left(1- frac 15 4 sin ^ 2 iright) hat g - e_ g left(1- frac 5 4 sin ^ 2 iright) hat h right)

 

 

 

 

(4)

which can be expressed in terms of orbital elements as

Δ i   =   − 2 π  

J

3

μ  

p

3

 

3 2

  cos ⁡ i  

e

g

 

(

1 −

5 4

 

sin

2

⁡ i

)

displaystyle Delta i = -2pi frac J_ 3 mu p^ 3 frac 3 2 cos i e_ g left(1- frac 5 4 sin ^ 2 iright)

 

 

 

 

(5)

Δ Ω   =   2 π  

J

3

μ  

p

3

 

3 2

 

cos ⁡ i

sin ⁡ i

   

e

h

 

(

1 −

15 4

 

sin

2

⁡ i

)

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

 

 

 

 

(6)

In the same article the secular perturbation of the components of the eccentricity vector caused by the

J

2

displaystyle J_ 2 ,

is shown to be:

(

Δ

e

g

, Δ

e

h

)

  =  

− 2 π  

J

2

μ  

p

2

 

3 2

(

3 2

 

sin

2

⁡ i   −   1

)

  ( −

e

h

,

e

g

)   +   2 π  

J

2

μ  

p

2

 

3 2

 

cos

2

⁡ i

(

e

h

,

e

g

)

  =

− 2 π  

J

2

μ  

p

2

  3

(

5 4

 

sin

2

⁡ i   −   1

)

 

(

e

h

,

e

g

)

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

J

3

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

2 π  

J

3

μ  

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 2pi frac J_ 3 mu p^ 3 frac 3 2 sin i left( frac 5 4 sin ^ 2 i - 1right)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

Δ Ω

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

2 π  

J

3

μ  

p

3

 

3 2

  sin ⁡ i  

(

5 4

 

sin

2

⁡ i   −   1

)

 

g ^

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

 

 

 

 

(9)

Adding the

J

3

displaystyle J_ 3 ,

contribution

2 π  

J

3

μ  

p

3

 

3 2

  sin ⁡ i  

(

5 4

 

sin

2

⁡ i   −   1

)

  ( 1   ,   0 )

displaystyle 2pi frac J_ 3 mu p^ 3 frac 3 2 sin i left( frac 5 4 sin ^ 2 i - 1right) (1 , 0)

to (7) one gets

(

Δ

e

g

, Δ

e

h

)

  =   − 2 π  

J

2

μ  

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) = -2pi frac J_ 2 mu p^ 2 3left( frac 5 4 sin ^ 2 i - 1right) 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

(

  0   ,   −

J

3

  sin ⁡ i

J

2

  2   p

 

)

displaystyle left( 0 , - frac J_ 3 sin i J_ 2 2 p right),

; the polar argument of the eccentricity vector increases with

− 2 π  

J

2

μ  

p

2

  3

(

5 4

 

sin

2

⁡ i   −   1

)

displaystyle -2pi frac J_ 2 mu p^ 2 3left( frac 5 4 sin ^ 2 i - 1right),

radians between consecutive orbits. As

μ = 398600.440

 km

3

/

s

2

displaystyle mu =398600.440 text km ^ 3 /s^ 2 ,

J

2

  =   1.7555  

10

10

 km

5

/

s

2

displaystyle J_ 2 = 1.7555 10^ 10 text km ^ 5 /s^ 2 ,

J

3

  =   − 2.619  

10

11

 km

6

/

s

2

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

one gets for a polar orbit (

i = 90

 deg

displaystyle i=90 text deg ,

) with

p = 7200

 km

displaystyle p=7200 text km ,

that the centre of the circle is at

(   0   ,   0.001036   )

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

( 0   ,   0.001036 )

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

J

2

displaystyle J_ 2 ,

term given by (7) and of the

J

3

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:

e = −

J

3

  sin ⁡ i

J

2

  2   p

displaystyle e=- frac J_ 3 sin i J_ 2 2 p ,

ω =   90  

deg

displaystyle omega = 90 text deg ,

Modern theory[edit] 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

J

2

displaystyle J_ 2 ,

term is used to counteract all secular perturbations, not only those (dominating) caused by the

J

3

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 (

i = 90

 deg

displaystyle i=90 text deg ,

) with

p = 7200

 km

displaystyle p=7200 text km ,

ignoring all perturbing forces other than the

J

2

displaystyle J_ 2 ,

and the

J

3

displaystyle J_ 3 ,

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),

. When we also include the forces due to the higher zonal terms the optimal value changes to

( 0   ,   0.001285 )

displaystyle (0 , 0.001285),

. Assuming in addition a reasonable solar pressure (a "cross-sectional-area" of 0.05 square meters per 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),

which corresponds to :

ω =   87  

deg

displaystyle omega = 87 text deg ,

, i.e. the optimal value is not

ω =   90  

deg

displaystyle omega = 90 text deg

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[edit]

This section may require cleanup to meet's quality standards. The specific problem is: 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 main perturbing force to be counter-acted to have a frozen orbit is the "

J

3

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 "

J

3

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 longitudial plane with one component

F

r

 

r ^

displaystyle F_ r hat r ,

in the radial direction and one component

F

λ

 

λ ^

displaystyle F_ lambda hat lambda ,

with the unit vector

λ ^

displaystyle hat lambda ,

orthogonal to the radial direction towards north. These directions

r ^

displaystyle hat r ,

and

λ ^

displaystyle hat lambda ,

are illustrated in Figure 1.

Figure 1: The unit vectors

ϕ ^

  ,  

λ ^

  ,  

r ^

displaystyle hat phi , hat lambda , hat r

In the article Geopotential model it is shown that these force components caused by the

J

3

displaystyle J_ 3 ,

term are

F

r

=

J

3

 

1

r

5

  2   sin ⁡ λ  

(

5

sin

2

⁡ λ   −   3

)

F

λ

= −

J

3

 

1

r

5

 

3 2

  cos ⁡ λ  

(

5  

sin

2

⁡ λ   − 1

)

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

 

 

 

 

(11)

To be able to apply relations derived in the article Orbital perturbation analysis (spacecraft) the force component

F

λ

 

λ ^

displaystyle F_ lambda hat lambda ,

must be split into two orthogonal components

F

t

 

t ^

displaystyle F_ t hat t

and

F

z

 

z ^

displaystyle F_ z hat z

as illustrated in figure 2

Figure 2: The unit vector

t ^

displaystyle hat t ,

orthogonal to

r ^

displaystyle hat r ,

in the direction of motion and the orbital pole

z ^

displaystyle hat z ,

. The force component

F

λ

displaystyle F_ lambda

is marked as "F"

Let

a ^

  ,  

b ^

  ,  

n ^

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

n ^

displaystyle hat n ,

points in the direction north and such that

a ^

  ,  

b ^

displaystyle hat a , hat b ,

are in the equatorial plane of the Earth with

a ^

displaystyle hat a ,

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 ,

making up the local coordinate system (of which

t ^

  ,  

z ^

,

displaystyle hat t , hat z ,

are illustrated in figure 2), and expressing their relation with

a ^

  ,  

b ^

  ,  

n ^

displaystyle hat a , hat b , hat n ,

, are as follows:

r

a

= cos ⁡ u

displaystyle r_ a =cos u,

r

b

= cos ⁡ i   sin ⁡ u

displaystyle r_ b =cos i sin u,

r

n

= sin ⁡ i   sin ⁡ u

displaystyle r_ n =sin i sin u,

t

a

= − sin ⁡ u

displaystyle t_ a =-sin u,

t

b

= cos ⁡ i   cos ⁡ u

displaystyle t_ b =cos i cos u,

t

n

= sin ⁡ i   cos ⁡ u

displaystyle t_ n =sin i cos u,

z

a

= 0

displaystyle z_ a =0,

z

b

= − sin ⁡ i

displaystyle z_ b =-sin i,

z

n

= cos ⁡ i

displaystyle z_ n =cos i,

where

u

displaystyle u,

is the polar argument of

r ^

displaystyle hat r ,

relative the orthogonal unit vectors

g ^

=

a ^

displaystyle hat g = hat a ,

and

h ^

= cos ⁡ i  

b ^

  +   sin ⁡ i  

n ^

displaystyle hat h =cos i hat b + sin i hat n ,

in the orbital plane Firstly

sin ⁡ λ =  

r

n

  =   sin ⁡ i   sin ⁡ u

displaystyle sin lambda = r_ n = sin i sin u,

where

λ

displaystyle lambda ,

is the angle between the equator plane and

r ^

displaystyle hat r ,

(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(5sin ^ 2 i sin ^ 2 u - 3right)

 

 

 

 

(12)

Secondly the projection of direction north,

n ^

displaystyle hat n ,

, on the plane spanned by

t ^

  ,  

z ^

,

displaystyle hat t , hat z ,

is

sin ⁡ i   cos ⁡ u  

t ^

  +   cos ⁡ i  

z ^

displaystyle sin i cos u hat t + cos i hat z ,

and this projection is

cos ⁡ λ  

λ ^

displaystyle cos lambda hat lambda ,

where

λ ^

displaystyle hat lambda ,

is the unit vector

λ ^

displaystyle hat lambda

orthogonal to the radial direction towards north illustrated in figure 1. From equation (11) we see that

F

λ

 

λ ^

  = −

J

3

 

1

r

5

 

3 2

 

(

5  

sin

2

⁡ λ   − 1

)

  cos ⁡ λ  

λ ^

  =   −

J

3

 

1

r

5

 

3 2

 

(

5  

sin

2

⁡ λ   − 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 -1right) cos lambda hat lambda = -J_ 3 frac 1 r^ 5 frac 3 2 left(5 sin ^ 2 lambda -1right) (sin i cos u hat t + cos i hat z ),

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 -1right) sin i cos u

 

 

 

 

(13)

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 -1right) cos i

 

 

 

 

(14)

In the article Orbital perturbation analysis (spacecraft) it is further shown that the secular perturbation of the orbital pole

z ^

displaystyle hat z ,

is

Δ

z ^

  =  

1

μ p

[

g ^

0

2 π

F

z

r

3

cos ⁡ u   d u +  

h ^

0

2 π

F

z

r

3

sin ⁡ u   d u

]

×  

z ^

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

 

 

 

 

(15)

Introducing the expression for

F

z

displaystyle F_ z ,

of (14) in (15) one gets

Δ

z ^

  = −

J

3

μ  

p

3

 

3 2

  cos ⁡ i   ⋅

[

g ^

0

2 π

(

p r

)

2

(

5  

sin

2

⁡ i  

sin

2

⁡ u   − 1

)

cos ⁡ u   d u   +

h ^

0

2 π

(

p r

)

2

(

5  

sin

2

⁡ i  

sin

2

⁡ u   − 1

)

sin ⁡ u   d u

]

×  

z ^

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

 

 

 

 

(16)

The fraction

p r

displaystyle frac p r ,

is

p r

  =   1 + e ⋅ cos ⁡ θ   =   1 +

e

g

⋅ cos ⁡ u +

e

h

⋅ sin ⁡ u

displaystyle frac p r = 1+ecdot cos theta = 1+e_ g cdot cos u+e_ h cdot sin u

where

e

g

=   e   cos ⁡ ω

displaystyle e_ g = e cos omega

e

h

=   e   sin ⁡ ω

displaystyle e_ h = e sin omega

are the components of the eccentricity vector in the

g ^

  ,  

h ^

displaystyle hat g , hat h ,

coordinate system. As all integrals of type

0

2 π

cos

m

⁡ u  

sin

n

⁡ u   d u

displaystyle int limits _ 0 ^ 2pi cos ^ m u sin ^ n u du,

are zero if not both

n

displaystyle n,

and

m

displaystyle m,

are even, we see that

0

2 π

(

p r

)

2

(

5  

sin

2

⁡ i  

sin

2

⁡ u   − 1

)

cos ⁡ u   d u

=   2  

e

g

 

(

5  

sin

2

⁡ i  

0

2 π

sin

2

⁡ u

cos

2

⁡ u   d u   −

0

2 π

cos

2

⁡ u   d u

)

=   2 π  

e

g

  (

5 4

sin

2

⁡ i − 1 )

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

 

 

 

 

(17)

and

0

2 π

(

p r

)

2

(

5  

sin

2

⁡ i  

sin

2

⁡ u   − 1

)

sin ⁡ u   d u

=   2  

e

h

 

(

5  

sin

2

⁡ i  

0

2 π

sin

4

⁡ u   d u   −

0

2 π

sin

2

⁡ u   d u

)

=   2 π  

e

h

  (

15 4

sin

2

⁡ i − 1 )

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

 

 

 

 

(18)

It follows that

Δ

z ^

 

=   2 π  

J

3

μ  

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 π  

J

3

μ  

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 &= 2pi 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 \&= 2pi 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

g ^

displaystyle hat g ,

and

h ^

displaystyle hat h ,

are the base vectors of the rectangular coordinate system in the plane of the reference Kepler orbit with

g ^

displaystyle hat g ,

in the equatorial plane towards the ascending node and

u

displaystyle u,

is the polar argument relative this equatorial coordinate system

f

z

displaystyle f_ z ,

is the force component (per unit mass) in the direction of the orbit pole

z ^

displaystyle hat z ,

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

Δ

e ¯

  =

1 μ

 

0

2 π

(

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 ^ 2pi 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

r ^

  ,

t ^

displaystyle hat r , hat t ,

is the usual local coordinate system with unit vector

r ^

displaystyle hat r ,

directed away from the Earth

V

r

=

μ p

⋅ e ⋅ sin ⁡ θ

displaystyle V_ r = sqrt frac mu p cdot ecdot sin theta

- the velocity component in direction

r ^

displaystyle hat r ,

V

t

=

μ p

⋅ ( 1 + e ⋅ cos ⁡ θ )

displaystyle V_ t = sqrt frac mu p cdot (1+ecdot cos theta )

- the velocity component in direction

t ^

displaystyle hat t ,

Introducing the expression for

F

r

  ,  

F

t

displaystyle F_ r , F_ t ,

of (12) and (13) in (20) one gets

Δ

e ¯

  =

J

3

μ  

p

3

  sin ⁡ i   ⋅

0

2 π

(

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 &Delta bar e = frac J_ 3 mu p^ 3 sin i cdot \&int limits _ 0 ^ 2pi left(- hat t left( frac p r right) ^ 3 2 sin u, left(5sin ^ 2 i sin ^ 2 u - 3right) - 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 -1right) cos uright)duend aligned

 

 

 

 

(21)

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

the integral above can be split in 8 terms:

0

2 π

(

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 π

t ^

 

(

p r

)

3

 

sin

3

⁡ u   d u

+ 6  

0

2 π

t ^

 

(

p r

)

3

  sin ⁡ u   d u

− 15  

sin

2

⁡ i

0

2 π

r ^

 

(

p r

)

3

 

sin

2

⁡ u   cos ⁡ u   d u

+ 3  

0

2 π

r ^

 

(

p r

)

3

  cos ⁡ u   d u

+

15 2

sin

2

⁡ i  

e

g

0

2 π

t ^

 

(

p r

)

2

     

sin

3

⁡ u   cos ⁡ u   d u

15 2

sin

2

⁡ i  

e

h

 

0

2 π

t ^

 

(

p r

)

2

     

sin

2

⁡ u  

cos

2

⁡ u   d u

3 2

 

e

g

 

0

2 π

 

t ^

 

(

p r

)

2

    sin ⁡ u   cos ⁡ u   d u

+

3 2

 

e

h

 

0

2 π

t ^

 

(

p r

)

2

   

cos

2

⁡ u   d u

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

 

 

 

 

(22)

Given that

r ^

= cos ⁡ u  

g ^

  +   sin ⁡ u  

h ^

displaystyle hat r =cos u hat g + sin u hat h

t ^

= − sin ⁡ u  

g ^

  +   cos ⁡ u  

h ^

displaystyle hat t =-sin u hat g + cos u hat h

we obtain

p r

  =   1 + e ⋅ cos ⁡ θ   =   1 +

e

g

⋅ cos ⁡ u +

e

h

⋅ sin ⁡ u

displaystyle frac p r = 1+ecdot cos theta = 1+e_ g cdot cos u+e_ h cdot sin u

and that all integrals of type

0

2 π

cos

m

⁡ u  

sin

n

⁡ u   d u

displaystyle int limits _ 0 ^ 2pi cos ^ m u sin ^ n u du,

are zero if not both

n

displaystyle n,

and

m

displaystyle m,

are even: Term 1

0

2 π

t ^

 

(

p r

)

3

 

sin

3

⁡ u   d u   = −

g ^

 

0

2 π

 

(

p r

)

3

 

sin

4

⁡ u   d u   +

h ^

0

2 π

 

(

p r

)

3

 

sin

3

⁡ u   cos ⁡ u   d u   =

g ^

 

(

0

2 π

 

sin

4

⁡ u   d u   +   3  

e

g

2

 

0

2 π

 

cos

2

⁡ u  

sin

4

⁡ u   d u     +   3  

e

h

2

 

0

2 π

 

sin

6

⁡ u   d u  

)

+

h ^

  6  

e

g

 

e

h

 

0

2 π

 

cos

2

⁡ u  

sin

4

⁡ u   d u =

g ^

 

(

2 π

(

3 8

  +  

3 16

 

e

g

2

  +  

15 16

 

e

h

2

)

)

+

h ^

 

(

2 π

(

3 8

 

e

g

 

e

h

)

)

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

 

 

 

 

(23)

Term 2

0

2 π

t ^

 

(

p r

)

3

  sin ⁡ u   d u   = −

g ^

 

0

2 π

 

(

p r

)

3

 

sin

2

⁡ u   d u   +

h ^

0

2 π

 

(

p r

)

3

  sin ⁡ u   cos ⁡ u   d u   =

g ^

 

(

0

2 π

 

sin

2

⁡ u   d u   +   3  

e

g

2

 

0

2 π

 

cos

2

⁡ u  

sin

2

⁡ u   d u     +   3  

e

h

2

 

0

2 π

 

sin

6

⁡ u   d u  

)

+

h ^

  6  

e

g

 

e

h

 

0

2 π

 

cos

2

⁡ u  

sin

2

⁡ u   d u =

g ^

 

(

2 π

(

1 2

  +  

3 8

 

e

g

2

  +  

9 8

 

e

h

2

)

)

+

h ^

 

(

2 π

(

3 4

 

e

g

 

e

h

)

)

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

 

 

 

 

(24)

Term 3

0

2 π

r ^

 

(

p r

)

3

 

sin

2

⁡ u   cos ⁡ u   d u   =

g ^

 

0

2 π

 

(

p r

)

3

 

sin

2

⁡ u

cos

2

⁡ u   d u   +

h ^

0

2 π

 

(

p r

)

3

 

sin

3

⁡ u   cos ⁡ u   d u   =

g ^

 

(

0

2 π

 

sin

2

⁡ u

cos

2

⁡ u   d u   +   3  

e

g

2

 

0

2 π

 

cos

4

⁡ u  

sin

2

⁡ u   d u     +   3  

e

h

2

 

0

2 π

 

sin

4

⁡ u  

cos

2

⁡ u   d u  

)

+

h ^

  6  

e

g

 

e

h

 

0

2 π

 

cos

2

⁡ u  

sin

4

⁡ u   d u =

g ^

 

(

2 π

(

1 8

  +  

3 16

 

e

g

2

  +  

3 16

 

e

h

2

)

)

+

h ^

 

(

2 π

(

3 8

 

e

g

 

e

h

)

)

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

 

 

 

 

(25)

Term 4

0

2 π

r ^

 

(

p r

)

3

  cos ⁡ u   d u   =

g ^

 

0

2 π

 

(

p r

)

3

 

cos

2

⁡ u   d u   +

h ^

0

2 π

 

(

p r

)

3

  sin ⁡ u   cos ⁡ u   d u   =

g ^

 

(

0

2 π

cos

2

⁡ u   d u   +   3  

e

g

2

 

0

2 π

 

cos

4

⁡ u   d u     +   3  

e

h

2

 

0

2 π

 

sin

2

⁡ u  

cos

2

⁡ u   d u  

)

+

h ^

  6  

e

g

 

e

h

 

0

2 π

 

cos

2

⁡ u  

sin

2

⁡ u   d u =

g ^

 

(

2 π

(

1 2

  +  

9 8

 

e

g

2

  +  

3 8

 

e

h

2

)

)

+

h ^

 

(

2 π

(

3 4

 

e

g

 

e

h

)

)

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

 

 

 

 

(26)

Term 5

0

2 π

t ^

 

(

p r

)

2

 

sin

3

⁡ u   cos ⁡ u   d u   = −

g ^

 

0

2 π

 

(

p r

)

2

 

sin

4

⁡ u   cos ⁡ u   d u   +

h ^

0

2 π

 

(

p r

)

2

 

sin

3

⁡ u  

cos

2

⁡ u   d u   =

g ^

  2  

e

g

 

0

2 π

 

sin

4

⁡ u  

cos

2

⁡ u   d u +

h ^

  2  

e

h

 

0

2 π

 

sin

4

⁡ u  

cos

2

⁡ u   d u = −

g ^

 

(

2 π

1 8

 

e

g

)

+

h ^

 

(

2 π

1 8

 

e

h

)

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

 

 

 

 

(27)

Term 6

0

2 π

t ^

 

(

p r

)

2

 

sin

2

⁡ u  

cos

2

⁡ u   d u   = −

g ^

 

0

2 π

 

(

p r

)

2

 

sin

3

⁡ u  

cos

2

⁡ u   d u   +

h ^

0

2 π

 

(

p r

)

2

 

sin

2

⁡ u  

cos

3

⁡ u   d u   =

g ^

  2  

e

h

 

0

2 π

 

sin

4

⁡ u  

cos

2

⁡ u   d u +

h ^

  2  

e

g

 

0

2 π

 

sin

2

⁡ u  

cos

4

⁡ u   d u = −

g ^

 

(

2 π

1 8

 

e

h

)

+

h ^

 

(

2 π

1 8

 

e

g

)

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

 

 

 

 

(28)

Term 7

0

2 π

t ^

 

(

p r

)

2

  sin ⁡ u   cos ⁡ u   d u   = −

g ^

 

0

2 π

 

(

p r

)

2

 

sin

2

⁡ u   cos ⁡ u   d u   +

h ^

0

2 π

 

(

p r

)

2

  sin ⁡ u  

cos

2

⁡ u   d u   =

g ^

  2  

e

g

 

0

2 π

 

sin

2

⁡ u  

cos

2

⁡ u   d u +

h ^

  2  

e

h

 

0

2 π

 

sin

2

⁡ u  

cos

2

⁡ u   d u = −

g ^

 

(

2 π

1 4

 

e

g

)

+

h ^

 

(

2 π

1 4

 

e

h

)

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

 

 

 

 

(29)

Term 8

0

2 π

t ^

 

(

p r

)

2

 

cos

2

⁡ u   d u   = −

g ^

 

0

2 π

 

(

p r

)

2

  sin ⁡ u  

cos

2

⁡ u   d u   +

h ^

0

2 π

 

(

p r

)

2

 

cos

3

⁡ u   d u   =

g ^

  2  

e

h

 

0

2 π

 

sin

2

⁡ u  

cos

2

⁡ u   d u +

h ^

  2  

e

g

 

0

2 π

 

cos

4

⁡ u   d u = −

g ^

 

(

2 π

1 4

 

e

h

)

+

h ^

 

(

2 π

3 4

 

e

g

)

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

 

 

 

 

(30)

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 &-10sin ^ 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)\&-15sin ^ 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)\&+3left( 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 - 1right)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

Δ

e ¯

  =

J

3

μ  

p

3

  sin ⁡ i   ⋅

0

2 π

(

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 =

2 π  

J

3

μ  

p

3

  sin ⁡ i  

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 &Delta bar e = frac J_ 3 mu p^ 3 sin i cdot \&int limits _ 0 ^ 2pi left(- hat t left( frac p r right) ^ 3 2 sin u, left(5sin ^ 2 i sin ^ 2 u - 3right) - 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 -1right) cos uright)du=\&2pi frac J_ 3 mu p^ 3 sin i frac 3 2 left( frac 5 4 sin ^ 2 i - 1right)left((1- e_ g ^ 2 + 4 e_ h ^ 2 ) hat g - 5 e_ g e_ h hat h right)end aligned

 

 

 

 

(32)

References[edit]

^ Eagle, C. David. "Frozen Orbit Design" (PDF). Orbital Mechanics with Numerit. Archived from the original (PDF) on 21 November 2011. Retrieved 5 April 2012.  ^ Chobotov, Vladimir A (2002). Orbital Mechanics (3rd Edition). American Institute of Aeronautics and Astronautics. p. 221.  ^ Frozen Orbits About the Moon. 2003 ^ Bell, Trudy E. (November 6, 2006). Phillips, Tony, ed. "Bizarre Lunar Orbits". Science@NASA. NASA. Retrieved 2017-09-08.  ^ Dirk Brouwer: "Solution of the Problem of the Artificial Satellite Without Drag", Astronomical Journal, 64 (1959) ^ Mats Rosengren (1989). "Improved technique for Passive Eccentricity Control (AAS 89-155)". Advances in the Astronautical Sciences. 69. AAS/NASA. 

Further reading[edit]

STUDY OF ORBITAL ELEMENTS ON THE NEIGHBOURHOOD OF A FROZEN ORBIT. including atmospheric

.