6
6 Orbit determination
methods by Iteration
6.1 Iteration on the semi parameter
In this method the problem can be stated as determining an elliptical orbit
from two position vectors and the time interval. This can also be called as
P-iteration.
6.1.1
Obtaining the eccentricity and semi major axis
The known values are 𝒓𝟏 , 𝒓𝟐 , 𝑡1 , and 𝑡2 . Hence the required time of flight
between the two position vectors is,
𝜏 = 𝑘(𝑡2 − 𝑡1 )
(6-1)
The Unit vector along the given position vector is,
𝑢̂𝑖 =
𝒓𝒊
, ′ 𝑖 = 1,2
𝑟𝑖
(6-2)
The magnitude of the vector is,
𝑟𝑖 = √𝒓𝒊 . 𝒓𝒊 ,
𝑖 = 1,2
(6-3)
The angle between these two vectors is,
cos(𝜈2 − 𝜈1 ) = 𝑢̂1 . 𝑢̂2
(6-4)
The conics equation is,
𝑟=
𝑝
1 + 𝑒 cos 𝜈
Rearranging the above equation,
(6-5)
𝑝
− 1, 𝑖 = 1,2
𝑟𝑖
(6-6)
cos 𝜈2 = cos(𝜈2 − 𝜈1 + 𝜈1 )
(6-7)
cos 𝜈2 = cos 𝜈1 cos(𝜈2 − 𝜈1 ) − sin 𝜈1 𝑠𝑖𝑛(𝜈2 − 𝜈1 )
(6-8)
𝑒𝑐𝑜𝑠 𝜈i =
Expanding
i.e.,
Multiplying Equation 6-8 with e both sides and rearranging, we have
𝑒 cos 𝜈1 cos(𝜈2 − 𝜈1 ) + 𝑒 cos 𝜈2
sin(𝜈2 − 𝜈1 )
(6-9)
−𝑒 cos 𝜈2 cos(𝜈2 − 𝜈1 ) + 𝑒 cos 𝜈1
sin(𝜈2 − 𝜈1 )
(6-10)
𝑒 sin 𝜈1 =
𝑒 sin 𝜈2 =
𝑒 2 = (𝑒 cos νi )2 + (𝑒 sin 𝜈𝑖 )2 ,
𝑖 = 1,2
(6-11)
Hence the value of e is determined. Substituting the value of e in Equation
6.12 we get the value of the semi major axis
𝑎=
𝑝
1 − 𝑒2
(6-12)
Hence from Equations 6.11 and 6.12, we obtained both the eccentricity and
semi major axis.
6.1.2
Utilization of keplerian Equation
The relation between eccentricities and true anomalies is given by
𝑟𝑖
(cos 𝜈𝑖 + 𝑒)
𝑝
(6-13)
𝑟𝑖
√1 − 𝑒 2 sin 𝜈𝑖 , 𝑖 = 1,2
𝑝
(6-14)
cos 𝐸𝑖 =
sin 𝐸𝑖 =
Substituting 𝐸1 , 𝐸2 values in the keplerian equation, we have
𝑀2 − 𝑀1 = 𝐸2 − 𝐸1 + 𝑒(sin 𝐸1 − sin 𝐸2 )
(6-15)
𝑀2 − 𝑀1 = 𝑛(𝑡2 − 𝑡1 )
(6-16)
𝜏 = 𝑘(𝑡2 − 𝑡1 )
(6-17)
But
Also
Substituting Equation 6.17 in Equation 6.16 and rearranging, we have
𝐹=𝜏−
𝑘
( 𝑀2 − 𝑀1 )
𝑛
(6-18)
Hence if the value of 𝐹 approaches zero then the assumed value for 𝑝 is
true.
𝑎
[1 − cos(𝐸2 − 𝐸1 )]
𝑟1
(6-19)
𝑎3/2
[𝐸2 − 𝐸1 − sin(𝐸2 − 𝐸1 )]
õ
(6-20)
𝑓 = 1−
𝑔 =τ−
𝑟2 = 𝑓𝑟1 + 𝑔𝑟1̇
(6-21)
The values of 𝑟1 , 𝑟2 are given and the f and g values are obtained from
Equations 6.19 and 6.20. Substituting these values in Equations 6.21, we
have the value of the velocity which completely defines the orbit.
6.2 Iteration on the True Anomaly
In this method the problem can be stated as determining an elliptical orbit
from two position vectors and the time interval. This can also be called as
ν-iteration.
6.2.1
Obtaining the eccentricity and semi major axis
Consider, the equation of conics,
𝑟=
𝑝
1 + 𝑒 cos 𝜈
(6-22)
At times 𝑡1 and 𝑡2 , it is possible to equate
𝑟1 (1 + 𝑒 cos 𝜈1 ) = 𝑝 = 𝑟2 (1 + 𝑒 cos 𝜈2 )
(6-23)
Rearranging Equation (1.23) we have,
𝑒=
𝑟2 − 𝑟1
𝑟1 cos 𝜈1 − 𝑟2 cos 𝜈2
(6-24)
By assuming 𝜈, we can find the value of eccentricity from Equation 6.24.
Substituting the value of e in Equation 6.25, we have
𝑝 = 𝑎(1 − 𝑒 2 )
∴𝑎=
𝑟1 (1 + 𝑒 𝑐𝑜𝑠𝜈1 )
(1 − 𝑒 2 )
(6-25)
(6-26)
Hence the values of a and e are determined from the above equations.
6.2.2
Utilization of keplerian Equation
For elliptic orbits, by utilizing,
sin 𝐸𝑖 =
√1 − 𝑒 2 sin 𝜈𝑖
𝑖 = 1,2
1 + 𝑒 cos 𝜈𝑖
(6-27)
cos 𝜈𝑖 + 𝑒
𝑖 = 1,2
1 + 𝑒 cos 𝜈𝑖
(6-28)
cos 𝐸𝑖 =
The values of 𝐸1 𝑎𝑛𝑑 𝐸2 are known and since the time interval is given let
us define a function 𝐹 which is a function of time
𝑀𝑖 = 𝐸𝑖 − 𝑒 sin 𝐸𝑖 , 𝑖 = 1,2
(6-29)
=𝜏−
𝑘
( 𝑀2 − 𝑀1 )
𝑛
(6-30)
If the assumed values of the true anomalies are true then the values of
function 𝐹 will be zero.
𝑎
[1 − cos(𝐸2 − 𝐸1 )]
𝑟1
(6-31)
𝑎3/2
[𝐸2 − 𝐸1 − sin(𝐸2 − 𝐸1 )]
õ
(6-32)
𝑓 = 1−
𝑔=𝜏−
𝑟2 = 𝑓𝑟1 + 𝑔𝑟1̇
(6-33)
The values of 𝑟1 , 𝑟2 are given and the 𝑓 and 𝑔 values are obtained from
Equations 6.31 and 6.32. Substituting these values in Equation 6.33, we
have the value of the velocity which completely defines the orbit.
6.3 Iteration on the eccentricity
In this method the problem can be stated as determining an elliptical orbit
from two position vectors and the time interval. This can also be called as
e-iteration. In this method the value of e is assumed and the orbit is
determined.
6.3.1
Development of quadratic re solvent
Here a quadratic expression is obtained and the orbit values are
determined approximately.
𝑟=
𝑝
1 + 𝑒𝑐𝑜𝑠𝜈
(6-34)
Here an arbitrary value of true anomaly 𝜈0 is considered.
𝒑
𝟏 + 𝒆𝒄𝒐𝒔(𝝂 − 𝝂𝟎 + 𝝂𝟎 )
(6-35)
𝑝
1 + 𝐶𝑣 𝑐𝑜𝑠(𝜈 − 𝜈0 ) − 𝑆𝑣 sin(𝜈 − 𝜈0 )
(6-36)
𝒓=
𝑟=
Let
𝐶𝑣 ≡ 𝑒𝑐𝑜𝑠𝜈0 , 𝑆𝑣 ≡ 𝑒𝑠𝑖𝑛 𝜈0
(6-37)
𝑟2 [1 + 𝐶𝑣 cos(𝜈2 − 𝜈0 ) − 𝑆𝑣 sin(𝜈2 − 𝜈0 )]
= 𝑟1 [1 + 𝐶𝑣 cos(𝜈1 − 𝜈0 ) − 𝑆𝑣 sin(𝜈1 − 𝜈0 )]
(6-38)
(𝑟2 − 𝑟1 ) + [ 𝑟2 cos(𝜈2 − 𝜈0 ) − 𝑟1 cos(𝜈1 − 𝜈0 )]𝐶𝑣
= 𝑆𝑣
[𝑟2 sin(𝜈2 − 𝜈0 ) − 𝑟1 sin(𝜈1 − 𝜈0 )
(6-39)
Assume,
𝛽𝑟 = 𝑟2 − 𝑟1
(6-40)
𝛽𝑐 = 𝑟2 cos(𝜈2 − 𝜈0 ) − 𝑟1 cos(𝜈1 − 𝜈0 )
(6-41)
𝛽𝑠 = 𝑟2 sin(𝜈2 − 𝜈0 ) − 𝑟1 sin(𝜈1 − 𝜈0 )
(6-42)
Substituting Equations 6.40, 6.41 and 6.42 in Equation 6.39 we have,
𝛽𝑟 + 𝛽𝑐 𝐶𝑣
= 𝑆𝑣
𝛽𝑠
𝛽𝑟 2 + 2𝛽𝑟 𝛽𝑐 𝐶𝑣 + 𝛽𝑐 2 𝐶𝑣 2
𝛽𝑠
2
= 𝑆𝑣 2 = 𝑒 2 − 𝐶𝑣2
2
2 )𝐶 2
2
2 2
(𝛽𝑐𝑜
+ 𝛽𝑠𝑜
𝑣𝑜 + 2𝛽𝑟𝑜 𝛽𝑐𝑜 𝐶𝑣𝑜 + 𝛽𝑟𝑜 − 𝛽𝑠𝑜 𝑒 = 0
The solution of Equation 6.45 is for the assumed value of
𝑒 2 > 𝛽𝑟2 /(𝛽𝑐2 + 𝛽𝑠2 ) For epoch 𝜈0 = 𝜈1
6.3.2
Computational of time interval
Depending on the assumed value of e, if 𝑒 2 < 1 (elliptical) then,
(6-43)
(6-44)
(6-45)
𝑝 = 𝑟1 (1 + 𝐶𝑣 )
𝑎=
(6-46)
𝑝
1 − 𝑒2
1
(6-47)
−3
𝑛𝑒 = 𝑘µ(2) 𝑎( 2 )
(6-48)
cos(𝜈2 − 𝜈1 ) = (𝒓𝟏 . 𝒓𝟐 )/𝑟1 𝑟2
(6-49)
Note that,
sin(𝜈2 − 𝜈1 ) =
𝑥1 𝑦2 − 𝑥2 𝑦1
[1 − 𝑐𝑜𝑠 2 (𝜈2 − 𝜈1 )]1⁄2
|𝑥1 𝑦2 − 𝑥2 𝑦1 |
(6-50)
The auxiliary values 𝑆𝑒 , 𝐶𝑒 are found from,
𝑆𝑒 =
𝑟1
1
[1 − 𝑒 2 ] ⁄2 𝑆𝑣
𝑝
(6-51)
𝑟1 2
[𝑒 + 𝐶𝑣 ]
𝑝
(6-52)
𝐶𝑒 =
So that,
sin(𝐸2 − 𝐸1 ) =
𝑟2
[
1 sin(𝜈2
𝑎𝑝] ⁄2
cos(𝐸2 − 𝐸1 ) = 1 −
− 𝜈1 ) −
𝑟2
[1 − cos(𝜈2 − 𝜈1 )]𝑆𝑒
𝑝
𝑟2 𝑟1
[1 − cos(𝜈2 − 𝜈1 )]
𝑎𝑝
(6-53)
(6-54)
For parabolic track:
𝑛𝑝 = 𝑘[µ]
1⁄
2
𝐷1 = [2𝑞]
1⁄
2
𝑆𝑣 /(1 + 𝐶𝑣 )
(6-55)
(6-56)
𝐷2 − 𝐷1 =
𝑟1 𝑟2
3
(2𝑞 ⁄2 )
{[1 + 𝐶𝑣 ] sin(𝜈2 − 𝜈1 )
(6-57)
− 𝑆𝑣 [1 − cos(𝜈2 − 𝜈1 )]}
For a hyperbolic track, 𝑒 2 > 1 the mean motion is as follows,
𝑝 = 𝑟1 (1 + 𝐶𝑣 )
𝑎=
𝑝
1 − 𝑒2
𝑛ℎ = 𝑘(µ)
𝑆ℎ =
(6-58)
1⁄
2
(6-59)
(−𝑎)−
3⁄
2
(6-60)
𝑟1 2
1
[𝑒 − 1] ⁄2 𝑆𝑣
𝑝
𝐶𝑒 = 𝐶ℎ =
sinh(𝐹2 − 𝐹1 ) =
(6-61)
𝑟1 2
[ 𝑒 + 𝐶𝑣 ]
𝑝
(6-62)
𝑟2
)
1 sin(𝜈2 − 𝜈1
[ −𝑎𝑝] ⁄2
𝑟2
− [1 − cos(𝜈2 − 𝜈1 )]𝑆ℎ
𝑝
𝐹2 − 𝐹1 = log{sinh(𝐹2 − 𝐹1 ) + [𝑠𝑖𝑛ℎ2 (𝐹2 − 𝐹1 ) + 1]
(6-63)
1⁄
2}
(6-64)
Hence the mean anomalies for three different cases are as follows,
𝐸2 − 𝐸1
(𝑀2 − 𝑀1 )𝑒 = 𝐸2 − 𝐸1 + 2𝑆𝑒 𝑠𝑖𝑛2 (
) − 𝐶𝑒 sin(𝐸2 − 𝐸1 )
2
(6-65)
1
{(𝐷2 − 𝐷1 )3 + 3𝐷1 (𝐷2 − 𝐷1 )2 + 6𝑟1 (𝐷2 − 𝐷1 )}
6
(6-66)
(𝑀2 − 𝑀1 )𝑝 =
(𝑀2 − 𝑀1 )𝑓 = −(𝐹2 − 𝐹1 ) + 2𝑆ℎ 𝑠𝑖𝑛2 (
𝐹2 − 𝐹1
) − 𝐶ℎ sin(𝐹2 − 𝐹1 ) (6-67)
2
And the corresponding computational time is given by:
𝜏𝑐 = 𝑘(𝑀2 − 𝑀1 )𝑒 /𝑛𝑒
(6-68)
𝜏𝑐 = 𝑘(𝑀2 − 𝑀1 )𝑝 /𝑛𝑝
(6-69)
𝜏𝑐 = 𝑘(𝑀2 − 𝑀1 )ℎ /𝑛ℎ
(6-70)
If 𝐹 = |𝜏 − 𝜏𝑐 | < 𝜖
Where 𝜖 is the tolerance, then the orbit is accurately determined. If this is not,
then repeat the calculations form Equation 6.45, by changing the value of 𝑒 2 .
6.3.3
Computation of position and velocity vectors
In this section, a short note is given for determining the velocity vector values for
various orbit tacks.
For ellipse case:
𝐶 = 𝑎[1 − cos(𝐸2 − 𝐸1 )]
(6-71)
1⁄
2 sin(𝐸2
− 𝐸1 )
(6-72)
1⁄
2
(6-73)
𝑆=𝑎
𝐷1 = 𝑟1 𝑆𝑣 /𝑝
For parabolic case:
𝐶=
1
(𝐷 − 𝐷1 )2
2 2
𝑆 = 𝐷2 − 𝐷1
For hyperbolic case:
(6-74)
(6-75)
𝐶 = −𝑎[cosh(𝐹2 − 𝐹1 ) − 1]
1⁄
2 sinh(𝐹2
𝑆 = [−𝑎]
𝐷1 = 𝑟1 𝑆𝑣 /𝑝
𝑓 = 1−
𝑔=
1
1 (𝑟1 𝑆
µ ⁄2
𝒓̇ 𝟏 =
− 𝐹1 )
1⁄
2
(6-76)
(6-77)
(6-78)
𝐶
𝑟1
(6-79)
+ 𝐷1 𝐶)
(6-80)
𝒓𝟐 − 𝑓
𝑔
(6-81)
Hence the fundamental set 𝒓̇ 𝟏 and 𝒓𝟏 is completed and the orbit is
determined.
6.4 Utilizing the 𝐟 and 𝐠 series
In this method the fundamental differential equations are used. This
method is valid for elliptic, hyperbolic and parabolic motion.
𝒓̈ = −
6.4.1
µ𝒓
𝑟3
(6-82)
Development of quadratic re solvent
1
𝒓𝟏 = 𝒓𝟎 + 𝜏1 𝒓𝟎̇ + 𝜏12 𝒓𝟎̈ + ⋯
2
(6-83)
1
𝒓𝟐 = 𝒓𝟎 + 𝜏2 𝒓𝟎̇ + 𝜏22 𝒓𝟎̈ + ⋯
2
(6-84)
𝜏1 = 𝑘(𝑡1 − 𝑡0 )
(6-85)
where,
𝜏2 = 𝑘(𝑡2 − 𝑡0 )
(6-86)
Taking the value of 𝑡0
𝑡0 =
𝑡1 + 𝑡2
2
(6-87)
Introducing Equation 6.82 into Equation 6.83
µ𝜏12
𝒓𝟏 = 𝒓𝟎 (1 − 3 ) + 𝜏1 𝒓𝟎̇ + ⋯
2𝑟0
(6-88)
µ𝜏22
) + 𝜏2 𝒓𝟎̇ + ⋯
2𝑟03
(6-89)
𝒓𝟐 = 𝒓𝟎 (1 −
𝒓𝟏 = 𝐴𝒓𝟎 + 𝜏1 𝒓𝟎̇
(6-90)
𝒓𝟐 = 𝐵𝒓𝟎 + 𝜏2 𝒓𝟎̇
(6-91)
Where,
µ 𝜏12
)
2𝑟03
(6-92)
µ 𝜏22
𝐵 ≡ (1 − 3 )
2𝑟0
(6-93)
𝜏2
𝜏1
𝒓𝟎 = (
) 𝒓𝟏 − (
)𝒓
𝐴𝜏2 − 𝐵𝜏1
𝐴𝜏2 − 𝐵𝜏1 𝟐
(6-94)
𝐴 ≡ (1 −
𝑟0 =
𝑟1 + 𝑟2
2
𝐴
𝐵
𝒓𝟎̇ = (
) 𝒓𝟐 − (
)𝒓
𝐴𝜏2 − 𝐵𝜏1
𝐴𝜏2 − 𝐵𝜏1 𝟏
6.4.2
(6-95)
(6-96)
Improvement of the position and velocity vectors
Approximating 𝑟0 , 𝑟0̇ from Equations 6.94 and 6.96 we have,
𝑟0 = √𝒓𝟎 . 𝒓𝟎
(6-97)
𝑟0̇ = (𝒓𝟎 . 𝒓𝟎̇ )/𝑟0
(6-98)
𝑉0 = √𝒓𝟎̇ . 𝒓𝟎̇
(6-99)
1 2 𝑉02
= −
𝑎 𝑟0
µ
(6-100)
𝒓𝟏 = 𝑓1 𝒓𝟎 + 𝑔1 𝒓𝟎̇
(6-101)
𝒓𝟐 = 𝑓2 𝒓𝟎 + 𝑔2 𝒓𝟎̇
(6-102)
We know that:
Here 𝑓1 , 𝑔1 , 𝑓2 , 𝑔2 are the 𝑓 and 𝑔 series which can be evaluated by using
𝜏2 , 𝜏1
1
1
𝑓 = 1 − 𝑢0 𝜏 2 − 𝑢0̇ 𝜏 3
2
6
(6-103)
1
1
𝑔 = 𝜏 − 𝑢0 𝜏 3 −
𝑢 ̇ 𝜏4
6
12 0
(6-104)
From Equations 6.101 and 6.102 eliminating,
𝑔2
𝑔1
𝒓𝟎 = (
) 𝒓𝟏 − (
)𝒓
𝑓1 𝑔2 − 𝑓2 𝑔1
𝑓1 𝑔2 − 𝑓2 𝑔1 𝟐
𝑜𝑟
(6-105)
𝒓𝟎 = 𝐶1 𝒓𝟏 + 𝐶2 𝒓𝟐
(6-106)
𝑔2
𝐷
(6-107)
where,
𝐶1 ≡
𝐶2 ≡ −
𝑔1
𝐷
𝐷 = 𝑓1 𝑔2 − 𝑓2 𝑔1
(6-108)
(6-109)
Similarly eliminating 𝒓𝟎̇ from Equations 6.101 and 6.102:
𝒓𝟎̇ = 𝐶1̇ 𝒓𝟏 + 𝐶2̇ 𝒓𝟐
(6-110)
where:
𝐶1̇ ≡ −
𝑓2
𝐷
(6-111)
𝐶2̇ ≡ −
𝑓1
𝐷
(6-112)
This process is repeated until the error between the values of 𝒓𝟎 in
Equations 6.94 and 6.106 is within the tolerance limit, i.e.
|(𝑟0 )𝑛+1 − (𝑟0 )𝑛 | < 𝜖1
(6-113)
|(𝑟0̇ )𝑛+1 − (𝑟0̇ )𝑛 | < 𝜖2
(6-114)
|(𝑉0 )𝑛+1 − (𝑉0 )𝑛 | < 𝜖3
(6-115)
Similarly,
where, n=1, 2,…, q and 𝜖1 , 𝜖2 , 𝜖3 are the tolerances
Finding the values of 𝑓1 , 𝑔1 , 𝑓2 , 𝑔2 as,
𝑓1 = 𝑓(𝑉0 , 𝑟0 , 𝑟0̇ , 𝜏1 )
𝑔1 = 𝑔(𝑉0 , 𝑟0 , 𝑟0̇ , 𝜏1 )
𝑓2 = 𝑓(𝑉0 , 𝑟0 , 𝑟0̇ , 𝜏2 )
𝑔2 = 𝑓(𝑉0 , 𝑟0 , 𝑟0̇ , 𝜏2 )
The values of 𝐶1 , 𝐶2 , 𝐶1̇ , 𝐶2̇ can be calculated from Equations 6.107, 6.108,
6.111 and 6.112. Similarly the values of 𝑓1̇ , 𝑔1̇ are obtained. Hence,
𝒓𝟏̇ = 𝑓1̇ 𝒓𝟎 + 𝑔1̇ 𝒓𝟎̇
(6-116)
⃗ is completed and the orbit is determined.
Hence the fundamental set 𝒓𝟏̇ , 𝒓
6.5
Gauss Orbit determination Method
This is the most approximate method which gives exact solution.
6.5.1
Ratio of sector to triangle
The ratio of sector ABC to triangle ABC is y. The orbit equation in terms of
semi parameter is
Fig. 6-1.
𝑝=
ℎ2
=> ℎ = √µ𝑝
µ
(6-117)
The area of the sector ABC is
1 𝜏
1
𝐴𝑠 = ∫ √µ𝑝 𝑑𝜏 = √µ𝑝 𝜏
2 0
2
(6-118)
The area of the triangle ABC is
1
𝑟 𝑟 sin(𝜈2 − 𝜈1 )
2 21
(6-119)
𝐴𝑠
√µ𝑝𝜏
=
𝐴𝑡 𝑟2 𝑟1 sin(𝜈2 − 𝜈1 )
(6-120)
𝐴𝑡 =
Ratio of sector to triangle
𝑦=
The conic equation in polar form
𝑟=
For times 𝑡1 and 𝑡2
𝑝
𝑝
=> = 1 + 𝑒𝑐𝑜𝑠𝜈
1 + 𝑒𝑐𝑜𝑠𝜈
𝑟
(6-121)
1 1
𝑝 ( + ) = 2 + 𝑒(cos 𝜈1 + cos 𝜈2 )
𝑟1 𝑟2
(6-122)
1 1
𝜈1 + 𝜈2
𝜈1 − 𝜈2
𝑝 ( + ) = 2 + 2𝑒 cos(
) cos(
)
𝑟1 𝑟2
2
2
(6-123)
The value of 𝑒 cos(
𝜈1 +𝜈2
2
) is unknown, hence for determining the value we
have,
√𝑟 𝑐𝑜𝑠
𝜈
𝑟(1 + 𝑐𝑜𝑠𝜈)
𝜈
𝑟(1 − 𝑐𝑜𝑠𝜈)
= ±√
; √𝑟 𝑠𝑖𝑛 = ±√
2
2
2
2
(6-124)
Introducing:
𝑟 = 𝑎(1 − 𝑒𝑐𝑜𝑠𝐸) ;
𝑟𝑐𝑜𝑠𝜈 = 𝑎(𝑐𝑜𝑠𝐸 − 𝑒)
(6-125)
𝑟(1 + 𝑐𝑜𝑠𝜈) = 𝑎(cos 𝐸 − 𝑒) + 𝑎(1 − 𝑒𝑐𝑜𝑠 𝐸)
= 𝑎(1 + 𝑐𝑜𝑠𝐸 − 𝑒(1 + cos 𝐸))
(6-126)
𝑟(1 + 𝑐𝑜𝑠𝜈) = 𝑎(1 − 𝑒)(1 + cos 𝐸) = 2𝑎(1 − 𝑒)𝑐𝑜𝑠 2
𝐸
2
(6-127)
𝐸
2
(6-128)
Similarly,
𝑟(1 − 𝑐𝑜𝑠𝜈) = 𝑎(1 − 𝑒)(1 − cos 𝐸) = 2𝑎(1 − 𝑒)𝑠𝑖𝑛2
Substituting Equation 6.124 in Equation 6.128,
𝜈
𝐸
= ±√𝑎(1 − 𝑒) 𝑐𝑜𝑠 ;
2
2
𝜈
𝐸
√𝑟 𝑠𝑖𝑛 = ±√𝑎(1 − 𝑒) 𝑠𝑖𝑛
2
2
∴ √𝑟 𝑐𝑜𝑠
Considering at two different times 𝑡1 and 𝑡2 ,
(6-129)
𝜈1 ± 𝜈2
)
√𝑟2 𝑟1 cos (
2
𝜈2
𝜈1
= [√𝑟2 cos ] [√𝑟1 cos ]
2
2
𝜈2
𝜈1
∓ [√𝑟2 sin ][√𝑟1 sin ]
2
2
(6-130)
Substituting Equation 6.129 in Equation 6.130,
𝜈1 ± 𝜈2
∴ √𝑟2 𝑟1 cos (
)
2
𝐸1
𝐸2
𝑐𝑜𝑠
2
2
𝐸1
𝐸2
− 𝑎(1 − 𝑒)𝑠𝑖𝑛 𝑠𝑖𝑛
2
2
= 𝑎(1 − 𝑒)𝑐𝑜𝑠
(6-131)
Rearranging ,we have
𝜈1 − 𝜈2
𝐸1 − 𝐸2
𝐸1 + 𝐸2
) = acos (
) − 𝑎𝑒 cos (
)
2
2
2
(6-132)
𝜈1 + 𝜈2
𝐸1 + 𝐸2
𝐸2 − 𝐸1
) = acos (
) − 𝑎𝑒 cos (
)
√𝑟2 𝑟1 cos (
2
2
2
(6-133)
√𝑟2 𝑟1 cos (
Similarly,
Multiplying Equation 6.133) with e and dividing it by √𝑟2 𝑟1 we have,
ecos (
𝜈1 + 𝜈2
)
2
1
𝐸2 −𝐸1
{a(1 − e2 )cos (
)
2
√𝑟2 𝑟1
𝜈2 − 𝜈1
− √𝑟2 𝑟1 cos (
)}
2
=
𝜈1 + 𝜈2
𝑝
𝐸2 −𝐸1
𝜈2 − 𝜈1
ecos (
)=
{cos (
) − cos (
)}
2
2
2
√𝑟2 𝑟1
Substituting Equation 6.135 in Equation 6.122, we have
(6-134)
(6-135)
1
1
2𝑝
2
√𝑟2 𝑟1
𝑝 (𝑟 + 𝑟 ) = 2 +
1
𝑝(
{cos (
𝐸2 −𝐸1
2
𝜈2 −𝜈1
) − cos (
2
𝜈2 −𝜈1
)} cos (
2
(6-136)
1 1
+ )=2
𝑟1 𝑟2
2𝑝
𝐸2 −𝐸1
{cos (
)
2
√𝑟2 𝑟1
𝜈2 − 𝜈1
𝜈2 − 𝜈1
− cos (
)} cos (
)
2
2
+
𝑝(
)
1 1
2𝑝
𝐸2 −𝐸1
𝜈2 − 𝜈1
+ )= 2+
cos (
) cos (
)
𝑟1 𝑟2
2
2
√𝑟2 𝑟1
𝜈2 − 𝜈1
− 2 cos 2 (
)
2
(6-137)
(6-138)
But we know,
µ𝑝𝜏 2
√µ𝑝 𝜏
2
𝑦=
=> 𝑦 = 2 2 2
𝑟2 𝑟1 sin(𝜈2 − 𝜈1 )
𝑟2 𝑟1 sin (𝜈2 − 𝜈1 )
(6-139)
Hence from Equation 6.139) we have,
2
𝑦 2 𝑟2 𝑟1 2 sin2 (𝜈2 − 𝜈1 )
𝑝=
µ𝜏 2
(6-140)
Substituting Equation 6.140 in Equation 6.138,
2
𝑦 2 𝑟2 𝑟1 2 sin2(𝜈2 − 𝜈1 ) 𝑟1 + 𝑟2
(
)
µ𝜏 2
𝑟1 𝑟2
2
𝑦 2 𝑟2 𝑟1 2 sin2 (𝜈2 − 𝜈1 )
𝐸2 −𝐸1
𝜈2 − 𝜈1
µ𝜏 2
=2+2
cos (
) cos (
)
2
2
√𝑟1 𝑟2
𝜈2 − 𝜈1
− 2 cos2 (
)
2
(6-141)
y 2 r2 r1 sin2 (𝜈2 − 𝜈1 )
(𝑟1 + 𝑟2 )
µ𝜏 2
2
𝑦 2 𝑟2 𝑟1 2 sin2 (𝜈2 − 𝜈1 )
𝐸2 −𝐸1
𝜈2 − 𝜈1
=2+2
cos
(
)
cos
(
)
2
2
µ𝜏 2 √𝑟1 𝑟2
𝜈2 − 𝜈1
− 2 cos2 (
)
2
(6-142)
y 2 r2 r1 sin2 (𝜈2 − 𝜈1 )
{(𝑟1 + 𝑟2 )
µ𝜏 2
𝐸2 −𝐸1
𝜈2 − 𝜈1
− 2 √𝑟1 𝑟2 cos (
) cos (
)}
2
2
𝜈2 − 𝜈1
= 2 (1 − cos 2 (
))
2
(6-143)
y 2 r2 r1 sin2 (𝜈2 − 𝜈1 )
{(𝑟1 + 𝑟2 )
µ𝜏 2
𝐸2 −𝐸1
𝜈2 − 𝜈1
− 2 √𝑟1 𝑟2 cos (
) cos (
)}
2
2
𝜈2 − 𝜈1
= 2 (sin2 (
))
2
(6-144)
y 2 r2 r1
𝐸2 −𝐸1
𝜈2 − 𝜈1
{(𝑟1 + 𝑟2 ) − 2 √𝑟1 𝑟2 cos (
) cos (
)}
2
µ𝜏
2
2
𝜈 −𝜈
2 (sin2 ( 2 2 1 ))
=
𝜈 −𝜈
𝜈 −𝜈
4 sin2 ( 2 2 1 ) cos2 ( 2 2 1 )
(6-145)
y 2 r2 r1
𝐸2 −𝐸1
𝜈2 − 𝜈1
)
{(𝑟
+
𝑟
−
2
𝑟
𝑟
cos
(
)
cos
(
)}
√
1
2
1
2
µ𝜏 2
2
2
1
=
𝜈 −𝜈
2 cos 2 ( 2 2 1 )
(6-146)
The ratio of sector to traiangle is,
𝜈2 − 𝜈1
2 )
𝑦 =
𝐸 −𝐸
𝜈 −𝜈
2𝑟1 𝑟2 [(𝑟1 + 𝑟2 ) − 2 √𝑟1 𝑟2 cos ( 2 2 1 ) cos ( 2 2 1 )]
2
µ𝜏 2 sec 2 (
(6-147)
µ𝜏 2
𝜈 −𝜈
sec 2 ( 2 2 1 )
2√𝑟1 𝑟2
𝑦2 =
𝐸2 −𝐸1
𝜈2 − 𝜈1
√𝑟1 𝑟2 [(𝑟1 + 𝑟2 ) − 2 cos ( 2 ) cos ( 2 )]
(6-148)
µ𝜏 2
𝜈 −𝜈
sec 3 ( 2 2 1 )
2√𝑟1 𝑟2
𝑦2 =
𝜈 −𝜈
𝐸 −𝐸
[√𝑟1 𝑟2 (𝑟1 + 𝑟2 ) sec ( 2 2 1 ) − 2 cos ( 2 2 1 )]
(6-149)
µ𝜏 2
𝜈 − 𝜈1
sec 3 ( 2
)
2
2𝑟1 𝑟2 √𝑟1 𝑟2
2
𝑦 =
𝜈 − 𝜈1
𝐸 −𝐸
(𝑟1 + 𝑟2 ) sec ( 2
) cos ( 2 1 )
2
2
4[
−
]
2
𝑟
𝑟
√1 2
(6-150)
µ𝜏 2
𝜈 −𝜈
sec 3 ( 2 2 1 )
8𝑟1 𝑟2 √𝑟1 𝑟2
𝑦2 =
𝜈 − 𝜈1
𝐸2 −𝐸1
(𝑟1 + 𝑟2 ) sec ( 2
)
cos
(
1
1
1
2
2 )]
[
−2+2−2
2
√𝑟1 𝑟2
(6-151)
µ𝜏 2
𝜈 −𝜈
sec 3 ( 2 2 1 )
8𝑟1 𝑟2 √𝑟1 𝑟2
𝑦2 =
𝜈 − 𝜈1
𝐸2 −𝐸1
(𝑟1 + 𝑟2 ) sec ( 2
)
cos
(
1
1
2
2 )]
[
− 2] + 2 [ 1 −
2
√𝑟1 𝑟2
≅
(6-152)
𝑚
𝑙+𝑥
𝑚
Hence the first gauss eqaution reduces to, 𝑦 2 = 𝑙+𝑥 , this is the relation
between ratio of sector to triangle and the differnce between eccentric
anamolies. The parameters 𝑚 and 𝑙 can be determined in further section.
6.5.2
Kepler’s Equation
Considering the kepler’s equation,
√µ 𝑘
𝑎
3⁄
2
(𝑡 − 𝜏) = 𝐸 − 𝑒𝑠𝑖𝑛𝐸
(6-153)
√µ 𝑘
𝑎
√µ 𝑘
3⁄
2
(𝑡2 − 𝑡1 ) = 𝐸2 − 𝐸1 − 𝑒(𝑠𝑖𝑛𝐸2 − 𝑠𝑖𝑛𝐸1 )
(6-154)
𝐸2 − 𝐸1
𝐸2 + 𝐸1
) cos (
)
2
2
(6-155)
𝜈2 − 𝜈1
𝐸2 − 𝐸1
𝐸2 + 𝐸1
) = a cos (
) − 𝑎𝑒 cos (
)
√𝑟2 𝑟1 cos (
2
2
2
(6-156)
𝐸2 + 𝐸1
𝐸2 − 𝐸1
𝜈2 − 𝜈1
√𝑟2 𝑟1
𝑒 cos (
) = cos (
)−
cos (
)
2
2
𝑎
2
(6-157)
3
𝑎 ⁄2
(𝑡2 − 𝑡1 ) = 𝐸2 − 𝐸1 − 2𝑒 sin (
From Equation 6.132,
Substituting Equation 6.157 in Equation 6.155,
√µ 𝑘
𝑎
3⁄
2
(𝑡2 − 𝑡1 ) = 𝐸2 − 𝐸1
𝐸2 − 𝐸1
𝐸2 − 𝐸1
− 2sin (
) {cos (
)
2
2
𝜈2 − 𝜈1
√𝑟2 𝑟1
−
cos (
)}
𝑎
2
√µ 𝑘
𝑎
3⁄
2
(𝑡2 − 𝑡1 ) =
(6-158)
𝐸2 − 𝐸1
− sin(𝐸2 − 𝐸1 )
𝐸2 − 𝐸1
𝜈2 − 𝜈1
√𝑟2 𝑟1
+2
sin (
) cos (
)
𝑎
2
2
(6-159)
But we know the following:
𝐸2 − 𝐸1
𝐸2
𝐸1
𝐸2
𝐸1
) = sin
cos − cos
sin
2
2
2
2
2
(6-160)
𝐸2 − 𝐸1
𝜈2
𝜈1
𝜈2
𝜈1
√𝑟2 𝑟1
)=
(sin cos − cos sin )
2
2
2
2
2
√𝑎𝑝
(6-161)
sin (
sin (
𝐸2 − 𝐸1
𝜈2 − 𝜈1
√𝑟2 𝑟1
sin (
)=
sin (
)
2
2
√𝑎𝑝
(6-162)
Substituting Equation 6.162 in Equation 6.159,
∴
√µ 𝑘
𝑎
3⁄
2
(𝑡2 − 𝑡1 )
=
𝐸2 − 𝐸1
− sin(𝐸2 − 𝐸1 )
𝜈2 − 𝜈1 √𝑟2 𝑟1
𝜈2 − 𝜈1
√𝑟2 𝑟1
+2
sin (
)
cos (
)
2
𝑎
2
√𝑎𝑝
(6-163)
Rearranging above equation we have,
𝜏=
𝑎
3⁄
2
õ
{𝐸2 − 𝐸1
− sin(𝐸2 − 𝐸1 )
𝜈2 − 𝜈1 √𝑟2 𝑟1
𝜈2 − 𝜈1
√𝑟2 𝑟1
+2
sin (
)
cos (
)}
2
𝑎
2
√𝑎𝑝
𝜏=
𝑎
3⁄
2
õ
{𝐸2 − 𝐸1 − sin(𝐸2 − 𝐸1 )} +
𝑟2 𝑟1
√µ𝑝
sin(𝜈2 − 𝜈1 )
(6-164)
(6-165)
But
𝑦=
1 𝑟2 𝑟1 𝑠𝑖𝑛(𝜈2 − 𝜈1 )
√µ𝑝 𝜏
=> =
𝑟2 𝑟1 𝑠𝑖𝑛(𝜈2 − 𝜈1 )
𝑦
√µ𝑝 𝜏
(6-166)
3
1 𝑎 ⁄2
1− =
[ 𝐸2 − 𝐸1 − sin(𝐸2 − 𝐸1 )]
𝑦
õ
(6-167)
But
sin(𝜈2 − 𝜈1 ) =
∴𝑦=
2√𝑎𝑝
𝐸2 − 𝐸1
𝜈2 − 𝜈1
sin (
) cos (
)
2
2
√𝑟2 𝑟1
√µ 𝜏
𝐸 −𝐸
𝜈 −𝜈
2√𝑎√𝑟2 𝑟1 sin ( 2 2 1 ) cos ( 2 2 1 )
(6-168)
(6-169)
1
∴ 𝑦 3 (1 − )
𝑦
=
𝐸2 − 𝐸1 − sin(𝐸2 − 𝐸1 )
√µ 𝜏
[
] ≅ 𝑚𝑋
3
𝐸2 − 𝐸1
𝜈2 − 𝜈1
3
sin ( 2 )
[2√𝑟2 𝑟1 cos ( 2 )]
𝑦 2 (𝑦 − 1) = 𝑚𝑋
(6-170)
(6-171)
But
𝑦2 =
𝑚
𝑙+𝑥
𝑦 = 1 + 𝑋(𝑙 + 𝑥)
(6-172)
(6-173)
Equations 6.172 and 6.173 are the two more possible ways to write a
compact eqaution. We have three equations now: 6.152, 6.172 and 6.173
Hence, we can find the solution to this problem.
6.5.3
Using 𝐟 and 𝐠 functions
𝑟2 = 𝑓𝑟1 + 𝑔𝑟1̇
(6-174)
𝑟2 = 𝑥𝑤 𝑃 + 𝑥𝑤 𝑄
(6-175)
𝑟1 = 𝑥𝑤𝑜 𝑃 + 𝑥𝑤𝑜 𝑄
(6-176)
Cross multiplying r2 with 𝑟1̇
̂
𝑟2 × 𝑟1̇ = 𝑓( 𝑟1 × 𝑟1̇ ) = 𝑓ℎ = 𝑓√µ𝑝 𝑊
(6-177)
Similarly multiplyng r2 with r1
̂
𝑟2 × 𝑟1 = −𝑔( 𝑟1 × 𝑟1̇ ) = −𝑔ℎ = −𝑔√µ𝑝 𝑊
(6-178)
∴𝑓=
𝑥𝑤 𝑦𝑤𝑜
̇ − 𝑦𝑤 𝑥𝑤𝑜
̇
√µ𝑝
𝑔=
𝑥𝑤𝑜 𝑦𝑤 − 𝑦𝑤𝑜 𝑥𝑤
√µ𝑝
(6-179)
Also,
𝑟 = 𝑎(1 − 𝑒𝑐𝑜𝑠 𝐸) => 𝑟̇ = 𝑎𝑒𝐸̇ sin 𝐸
(6-180)
𝑥 = 𝑟𝑐𝑜𝑠𝜈 = 𝑎(𝑐𝑜𝑠𝐸 − 𝑒)
(6-181)
𝑥̇ = 𝑟̇ 𝑐𝑜𝑠𝜈 − 𝑟𝑠𝑖𝑛𝜈𝜈̇ = −𝑎𝐸̇ sin 𝐸
(6-182)
𝑦 = 𝑟𝑠𝑖𝑛𝜈 = 𝑎√1 − 𝑒 2 𝑠𝑖𝑛𝐸
(6-183)
𝑦̇ = 𝑟̇ 𝑠𝑖𝑛𝜈 − 𝑟𝑐𝑜𝑠𝜈𝜈̇ = 𝑎𝐸̇ √1 − 𝑒 2 𝑐𝑜𝑠𝐸
(6-184)
𝑀 = 𝐸 − 𝑒𝑐𝑜𝑠𝐸
(6-185)
𝑀̇ = 𝐸̇ − 𝐸̇ 𝑒𝑐𝑜𝑠 𝐸
(6-186)
1 µ
√ = (1 − 𝑒𝑐𝑜𝑠𝐸)𝐸̇
𝑎 𝑎
(6-187)
1 µ
𝐸̇ = √
𝑟 𝑎
(6-188)
Also,
Hence,
𝑥̇ = −
𝑓=
√µ𝑝
√µ𝑎
𝑠𝑖𝑛𝐸 𝑦̇ =
𝑐𝑜𝑠𝐸
𝑟
𝑟
𝐸
𝑎 (cos 2 − 𝑒) √µ𝑝 cos 𝐸1 + 𝑎√µ𝑝 sin 𝐸2 sin 𝐸1
𝑟1 √µ𝑝
(6-189)
(6-190)
𝑎
𝑓 = 1 − [1 − cos(𝐸2 − 𝐸1 )]
𝑟
𝑔=
(6-191)
𝑎2 √1 − 𝑒 2 (cos 𝐸1 − 𝑒) sin 𝐸2 − 𝑎2 √1 − 𝑒 2 (cos 𝐸2 − 𝑒) sin 𝐸1
(6-192)
√µ𝑝
𝑔=𝜏−
𝑎
3⁄
2
õ
[(𝐸2 − 𝐸1 ) − sin(𝐸2 − 𝐸1 )]
(6-193)
𝑟2 = 𝑓𝑟1 + 𝑔𝑟1̇
(6-194)
The values of 𝑟1 , 𝑟2 are given and the 𝑓 and 𝑔 values are obtained from
Equations 6.191 and 6.193. Substituting these values in Equation 6.194 we
have the value of the velocity which completely defines the orbit.
6.6 Lambert-Euler Orbit Determination
6.6.1
Chord length as function of Eccentric anomalies
𝑐 2 = (𝒓𝟏 − 𝒓𝟐 ). (𝒓𝟏 − 𝒓𝟐 ) = 𝑟22 + 𝑟12 − 2(𝒓𝟏 . 𝒓𝟐 )
(6-195)
The chord length c between radii 𝒓𝟏 , 𝒓𝟐
𝒓𝟐 . 𝒓𝟏 = 𝑟2 𝑟1 𝑐𝑜𝑠(𝜈2 − 𝜈1 )
(6-196)
Substituting Equation 6.196 in Equation 6.195
𝑐 2 = 𝑟22 − 2𝑟2 𝑟1 + 𝑟12 + 2𝑟2 𝑟1 − 2𝑟2 𝑟1 𝑐𝑜𝑠(𝜈2 − 𝜈1 )
𝑐 2 = ( 𝑟2 − 𝑟1 )2 + 4𝑟2 𝑟1 𝑠𝑖𝑛2 (
𝜈2 − 𝜈1
)
2
(6-197)
(6-198)
But we know,
𝑟 = 𝑎(1 − 𝑒 𝑐𝑜𝑠 𝐸)
Hence,
(6-199)
𝑟2 − 𝑟1 = 𝑎𝑒(𝑐𝑜𝑠 𝐸1 − 𝑐𝑜𝑠 𝐸2 )
𝑟2 − 𝑟1 = 2𝑎𝑒 𝑠𝑖𝑛 (
𝐸2 + 𝐸1
𝐸2 − 𝐸1
) 𝑠𝑖𝑛 (
)
2
2
𝐸2 + 𝐸1
𝐸2 − 𝐸1
𝑟2 + 𝑟1 = 2𝑎 [1 − 𝑒 𝑐𝑜𝑠 (
) 𝑐𝑜𝑠 (
)]
2
2
(6-200)
(6-201)
(6-202)
Now, since
𝑟 𝑠𝑖𝑛 𝜈 = √𝑎𝑝 𝑠𝑖𝑛 𝐸
(6-203)
𝑟 𝑐𝑜𝑠 𝜈 = 𝑎(𝑐𝑜𝑠 𝐸 − 𝑒)
(6-204)
√𝑟 𝑠𝑖𝑛
𝜈
𝑟(1 − 𝑐𝑜𝑠 𝜈)
=√
2
2
(6-205)
√𝑟 𝑐𝑜𝑠
𝜈
𝑟(1 + 𝑐𝑜𝑠 𝜈)
=√
2
2
(6-206)
From Equation 6.206 and Equation 6.199 we have,
√𝑟 𝑠𝑖𝑛
𝜈
𝐸
= √𝑎(1 + 𝑒) 𝑠𝑖𝑛
2
2
(6-207)
√𝑟 𝑐𝑜𝑠
𝜈
𝐸
= √𝑎(1 − 𝑒) 𝑐𝑜𝑠
2
2
(6-208)
Hence the factor in Equation 6.209 can be formed.
𝜈2 − 𝜈1
)
√𝑟2 𝑟1 𝑠𝑖𝑛 (
2
𝜈2
𝜈1
] [√𝑟1 𝑐𝑜𝑠 ]
2
2
𝜈1
𝜈2
− [√𝑟1 𝑠𝑖𝑛 ] [√𝑟2 𝑐𝑜𝑠 ]
2
2
= [√𝑟2 𝑠𝑖𝑛
(6-209)
= √𝑎2 (1 − 𝑒 2 ) [𝑠𝑖𝑛
𝐸2
𝐸1
𝐸1
𝐸2
𝑐𝑜𝑠 − 𝑠𝑖𝑛 𝑐𝑜𝑠 ]
2
2
2
2
(6-210)
𝐸2 − 𝐸1
]
2
(6-211)
= √𝑎2 (1 − 𝑒 2 ) [𝑠𝑖𝑛
Multiplying Equation 6.211 with 4 and squaring the equation we have,
𝜈2 − 𝜈1
𝐸2 − 𝐸1
) = 4𝑎2 (1 − 𝑒 2 )𝑠𝑖𝑛2 (
)
2
2
4𝑟2 𝑟1 𝑠𝑖𝑛2 (
(6-212)
Substituting the value in Equation 6.198 we have,
𝑐 = 2𝑎√1 − 𝑒 2 𝑐𝑜𝑠 2 (
6.6.2
𝐸2 − 𝐸1
𝐸2 − 𝐸1
) 𝑠𝑖𝑛 (
)
2
2
(6-213)
Transformation of Variables
In the Equation 6.213 the value of 𝑒 𝑐𝑜𝑠 (
𝐸2 +𝐸1
2
) is not know.
Let,
𝐸2 + 𝐸1
𝑒 𝑐𝑜𝑠 (
) ≡ 𝑐𝑜𝑠 𝜈
2
(6-214)
Hence,
𝑐 = 2𝑎 𝑠𝑖𝑛 𝜈 𝑠𝑖𝑛 (
𝐸2 − 𝐸1
)
2
(6-215)
The sum of the radii is,
𝐸2 − 𝐸1
𝑟2 + 𝑟1 = 2𝑎 [ 1 − 𝑐𝑜𝑠 𝜈 𝑐𝑜𝑠 (
)]
2
(6-216)
Summing Equation 6.216 and Equation 6.215 we have,
𝑟2 + 𝑟1 + 𝑐 = 2𝑎 [ 1
𝐸2 − 𝐸1
𝐸2 − 𝐸1
− 𝑐𝑜𝑠 𝜈 𝑐𝑜𝑠 (
) + 𝑠𝑖𝑛 𝜈 𝑠𝑖𝑛 (
)]
2
2
(6-217)
Similarly subtracting Equation 6.216 and Equation 6.215 we have,
𝑟2 + 𝑟1 − 𝑐 = 2𝑎 [ 1
𝐸2 − 𝐸1
𝐸2 − 𝐸1
− 𝑐𝑜𝑠 𝜈 𝑐𝑜𝑠 (
) − 𝑠𝑖𝑛 𝜈 𝑠𝑖𝑛 (
)]
2
2
(6-218)
Let,
𝐸2 − 𝐸1
𝜖 ≡𝜈+(
)
2
(6-219)
𝐸2 − 𝐸1
𝛿 ≡𝜈−(
)
2
(6-220)
From Equation 6.216, 6.217, 6.219 and 6.220 we have,
2𝑎[1 − 𝑐𝑜𝑠 𝜖] = 𝑟2 + 𝑟1 + 𝑐
(6-221)
2𝑎[1 − 𝑐𝑜𝑠 𝛿] = 𝑟2 + 𝑟1 − 𝑐
(6-222)
Rearranging the above equations we have,
𝑐𝑜𝑠 𝜖 = 1 −
1
[𝑟 + 𝑟1 + 𝑐]
2𝑎 2
(6-223)
𝑐𝑜𝑠 𝛿 = 1 −
1
[𝑟 + 𝑟1 − 𝑐]
2𝑎 2
(6-224)
Transforming into half angle relationship, we have
𝑠𝑖𝑛
𝜖
1
= ± √ (𝑟2 + 𝑟1 + 𝑐)
2
4𝑎
(6-225)
𝑠𝑖𝑛
𝛿
1
= ± √ (𝑟2 + 𝑟1 − 𝑐)
2
4𝑎
(6-226)
6.6.3
Auxiliary angles determination
𝜈2 − 𝜈1
)
2
√𝑟2 𝑟1 𝑐𝑜𝑠 (
𝜈2
𝜈1
] [√𝑟1 𝑐𝑜𝑠 ]
2
2
𝜈1
𝜈2
+ [√𝑟1 𝑠𝑖𝑛 ] [√𝑟2 𝑐𝑜𝑠 ]
2
2
= [√𝑟2 𝑐𝑜𝑠
(6-227)
From Equation 6.203, 6.204, 6.205 and 6.206 we have,
√𝑟2 𝑟1 cos (
𝜈2 − 𝜈1
𝐸2 − 𝐸1
𝐸1 + 𝐸2
) = 𝑎 cos (
) − 𝑎𝑒 cos (
)
2
2
2
(6-228)
𝜈2 − 𝜈1
𝐸2 − 𝐸1
) = 𝑎 [𝑐𝑜𝑠 (
) − cos 𝜈]
2
2
(6-229)
𝜈2 − 𝜈1
𝜖
𝛿
) = 2 a sin sin
2
2
2
(6-230)
√𝑟2 𝑟1 cos (
√𝑟2 𝑟1 cos (
𝜈2 − 𝜈1
𝛿 √𝑟2 𝑟1 cos ( 2 )
sin =
𝜖
2
2𝑎 a sin 2
(6-231)
Hence the relation is obtained.
6.6.4
Lambert-Euler equation
The Kepler’s equation is,
√µ 𝑘( 𝑡2 − 𝑡1 )
𝑎
3⁄
2
= 𝐸2 − 𝐸1 + 𝑒(sin 𝐸1 − sin 𝐸2 )
(6-232)
𝜖 − 𝛿 = 𝐸2 − 𝐸1
(6-233)
But,
Hence,
√µ 𝜏
𝑎
3⁄
2
= 𝜖 − 𝛿 + 𝑒(𝑠𝑖𝑛 𝐸1 − 𝑠𝑖𝑛 𝐸2 )
(6-234)
√µ 𝜏
𝑎
3⁄
2
= 𝜖 − 𝛿 + 2𝑒 𝑐𝑜𝑠
𝐸1 + 𝐸2
𝐸2 − 𝐸1
𝑠𝑖𝑛
2
2
(6-235)
Introducing the parameters, we have
√µ 𝜏
3
𝑎 ⁄2
= 𝜖 − 𝛿 + 2𝑒 𝑐𝑜𝑠 𝜈 𝑠𝑖𝑛
𝜈=
√µ 𝜏
3
𝑎 ⁄2
𝜖+𝛿
2
= 𝜖 − 𝛿 + 2𝑒 𝑐𝑜𝑠
√ µ 𝜏𝑐
𝑎
3⁄
2
𝜖−𝛿
2
𝜖−𝛿
𝜖−𝛿
𝑠𝑖𝑛
2
2
= 𝜖 − 𝑠𝑖𝑛 𝜖 − (𝛿 − 𝑠𝑖𝑛 𝛿)
(6-236)
(6-237)
(6-238)
(6-239)
µ 𝜏
Now we have the value of √ 3⁄ 𝑐 to substitute in g, hence
𝑎 2
𝑎
𝑓 = 1 − [1 − 𝑐𝑜𝑠(𝐸2 − 𝐸1 )]
𝑟
𝑔=𝜏−
𝑎
3⁄
2
õ
(6-240)
[(𝐸2 − 𝐸1 ) − 𝑠𝑖𝑛(𝐸2 − 𝐸1 )]
(6-241)
𝑟2 = 𝑓𝑟1 + 𝑔𝑟1̇
(6-242)
The values of r1 , r2 are given and the f and g values are obtained from
Equations 6.240 and 6.241. Substituting these values in Equation 6.242 we
have the value of the velocity which completely defines the orbit.
6.7 Two Position Vectors and Time
(Lambert’s Problem)
We will discuss Lambert’s original geometrical formulation developed in
1761. Two paths satisfy the requirement to find an orbit between the two
vectors, in a given time. The two vectors identify the orbit plane. If ∆𝜈 =
180𝑜 , then many orbit planes can be used.
Fig. 6-2 .
𝑐𝑜𝑠(∆𝜈) = 𝑐𝑜𝑠(∆Ѳ) =
𝒓𝒐 . 𝒓
𝑟𝑜 𝑟
𝑠𝑖𝑛(∆𝜈) = 𝑡𝑚 √1 − 𝑐𝑜𝑠 2 (∆𝜈)
6.7.1
(6-243)
(6-244)
Minimum Energy Solution
Lambert’s theorem: “The orbit transfer time depends only upon the semi
major axis, the sum of the distances of the initial and final points of the arc
from the centre of force and the length of the chord joining these points.”
𝑐 = √𝑟𝑜2 + 𝑟 2 − 2𝑟𝑟𝑜 𝑐𝑜𝑠(∆𝜈)
Define the semi perimeter,
(6-245)
𝑠=
1
(𝑟 + 𝑟𝑜 + 𝑐)
2
(6-246)
Fig. 6-3 .
Note that sum of the distances from any point on ellipse to foci on ellipse
to the foci is constant and equal 2𝑎
2𝑎 = 𝑟𝑜 + (2𝑎 − 𝑟𝑜 ) = 𝑟 + (2𝑎 − 𝑟)
(6-247)
So, the secondary focus F’ is the intersection of the two circles shown. The
radius of the setwo circles depend on 𝑎 or for each 𝑎 we can draw two
circles. The locus of F’ lies on the two branches of a hyperbola with
eccentricity 1⁄𝑒𝑓 , where 𝑒𝑓 is the minimum eccentricity orbit.
𝑒𝑓 = 𝑟 − 𝑟𝑜 ⁄𝑐
(6-248)
F’ whose closer to F has an orbit with smaller e.
A minimum Energy orbit is the solution orbit that has minimum energy
𝜉=−
𝜇
2𝑎
(6-249)
And so, it has minimum 𝑎 ≥ 𝑎𝑚𝑖𝑛
For minimum Energy Orbit, F’ has only one possible location, when the two
circles touch.
∴ (2𝑎𝑚𝑖𝑛 − 𝑟𝑜 ) + (2𝑎𝑚𝑖𝑛 − 𝑟) = 𝑐
∴ 𝑎𝑚𝑖𝑛 =
𝑟𝑜 + 𝑟 + 𝑐 𝑠
=
4
2
(6-250)
(6-251)
Note: for minimum Eccentricity Solution Orbit, the major axis is parallel to
the chord; this solution is called the fundamental ellipse.
To find eccentricity, consider the above triangle
Note:
2𝑎 − 𝑟 =
𝑐 + 𝑟 + 𝑟𝑜
𝑐 + 𝑟𝑜 − 𝑟
−𝑟 =
=𝑠−𝑟
2
2
(6-252)
(2𝑎𝑚𝑖𝑛 𝑒𝑚𝑖𝑛 )2 = {(𝑠 − 𝑟) 𝑠𝑖𝑛 𝛼}2 + {𝑟 − (𝑠 − 𝑟) 𝑐𝑜𝑠 𝛼}2
(6-253)
2
2
4𝑎𝑚𝑖𝑛
𝑒𝑚𝑖𝑛
= (𝑠 − 𝑟)2 (1 − 𝑐𝑜𝑠 2 𝛼) + 𝑟 2 + (𝑠 − 𝑟)2 𝑐𝑜𝑠 2 𝛼
− 2𝑟(𝑠 − 𝑟) 𝑐𝑜𝑠 𝛼
(6-254)
𝑟𝑜2 = 𝑟 2 + 𝑐 2 − 2𝑟𝑐 cos α
(6-255)
𝑟 2 + 𝑐 2 − 𝑟02
𝑐𝑜𝑠 𝛼 =
2𝑟𝑐
(6-256)
But,
And note that:
2𝑠(𝑠 − 𝑟𝑜 ) − 𝑟𝑐 = 2 (
𝑟 + 𝑟𝑜 + 𝑐 𝑟𝑜 + 𝑟 + 𝑐
)(
− 𝑟𝑜 ) − 𝑟𝑐
2
2
(6-257)
−𝑟𝑜2 + 𝑟 2 + 𝑐 2
=
2
(6-258)
Compare Equation 6.256 and Equation 6.258:
∴ 𝑐𝑜𝑠 𝛼 =
2𝑠(𝑠 − 𝑟𝑜 )
−1
𝑟𝑐
(6-259)
Substitute in Equation 6.254,
2
2
∴ 4𝑎𝑚𝑖𝑛
𝑒𝑚𝑖𝑛
= 𝑠2 −
4𝑠
(𝑠 − 𝑟0 )(𝑠 − 𝑟)
𝑐
(6-260)
Solve for 𝑒𝑚𝑖𝑛 , also note that we can show that:
∴ 𝑃𝑚𝑖𝑛 =
2
2
4𝑎𝑚𝑖𝑛
𝑒𝑚𝑖𝑛
= 𝑠 2 − 2𝑠𝑃𝑚𝑖𝑛
(6-261)
2
𝑟𝑟𝑜
(𝑠 − 𝑟𝑜 )(𝑠 − 𝑟) =
(1 − 𝑐𝑜𝑠 ∆𝜈)
𝑐
𝑐
(6-262)
And we can show that
∴ 𝑒𝑚𝑖𝑛 = √1 −
2𝑃𝑚𝑖𝑛
𝑠
(6-263)
To calculate the time,
𝛼𝑒
𝑟𝑜 + 𝑟 + 𝑐
𝑠
)=√
=√
2
4𝑎
2𝑎
𝑠𝑖𝑛 (
(6-264)
𝛽𝑒
𝑟𝑜 + 𝑟 + 𝑐
𝑠−𝑐
𝑠𝑖𝑛 ( ) = √
=√
2
4𝑎
2𝑎
(6-265)
𝑠𝑖𝑛ℎ(
𝛼ℎ
𝑟𝑜 + 𝑟 + 𝑐
𝑠
)=√
=√
2
−4𝑎
−2𝑎
(6-266)
𝑠𝑖𝑛ℎ(
𝛽ℎ
𝑟𝑜 + 𝑟 − 𝑐
𝑠−𝑐
)=√
=√
2
−4𝑎
−2𝑎
(6-267)
Kaplan shows that the general time of flight, sometimes called Lambert’s
equation, is
𝑎3
∆𝑡 = √ 𝜇 [2𝜋𝑛𝑟𝑒𝑣 + 𝛼𝑒 − 𝑠𝑖𝑛 𝛼𝑒 ∓ (𝛽𝑒 − 𝑠𝑖𝑛 𝛽𝑒 )]
−𝑎3
=√
6.7.2
𝜇
[𝑠𝑖𝑛ℎ 𝛼ℎ − 𝛼ℎ ∓ (𝑠𝑖𝑛ℎ 𝛽ℎ − 𝛽ℎ )]
(6-268)
(6-269)
Gibbs method
𝒓𝟏
𝑟1
(6-270)
𝒓𝟐 × 𝒓𝟑
‖𝒓𝟐 × 𝒓𝟑 ‖
(6-271)
̂𝑟 =
𝑈
1
𝐶̂23 =
̂𝑟 . 𝐶̂23 = 0
∴𝑈
1
(6-272)
𝒓𝟐 = 𝐶1 𝒓𝟏 + 𝐶3 𝒓𝟑
(6-273)
It is easy to show for any 𝒓:
𝒓
𝑽 × 𝒉 = 𝜇 [ + 𝒆]
𝑟
(6-274)
Multiplying the equation with 𝒉,
𝒓
𝒉 × (𝑽 × 𝒉) = µ [𝒉 × + 𝒉 × 𝒆]
r
(6-275)
𝒉 × (𝑽 × 𝒉) = 𝑽(𝒉. 𝒉) − 𝒉(𝒉. 𝑽) = ℎ2 𝑽 − 0 = ℎ2 𝑽
(6-276)
But,
∴𝑽=
𝜇 𝒉×𝒓
[
+ 𝒉 × 𝒆]
ℎ2
𝑟
(6-277)
𝒆 = ep̂ ; 𝒉 = hw
̂; w
̂ × p̂ = q̂
(6-278)
µ w
̂ ∗𝒓
∴𝑽= (
+ eq̂)
h
r
(6-279)
If we use 𝐫𝟏 , 𝐫𝟐 , 𝐫𝟑 to find e, h, q̂, w
̂ , then we can compute 𝐫𝟏 , 𝐫𝟐 , 𝐫𝟑
𝒆. 𝐫𝟐 = 𝒆. (C1 𝐫𝟏 + C3 𝐫𝟑 )
(6-280)
From orbit equations:
𝒓𝟏 . 𝒆 =
ℎ2
− 𝑟1 ;
𝜇
𝒓𝟐 . 𝒆 =
ℎ2
ℎ2
− 𝑟2 ; 𝒓𝟑 . 𝒆 =
− 𝑟3
𝜇
𝜇
(6-281)
Substitute into the above equations;
ℎ2
ℎ2
ℎ2
( − 𝑟2 ) = 𝐶1 ( − 𝑟1 ) + 𝐶3 ( − 𝑟3 )
𝜇
𝜇
𝜇
(6-282)
ℎ2
1
𝜇 1 + 𝑒 𝑐𝑜𝑠 𝜃
(6-283)
ℎ2
𝜇
(6-284)
ℎ2
𝑟1 𝑒 𝑐𝑜𝑠 𝜃 =
− 𝑟1
𝜇
(6-285)
𝒓𝟏 . e⃗ = r1 e cos Ѳ
(6-286)
𝑟1 =
𝑟1 + 𝑟1 𝑒 𝑐𝑜𝑠 Ѳ =
But
∴ 𝒓𝟏 . e⃗ =
h2
− r1
µ
(𝒓𝟐 = 𝐶1 𝒓𝟏 + 𝐶3 𝒓𝟑 ) × 𝒓𝟏 ⇒ 𝒓𝟐 × 𝒓𝟏 = 𝐶3 ( 𝒓𝟑 × 𝒓𝟏 )
(6-287)
(6-288)
(𝒓𝟐 = 𝐶1 𝒓𝟏 + 𝐶3 𝒓𝟑 ) × 𝒓𝟑 ⇒ 𝒓𝟐 × 𝒓𝟑 = −𝐶1 (𝒓𝟑 × 𝒓𝟏 )
(6-289)
2 (r⃗⃗⃗3 × ⃗⃗⃗
r1 ) to eliminate C2 , C3 ∶
ℎ2
( 𝒓𝟏 × 𝒓𝟐 + 𝒓𝟐 × 𝒓𝟑 + 𝒓𝟑 × 𝒓𝟏 )
𝜇
= 𝑟1 (𝒓𝟐 × 𝒓𝟑 ) + 𝑟2 (𝒓𝟑 × 𝒓𝟏 ) + 𝑟3 (𝒓𝟏 × 𝒓𝟐 )
(6-290)
𝑵 = 𝑟1 (𝒓𝟐 × 𝒓𝟑 ) + 𝑟2 (𝒓𝟑 × 𝒓𝟏 ) + 𝑟3 (𝒓𝟏 × 𝒓𝟐 )
(6-291)
𝑫 = ( 𝒓𝟏 × 𝒓𝟐 + 𝒓𝟐 × 𝒓𝟑 + 𝒓𝟏 × 𝒓𝟏 )
(6-292)
𝑵=
h2
𝑫
µ
(6-293)
𝑵
𝑫
(6-294)
ℎ = √𝜇
Note: 𝑫 is perpendicular to orbit plane
𝑫
∴𝑤
̂ = ‖𝐷‖
(6-295)
1
𝑺
𝐷𝑒
(6-296)
Reading:
𝑞̂ =
Where,
𝑺 = 𝑟1 (𝒓𝟐 − 𝒓𝟑 ) + 𝑟2 (𝒓𝟑 − 𝒓𝟏 ) + 𝑟3 (𝒓𝟏 − 𝒓𝟐 )
(6-297)
𝜇 𝑤
̂ ∗𝒓
(
+ 𝑒𝑞̂)
ℎ
𝑟
(6-298)
𝜇 𝑫×𝒓
(
+ 𝑺)
𝑵𝑫
𝑟
(6-299)
∴𝑽=
𝑽=√
Algorithm:
Given 𝐫𝟏 , 𝐫𝟐 , 𝐫𝟑
Calculate r1 , r2, r3 , C12 , C23 , C31
̂𝑟 . 𝐶̂23 = 0
Verify: 𝑈
1
Calculate ∶ 𝑵, 𝑫, 𝑺, 𝑽𝟐
6.8 Angles-Only Orbit Determination
6.8.1
Gauss’s Method
𝒓𝟏 = 𝑹𝟏 + 𝜌𝑖 𝜌̂𝑖 ; 𝑖 = 1,2,3
(6-300)
𝒓𝟐 = 𝐶1 𝒓𝟏 + 𝐶3 𝒓𝟑
(6-301)
𝒓𝟏 = 𝑓1 𝒓𝟐 + 𝑔1 𝑽𝟐
(6-302)
𝒓𝟑 = 𝑓3 𝒓𝟐 + 𝑔3 𝑽𝟐
(6-303)
But,
Where 𝑓𝑖 , 𝑔𝑖 are lagrange coefficents evaluated at time 𝑡𝑖
Equation 6-301 × ⃗⃗⃗
r3 :
𝒓𝟐 × 𝒓𝟑 = 𝐶1 ( 𝒓𝟏 × 𝒓𝟑 )
∴ 𝐶1 =
(𝒓𝟐 × 𝒓𝟑 ). (𝒓𝟏 × 𝒓𝟑 )
‖𝒓𝟐 × 𝒓𝟑 ‖2
(6-304)
(6-305)
Similarly,
𝐶3 =
(𝒓𝟐 × 𝒓𝟏 ). (𝒓𝟑 × 𝒓𝟏 )
‖𝒓𝟏 × 𝒓𝟑 ‖2
From Equation 6-302 and 6-303:
(6-306)
𝒓𝟏 × 𝒓𝟑 = (𝑓1 𝑔3 − 𝑓3 𝑔1 )𝒉
(6-307)
Where 𝒉 = 𝒓𝟐 × 𝑽𝟐
‖𝒓𝟏 × 𝒓𝟑 ‖2 = (𝑓1 𝑔3 − 𝑓3 𝑔1 )2 ℎ2
(6-308)
𝒓𝟐 × 𝒓𝟑 = 𝒓𝟐 × (𝑓3 𝒓𝟐 + 𝑔3 𝑽𝟐 ) = 𝑔3 𝒉
(6-309)
𝒓𝟐 × 𝒓𝟏 = 𝑔1 𝒉
(6-310)
Similarly,
Substitute from Equations 6-308, 6-309, and 6-310 into Equations 6-306
and 6-307,
∴ 𝐶1 =
𝑔3
𝑓1 𝑔3 − 𝑓3 𝑔1
(6-311)
𝐶3 = −
𝑔1
𝑓1 𝑔3 − 𝑓3 𝑔1
(6-312)
Assume times between observations are small and let 𝜏𝑖 = 𝑡𝑖 − 𝑡2
∴ 𝑓1 ≈ 1 −
1𝜇 2
𝜏
2 𝑟23 1
(6-313)
𝑓3 ≈ 1 −
1𝜇 2
𝜏
2 𝑟23 3
(6-314)
𝑔1 ≈ 𝜏3 −
1𝜇 3
𝜏
6 𝑟23 1
(6-315)
𝑔3 ≈ 𝜏3 −
1𝜇 3
𝜏
6 𝑟23 3
(6-316)
Substitute into Equations 6-311 and 6-312
−1
𝜏3
1𝜇
1𝜇
𝐶1 ≈ ( 1 − 3 𝜏32 ) . (1 − 3 𝜏 2 )
𝜏
6 𝑟2
6 𝑟2
(6-317)
Where 𝜏 = 𝜏3 − 𝜏1
Also, Note that
−1
1𝜇
(1 − 3 𝜏 2 )
6 𝑟2
1𝜇 2
𝜏
6 𝑟23
(6-318)
∴ 𝐶1 ≈
𝜏3
1𝜇
[1 + 3 (𝜏 2 − 𝜏32 )]
𝜏
6 𝑟2
(6-319)
𝐶3 ≈ −
𝜏1
1𝜇
[1 + 3 (𝜏 2 − 𝜏12 )]
𝜏
6 𝑟2
(6-320)
≅1+
Similarly,
Recall that
𝐑 𝟐 + ρ2 ρ
̂2 = C1 (𝐑 𝟏 + ρ1 ρ
̂)
̂)
1 + C3 (𝐑 𝟑 + ρ3 ρ
3
(6-321)
Rearrange:
∴ 𝐶1 𝜌1 𝜌̂ − 𝜌2 𝜌
̂2 + 𝐶3 𝜌3 𝜌
̂3 = −𝐶1 𝑹𝟏 + 𝑹𝟐 − 𝐶3 𝑹𝟑
(6-322)
In Equation 6-265, Ci are functions of 𝑟2 ; So, we can solve for 𝜌𝑖 as
functions of 𝑟2 only. This can be done as follows:
∴ 𝐶1 𝜌1 𝜌
̂1 . (𝜌
̂2 ∗ 𝜌
̂)
̂2 ∗ 𝜌
̂)
3 = (−𝐶1 𝑹𝟏 + 𝑹𝟐 − 𝐶3 𝑹𝟑 ). (𝜌
3
(6-323)
𝐷𝑜 = 𝜌
̂.
̂2 ∗ 𝜌
̂)
1 (𝜌
3
(6-324)
Let
∴ 𝜌1 =
Where
1
1
𝐶3
(−𝐷11 + 𝐷21 − 𝐷31 )
𝐷𝑜
𝐶1
𝐶1
(6-325)
𝐷𝑖1 = 𝑹𝟏 . (𝜌
̂2 ∗ 𝜌
̂)
3
(6-326)
Similarly,
𝜌2 =
1
(−𝐶1 𝐷12 + 𝐷22 − 𝐶3 𝐷32 )
𝐷0
(6-327)
Where
Di2 = 𝑹𝟏 . (ρ
̂1 ∗ ρ
̂)
3
𝜌2 =
(6-328)
1
𝐶1
1
(− 𝐷𝑖3 + 𝐷23 − 𝐷33 )
𝐷0
𝐶3
𝐶3
Where,
𝐷𝑖3 = 𝑹𝟏 . (𝜌
̂1 ∗ 𝜌
̂)
2
(6-329)
Substitute for C1 , C3 into Equation 6-327
∴ 𝜌2 = 𝐴 +
𝜇𝐵
𝑟23
(6-330)
Where,
1
𝜏3
𝜏1
(−𝐷12 + 𝐷22 + 𝐷32 )
𝐷0
𝜏
𝜏
(6-331)
1
𝜏3
𝜏1
[𝐷12 (𝜏32 − 𝜏 2 ) + 𝐷32 (𝜏 2 − 𝜏12 ). ]
6𝐷0
𝜏
𝜏
(6-332)
𝐴=
𝐵=
Recall,
𝒓𝟐 . 𝒓𝟐 = (𝑹𝟐 + 𝜌2 𝜌
̂).
̂)
2 (𝑹𝟐 + 𝜌2 𝜌
2
𝑜𝑟
2
𝑟22 = 𝜌22 + 2(𝑹𝟐 . 𝜌
̂)𝜌
2 2 + 𝑅2
(6-333)
(6-334)
Equations 6-330 and 6-334 are two equations in two unknowns ρ2 , r2 .
Substitute from Equation 6-330 into Equation 6-334
2
∴
𝑟22
𝜇𝐵
𝜇𝐵
= (𝐴 + 3 ) + 2𝐸 (𝐴 + 3 ) + 𝑅22
𝑟2
𝑟2
(6-335)
where E = ⃗⃗⃗⃗
R2. ρ
̂2
Rearranging and Let = 𝑟2
𝛼 = −(𝐴2 + 2𝐴𝐸 + 𝑅22 )
(6-336)
𝑏 = −2𝜇𝐵(𝐴 + 𝐸)
(6-337)
𝑐 = −𝜇 2 𝐵 2
(6-338)
∴ 𝑥 8 + 𝑎𝑥 6 + 𝑏𝑥 3 + 𝑐 = 0
(6-339)
Solve numerically for 𝑥, then substitute 𝑅2 into 6-330 to get 𝜌2 . From
Equation 6-265 we can get 𝜌1 and 𝜌3 , similar to 𝜌2
∴ 𝒓𝟏 = 𝑹𝟏 + ρi ρ̂i i = 1,2,3
(6-340)
We can find 𝑽𝟐 From Equation 6-302 as follows:
𝒓𝟐 =
1
𝑔1
𝒓 𝟏 − 𝑽𝟐
𝑓1
𝑓1
(6-341)
Solve for 𝑽𝟐
∴ 𝑽𝟐 =
1
(−𝑓3 𝒓𝟏 + 𝑓1 𝒓𝟑 )
𝑓1 𝑔3 − 𝑓3 𝑔1
(6-342)