6 Case I: Radial, Transverse, and Orthogonal Components
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274
11 General Perturbations Theory
can be made as follows. If cj represents any one of the elements, then
@R
@r
D rR
; j D 1; : : : ; 6
@cj
@cj
(11.119)
O So we need only evaluate
where r D xOi C yOj C zk.
@r
@R
DF
@cj
@cj
(11.120)
in terms of the force components R0 ; S0 ; W 0 , and use these partial derivatives in
Eqs. (11.108)–(11.114). As an example, we consider @R=@a. Using the notation
O are unit vectors of a perifocal frame (see Sect. 5.2),
where PO and Q
O
r D xOi C yOj C zkO D PO C ÁQ
(11.121)
@
@r
O @Á
D PO
CQ
@a
@a
@a
(11.122)
@
@R
O @Á
D F PO
CF Q
@a
@a
@a
(11.123a)
Thus,
Equation (11.120) yields
where by Eq. (11.84)
@
D cos E e D
@a
a
p
@Á
Á
D sin E 1 e2 D
@a
a
(11.124a)
(11.124b)
In Eq. (11.115) we defined
F D R0 uO r C S0 uO Â C W 0 uO W
(11.125)
O Q
O are related as shown in Fig. 11.2. So
The unit vectors uO r ; uO Â and P;
F PO D R0 cos f
S0 sin f
(11.126)
O D R0 sin f C S0 cos f
F Q
(11.127)
11.6 Case I: Radial, Transverse, and Orthogonal Components
275
Fig. 11.2 Relation of
perifocal and polar unit
vectors
Also,
D r cos f ; Á D r sin f , so Eq. (11.126) becomes
1
F PO D ŒR0
S0 Á
r
O D 1 ŒR0 Á C S0
F Q
r
(11.128a)
(11.128b)
Substituting Eqs. (11.128) and (11.128) into (11.123) we have
1
@R
D R0 .
@a
ra
2
C Á2 / D
r 0
R
a
(11.129)
This is the partial derivative for the force function postulated. In a similar manner,
the other derivatives can be found. These are
r
@R
D R0
@a
a
Ä
@R
r
D R0 a cos f C S0 a sin f 1 C
@e
a.1 e2 /
p
R0 ea sin f
@R
S 0 a 2 1 e2
D p
C
@
r
1 e2
@R
D S0 r cos i W 0 r sin i cos u
@
@R
D S0 r
@!
@R
D W 0 r sin u
@i
(11.130a)
(11.130b)
(11.130c)
(11.130d)
(11.130e)
(11.130f)
276
11 General Perturbations Theory
Substituting Eqs. (11.130) into Eqs. (11.108), we have the element rates
2e sin f 0 2aÁ 0
R C
S
nÁ
rn
Ä 2 2
Á sin f 0
Á
aÁ
r2 0
eP D
R C 2
S
na
a ne
r
Ä 2
Ä
Á cos f
2r
r
Á2 sin f
0
P D
1 C 2 S0
R
2
2
ane
na
a ne
aÁ
aP D
P D
!P D
r sin u
W0
a2 nÁ sin i
Ä
Á cos f 0 Á sin f
r
R C
1 C 2 S0
ane
ane
aÁ
r sin u cot i 0
W
a2 nÁ
r cos u 0
di
D 2
W
dt
a nÁ
(11.131a)
(11.131b)
(11.131c)
(11.131d)
(11.131e)
(11.131f)
(11.131g)
p
where Á , 1 e2 . Equations (11.131) are the Gauss variational equations. In a
P and aP , independent of , can be transformed. We
similar manner the equations for M
can see clearly from Eqs. (11.131d) and (11.131g) that the spatial orientation of the
orbit will change only when there is a component of specific force W 0 (McCuskey
1963, pp. 144–147).
11.7 Case II: Tangential, Normal, and Orthogonal
Components
We wish to express the radial and transverse components of Case I in terms of the
tangential and normal components. Let
G D T 0 uO T C N 0 uO N C W 0 uO W
(11.132)
define the force. uO T ; uO N ; uO W are unit vectors along the tangent in the direction of
motion, along the normal to the orbit directed toward its concave side (i.e., towards
the attraction center), and perpendicular to the orbit in the sense uO W D uO T uO N ,
respectively. So W 0 is the same as W 0 of Case I. This triad forms a coordinate system
which is sometimes referred to as an NTW frame.
11.7 Case II: Tangential, Normal, and Orthogonal Components
277
We are concerned with transforming from R0 ; S0 to T 0 ; N 0 . The radial direction
component of G is
R0 Á G uO r D T 0 .uO T uO r / C N 0 .uO N uO r /
(11.133)
and in the transverse direction it is
S0 Á G uO Â D T 0 .uO T uO Â / C N 0 .uO N uO Â /
(11.134)
The scalar products are evaluated knowing that uO T and uO N are along and
perpendicular, respectively, to the velocity vector. So we may write
v D rP uO r C rfP uO Â D v uO r
(11.135)
rP
D uO r uO T D uO Â uO N
v
(11.136a)
Then the scalar products
rfP
D uO Â uO r D
v
uO r uO N
(11.136b)
The first parts of Eq. (11.136) can be seen from dot products of Eq. (11.135) by uO r
and uO Â , respectively. The second parts are from the unit vectors geometry. From the
elliptic motion we have
a.1 e2 /
1 C e cos f
p
r2 fP D h D k2 Ma.1
rD
(11.137a)
e2 /
(11.137b)
Evaluation of rP from the first and elimination of fP by means of the second yields
p
k Me sin f
rP D p
a.1 e2 /
p
k M.1 C e cos f /
rfP D
p
a.1 e2
p
k M.1 C e2 C 2e cos f /1=2
vD
p
a.1 e2 /
(11.138)
(11.139)
(11.140)