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6 Case I: Radial, Transverse, and Orthogonal Components

# 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

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

rn

Ä 2 2

Á sin f 0

Á

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

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

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)

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6 Case I: Radial, Transverse, and Orthogonal Components

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