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A I LONG-TERM MOTION OF A LUNAR SATELLITE Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland https://ntrs.nasa.gov/search.jsp?R=19660007958 2018-09-14T09:36:28+00:00Z
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Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

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Page 1: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

A

I

LONG-TERM MOTION OF A LUNAR SATELLITE

Frances A. Frost

October 1965

Goddard Space Flight Center Greenbelt, Maryland

https://ntrs.nasa.gov/search.jsp?R=19660007958 2018-09-14T09:36:28+00:00Z

Page 2: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

P

CONTENTS

Page - . -

I. INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

11. THE HARMONIC ANALYSIS METHOD . . . . . . . . . . . . . . . . . . . . . 4 s * -

1x1. THE SOLUTION WITH ELLIPTIC INTEGRALS . . . . . . . . . . . . . . . 15

IV. CONCLUSIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

iii

Page 3: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

LONG-TERM MOTION OF A LUNAR SATELLITE

I. INTRODUCTION

The purpose of this report is to examine two special methods of studying the long-term motion of a lunar orbiter. The Hamiltonian used is the long- period Hamiltonian (in which the argument of pericenter and inclination of the satellite zppzr kt other argdar v$r.i&Les have been eliminated) as derived by Kozai and Giacaglia, et. al., (References 2, 3). It includes the perturbation of the earth as a point mass and the principal part of the oblateness of the moon.

This long-period Hamiltonian is

where + = 1 - e Z

e = eccentricity of the satellite's orbit

p = gravitational constant times the mass of the moon = 3.6601891 x decamegameters3/centiday2

a = semimajor axis of the satellite's orbit (decamegameters)

i = inclination of satellite orbit plane to the moon's equatorial plane

g = argument of pericenter of the satellite

1

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L

3 4 K, = - (centiday - ')

8~ n

ne = mean motion of the moon = 2.2802713 x (radians/centiday)

n = i . ~ 2 L - 3 = mean motion of the satellite

E = ratio of sum of masses of earth and moon to the mass of the earth = 1.0123001

3 K = - J b2 n2 (decamegameters 2/centiday2) 2 2 4 J = principal part of the oblateness of the moon

2 = 2.41 x (Reference 2)

b = mean radius of the moon = 0.1738 decamegameters

Since the Hamiltonian (1) is not an explicit function of time, i t has a constant value, F = 1/6 C .

Special methods are needed for the problem of the long-term motion since the standard procedures - von Zeipel's method and the method of successive approximations - break down at this point for the lunar satellite (e.g. see Ref- erence 2). The special methods, which involve harmonic analysis and elliptic integrals, t o be considered in this report were suggested many years ago by E . W. Brown in his paper "On the Stellar Problem of Three Bodies'' (Reference 1).

In the method involving harmonic analysis it is assumed that T~ and d t /dg may be represented by cosine ser ies with the argument of pericenter g as the variable. Integrating dt/dg gives g as a sum of a linear expression of time and a sine series in g . On reversing this expression we have g as a function of time. Then ~2 is approximated as a cosine series in time. In the second of these methods proposed by Brown,

2

. .

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4

and an approximation of cos 2g found by neglecting the oblateness of the moon a r e used to give dT'/dt and dg/dT2 as functions of 77' alone. These result in elliptic integrals of the first and third kinds a s the solutions for qi as a function of time and g as a function of q2 (and therefore of time).

Fo r the harmonic analysis method it is necessary to solve the Hamiltonian for '1' corresponding to chosen values of the argument of pericenter. To be able to compare the values of 7' and g as obtained from harmonic analysis and elliptic integrals the portion of the Hamiltonian containing the oblateness of the moon was neglected, both in solving for q' and evaluating d t i d g . Setting J z = 0 results in the following quadratic equation in x = 7' :

3 ( 1 - 5 c o s 2 g ) x ' - [ 5 + 9V2-15cos2g(l+v') tC'1 x -t 1 5 v 2 (1 - c o s 2g) = o

where

and v = H/L are constants. 77: is the value of 7' when g = 0. If the oblate- ness of the moon is included in the Hamiltonian, however, an approximation used in this report is to represent ( 1 - e2)-3/2 by its Maclaurin series terminating the series to give the quadratic equation in e2 (for small to moderate values of e )

where

3 4

a = 3K1 L t - % (- 1 + 15 v') - 1 SKILcos 2g

P = (KIL t Kz) (-1 t 9v2) + C + 15 K,L(1 - v') C O S 2g

Y = 2(K1L + K2) ( - 1 t 3v') - C.

3

Page 6: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

This quadratic is then solved for T~ = 1 - e,, given specific values of cos 2g:

(the sign is chosen so that T~ lies between zero and one).

11. THE HARMONIC ANALYSIS METHOD

Given the initial conditions

L = 0.06 C = 0.055 H = 0.05 g = o

we find

e = 0.39965264 a = 0.98 355573 decamegameters i = 24.619974O

K, = 3.105573 X

K, = 2.1003149 X 10-8 C = 7.6893300x

We assume

q2 = % c o s 2kg and - d t - - c b k c o s 2kg, k > 0 dg k > O

terminating the series with c o s 8 g so that

77' = a. t al cos 2g t a, c o s 4g t a3 c o s 6 g t a, cos 8 g

- bo + b, cos 2g t b, C O S 4g + b, c o s 6 g t b, C O S 8g. d t

d g - -

4

Page 7: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

4

T - 2

2g 0 77

q 2 , K, = 0 ,84027778 .77996317 .80665828

, K, f 0 -84027778 .78055119 .80702653

Solving the Hamiltonian for q2 given the five values 2g =0, T , 771'2, 7 ~ / 3 , 2 ~ / 3 , we have the following results:

277 3 3

- 77 -

.82299779 .79219924

-82317074 .79270692

77 - 77 - 77 0 2 3 2g

d t - 3 K, = 0 8781.8656 6732.7808 8268.1582 8735.5882 dg

dt -, K, f 0 8663.7170 6662.5543 8155.7627 8611.8994 dg

Corresponding to each of the five special values of 2g with the associated values of q2, d t / d g is evaluated using

27r 3

7538.4678

-

7446.8692

Then the following table may be used to solve for the coefficients in the series for dt /dg :

Letting f, denote the kfh value of in the table, the equations

f = a. t al t a , t a3 t a4

f , = a0 -a l t a , - a3 + a 4

f, = a0 - + a4

1 1 1 f , = a,, +- al 2

1 a, t a3 -- as 1 1 f = a. - - a l - - - z - a3 -Ta4

5 2 2 2

5

Page 8: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

c

may be solved for the coefficients ak in the terminated series expansion for 7 , . For K, = 0 we then have

a. = - 1 ( f l t f , ) t- 1 ( f , t f5) = .80843917 6 3

a l =- 1 ( f l - f , t f, - fs) = .03637106 3

= .00173110 1 1 4 2

1 1

a, = - ( f l t f , ) -- f 3

a3 - 6 5

a4 = f , - a. t a, _ - - .00004979.

- _ ( f l - f , ) -5(f4 - f ) = - .00021374

Thus, if K, = 0,

q2 = .80843917 t .03037106 COS^^ t .00173110 C O S 4g ( 2 )

- .00021374cos 6 g - . 0 0 0 0 4 9 7 9 ~ 0 s 8 g

while, if K, # 0,

r12 = ,80876405 + .03006347 C O S 2g t .00169398 C O S 4g

Similarly, the five coefficients bk in the terminated cosine series for d t /dg may be evaluated,so that if K, = 0,

d t - = 8010.4597 t 1082.0684 C O S 2g - 255.4175 C O S 4g dg

- 57.5260 c o s 6 g t 2.2810 c o s 8 g

6

a, = - 1 ( f l t f , ) -- 1 f 3 = .00173110 4 2

a3 = - 1 ( f l - f , ) -5(f4 1 - f5) = - .00021374 6

a4 = f , - a. t a, = - .00004979.

Thus, if K, = 0,

( 2 ) q2 = .80843917 t .03037106 c 0 ~ 2 g t .00173110 C O S 4g

- . 0 0 0 2 1 3 7 4 ~ 0 ~ 6 g - . O 0 0 0 4 9 7 9 C 0 ~ 8 g

while, if K, # 0,

r12 = ,80876405 + .03006347 C O S 2g t .00169398 C O S 4g

Similarly, the five coefficients bk in the terminated cosine series for d t /dg may be evaluated,so that if K, = 0,

d t - = 8010.4597 t 1082.0684 C O S 2g - 255.4175 C O S 4g dg

- 57.5260 c o s 6 g t 2.2810 c o s 8 g

Page 9: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

and, if K, # 0 ,

dt - = 7907.3015 + 1055.3976 COS 2g - 246.3136 C O S 4g dg

- 54.8163 C O S 6g t 2.1476 C O S 8g.

Integrating this expression for d t /dg gives

t = bog t- bl sin 2g + - b2 sin 4g t- b3 sin 6g t - b4 sin 8g 2 4 6 8

(the constant of integration i s zero in this example because of the initial condi- tion g = 0). Then

In particular ,

(K2 = 0) g = .12483678 x 1O-j t - .06754097 sin 2g

t -00797137 sin 4g t .00119689 sin 6g - .00003559 sin 8 g

(K, # 0) g = .12646539 x t - .06673563 sin 2g t .00778754 sin 4g

t .00115539 sin 6g - .00003395 sin 8 g .

Since we may write g = 7 + cp(g) , where 7 is a linear function of time and (p( g) i s a trigonometric function in g , Lagrange's expansion theorem may be applied to reverse the equation and give g as a function in time. By this theorem, if

Page 10: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

then a series may be used to express y as a function of X:

In this case, only the first four terms of this series were taken and all terms after si n 8 g were eliminated.

Hence,

(C + J 2 6 t!!) s i n 6 7 t (d +;+:) s i n

where

k = O 2

a = -.067540965 b = .0079713748 C .0011968 934 d -.000035595648

A = - 2(ab + b c + cd)

q = 2(a2 - 2ac - 2bd)

k, f 0

-. 06673 563 .007 78 7 54 .00115539

- .000033 9 5

J = 6 a ( b - d )

D = 4(2ac t b2)

8

Page 11: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

P

n - 2

,77996317

7r - 2

Y

.71946556 .47651148 .97849565

.81080056 .82570548 .79556239

.72021381 .47693763 .97942282

ab2 a3 - b2c - bcd --- d2c - adb + - - a 2 c -- C 6

.78055119

- 2abc - 2b2d - bc2 - 2a2d - d3 - 2dc2

.81107413 ,82582292 .79599495

Thus, for K, = 0, b

g = r - .06685092 s in 27 t -01255112 sin 47

- .00091832 s i n 6 7 t .00007192 s in 87 (3)

where r = 0.12483678 X t ( t in centidays), while i f K, # 0,

g = 7 - .06606959 s i n 27 t .01225829 s i n 4 7 - .00088524 sin 6 7

t .00006836 s i n 87. (3')

( 7 = .12646539 X t, t in centidays).

Letting 27 take on the five special values 0, n, n / 2 , n / 3 , 2 n i 3 , these equations for g give corresponding values of 2 g ; from these we may evaluate '1, using the terminated cosine series (2) and (2'). This gives the following results:

l o g , K, = O

I 77, , K, = 0 34027778 I L

9

Page 12: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

a

Assuming q2 may be written as a cosine ser ies in 7 and taking the first five terms of the series, we may evaluate the coefficients of these te rms as above. Thus, we have for K, = 0,

q2 = .81046278 t .03015257 cos 27 - .00034004 c o s 47

t .00000474 C O S 67 - .OOOOO226 C O S 8 7

(7 = .12483678 x t )

and for K, f 0 ,

v2 = .81074412 t .02985152 cos 27- - .00032982 cos 47

t .00001177 COS 67 + .00000019 c o s 87. (4')

( 7 = .12646539 x t)

In this way both g and q2 may be found as functions of time.

Tables 1 and 2 give the values of T* and g . Equations (4) and (4') a r e plotted against time in Figure 1. This shows the shift that occurs when the oblateness of the moon is considered in the long-period Hamiltonian. Figure 2 shows similar results for g as given by Equations (3) and (3').

As may be seen from these figures, the effect of the moon's oblateness on a satellite at this distance from the moon (5.659 moon radii) is not great. For a satellite near the surface of the moon, the effect of the moon's oblateness on the eccentricity and argument of pericenter of the satellite would be much more pronounced.

10 I

Page 13: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

0 20 40 60 80

100 120 14 0 160 180

200 220 240 260 280

300 32 0 34 0 360 380

400 42 0 440 46 0 480

500 52 0 54 0

.84027782

.83675070 A2692477 .81298783 .79215000

.78627455

.78029692

.78191710

.79068695

.8042432 7

.81911010

.83167979

.83906558

.8396547 8

.83332190

.82144059

.80671501

.79268607

.78291847

.78002!546

.7 8480573

.79594596

.81050099

.82482343 A3551665

.84017939

.83780875

.82891087

Table 1 +, g Harmonic Analysis (Kz = 0)

g Radians

0 .2273639 .4543801 .a350423 -92 871 34

1.1952296 1.4843190 1.7799106 2.0606739 2.3168499

2.5537270 2.7820046 3.0090020 3.2365252 3.4636120

3.6913262 3.9260582 4.17 83800 4.4552836 4.7497567

5 -0413928 5.3120934 5.5589943 5.7910522 6.0182398

6.2454964 .1897622 .4167136

g Degrees

0 13.02699 26.03406 39.25093 53.21135

68.48161 85.04521 101.98136 118.06792 134.33726

146.31778 159.39712 172.40312 185.43923 198.45035

211.49741 224.94656 239.40354 255.26 894 272.14101

288.85053 304.36053 318.50691 331.80284 344.81974

357.84058 10.87257 23.87 593

F x

.7 60 86 524

.76086623

.76086372 - 7 60 86474 .76086799

.76085606

.76086390 -7 60058 84 .76086051 .76086867

.7 6086276

.76086533

.76086577

.76086555

.76086267

.76086784

.76086301

.76086301

.760857 03

.76086506

.76085565

.7 6086641

.7 6086596

.76086308

.76086616

.76086530

.76086611

.76086439

F - C x 10-10

0 .089812 -.160298 -.059685 .266027

-.927684 -.143245 -.649720 -.4 81463 .334239

-.2575 0

.043769

.022168 -051159

-.266027 .250679 -.232489 -. 829345 -.027284

-.967474 .lo7434 .062527 -.225099 .082991

-. 0034 10 .077875 -,093791

11

Page 14: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

Table 2 T', g: Harmonic Analysis (Kz # 0)

0 20 40 60 80

100 120 140 160 180

200 220 240 260 280

300 320 340 360 380

400 420 440 460 480

500 520 540

.84027778

.83668256

.82670981

.81265809

.79790366

.78620654

.78072811 .78300817 .79240119 .80631778

.8 2111 800

.83316153

.83963119

.83905608

.83156597

.81886887

.80395451

.79050779

.78212960

.78109380

.78771252

.80013334

.81502776

.82865910

.83771752

.84018579

.83 548432

.82467794

g Radians

g Degrees

0 .23048543 .46072723 .69497792 .94291820

1.2142525 1.5074170 1.8058868 2.087721 8 2.3446295

2.5832362 2.8143461 3.04451 64 3.2751 774 3.5053372

3.7369943 3.9777790 4.2384158 4.5239208 4.8229269

5 S136422 5.3809400 5.625948 5.8590324 6.0893168

.03656091

.26 6 9 5 049

.497 3 894 6

0 13.205842 26.397726 39.819302 54.025233

69.571 544 86.383185

103.46969 11 9.61 765 134.33738

148.00853 161.2501 6 174.43794 187.65384 200.84103

214.1 1400 227.90995 242.84334 259.20157 276.33336

292.99012 308.30515 322.34308 335.69783 348.89215

2.094786 15.295136 28.49 8317

12

F x 10-5

-76393302 .76898250 .76906286 .76911933 .76917259

.76921679

.76926319 .76923517 .76919183 .76914220

.76908830

.76901856

.76 894303

.76895149

.76903174

.7690969 5

.76915085

.76919868

.76924 15 5

.76924995

.7 69209 82

.76916465

.76911089

.76905185

.76896972

.7 6 893447

.76899601

.76907304

0 .0495106 .1298758 .1863440 .2395950

.28379 64

.3201989

.3021796

.2588421

.2092065

.1553075

.0855607

.0100385

.0184968

.09 874 83

.1639591

.2178580

.2656861

.3085574

.3169589

.2768274

.2316596

.1778971

.1188595

.0367322

.0014779

.0630166

.1400508

Page 15: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

. I

~

. .

..

- _

N c 0 C

9'

'0s v113

13

Page 16: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

I I I I \ \ I I I I I I I I 1 I 1

I I 1 IVI I I I I 1 I I I I 0 I l l I U l I I I I I 1 I I Y) m

a s 0 N

0 C VI

c

.- cy

Y

v) .- cy

Y

14

Page 17: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

. III. THE SOLUTION WITH ELLIPTIC INTEGRALS

The second method suggested by Brown, starting with 4 and g, results in elliptic integral expressions for 77, and g, but for this solution the oblateness portion of the Hamiltonian is neglected.

From the long-period ami l ton ian (1) and the definition of the Delaunay variables, we have

d77 - 1 dC - 1 3F - d t L d t L a g - _ - - _- - -

Setting K, = 0 in the Hamiltonian gives

where the constant W is evaluated for g = 0. 77: is the value of q2 when g = 0. On solving the Hamiltonian (6) for cos 2g, we have

Then, w = 2(5 - 677: t 3v2) where

which may be substituted in (6) f o r s i n 2 g = [ l - cos2 2eJ112 to give

- _ dx - 2773 = T 4&K1 [(x - xl)(x - x,)(x - x3)I1I2, d t

where x = T ~ , x1 = 77: and the roots x2,x3 of the polynomial are defined by the equations

5 x2x3 =3 v2

1 3

x, t x3 =- (5 t 5v2 - 2x1).

In this expression for dx/dt the minus sign is used when sin 2g is positive and the plus sign when s i n 2g is negative.

15

Page 18: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

If the roots of the polynomial are relabeled to give a strict ordering xi < xk < xi with the relationship x' 1 - < x - < xi, we have

= T 4 K , v'6 dt dx

J (x - x i ) (x - x;, (x - xi)

Integrating both sides gives

,.X rt

dx = T 4 K , V'K J dt J (x - x;, (x - x;, ( x - x;, ' x o

where xo is the initial value of x and to may be taken as zero.

Set

x; - x; y 2 x =

1 - y2

and we have

dY Y 2 7 4 K 1 tv% = rn I, J (1 - k2 y2) (1 - y2)

16

Page 19: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

d

L

or

. .

,.

dY = T 2 K 1 t ,/- JY (1 - k2 y2) (1 - y2) y o

Let us define

v = 2K1 t J ~ ( x ; - xi)

Then

, / x x - dx

2 J(x - xi) (x - xi) (x - xi)

dx dx

- xi)(. - xi) (x - x‘) 3 (x - x’) (x - x’) (x - XI) 4: J 1 2 3

where

17

Page 20: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

c

so that

dY J(l - y2)(1 - k2y2)

Thus we may solve for x as a function of time, using the Jacobi elliptic function of u and v:

x = x l l t (x2' - xI1) sn2 (u i v)

s n u c n v dn v f s n v cn u dn u

1 - k2 sn2 u s n 2 v = X l l t (x21 - XI')

For the given initial conditions we have

xi = .77996317

xi = .84027778

xi = 1.48392569

k2 = .08567872

sn u = 1

c n u = O

cn v dn v 1 - .08567872 s n 2 v

X = .77996317 t .06031461

18

Page 21: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

To evaluate g , the Hamiltonian is solved for cos 2g, giving I

2K, 1 - 3u2x-1 t (5- 3x) (1 - 31. x3/2

This expression is substituted for cos 2g in

Since

* dg dx dg dx d t dx

g = - * - = 4& K, - ,

where Q(x) = (x - xl)(x - x2) (x - x3), this substitution gives

Integrating this expression, we have

?4&Klg=-[- 1 c t2(1-3v2)K1] f dx

3 L xO (1 - X ) r n

--[- u2 t 2(3u2- 5 ) K j I dx

(x - u2) +) 0

3 L

19

Page 22: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

- -[- 1 c t 2 ( 1 t 3v2) Kl] [ dx

3 L fm-)

2 K 2 dx dx 0 .

+- - (1- 372) 3 L

dX .L- - K2 (v2 - 3 v 4 - 2 )

- - - K 2 ( 7 t 6 v 2 )

3 v2L

dx dx 3 L x 3 v Q o e

In actually evaluating g only the first three summands of the preceding were used since the oblateness of the moon was omitted from the Hamiltonian. The elliptic functions were evaluated using Reference 6.

IV. CONCLUSIONS

The results of these two special methods involving harmonic analysis and elliptic integrals are compared using the given initial conditions and the harmonic analysis solutions (3) and (4) for '12 and g when K, = 0. These results a re tabulated in table 3 together with the differences between the values obtained by the two methods .

In table 2 are the values of T~ and g when the oblateness of the moon is

The value of the Hamiltonian (1) was computed for ^r12 and g retained in the Hamiltonian and a Maclaurin ser ies is used to give a quadratic equation in . e2 . (at 20 day intervals) and compared to the initial value ( T~ = .84027775, g = 0) C = .76893300 x 10-5.

Although g is periodic in this example and all values are admissible, there is the possibility that a lunar satellite may have a stable orbit but that for some I

value go of the argument of pericenter, dg/d t will be zero so that d t,/dg will be undefined. In such an event the special values of 2g used to evaluate the coeffi- cients in the cosine series representation of 772 and dt/dg must be restricted to the regionof admissiblevalues, i.e., go < g < (n - go ) .

20

Page 23: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

a

.

t (days)

0 20 40 60 80

100 120 14 0 160 180

2 00 220 240 260 280

300 320 340 360 3 80

400 420 440 460 480

500 520 540

Table 3 +, g ; Elliptic Integrals and Harmonic Analysis (J2 = 0)

TI2

Integrals Elliptic

.&IO27778

.03U I -x I I 4

.82691973

.81298180

.79820373

.78626473

.78029580

.78191519

.79068037

.80424132

A1911161 A3167986 .83906625 A3965270 .83331311

.82142913

.80670000

.79266806

.78290935 ,78002606

.78480504

.7 95946 5 8

.81050828 A2483048 .83552045

.84018034

.83 7 801 04

.82889504

noCl7Avv

r12 Harmonic Analysis

.8402 77 7 8

.836750?0

.82692477

.E31298783

.79821500

.7 8627455

.78029692

.78191710

.79068695

.80424327

.81911010 A3167979 A3906558 A3965478 A3332190

.82144059

.80671501

.79268607

.78291847

.78002546

.78480573

.79594596 ,81050099 .82482343 ,83551665

A4017939 .83780875 .82891087

I b 2 1

0 x 10'5 2 9 6 .504 .603 1.127

.982

.112

.191

.658

.195

.151

.007

.067

.208

.879

1.146 1.501 1.801 .912 .060

.069

.062

.729

.705

.380

.895

.771 1.583

(radians)

0 .22747356 .45445976 .68513324 .92 893 74 0

1.1954244 1.4844038 1.7799258 2.0605636 2 ,3167928

2.5537699 2,7820064 3.0090287 3.2366909 3.4638279

3.6914969 3.9263168 4.1787443 4.4555575 4.7499516

5.0414 831 5.3120974 5.5591051 5.7911936 6.0183377

6.2456836 .19007482 .41701462

0 .2273639 .4543801 .6 85 042 3 .9287134

1.1952296 1.4843190 1.7799106 2.0606739 2.3168499

2.5537270 2.7820046 3.0090020 3.2365252 3.4636120

3.6913262 3.9260582 4.1783300 4.4552326 4.7497667

5.041392 8 5.3 12 0934 5.5589943 5.7910822 6.0182398

6.24 549 64 .1897622 .4167136

~

0 x 10- 1.0966 .7966 .9094

2.240

1.946 .848 .152 1.103 .571

.429 ,018 .267 1.657 2.159

1.707 2.586 4.143 3.249 1.909

.903

.040 1.111 1,314 .979

1.872 3.1262 3.0102

21

Page 24: Frances A. Frost - NASA · Frances A. Frost October 1965 Goddard Space Flight Center Greenbelt, Maryland ... q = 2(a2 - 2ac - 2bd) k, f 0 -. 06673 563 .007 78 7 54 .00115539 - .000033

The harmonic analysis method makes it possible to include the oblateness of the moon when finding T~ and g as functions of time. Greater accuracy may be achieved by the retention of terms of the series after 8 g or 8$. An advantage of this method is the ability to include the oblateness of the moon. This is espe- cially advantageous i f the Orbiter is to be near the moon where the effect of the moon's oblateness will be greater.

1.

2.

3.

4.

5.

6 .

REFERENCES

Brown, E . W. "The Stellar Problem of Three Bodies," Monthly Notices of the Royal Astronomical Society, Vol 97, 56-66, 116-127, 388-395 (1936).

Giacaglia, Georgio E . 0. et al., The Motion of a Satellite of the Moon. GSFC X-547-65-218, June 1965.

Kozai, Y. "Motion of a Lunar Orbiter," Journal of the Astronomical Society of Japan, Vol 15, 301-312 (1963).

Byrd, P., and Friedman, M., Handbook of Elliptic Integrals for Engineers and Physicists, Springer-Verlag, 1954.

Burington, R.S., Handbook of Mathematical Tables and Formulas, Handbook Publishers, Inc., Sandusky, Ohio.

Velez, Carmelo E., Subroutines for Elliptic Functions, F 178, 179, 180. June 1965.

' .

22