Smithsonian

Contributions to Astrophysics

VOLUME 5, NUMBER 9

ROTATION OF AN EARTH SATELLITE

IN FLIGHT ALONG ITS ORBIT

By YUSUKE HAGIHARA

SMITHSONIAN INSTITUTION

Washington, D.C.

1961
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Contents

Pace

Introduction 113

Equations of motion 113

Equatorial circular orbits 119

Inclined circular orbits 124

Eccentric orbits 134

Summary 139

Keferences 142

Abstract 143

in

Rotation of an Earth Satellite

in Flight along its Orbit

By YUSUKE HAGIHARA

 1

Introduction

The translational motion of an earth satellite as affected by radiation pressure (Musen, 1960), the

electrostatic charge induced in the satellite body by the charge distribution in space, the electric

current induced by the satellite's motion through the geomagnetic field (Jastrow and Pearse, 1957;

Wyatt, 1960; Beard and Johnson, 1960), and the induced magnetization caused by the geomagnetic

field (La Paz, 1960) have been discussed recently. The rotational motion of an earth satellite,

however, has not yet been worked out in detail, except for the work of Roberson (1958) and others on

the gravitational torque.

In the present paper I propose to discuss the rotational motion of an earth satellite during the

period of its flight along its orbit around the earth's center, and shall base the discussion on Euler's

equations of motion for the rotation of a rigid body around its center of mass. The body of the

satellite is supposed to be symmetrical, not only in its external shape but also in its dynamical

properties about its axis of symmetry; that is, if we denote the principal moments of inertia by

A, B, C, then we have A—B. Although it is assumed that C<CA in the present discussion, which

applies to a body of a long prolate spheroidal shape, the result can easily be applied to a body of a

flat oblate spheroidal shape, merely by changing the sign of A— C, at least for the tidal torque.

The effect of tidal torque is fully discussed in the present paper for the cases of a circular and an

elliptic orbit. The figure of the earth is assumed to be spherical. The instantaneous orientation of

the satellite body is referred to a coordinate system that is fixed in space. It will be shown that this

frame should be such that its Z-axis is directed to the north pole of the orbital plane of the satellite,

and that the body makes a precessional motion of long period around this Z-axis; on this motion is

superimposed the nutation of short period, of the order of the satellite's orbital period.

The hydrodynamic torque is considered on the basis of an irrotational motion for an incompressible in viscid fluid, as a rough approximation until a rigorous aerodynamic treatment is available.

The solution of the equations of rotational motion, including the hydrodynamic torque, is highly

complicated, because the air density varies with height above the earth's sea-level; the integration

is thus carried out by expanding in powers of the orbital eccentricity.

The effect of the magnetic torque due to the induced magnetization caused by the geomagnetic

field is considered. The geomagnetic field is assumed to be a dipole whose pole coincides with the

earth's pole.

The radiation pressure may produce a torque on the satellite. However, if its external shape is

symmetrical, the resultant of the radiation pressure acts at the center of mass of the satellite and

we can expect scarcely any torque due to the radiation pressure, although the effect on the translational motion is evidently very important. The electrostatic charge induced by the charge distribution in space, and the distribution of electric current inside the satellite induced by its motion through

the geomagnetic field, may also produce torques. However, if we assume a symmetrical structure

of the satellite body, then these effects are considered to be insignificant. Hence we shall neglect

1

 Smithsonian Astro physical Observatory.
114 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS

here the torques due to radiation pressure, electrostatic charge, and electric current distribution

inside the body.

The solution given in the present paper can be tested by photometric observations of the

variation in light of an earth satellite or by ultrahigh-frequency radio echoes from an earth satellite,

when we have numerical values available for the figure and structure of the satellite and its orbit,

and for the variation of the orbital elements.

Equations of motion

Consider an earth satellite of long cylindrical shape whose principal moments of inertia are A, B,

and C; A=B and C<C.A, the C-axis being taken along the longest axis of the cylinder. The satellite is supposed to move in a field of force represented by the force function U for the torque. Denote the angular velocities around the principal axes of inertia by al} o>2, and «3. Then Euler's

equations of motion for the rotation of the satellite around its center of mass (see, for example,

Klein and Sommerfeld, 1923; Whittaker, 1927; Charbonnier, 1927; Deimel, 1950) are:

(1)

where Kia1} K2O2, K3O}3 represent the rate of retardation of the angular velocities around the three

principal axes of inertia due to any cause.

Take coordinate axes of fixed direction in space with the center of mass of the satellite as the

origin; for example, the equatorial axes for which the Z-axis is directed parallel to the earth's north

FIGUU 1.—Relation between the coordinate system (XYZ) fixed in space and the routing coordinate system (xyx)

attached to the satellite, with the Eulerian angle
NO. » ROTATION OF AN EABTH SATELLITE 11 5

pole, the JY-axis in the plane of the earth's equator and directed to the vernal equinox, and the Yaxis in the plane of the earth's equator and perpendicular to the JST-axis. Denote the Eulerian

angles of the principal axes of inertia of the satellite referred to these (-XTZ)-axes by 0, <p, and ^.

Then we have, as shown in table 1 (p. 142), the direction cosines of these fixed (XYZ)-axes and the

rotating axes (zyz) that are attached to and carried by the body of the satellite. The x-, y-, 2-axes,

which^are, respectively, in the direction of the A-, B-, C-axes of the principal moments of inertia

of the satellite, are illustrated in figure 1.

Then we have the relations:

dB . , d<p . n , dd . . d<p . . . , d\p, d<p . ,_,.

a>i=-j-sin^—-j*rsin0cos^, a>2=-^cos^+-^-sin0sin^, °*==?t^~dtC0S ' ^

or

wi sin ip-\- w2 cos \p =-n> — «i cos ^+a>2 sin $=jf: sin 0. (2a)

^ The force function for the tide-generating action on the satellite is expressed (Plummer, 1918)

by the equation

I^^'-]®' (3)

where m

r

 denotes the mass of the earth, G the constant of universal gravitation, a the semimajor

axis of the satellite's orbit, r the radius vector of the satellite, both counted from the earth's center,

and z is the rectangular coordinate in the direction of the principal axis C of the satellite. Write

C-A 3m'GA E

A —° 2a3 — e»

then we have,

Here the (xyz) -axes are the moving coordinates attached to and carried by the satellite, and their

directions coincide with its principal axes of inertia.

By substituting equation (2) in equation (1) and noting that

. . bU , . W W . ZU . , W W VU W sinv^—rco s ^-^—=^—> cos^ ^ sin ^-5—=—5—1 3r~:saE

Y 3rr»

 dx dy d6 dx dy &p dz d^

we get the equations of motion (1) in the form:

A

 ^A (W - • »• " + c l - > (?M •»»)

=—KI sin ^ (-r: sin \p— ~? sin 0 cos ^ j—JC2 COS ^(JI cos ^+^7 sin 0 sin ^)+-^»

. ,fdd ,.d<p . . . A. I dZ7

K2 sin ^ ( j - cos ^+ JT sin 0 sin \f/)-]—:—- -=—»

\rf< rf< / sin 0 5^

^ /rfV , d?tp n d<pdB . A /^ , ^
SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS TOL. 5

Torque due to resistance.—At first we consider the general behavior of the resisting retardation of

the angular velocities due to the terms — Ki«i, — K2OJ2, — K3W3 of the equations (1). As we have supposed C<CA, we can assume that K! = K2>K3. From equations (1) we have, by neglecting the force

function U,

From the third equation of (5) we get

io3=a exp*j — -gt >, (6)

where a is an integration constant. Inserting this expression in (5) we have

From these we obtain the equations:

Integrating these, we have

" | ^ « y ^t[ ^ exp | -

with two integration constants 5 and i£. From these equations we can see that, if initially a>i=a>2

= 0 , then ajj and o>2 are always zero, and that o>3 varies more rapidly than o>i and OJ2, as C<C^A. We

obtain from these integrals the equations

= cos [^! exp /-g « j-+if] • 5 exp /- | ( j- - (7)

The position in space {0s,<P^) of the axis of instantaneous rotation referred to the (XYZ)-axes

is obtained from table 1 (p. 142) :

sin 6, cos y>s sin 8S sm ^>s cos 03

By substituting equations (6) and (7) in these ratios we get

« , (8)

tan 0,=^ exp /(_|+£ ) ^ . (9)

We can see from these that the polar angle 6, of the instantaneous axis tends to TT/2 for f->°° if

(K,/.A)< (K3/<7) ; and tends to 0 for t-^co if (KI/A)XK3/C). The azimuth angle <p, of the instantaneous

axis decreases constantly towards the value K, which is determined by the initial value of coi/o?2.

Hydrodynamic torque.—Consider an ellipse of axes c>a immersed in an incompressible in viscid

fluid of density p. In a two-dimensional irrotational motion of the fluid across the ellipse with
ROTATION OF AN EARTH SATELLITE 117

FIGURE 2.—Two-dimensional stream lines around an ellipse immersed in an incompressible fluid.

speed V at infinity, while making an angle 5 with the major axis of the ellipse, the torque M exerted

to the ellipse is, according to Cisotti (Lamb, 1932; Milne-Thomson, 1955),

in the sense of increasing 8, or

\M\=irP(<*—a

2

) Vs sin 8 cos 8,

\M\=M sin 5 cos 8, M=TTP{&—a

2

) V2.

(10)

(11)

We extend this result to a three-dimensional motion for an ellipsoid of revolution c>a=b. Then

the same formula (11) can be expected to represent the torque exerted on the ellipsoid moving with

speed V relative to the fluid by making an angle 8 with the major axis c of the ellipsoid.

In the high atmosphere the air density is so low that this hydrodynamic approximation is

certainly not valid. Until a rigorous aerodynamical treatment is available, based on a consideration

of the detailed mechanisms involved when particles of the atmosphere impinge on the surface of

the satellite, we shall tentatively assume the formula (11), with a correcting numerical factor (Rand

Symposium, 1959) which I shall omit.2

There may be a torque due to radiation pressure. If the resultant R of the solar radiation

pressure reacts at a point at a distance D from the center of mass of the satellite in the direction of

its longest axis, which makes an angle <r with the direction towards the sun, then a torque of magnitude RD sin <r will result. However, if the structure of the satellite is symmetrical, then the

resultant of the radiation pressure acts at the center of mass of the satellite and D—0; hence the

torque vanishes, although the effect on the translational motion of the satellite is very important

(Musen, 1960).

The electrostatic charge induced by the charge distribution in space also causes a drag on the

motion of a satellite. The symmetrically induced charge has little effect. The induced current

due to the motion across the geomagnetic field also affects the drag (Jastrow and Pearse, 1957;

Wyatt, 1960; Beard and Johnson, 1960). If the satellite is symmetrically constructed, then these

1

 Personal communication from Dr. C. A. Whitney. The numerical factor is not yet available.

581553—61 2
1 1 8 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS VOL. S

effects produce an insignificant amount of torque on the satellite (Spitzer, 1960). Hence we neglect

all these effects in the present paper.

Magnetic torque.—The effect of magnetic damping on the motion of a satellite has been considered by La Paz (1960) and others. I shall now consider the effect of this magnetic induction

on torque.

If we denote the mean flux of the magnetic induction by B, the dipole moment of a ferromagnetic material placed in a magnetic field H is represented (Spitzer, 1960) by

IT , HX (B—H) • (volume of the satellite)

|L| ^ ,

or

i,, IHI • IBl • (volume of the satellite)

|L|««J—!——•—* -: • sin y,

where y denotes the angle between the magnetic field H and the magnetic induction B, where B=H +

4x1=/xH inside a ferromagnetic material (Jeans, 1940). Here I is the magnetization and n the

magnetic permeability. We assume that the induction is in the direction of the 2-axis of the satellite,

so that we have, with a constant field Ho which will be defined later,

I

B

!=M(J|)CO S y-H0. (12)

With these assumptions we have

.T , T /H V . T Hln • (volume of the satellite) „ oN

' ' = °\H~) sin y cos7t °*" TH- ^ '

the torque being in the direction of turning B towards H, that is, of decreasing y.

The earth is supposed to be a uniformly magnetized sphere of magnetic moment Ms and magnetization Is, where Mg= (4T/3)/*. Then at a point with cylindrical coordinates r and 6 referred

to the earth's center, the radial and the tangential components of the force are poloidal and are

represented by the equations

where 0 is the angular distance from the geomagnetic north pole. Here R is counted positive for the

outward direction and S is positive for the direction of increasing 0, and r0 denotes the earth's radius

(Chapman and Bartels, 1951). Then we have

Assume that the geomagnetic pole coincides with the earth's north pole. The components of the

geomagnetic field in the directions of the (XFZ)-axes at a point of longitude X and latitude fi are

Hx 3 sin/S cos 0 H r 3 sin 0 cos 0 . . Hz 1—3 sin2/3 ,1it*

7—7= — COS X. 7TFF = 1 = Sin X, rrf, = -== (14)

iHl Vl+3sin2/3 |Hj Vl+3i s3 ' |H| Vl+3i 23

In fact,

HZ=S sin B-R cos d=H0 (^J (1-3 sin2 0),

Hs=— S cos d-R sin 0=—Ho ft£\ - 3 sin 0 cos ft
HO. • ROTATION OP A N EARTH SATELLITE 119

where R is the radial direction perpendicular to the Z-axis and X, /3 denote respectively the geographic

longitude and latitude of the point in question at a distance r from the earth's center, under our

assumption.

The first approximation.—We suppose that «3 is very large initially, because of the procedure in

launching the satellite. We put

The value of n is supposed to be very large compared with «2 and «3. Thus we can treat n as a

constant and set d£7/5^=0, as our first approximation. We are left with two equations:

Ad

2d A Sd<p\2

 . . n . „ d<p . o bU d / A < b . . o . „ A bU /1CX A -jji—A I ~ ) sin 6 cos 0+Cn -£ sin 6=^-rt -j-± [ A ^ sin2

 d-\-Cn cos B )=-5r— (16)

dt2 \dt / dt bd dt\ dt / b<p

Observations of earth satellites show that n is of the order of a few seconds and that the precessional

motion for dd/dt and d<p/dt is of the order of a few weeks (P. Zadunaisky, personal communication). Hence we assume that dd/dt and d<p/dt are very small compared with n; we take only

those terms that contain n as a factor and neglect the remaining terms. We take thus

dip^ 1 bU d6=i 1 bU

dt^Cn sine bd9 dt Cn sin 0 b<p'

where

as the equations for determining our first approximation to the solution.

As a first approximation we solve equations (17) and then substitute these first approximate

solutions into the left-hand sides of equations (16) in the terms that are not multiplied by n; we

solve again for 6 and <p, which appear in the terms multiplied by n, as the two new unknown functions in our second approximation. We obtain the change of n from (15) by substituting in it these

second approximate values of 6 and tp.

We are principally concerned with the precessional motion of the satellite, so that we eliminate

short-period variations with periods of the order of the orbital period of revolution by averaging

the quantities in the equations over this short period.

Equatorial circular orbits

Tidal torque.—In a circular orbit we have r=a and from equation (18) we obtain

(19)

The coordinates of the earth's center relative to the center of mass of the satellite are:

Xg=— a cos (o+w), YB=— a sin (H-u), Zs=0, (20)

where v is the true anomaly and w is the longitude of the pericenter from the node which is now taken

as the .X-axis. By referring to table 1 (p. 142), we have

z=XB sin 6 cos <p-\-YB sin 6 sin <p+Zg cos 0. (21)

Hence:

(-) =cos* (v+w) sin2 d cos* v?+sin! (v+a) sin* 6 sin1 y>+2 sin (v+w) cos (»+«) sin1 d sin <p cos <p.
1 2 0 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS VOL. 5

In order to obtain the precessional motion by eliminating short-period terms we take the mean

value of U with respect to time t:

where M denotes the mean motion in the orbital revolution. Hence the equations of motion (17)

become

where E^=~> and #i> 0 for C<A, #!<0 for C>A.

The solution of (22) is

0=0O= constant, p—<p0=EiCos60' (t—10), (23)

where we have taken (p=<po at t=t0. Thus the 2-axis of the satellite turns around the north pole

by making with it a constant angle 0O with the angular speed Ex cos 0O in the direct sense, that is, in

the sense of the orbital revolution of the satellite. This is precession.

The complete equations of motion, before we take the mean value of U, are

=E1 cos0 • [cos2(H-«-*>) + l], (24)

where v=ix (t—tx), if we denote the mean motion by n and the epoch of the pericenter passage by

t\. The 2-axis of the satellite makes a nutation with the period of the orbital revolution of the

satellite superimposed on this precessional motion.

Hydrodynamic torque.—According to equation (11) the hydrodynamic torque M is represented

by |M|=M sin 8 cos 5 in the sense of increasing 5. We first compute the angle 8. For a circular

motion of the satellite we have, from equation (20),

=an sin (w-K>), -^-=—a^ cos

By referring to table 1, we find the angle 5 between the 2-axis of the satellite and the direction of the

relative velocity V of the atmospheric gas and the satellite, with V=|V|:

cos

as

5= —77^ sin 0 cos <p-\—if sin 0 sin v? /F=si n 0 sin (v+«—<p), (26)

Next we compute the direction cosines (I, m, n) of the vector M, which is perpendicular to both

the 2-axis of the satellite and the velocity vector V. They are obtained from the relation,

IV mV nV

dYg y dZg v dZg -y. dXg 7 dXg v dYg v

dt Zi~ dt Xt ~lt*t~~dTZj* ~dTYz~~dTXt

where Xt, Yt, Z, denote the direction cosines of the 2-axis of the satellite referred to the (XYZ)-&xes,

which are fixed in space. The result is

l=—cos(w-K>)cos0, m—— sin (o+w) cos0, n=sin (t?+oj)sin 0sin^+cos (»+w)sin0 cos^>. (28)
NO. » ROTATION OF A N EARTH SATELLITE 121

The directions of the vector M referred to the (xyz)-&xes attached to the body of the satellite are, from

table 1 (p. 142),

\ = l(x)X)-\-m(x,Y)-\-n(x,Z) = —cos (y+o>)(cos tp cos ^—cos 0 sin <p sin $

—sin (v+a>) (sin <p cos ^+co s 6 cos <p sin >p),

My/\M\=l(y,X)-{-m(y,Y)+n(y,Z)=cos (y+a>) (cos <p sin ^+cos 0 sin <p cos #)

+sin (y-fw) (sin ip sin if/— cos 0 cos (p cos $ ,

where (A, B) denotes the direction cosine of the angle between the axis represented by A and the

axis represented by B. The components of M referred to the axes (0, <p, ^) are :

= (MJ; sin \I/+MV cos ^)/|M|=cos (y+w) cos 0 sin <p— sin (y+co) cos 0 cos p,

= (— Mz cos ^+MW sin ^)/(|M|sin 0)=cos (o+w) cos ^J+sin (#+&>) sin ^>,

Thus we have the part due to the hydrodynamic torque in the equations of motion by substituting

these expressions together with (11) and (26) in equation (17):

(-77 ) =—M sin 0 cos 9 sin2 (y+w—<p) Vl—sin2 0 sin2 (v+u—cp),

jjZj = M sin (w+o)—<p) cos (y+w—(p) Vl—sin2 0 sin2 (v+w—<p). (29)

If we put together the part due to the tidal torque and that due to the hydrodynamic torque,

we get the equations:

-=—Ei sin 0 sin 2(y+co—<p)— M sin 0 cos 0 sin2 (y+&>—<p) Vl—sin2 0 sin2

at

~r=Ei cos 0 • [cos 2(y-f co—^>)+l]+M sin (y+co—<p) cos (y-f co—^>) Vl—sin1 0 sin2 (y+co—p). (30)

at

Take the mean values of the right-hand members with regard to t, that is, with regard to v from 0 to

2TT, since we are dealing with the case of circular motion; then

(sin 2(y+co—<p))=0,

(cos 2(y-fto—<p))=0,

. 2 sin3 0 1 cos8 6

(sin2 {v+a—<p) Vl—sin20 sin8 (y+co—<p))= . V 2 • E (sin 0) + ^, • 2 a F (sin 0),

; Q

(sin (y+w—v?) cos (y+w—y>) Vl—sin2 0 sin2 (y+w—^))=0,

where F and E, respectively, denote the complete elliptic integrals of the first and the second kinds

(de Haan, 1858):

/»r/2 fr/2

F (sin 0) = Vl-sin2 0 sin2 x • dx, E (sin 0)= ,

Jo Jo VI—a

2

 0 sin2 z
1 2 2 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS

Thus we get the equations of motion:

do ,., • n n f2sin20— MsinecoBd\

(8in

 w * /

^ (31)

In order to solve these equations we expand the elliptic integrals in powers of sin2 0, by assuming

that sin 6 < 1:

E

 o*«-i

The first equation of (31) takes the form

dd

sin 5 cos 5

ri i . ,. , \

Mdt \2 4 • /

After integration for a small sin2 0 we get approximately,

tan0=tan0o -exp/ —^-(t-t0) \ , (32)

where we suppose that 0=0O for t=to. By substituting (32) in the second equation of (31) and

integrating it we get, approximately,

^ ( - . . .) (t-Q, (33)

where we take <p=<po for 1=1$.

The precessional motion of the z-axis of the satellite due to the tidal torque alone is modified

by the addition of the hydrodynamic torque in such a way that the tangent of the angle 9

decreases exponentially and the speed of the rotation of the plane (Zz) changes from Et cos 0O to J5i

( 1—- tan2 0O+ • • • )• Thus the satellite finally tends to direct its longest axis towards a direction

perpendicular to the velocity of the satellite relative to the earth's atmosphere.

Magnetic torque.—As the orbital plane is supposed to be the equator, the direction of the

geomagnetic field is always in the Z-direction and hence 7=0. We have

L=— Lo ( — ) sin 0 cos 0, L*= — |L| sin 0 sin ^,

I*=—|L| sin 0 cos ^, Lf= — |L| sin 0 cos 0 sin (X—<p),

and

L#= — |L| sin 0, L,=0, L* = — |L| sin 0 sin (X—<p).

Hence

D (2)' (M) (^(^'-,) . (34)
NO. 9 ROTATION OF A N EARTH SATELLITE 123

The solution is:

<P= po=constant,

1 1, /l+sin0 o \, 1 T Sro\

6

, x

The angle 0 decreases from 0O for t=t0, while <p is kept constant.

When we consider together the tidal, the hydrodynamical, and the magnetic torques, we

obtain, by averaging over the short period,

d9 ** • a /. f 2 sin2 0—1 „ . . _N . cos2 0 „ , . M"\ T /ro\

6

 • , n n

Tr-Msm 6 cos g | 3s . n 2 g E (sin ^+3^ ^ F (sin ^)|-X 0 (£) sin' 0 cos 5,

jj=Ei cos 0. (36)

If we suppose that sin 0-C1, then equation (36) reduces to

^=- M sin 6 cos 0 • (i- | sin2 0)-Lo (^Y sin2 6 cos 0,

^=^ l C os0 .

at

The solution is, approximately:

tan0=tan0o-exp-< —wit—to) ft

<p-<po=E1(l~ tan2 0O+ . . .) (t-to)+ . . ., (37)

and the effect of the magnetic torque is small compared with that of the hydrodynamical torque,

if M and i0(r0/a)6 are of the same order of magnitude.

Choice of the reference axes.—It is questionable whether we may take our reference axes (XYZ)

fixed in space in such a way that the Z-axis is directed towards the pole of the equator, because

there exists no invariable plane in the rotation of a satellite that is not in a free rotation; therefore,

we now examine the question by taking the Z-axis to be in the direction of the initial angular

momentum of the satellite (Beletskii, 1959). Let the Z-axis be in the direction of the velocity of the

satellite at the pericenter v=0. For convenience, we take «=0. Then

^=0 , YB=—a cos v, ZB=—a sin v,

and

z——a cos v sin 0 sin <p—a sin v cos 0, \(~ ) /~o

 sm 2 G sm* ^ 2 cos ! & in 0 sin <p—a sin v cos 0, \(~ ) /~ o '

 sm 2 G

The equations of motion take the form

dd T-, . . . d<p ,-, „ ,

-T-=—J^i sin 0 sin <p cos <p, -jr——^1 cos 6 cos 1 ^>.

(It Clt

In order to solve these equations we suppose that sin 0^0, cos <p^0. Then we get

cos 0 d6 sin <p djp

sin 0 cfc cos ^> dt

or, sin 0 • cos <p=constant.
1 2 4 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS VOL. S

This integral shows that the angle between the equatorial pole, which is now the X-axis,

and the 2-axis of the satellite is constant, in accord with the result obtained on pages

120-122; that is, the equatorial pole is the pole of the precession. Hence we should take the north

pole of the equator as the fixed Z-axis in space to which the rotational motion of the satellite moving in an equatorial orbit should be referred. It will be shown in the next section that the pole of

the orbital plane should be taken as the Z-axis fixed in space to which the rotational motion of the

satellite is to be referred, and it is the pole of the precessional motion.

Inclined circular orbits

Tidal torque.—Let the inclination of the orbital plane of the satellite to the earth's equator be /

and the longitude of the node be ft. The coordinates (XYZ), referred to the axes parallel to the

earth's north pole and its equator, of the center of the earth relative to the center of mass of the

satellite are:

XE=—a{cos (0+00) cos ft — sin (v-\-w) sin ft cos /} ,

YB = — a{cos (tf+a>) sin ft+sin (#+«) cos ft cos /} ,

ZB = —a sin (v-\-w) sin /. (38)

Thus we have as before,

z=XE sin 0 cos <p-\- YB sin 0 sin <p-\-Zs cos 0,

The mean value of (2/a)2 is obtained as follows:

\(-) /—o (cosS &+sin2 ft cos21) sin2 0 cosV+<> iTM3* & cos 2 J+sin2 ft) sin20 sin2 <p

+ - sin2 / cos2 0+sin2 / sin ft cos ft sin2 0 sin <p cos <p

+sin / cos 7 cos ft sin 0 cos 0 sin <p—sin / cos / sin ft sin 0 cos 0 cos <p

By substituting this expression in equation (17), and after some algebraic computation, we get

—-ET J7= O sin2 7 sin (2 ft—2^) sin 0-+-sin I cos I cos (ft — <p) cos 0,

EJ\ at z

—-pr -n — —cos21 cos 0-f-sin2 7 sin2 (ft —<p) cos 0+sin / cos /sin (ft — <p) • (cos2 0- sin2 0)/sin 0. (39)

Now we can carry out the operation,

[—cos / sin 0+sin / cos 0 sin (ft — <p)} -y— sin 7sin 0 cos (ft — <p) ^>

by substituting the values of d0/dt and d<p/dt as represented on the right-hand sides of the above

differential equations; the result is zero. Hence we have an integral,

cos /cos 0+sin /si n 0 sin (ft — tp)= constant.

By referring to the spherical triangle ZPz shown in figure 3, in which P is the pole of the orbital

plane, Z is the earth's north pole, and z is the direction of the 2-axis of the satellite, we see that

cos P2=co9 / cos 0+sin /si n 0 sin (ft— <P)=constant. (40)
ROTATION OF AN EARTH SATELLITE 125

FIGURE 3.—Relation among the coordinate system (XYZ), fixed in space, the rotating coordinate system

attached to the satellite, and the orbital plane NR and its pole P.

This equation shows that the pole of the orbital plane should be taken as the X-axis fixed in space to

which the rotational motion of the satellite is to be referred, and that it is indeed the pole of the precessional

motion.

As a special case let us consider a polar orbit; that is, an orbit with 7=90°. Then we have

XE=—a cos (t?+w) cos Q,, YB=—a sin (fl+oj) sin Q>, ZB=—a sin (v-\-ui),

2 \ 2 \ 11 1

- I )= - cos 2 Q> sin2 0 cos 2 <p-\-x sin2 Q> sin2 8 sin2 <p+-z cos 2 0+si n & cos Q, sin2 6 sin <p cos <p.

a/ / £ Z L

For simplicity, take £1=0; thftn the equations of motion are

1 de . a . •jr 3-=sin 0 sin <p cos <p, n,i at

1

 d<p a . , -FT -37=—cos 0 sin1

 <p. Hi at

We get, as before, the integral: sin 0 sin <p= constant. It shows that the angle between the z-axis

of the satellite and the F-axis, which is now the pole of the orbital plane, is constant. Again, we

591553—61 3
126 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS vox,. 5

have seen that the pole of the orbital plane should be taken as the Z-axis fixed in space to which the

rotational motion of the satellite is to be referred, and that it is indeed the pole of the precessional

motion.

Hydrodynamic torque.—We take the earth's north pole as the Z-axis. We have

—Xg/a=cos (fl+w) cos ft — sin (v-\-a) sin ft cos /,

— YE/a=±cos (v+«) sin ft+sin (y+w) cos ft cos /,

—ZBja=s\n (v+u) sin /, (41)

and

1 dJT

rr=—sin (v-\-cS) cos ft—cos (fl+u) sin ft cos I, ita at

1 flV

rr=—sin (#+«) sin ft+cos (v-\-o>) cos ft cos /,

na at

1 d ZB / 1 \ • r

5— =cos (v+w) sin /,

p.a at

We get from equation (27), page 120, the equations:

Z={sin (v-\-w) sin ft—cos (#+«) cos ft cos /}cos 0+cos (w+w) sin /sin 6 sin <p,

m=—cos (t?+w) sin /si n 6 cos <p—{sin (v-\-u) cos ft+cos (f+co) sin ft cos /} cos 6,

n.={sin (fl-f to) cos ft+cos (u+o?) sin ft cos /}sin 0 sin ^>

—{sin fy+co) sin ft—cos (tf+co) cos ft cos /} sin 6 cos <p, (43)

and from equation (26) we obtain

cos 5=sin (p-fw) cos (ft — <p) sin 0+cos (»+w){sin (ft —^) sin 0 cos /—cos 0 sin /} . (44)

We proceed in the same way as in pages 120 to 122.

Mz

jjT^=sin (fl+w) {sin ft (cos <p cos \p—sin <p sin ^ cos 0)—cos ft (sin <p cos ^+cos <p sin yp cos 0)}

+cos (»+«) {—sin ft (sin <p cos ^+cos <p sin ^ cos 0)+cos ft (cos <p cos ^—sin <p sin ^ cos 0)} cos /

—cos (c+w) sin 0 sin ^sin /,

M

nr^j=sin (»+w) {—sin ft (cos> sin ^+sin <p cos ^ cos 0)+cos ft (sin <p sin $—cos v? cos ^ cos 0)}

+cos (»+w){sin ft (sin <p sin if/—cos ^ cos ^ cos 0)+cos ft (cos <p sin ^+sin <p cos ^ cos 0)) cos /

—cos (»+«) sin 0 cos ^ sin /,

=sin (»+w) sin (ft — <p) sin 0 cos 0,
NO. B ROTATION OF AN EARTH SATELLITE 127

and

Me =—sin (w+o>) cos (ft—v?) cos 0—cos (v+a>){sin (ft— <p) cos 0 cos 7+sin 0 sin /} ,

| XT * |

M

j^|j={—sin (w+w) sin (ft— <p)+cos (v+u) cos (ft— <p) cos 7}/sin 0,

7=r^=sin (v-\-cS) sin (ft — <p) sin 0 cos 0.

Hence we get

-JT ) =—M sin 5 cos 5 • [—sin (w+w) cos (ft — <p) cos 0—cos (w+w) {sin (ft — <p) cos 0 cos 7+sin 0 sin 7}],

/d\p\ , , sin 5 cos 8 r , , v . , _ . . /i \ / ^ \ n f jj I = M ;—-— [—sm (fl-f-tt) sin (ft —^>)+cos (»+w) cos (ft—v) cos/],

\dt / 2 sin 0

(-J7 ) = M sin 5 cos 5 • sin (u+a>) sin (ft — <p) sin 0 cos 0. (45)

Since

cos P2=cos 7 cos 0+sin 7 sin 0 sin (ft — <p),

sin P2 cos ZZP2=sin 7 cos 0—cos 7 sin 0 sin (ft — </>),

sin P2 sin ZZPz=sin 0 cos (ft —*?),

we see, by referring to figure 3, that

cos 8=—sin Pz cos (#+&>+ /ZPz)

and that

-~=—Ei sin 0i • sin 2(tf+a>—^)— Msin 0i cos0i sin2 (w+w—^) Vl~sin20! sin2

^=Ey cos 0i • [cos 2(fl+u—^>i)+l]+Msin (p-f^—^j) co s (o+w—<pj)^l—sin20i sin2 (©+«—^),

(46)

where Oi=Pz and ^>i= ^ZPz. Hence the equations can be treated in the same way as in (30), by

taking the pole of the orbital plane as the new Z-axis fixed in space for referring the rotational

motion of the satellite.

If an inclined orbit is given, we obtain the pole of the orbital plane by the equations,

XP=sin /si n ft, YP=— sin 7cos ft, ZP=cos 7, (47)

and we study the rotational motion of the satellite by referring to the orbital plane and its pole.

Motion of the orbital plane.—We now examine the effect of the motion of the orbital plane on

the rotation of the satellite. It is known that the disturbing action causes the orbital plane of the

satellite to rotate around the pole of the equator with a period of a few months (Y. Kozai, personal communication), while the orbital inclination is kept constant.
128 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS VOL. 5

From the spherical triangle PZz shown in figure 3, where we write the angle /.ZPz=<px and

the angle ZPZz=QQ°— & + <p, we obtain:

cos 0=cos / cos 0!+sin / sin 0X sin <pu

sin 0 sin (& — <p) = — sin 0t cos /cos px + cos 0i sin /,

sin 0 cos (& — <p)=sin 0, sin ipx. (48)

Suppose that we have studied the rotational motion of the satellite by means of the angles 0t

and tpx referred to the orbital plane and its pole by the theorem given on page 125, and that we have

obtained the solution in the forms of equations (32) and (33):

tan 0i=tan 0iO • exp-l —^-(t—t0) >> (32a)

<Pi—<pio=E- (t—t0), (33a)

where

Here we have changed the notation from 0 and <p of equations (32) and (33) to dx and <px referred to

the orbital plane and its pole. If we substitute these expressions in equations (48) and if we

assume that the variation, caused by the perturbations of the translational motion of the satellite's

orbital plane referred to the equator and its pole, is expressed by the relations

I=I0, &= &o+vt, (49)

then we obtain the equations:

cos 0=cos 70 cos 0io+sin 7O sin 0io sin (<pl0-\-Et) • exp-< —M— (t—10) V>

sin 0 sin (<p—Q,0—vt)=sin 01O cos 70 cos (<fno+Et) • expx —5- (t—10) V+cos 0iO sin 70,

sin 0 cos (v>— Q,o— vt)=sin 0iO sin (piO+2?it) • exp« — ^ (£—10) VBy solving these equations we obtain the angles 0 and <p, which show the orientation of the z-axis

of the satellite referred to the equational coordinate axes (X,Y,Z), in the form:

f ]\f ^

cos 0=cos 70 cos 0io+sin 70 sin 0iO sin (<pio-\-Et) • expx — —- (t— to) [>

r — r M 1 1 «

tan dp— 0,0— vt)— sin 0iocos70 cos (y>io+£'O + cos0iosin7o-exp-< — (t— t0) > /[sin0iosin (<piO-\-Et)].

(50)

The results shown in equations (32) and (33) are obtained under the assumption that sin 0-C1.

There is no restriction on the value of 70 in these formulas. From this result we see that the motion

of the orbital plane affects only the azimuth angle <p, and not the polar angle 0, when both angles

are referred to the equator and its pole.
NO. 9 ROTATION OF AN EARTH SATELLITE 129

Magnetic torque.—By combining the direction cosines of the z-axis of the satellite,

X,=sin 0 cos <p, F*=sin 9 sin <p, Zt=cos 9,

with the direction cosines of the magnetic field H of equation (14) (p. 118) we obtain:

3 sin jS cos ft . . A . . 1—3 sin2 ft . ,„ .

cos 7=— , = sin 0 cos (X—<p) -\—. •=. cos 0. (51)

VI+3 sin2 0 Vl+3 sin2 ft Further, we obtain:

Lx_KTZZ—HZYZ_—3 sin ft cos ft sin X cos 0— (1—3 sin2 ft) sin 0 sin <p

\L\~ [H| ~"~ Vl+3 sin2 ft

L r KzXz—JixZz_(l—S sin2 ft) sin 0 cos y>+3 sin ft cos ft cos X cos 0

jEj [Hi "" Vl+3 sin2 ft " '

Lz HXT^—HrXz_3 sin ft cos ft sin 9 sin (X—y).

|L| |H| ~~ Vl+3 sin2 ft

U 3sinftcosft m ( } co g o g cQ s (x _ } s b l-3sin2ft= g m

|L| Vl+3 sin2 ft Vl+3 sin2 ft

Lj,==3smft_cosft [gi n (x _ ^ g m +co g fl (x _ ^ cQs ] + 1-3 sm2£ g m co g

|L| Vl+3 sin2 ft Vl+3 sin2 ft

Lz 1—3 sin2 ft . . - • /v \

7T1 = -T =

 s m

 »

 co s

 ^ si n (X—Wi

|L| Vl+3 sin2 ft

Le 3 sin ft cos ft . ^ N , 1—3 sin2 ft . 4 =

 M cos 0 cos (X—y)+-^ sm 0,

Vl+3 i 2 ft IM Vl+3 sin2

 ft

L»_3 sin ft cos ft sin (X—<p)

|L|~Vl+3sin2ft sin 0

L* 1—3 sin2 ft .

= F

 8 i

Vl+3 sin2 ft

8 i n

 0 co s * si n (^~*»)i

Hence the equations of the rotational motion of the satellite are:

^ \ =_ Lo (-} Vl+3 sin2 ft • sin y cos 7X {3 sin ft cos ft cos 9 cos (X—v) + (l— 3 sin2 ft) sin 9},

^ \ =—L0 (-} Vl+3 sin2 ft • sin 7 cos 7X3 sin ft cos ft sin (X—<p)/sin 9, dtjz \a/

QF\ =—L0 (-} Vl+3 sin2 ft • sin 7 cos 7X (1 - 3 sin2 ft) sin 9 cos 9 sin (X-*>). (52)

dty3 \aj

Because of the orbital motion of the satellite the longitude X and the latitude ft, indicating the

position of the satellite, vary according to the relations,

sin ft=sin /sin (c-j-w), tan X=cos /ta n (#+«)• (53)
130 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS VOL. B

Put

. „ _. 3 sin/3 cos/3 . , „ „. 1—3 sin2/3 v ....

cos (H, R)=—. -=sin Xi, cos (H, Z)= .

 K =cos Xi. (54)

V

 Vl+3sin2^ Vl + 3sin28

Denote the pole of the instantaneous direction of the vector H by H as shown in figure 4. Then

ZH—xi, Hz=A, Zz=6. Further denote the angles AZHz and ZHzZ by 0 and X, respectively,

the angle ZHZz being 180°— (X— <p). Then for the spherical triangle ZHz shown in figure 4 we

have the following relations:

cos vl=sin xi sin 8 cos (X— <p)-\-cos xi cos 6,

sin A cos X=—sin xi cos 8 cos (X— p)+cos xi sin 8,

sin A sin X=sin xi sin (X—<p),

sin A cos 0=—cos xi sin 8 cos (X—y>)+sin xi cos 8,

sin A sin 6=sin 8 sin (X— <p). (55)

By'these formulas we see that y=A from equation (51) and that the equations (52) are transformed

into

^L\ =—L0 (- Y (1+3 sin2 0) sin A cos A • sin A cos X,

^ \ =La (j*\ (1+3 sin2 0) sin A cos 4 • sin A sin X/sin 0. (56)

From equation (55), by substituting (56) we obtain:

d' cos A . _ dXi dp , _ d8 • n • A • -o- (d\ d<p\ jr =—sm A cos 9 • -JT TT—sin A cos X • -77—sm 8 sm A sm X • ( -77—-77 h dt dp dt dt \dt dt/

sin2 A • -3-=sin A sin X • -77+sin 8 sin A cos X • -T;+COS A sin 9 • -777 -77—sin 8 sin ^1 cos X • -57*

at dt dt dp dt dt

sin2 dX T /r o \

6

/1 1 o • *a\ • a A A • vw / A -o- 1 cos 9 sin xA jr=^o ( — ) (l+3sin2

j8)sin3 A cos >4 sin XX( cos .A cos XH r—-—-)

+sin ^1 sin 0 • -jr- -77—sm Xi sin A cos 9 • -JT;

dp dt dt

or

^=~L o (-Y (1+3 sin2 /3) sin2 ^1 cos yl+cos 9 • ^ ^+sin 8 sin X • § , (57)

at \a / dp at dt

. , A dQ A • _ dX\ dp . . . A _ d\ ,__x

sin2 A • -Tr=cos -4 sm 0 • -J-T- -37—sm 8 sin A cos X • -r:» (58)

dt dp dt dt

Now we can divide the satellite's motion into two parts:

I. The part of the motion around the direction H of the geomagnetic field at the position

(X,/9) of the satellite, when we suppose that H and thus the satellite are instantaneously at rest;

that is, when d\/dt=d&/dt=O.

II. The part due to the motion of H caused by the translational motion of the satellite in

longitude and latitude; that is, when dXfdt^O,
NO. 9 ROTATION OP AN EARTH SATELLITE 13 1

Accordingly, the equations (57) and (58) split into two parts,

dt \dt A+\dt Jn dt KdtA^KdtJn

such that

0 =-L0 P£j (1+3 sin2/S) sinM cos A,

fw/ i =0; (59)

( -y- ) =cos A sin 6 • -rx -r:—sin 6 sin A cos X • -y' (60)

\dt/n dp at dt

At first we integrate the equations (59). The solution, obtained in the same way as in pages

122 to 123, is:

1 _ 1 \ _ M V

mA sin AQ/ \aj v / v w 4 *

—sin A V+o /

6=0 O = constant. (61)

Or, if sin A<g.l, the solution is, approximately,

tan ^4=tan A* • exp /— Lo ft£\ (1+3 sin2 0)(t-to) \> e=9o. (61a)

Hence the first part of the motion consists hi an exponential shortening of the arc A, that is, the

angular distance of the z-axis of the satellite from the instantaneous direction of the geomagnetic

field for the point (X,/3), while the arc A is kept in the same direction fixed in space, as G=0O (see

fig. 4). The integration of the second part of the equations is difficult and I do not go into it at

present.

As the evaluation of the mean values with respect to t of the right-hand members of equation

(56) is also difficult, I am now compelled to assume that sin 0<S1. We then have, up to the

second order terms in sin I, the expressions:

sin Xt=—3 sin / sin v,

9

cos Xi=l—x sin21 sin2 v,

sin X=sin »—- sin* / sin v-\-^ sin* I sin8 v,

cos X=cos p+x sin2 / sin2 v cos v,
132 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS VOL. 5

and

cos ^4=sin 0 cos <p sin Xi cos X+sin 0 sin <p sin Xi sin X-f cos 0 cos Xi

=cos 0 • 1—3 sin I tan 0(cos <p sin v cos fl+sin <p sin2 v)— -? sin2 / sin2 v \>

sin A=cos 0 • 1 + 3 sin I cot 0 • (cos <p sin 0 cos p+sin <p sin2 «)

/ 9 3 \ 9 • "1

—sin2 / • I ^+ T cot2 0 j(cos <psir\ v cos v+sin ^> sin2 v)z-\-- sin2 / cot2 0 sin2 v \>

sin ^1 cos A=sin d cos 0 • 1+3 sin / • (cot 6— tan 6) (cos <p sin w cos v+sin <p sin2 v)

/27 3 \ 9 . ~|

—sin2 / • (-K-+T cot2 0 J(cos ^> sin v cos «+sin p sin2 v)2+^ sin2 / • (cot2 0—1) sin2 v \>

sin A cos X=sin 6 • 1+3 sin / cot 0 • (cos <p sin v cos w+sin <p sin2 «)—^ sin2 / sin2 v U

sin A sin X=—3 sin / • [cos <p sin2 0+sin <p sin «; cos p]+0Xsin2 /.

By substituting these on the right-hand sides of equation (56) and by taking the mean values, we

get:

-TT) =—L0 (~J sin2 d cos 0 • 1+ - sin / • (2 cot 9— tan 6) sin <p

- | sin2 / • (cot2 0-2)-sin2 / • (y~f | cot2 0) (l+ | sin2 v>Yl

^f) ~~~^° \ ) cos 0 • Ksin /cos H~9 sin2 / • (cot 0—tan 0)Qsin <p cos <p\ • (62)

At first we solve equation (62) by putting sin /=0 . Then we have

p=Vo=constant, -77=—Lo( —) sin2 0 cos 0;

or, by integrating,

l+sin0 \ 1. /1+sin 0O\ 1, 1 T AoV /. ,N /PM

2 '°

where we have taken 0=0O at t=tQ, as before. Let us assume further that sin 0-C1; then we get

sin 0=sin 0o—sin20o*Zo(-) (t—%)+... (63a)

Now we turn to the second equation of (62) and substitute (63a) for sin 0; then we have

dp 3 r /r 0V • T

-j7=—7i Lo I — ) sin / • cos 0O cos <p;
NO. g ROTATION OF AN EARTH SATELLITE

or, integrating from the initial value <p=<p0 for t=t0, we Lave

133

log

l+tan |

1—tan ^

—log

l+tan |

1 — tan —

=—^Lo ( — ) sin/ • cos 0O • (t—t0), (64)

or,

or,

tan £-tan § + ... =-\ Lo (^Y sin / cos 0O • (t-t0),

3 fv \6

<P=<Po—T i o (— j sin / cos d0 • (t—t0). (64a)

We return to the first equation of (62) with these values of 0 and <p given by equations (63a),

(64a). Then we have the equations for the second approximation:

-jjj =—LQ (—} sin2 0 cos 0X 1+3 sin / cot 0C sin <p0

X -j 1—Lo(—J (-sin I cos 0O cos <p0—sin 0O J(t—tQ) V +

By integrating this equation we get

/r \ 6

sin 0=sin 0O—Lo I— J (sin2 0o+3 sin /sin 0O cos 0O sin ^0)(^—*o)+ • • • (65)

The formulas (64a) and (65) give the motion in 0 and <p to our degree of approximation.

FIGURE 4.—Relation between the direction H of the geomagnetic field at a point (XjS) on its orbit and

the fixed equatorial axes (XYZ).
1 3 4 SMITHSONIAN CONTRTBTJTIONS TO ASTROPHYSICS VOL. a

The formulas for 0 and <p referred to the coordinate axes {XYZ) at rest in space are transformed to the coordinate axes referred to the orbital plane and its pole P by the formulas:

cos 0i=cos /cos 0+sin /si n 0 sin (ft — <p),

sin 0i cos <pi=— sin 0 cos /sin (ft — <p)-\-cos 0 sin /,

sin 0i sin ^i=sin 0 cos (ft — <p). (66)

Eccentric orbits

Tidal torque.—According to the theorem which we have proved earlier (pp. 124-125), we can take the

orbital plane and its pole as our reference frame (XYZ). The coordinates of the earth's center,

relative to the center of mass of the satellite, are:

XB= —r cos («?+«), YB— — rsin (o+w), ZB=0. (67)

The force function for the tidal torque is, from equation (3a), expressed in the form:

We have

- ) =cos2 (o+«) sin20 cosV+sin2 (v-\-u>) sin20sinV+2 sin (»+«) cos (v-\-u) sin20 sin <p cos <p,

and the equations of motion are

d0= 1 bU dv 1 W fRQ.

dt Cn sin 0 5^' dt Cn sin 0 50* ^ ;

FIGURE S.—Spherical triangle formed by the pole P of the orbital

plane, the north pole Z of the equator, and the longest axis x of

the satellite.
NO. 9 ROTATION OF A N EARTH SATELLITE 135

In order to eliminate the effects of short-period terms we take the mean values with respect to

the mean anomaly M, by noting that

(j 1^ * cos*). (70)

We get

1 H'/aVdMj 1 f2' 1 /a\ , 1 C2*l+ecosv, 1

aY\ 1 f2*/2Vl+ecosz> 11 /I

 /I . 1 . . \ sin2 0

(

s m

 *cos ^+2sin * sm "rw^w

ni 2(l-«2

)

3

'

2

*

Hence the equations for the precessional motion are :

( T ) ~0>

The solution is easily found:

Hence, for an eccentric orbit, the only modification is of the factor 1/(1—e*)3n in the rate of increase

of <p.

Hydrodynamic torque.—The equation (11)

|M| =M sin 5 cos 5, M=*-p(c2-a2) V2,

is transformed to

| M | =M0 — (¥• Y sin 5 cos 5, (73)

PO \ '0 /

where

M0=,rp0(c2-a2)F0

2

, p/p o =exp/--\, (F/Fo)2=(a/r)». (1-e1 cos'

if r0 denotes the radius of the earth, po the air-density on the earth's surface, Vo the velocity of the

satellite at the pericenter, and u the eccentric anomaly in the orbital motion of the satellite. Thus

we have

|M| =M0 • exp / ~ \ - (*\ (1 -e 2 cos1 u) sin 5 cos d. (74)

We propose to express the quantities in this formula by u and to take the mean values with respect

to the mean anomaly M.

For simplicity we take w=0 by rotating the XF-axes by an angle w; the X-axis is thus directed

to the pericenter of the orbit. Then we have

XH=— r coso=— a(cos u— e), YB= — r sin v = — aVl—«2• sin u, ZM=0, (75)

and

dXR fta? • dYB ixa2 n y dZs _

= sm u —~ V1 —« • cos u, —n-=0,

r dt

rr= sm u, —rr—~ V1 —« • cos u, n

dt r ' dt r dt
1 3 6 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS

The direction cosines of the angle between the velocity vector V and the (XY)-axes are

,__„ sinw ,vxr. Vl—<?2-cosu.

cos (X,V) = -j= o o > cos (F,V) = — >

Vl

 2 2

= o o > cos (F,V) = ,

V — e2cos2w V1 —« cos2

thus we get the values for the direction cosines I, m, n of equation (27):

, -Jl — e

2

- co s w „ /=—j L

?== cos 0,

•y/l—e2 cos2 it

sin w .

—e2 cos 0, cos2

. . Ji — e2-cosw . .

2 2 sin ^ sin v+ , = sin 0 cos <p. 2

 cos

2

 u Vl

 2 VI — e ''cos

 i

u •yjl — e

2

cos

2

u

Then we have

'" 'x ——, ~

e

 'cosu (cos <p cos ^—cos 6 sin <p sin \p)—, = (sin <p cos ^+co s 0 cos <p sin ip),

•yjl—e2cos

2

u Vl — «

2

cos2w

"~ (cos ^ sin iH~ cos 0 sin ^» cos \J/) 4-7===== = (sin ^» sin \f/— cos 0 cos <p cos ^),

I—g2 cos2 u VI — fi cos u

\M\~V'

and

Me COS

= • (Vl—£2 • cos u sin <p—sin u cos <p),

I

M

I Vl—e 2cos 2w

i^=-p^ (Vl —e2 • cos w cos ^>+sin w sin v?),

cos g= (sin w cos «p—^l—e2- cos « sin <p).

Vl — «

2

 cos2 u

Hence our differential equations are

f JT j =—Mo exps (l-ecos« ) >-(l—e2 cos2 w)-(l—g cosw)~*

v, • /, «/sin wcos(o— Vl — e2• cos«sin«?V f /sinwcosv>—ijl—e2-coswsiny>V\

Xsuiecosei , -)<l—\ / I r

\ Vl—c2coslu / V. \ Vl—c2cos2w / J

(-jjj=Moexp'l (1— g cos w) Wl— c*cos2 u)-(l—e cosw)~2

,,/sin usin y+Vl—e2-cosucosy?\ /sin it cos y—Vl—e'-cosusinyX

\ Vl—e

2

cos

2

u J\ Vl—e2cos2w /

, /si n u cos <p— Vl—e 2 • cos -M sin y?

V V

ln
ROTATION OF AN EARTH SATELLITE 137

In order to eliminate the effect of short-period terms we take the mean values of the right-hand

members with respect to M by noting that

We make the transformation,

, , , dM , r , ,, . ,

dM=-j— du=- du= (l—e cos tt)cra.

sin u . -Jl—e2-cosu

- SlU A, .- _—CU b A,

i—e2 cos 2 w

so that

du=dx, 1 cos 2

 X .. .

+sin2 —e* x. cos2

 u 1—e

2

 cos2 M 1—e

2

By integrating this relation between du and dx we obtain the equation,

1—e

l+e tan"

u

tan -

E*

l +e

"l—e

tan' =X+constant,

Or, by taking the constant equal to 0, we have

/ e2 \ e2

X=w ( 1+—-}- • • • )' cos

 IM=COS X+-p sin2

 x+ . . ..

\ 4 / 4

Further, we have

sin u cos y— VI—e2 • cos u sin y>

Vl—e2 cos2 w

sin u sin ^+yi—e

2

 • cos w cos <p

—e2 cos2

cos2 u

=cos (x—<p).

Hence we get

d6\ Mo . a a J a\ f2T fa{ . e3 . . .

JT ) = , sin 6 cos 5 • exp -< U I exp -< —( e cos x+77 sm2x+ . . .

(78)

(79)

(79a)

• sin2 (x—^o)Vl— sin25 sin2 (X—^>) • <fx,

exp-< — ( ecosx+77sin2x+ . . .) fX(-j =+sin2x)

x(l- « cos X—j sin2 X+ . . .J • sin (x—<p) cos (x—<p)y\—sin2flsin2 (x—<p) -dx. (80)

exp

Now we have (see p. 122)

sin2 (X-^) Vl-sin2 flsin* (x-y) •^" ?"^7 1 E(sin

1 f2ir

— I cos (X—ip) sin2 (X—(p) VI—sin2 6 sin2 (x—<p) • dx=O,

I* Jo

~ f2' sin3 (x-<p) Vl-sin20si

41T Jo

F(si n

 ^ '

- ^ ) • dx=O.
138 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS

But we write

i n 3 .. . 15 . .

•••) •

- J* sin< ,. VI=S^i ^ • *-| . £ (-1)- -I • |^ | • |^±i. • • • -I (•) (sin'

E(sin «>-Jl (l-sin« , sin' *)««fc=f • S (-1)"

 2-|=i • g£ | I. ( *) (sin' «-

F(sin ,)= Jf (1-Sin> • sin= *)-« <fa=|. fj (-1)- ^ • g= | I • ("*) (sin' V

=|(l+ i sin1 e+gsin< «+...). (82)

Then we obtain the equations of the precessional motion of the satellite up to the second order of

the orbital eccentricity e in the form:

/d6\ 2M0(l-e*Y/2 f a\ . a OK.f J2 sin2 6-1 „ , . M . 1-sin2 $Vf . o.

( 37 ) = ^^ — exp-< \- sm 6 cos 0X "S o • , , E(sm g)+ . 2 F(sm (9)

\.<«/2 7T I. ''o / L I. 3 sin2 0 v 7 3 sm2 9 v y

{ _£} . [_, (!_|i) sin, cos,. 7 (8V) ] , t f r exp (83)

where the right-hand members are also expanded in powers of a=a—r0, the height above the sealevel of the earth, up to the second order.

If we again suppose that sin 0<C 1, then, by substituting the values in the first approximation for

6 and tp from equations (32) and (33) in the terms multiplied by c2, we have

_ I tan' 0Q • exp{-M(t-tQ)}]

{~}-[-?(l-D]-

 (83a)
NO. 9 ROTATION OF AN EARTH SATELLITE 139

By integrating these equations we obtain an approximate solution:

tanfl r/ 1 Mn f \ (, O\ ,/3 \ /

L

M(12)3/ 2

 { j {0h 2( ) (

 C°S

a \ (/, sin2 0O\ ,/3 52\/5 1

- sin2 &0-4 *2 sin 2^& • ^gjj ) —j~ f («-« J

L till Y' I "TTt i •• ' , Q st \ / j J\

 A 91 / /« 9\ 4 / 9 J * *

L—x tan2 0O ) • (t—to)—7 e^iVioCl—e

2

)

i/2

tan <po * _ ^ exp -_< _

(84)

Variations of the orbital form.—The values of the orbital semimajor axis a and the orbital eccentricity e vary slightly because of perturbations due to the figure of the earth, lunar and solar gravitational action, atmospheric drag, radiation pressure, and so forth. Suppose that the variations are

expressed by the equations,

e=eo+ei(t—to) + e2(t—to)

2+ . . .,

a=ao+a1(t-10)+02(1-10)*+ . . •• (85)

If we consider the effect on the tidal torque only, the effect is expressed, by use of equation (72),

by the relation,

cos d0 lt t N , Ei cos dp

If we consider the effect on the hydrodynamic torque, the effect will be obtained by computing by use of equation (84) the increments of 0 and <p due to the increments of a and e shown by

equation (85).

Summary

The rotational motion of an earth satellite during its orbital flight around the earth's center is

discussed on the basis of Euler's equation of motion (1) for the rotation of a rigid body around its

center of mass. The body of the satellite is supposed to be symmetrical, both geometrically and

dynamically, that is, with regard to its principal moments of inertia, so that A—B and Cj^A.

Especially the satellite body is supposed to have a long prolate spherical shape.

The effects are discussed of the tidal or the gravitational torque due to the action of the earth,

which is supposed to be spherical; of the hydrodynamical torque caused by the satellite's motion

through the earth's atmosphere; and of the magnetic torque due to the magnetic induction of the

satellite material by the geomagnetic field. The radiation pressure may exert a torque, but its

effect is neglected because of the symmetrical shape of the satellite body. The electrostatic charge,

induced by the charge distribution in space through which the satellite moves, and the electric

current distribution inside the satellite body, induced by its motion through the geomagnetic field,

produce effects that are neglected for the same reason.
1 40 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS VOL. 5

To study the precessional motion of the satellite, we eliminate the short-period motions, those

having the period of the orbital revolution of the satellite. We summarize the results separately

for the tidal torque, the hydrodynamic torque, and the magnetic torque.

1. The effect of the tidal torque is discussed on the basis of the tide-generating force exerted

by the spherical earth on the satellite, in the form shown in equation (3):

r 7 _ 3 m'G

2 a3

where m' denotes the earth's mass, G the constant of gravitation, a the semimajor axis of the satellite

orbit and r the radius vector of the satellite, both measured from the earth's center, and z the rectangular coordinate in the direction of the principal axis having the moment of inertia C of the

satellite.

Let the instantaneous angular velocities of the satellite around its principal axes of inertia be

wi, w2 and co3 respectively, a>3 being around the z-axis having the moment of inertia C. Let the

resistance to the rotation be supposed to be KI«I, K2O>2, K3CO3 around each of the principal axes of inertia.

We take reference axes (XYZ) fixed in space and refer the position of the principal axes of inertia

of the satellite by use of the Eulerian angles 0, <p, and \f/. The angle 0, the angle between the z-axis

that is attached to and rotating with the satellite and the Z-axis that is fixed in space, decreases

exponentially at the rate —K3/C, and u)i, w2 decrease exponentially at the rate —KI/A as shown in

equations (6) and (7). The position 0, of the instantaneous axis of rotation for f—»»tends to ir/2 if

KI/A<CK3/C, and tends to 0 if KI/A^-KZ/C. The value of <ps, the azimuth angle between the plane

(z,,Z) and the fixed reference plane (Z,F), of the instantaneous axis of rotation, decreases constantly

towards a constant value K determined by the initial value of «i/u2, as shown in equations (8)

and (9) (p. 116).

We suppose that co3 is very large initially and suppose, for the first approximation, that co3=n=

constant, just as in the case of the precession of the earth's rotation axis. As the (XYZ)-&xes fixed

in space we take the axes drawn from the satellite's center of mass parallel to the direction of the

vernal equinox, to that of its perpendicular on the earth's equator, and to that of the north pole of

the earth's equator. For an earth satellite with a circular equatorial orbit (see p. 120) we have:

0=0o=constant,

<P—<Po=Ei • cos 0O • (t—£Q),

u a n t , . 1 -a 3m'G(A—C) where 0=0O, v=vo for t=t0 and Ex— -^\~ -•

Thus the z-axis of the satellite rotates in the sense of orbital revolution of the satellite around

the pole of its orbital plane by describing a small circle of angular radius 0 with the pole. This is

precession. Short-period variations of the period of orbital revolution are superposed on this

precession. This is nutation.

If we take our reference axes (XYZ) to be such that the Z-axis is directed towards the direction

of the initial angular momentum of the satellite, then we get an integral showing that the angle

between the z-axis and the earth's north pole is constant (p. 124). Hence the earth's pole is seen to

be the pole of the precessional motion, and the earth's north pole should be taken as our fixed Z-axis

to which we refer the rotation of the satellite moving in an equatorial orbit. The same theorem is

proved for an inclined orbit; that is, the angle is constant between the z-axis attached to and rotating

with the satellite and the pole of the orbital plane, and the pole of the orbital plane is the center of

the precession. The theorem is proved also for a polar orbit (p. 125). Hence we see that the pole of

the orbital plane should be taken as the Z-axis fixed in space to which the rotational motion of the

satellite is to be referred, and it is indeed the center of the precessional motion of the satellite. This

fact is proved to be true, by a suitable transformation of the coordinate axes, even if we consider
NO. 9 ROTATION OF A N EARTH SATELLITE 141

the hydrodynamic torque (p. 127). Hence, if an inclined orbit is given, we compute the pole of the

orbital plane by equation (47) and study the rotational motion by referring to the orbital plane

and its pole.

The orbital plane of the satellite shifts because of the perturbations. If the motion of the

orbital plane is given by equation (49), then the effects are shown to be those expressed by equation

(50) (p. 128).

For an eccentric orbit the tidal torque causes the precession to have a minor change:

a-* ^ , -

E

i

 co s e° (f n (7o\

\i e )

The semimajor axis a and the eccentricity e of the satellite orbit vary, due to perturbations;

this variation is taken into account by supposing that the variation is that given by equation (85).

The result is shown in equation (86) (p. 139).

2. Suppose a prolate spheroid with axes c>a=b. An irrotational motion of an incompressible

inviscid fluid of density p, moving across the spheroid with speed V at infinity, exerts a hydrodynamic torque M making an angle 8 with the major axis c of the spheroid. This torque, according

to Cizotti (see Lamb, 1932; Milne-Thomson, 1955), is taken to be:

|M|=Msin 8 cos 5, M=7rp(c2-a2)F2, (11)

in the sense of increasing 8 (see fig. 2, p. 117). I assume this formula for the moment, as we have

no rigorous formula available for our actual case.

For an equatorial circular orbit the equations for the precessional motion of the satellite are

obtained in equation (31). The hydrodynamical torque affects the precession only in 0, but not in

ip. The equations are solved for sin 0-C1 in equations (32) and (33). The angle 0 decreases

exponentially and the motion in the angle <p is modified by only a small amount (p. 122). Finally, the

s-axis of the satellite tends to be directed parallel to the Z-axis, that is, perpendicular to the direction of the satellite's velocity relative to the earth's atmosphere.

It is shown that the effect of the hydrodynamical torque can be properly treated by referring

to the orbital plane and its pole as the fixed axes in space, and this pole is the center of the precession.

The effect on the satellite moving in an eccentric orbit is discussed (p. 135). The torque is

taken to be

|M| =M0 (—)(TT) sin 8 cos 8, (73)

\Po/ \Vo/

where

if r0 denotes the radius of the earth, p0 the air-density on the earth's surface, Vo the velocity of the

satellite at its pericenter, and u the eccentric anomaly in the orbital motion of the satellite. The

equations of motion (77) are transformed by (78) to (83) by the introduction of the complete elliptic

integrals (82) and the definite integrals (81). The equations are solved for the case sin 0<C1 by

expanding in power series of e up to the square in the form (84).

3. The magnetic torque due to the induced magnetization of the satellite is taken to be

I T| T /H\2

 . T HI (permeability) X (volume of the satellite) /1ON

' ' °VHV

 smycosy> ^o=—2

-^ —— ^ ~> 03)

in the sense of decreasing y, where i7=|H| denotes the magnetic field at the satellite position and

7 the angle between the geomagnetic field |H| and the z-axis attached to the satellite, of which the

latter is supposed to be the direction of the magnetic induction. The earth is assumed to be a

uniformly magnetized sphere and the geomagnetic pole is assumed to coincide with the earth's

pole. The geomagnetic field is then that of a dipole and is expressed by equation (14) (p. 119).
142 SMITHSONIAN CONTRIBUTIONS TO ASTROPHYSICS

For an equatorial circular orbit the magnetic torque affects only the rotational motion in 0

and yff, but not in <p, as shown in equation (34). Hence the azimuth angle <p is kept constant, while

6 decreases nearly linearly with t (p. 123). The effects of all three kinds of torques, the tidal, the

hydrodynamical and the magnetic, are described by equation (36). The solution (37) is carried

out for sin 0-C1. The effect of the magnetic torque is small, compared with that of the hydrodynamic torque, if M and Xo(ro/a)6

 are of the same order of magnitude.

The computation of the effect of the magnetic torque on a satellite with its orbital plane

inclined to the equator is very complicated. The equations of motion are deduced in the

form of equation (52). It is seen that the angle ^ZHz=Q is constant in the spherical triangle

formed by the equator's pole Z, the direction of the geomagnetic field H at the satellite, and the

0-axis attached to the satellite (see fig. 4); and that the angle between H and s decreases exponentially, if we disregard the orbital motion of the satellite. This circumstance, however, does not

help us in integrating our equations of motion.

We are thus compelled to assume that the inclination / is small. The equations for the precession are now in the form (62) and are solved by successive approximations (64) or (64a) and

(65). To include the magnetic torque, we use the following procedure. First, we study the precessional motion caused by the tidal and hydrodynamical torques, with reference to the orbital

plane and its pole, according to the theorem stated on page 125. Then we transform back to the

equatorial coordinates by using equation (66). To this motion we add the precession due to the

magnetic torque, supposing that the effect of the magnetic torque is sufficiently small.

TABLE 1.—Direction cosines of fixed and rotating axes

Rotating

axes

X

v

z

Fixed axes

X

cos 0 cos <p cos yp—sin <p sin ^

— cos 0 cos v sin \p—sin <p cos if>

sin 0 cos <p

Y

cos 0 sin <p cos ^-4-cos <p sin $

— cos 0 sin <p sin ^+cos <P COS \f>

sin 6 sin <p

Z

—sin 0 cos #

sin 0 sin t//

cos 9

References

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1960. Charge and magnetic field interaction with

satellites. Journ. Geophys. Res., vol. 65,

pp. 1-7.

BELETSKH, V. V.

1959. The motion of an artificial earth satellite

relative to its centre of mass. (In Russian.) Translated by J. W. Palmer,

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Ministry of Supply, London.

CHAPMAN, S., AND BARTELS, J.

1951. Geomagnetism, ed. 2, vol. 2, Clarendon

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1927. Trait6 de ballistique exterieure, vol. 2,

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DEIMEI. R. F.

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1858. Tables d'int4grales d6finies, 572 pp. Van

der Post, Amsterdam.

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Abstract

The rotational motion, especially the precessional motion, of an earth satellite around its center of mass during

its orbital flight around the earth's center is discussed on the basis of Euler's equations. The body of the satellite

is supposed to be symmetrical, both geometrically and dynamically. The effects of tidal and hydrodynamical torques

and also of magnetic torque due to the magnetic induction in the satellite by the geomagnetic field are considered.

It is proved that the pole of the orbital plane should be taken as the coordinate axis fixed in space to which the

rotational motion is to be referred, and that it is indeed the pole of the precessional motion of the satellite, if the

tidal and hydrodynamic torques are taken into account.

The paper begins by discussing in general the effect of torque due to resistance, and describes how the angular

velocity decreases and how the instantaneous axis of rotation changes with time. The initial angular velocity around

the longest principal axis of inertia is supposed to be very high, and the solution is carried out under this assumption.

It is shown that the hydrodynamic torque affects the precessional motion both in the Eulerian polar angular distance

8 and in the azimuth angle <p of the longest principal axis of inertia, but the tidal torque affects only <p and the magnetic torque affects only 0; and that the satellite, by performing a precessional motion around the pole of the orbital

plane in the sense of orbital revolution, finally tends to direct its longest principal axis of inertia to a direction perpendicular to its orbital velocity. The total effect of all the three kinds of torques is discussed for circular and eccentric

orbits of the motion of the satellite around the earth's center. For the case of an eccentric orbit, the solution is expanded in a series proceeding in powers of the orbital eccentricity. For the effect of the magnetic torque the solution

is carried out only for the case of small orbital inclinations. In some cases in which the integration seems difficult the

solution is obtained by expanding in powers of sin 0 by the assumption that sin 0<3C 1. The effect of the motion of the

satellite's orbital plane due to perturbations is also considered.

It is hoped that the results offered in the present paper will be tested by photometric observations on the

variation in brightness of satellites and by observations of ultrahigh-frequency radio echoes from satellites.