Astronomy 6570 – Physics of the Planets

Precession: Free and Forced

Planetary Precession

We have seen above how information concerning the distribution of density within a planet (in particular, the

polar moment of inertia factor, C/MR2) can be derived from measurements of the oblateness and J2 of a

rotating planet. In some cases, notably the Earth and Mars and probably Saturn in the relatively near future,

we can also obtain such information from studying the planet’s spin-axis precession. Precession may take

one (or both) of two forms – free or Eulerian precession, and forced precession.

Free precession

Let the principal moments of inertia of the planet (or satellite) be A B < C, and the

corresponding body-fixed principal axes be denoted 1, 2, and 3. The angular

momentum of the planet relative to its center of mass is

H = I i

= A1

x1+ B

2x

2+ C

3x

3 where I is the inertia tensor and x

i are the principal axes.

If there are no external torques onthe planet, then we have

dH

dt=

H

t+ H = 0

= I i + I i( )

In terms of Cartesian components,

A1+ C B( ) 2 3

= 0 (1)

B2+ A C( ) 3 1

= 0 (2)

C3+ B A( ) 1 2

= 0 (3)

These equations are know as Euler's equations. In the case of

a planet flattened by rotation, so that A = B < C, they have a

very simple solution:

(3) C3= 0

3= constant

d

dt(1) : A

1+ C A( ) 3 2

= 0

A1+ C A( )

22

3 1

A= 0 from (2)

i.e. 1+

C A

A3

2

1= 0

Writing C A

A3= , the general solution is

1= cos t t

0( )

2= +

A

C A3

sin t t0( )

= sin t t0( )

i.e., the instantaneous angular velocity vector, , precesses around the C axis at a rate , with a constant

angular displacement, , given by

tan =

(See diagram next slide)

Such a precessional motion (known as the "Chandler Wobble") is observed for the Earth, with a very small

amplitude of = 0 .2 = 1 10 6 rad. (The corresponding linear displacement of from the C axis at the

Earth's poles is R 6 meters.)

The observed period, however, is 2obs

~ 434 days, whereas the period predicted by the above solution is

2obs

=A

C Ai

2

3

306 days.

(see below)

This discrepancy was unresolved for many years, but was eventually shown by Simon Newcomb to be due to non-rigidity of the Earth’s

mantle. This led to one of the earliest estimates of the Earth’s elasticity by Lord Rayleigh*.

(Exercise left to the student: describe the motion, relative to inertial space, of the Earth’s C axis during a precessional cycle.)

Footnote: Free precession is sometimes referred to as “Eulerian nutation” in mechanics books. This term is confusing, as the Earth also

experiences an oscillatory component of its forced precession with an 18 year period which astronomers refer to as nutation.

*More recent studies have suggested that the excitation of the Chandler wobble is due to variations in salinity and temperature of the

ocean, as well as changes in ocean currents and atmospheric circulation. Other possible contributors include large earthquakes.

Forced Precession

Because of the rotational flattening of a planet’s figure, the sun and any large, non-equatorial satellites exert a

torque on the planet which attempts to align the planet’s spin axis with the normal to the orbit plane. The

actual effect of such a torque is to force a precession of the spin axis about the orbit normal, as follows:

n = orbit normal

= spin vector

i = inclination of orbit on planet's equator

T = satellite/solar torque

H = C = spin angular momentum

In time t, the angular momentum changes by H = T t.

the angular momentum vector rotates about n by an angle

=T t

H sin i

precession rate =d

dt=

T

H sin i

=T

C sin i

To calculate T, we consider the reaction torque exerted by the planet on the satellite, averaged around one orbit:

Orbit

Equatorial

plane

(xy plane)

S = satellite/sun

N = ascending node on Equator

NP =

SP = 2

C = planet's spin axis

NS = u

The planet's gravitational field is

VG

r,( )GM

r+

GMR2

r3

J2

P2

cos( )

which leads to an instantaneous torque on S of

T = r m VG( ), where m is the satellite mass

= mV

G ˆ = 3GMmR

2

r3

J2sin cos ˆ

Since T varies in both amplitude and direction as the satellite moves around its orbit,

we resolve T into cartesian componnets Tx (towards N ) and T

y, and average around one orbit:

ˆ = sin x + cos y

T

x= T

0sin cos sin

Ty= T

0sin cos cos

where T0

3GMmR2

r3

J2

From spherical trigonometry we derive the relations

cos = sin isinu

tan = cos i tanu,

while r is given by the equation of the orbit:

r 1=

1+ ecos u( )a 1 e

2( ).

We can simplify the algebra by assuming that

(i) e 0

(ii) i << 2

so that we can set r a, u, sin 1, and cos isinu.

We then have the approximate results:

Tx

T0i sin2 u

Ty

T0i sinu cosu

Upon averaging arounnd one orbit 0 u 2( ), Ty cancels and we have

T =1

2T

0i x

=3

2GMm

R2

r3J

2i x

A slightly more complicated analysis valid for all i yield

T =3

4GMm

R2

a3J

2sin2i x,

showing that T is zero for both i = 0 and i =2

, and a maximum for i =4

.

Returning to the precession rate formula, and noting that the torque exerted by the satellite on the planet is minus the above result, we have

d

dtplanet =

3

2

GMmR2

C a3J

2cos i

This expression can be further simplified by substituting

J2=

C A( )MR2

and using Kepler's 3rd law: n2a3= G M + m( ) :

d

dt=

3

2

C A

C

m

M + m

n2

cos i

The factor m

M + m is ~

m

M for satellite-induced precession, but ~1 for solar-induced precession.

Let us evaluate the solar and lunar contributions to the Earth’s precession rate:

Sun Moon

1.0 1/81.3

n 2 /365d 2 /27.3d

2 /1d 2 /1d

4.72 x 10-5 + 10.37 x 10-5 = 15.1 x 10-5 d-1

m

M + mi

n2

m

M + m

So we see that the lunar term is dominant, and contributes ~ 69% of the total. The observed precession rate of

the Earth’s spin axis is

d

dt= 50 .4 yr

1= 6.69 10

7d

1,

corresponding to a period of 25,600 yrs., and the inclination of both the sun’s and moon’s orbits to the

equator is i ~ 23°.5, from which we may calculate the quantity

for the Earth.

C A

C= 0.00328 = 1

305

Terrestrial forced precession

This quantity may be combined with the measured value of

To give the polar moment of inertia of the Earth:

Note that this direct determination of C/MR2 is in good agreement with that inferred indirectly from the Earth’s

rotational flattening using the Darwin-Radau approximation.

At present, no other planet but Mars has a measured forced precession rate (tracking of the 2 Viking landers on

Mars was precise enough to do this), so we cannot generally apply this technique to determine accurate

moments of inertia. In the future, however, such measurements may well be possible, at least Saturn.

J2

C A

MR2= 0.001083

C

MR2= 0.331

Examples of spin precession periods.

* Affected by solar torque on satellite orbits

Outer planet precession

Digression: mutual precession

We have discussed (a) the nodal precession rate of satellite orbits due to the planetary J2, and (b) the precession

of the planet’s spin axis due to the satellite torque on the equatorial bulge. How are these 2 different view

points to be reconciled, and what if anything, remains fixed in space? The answer, of course, is that (for an

isolated planet and satellite system) only the total annular momentum vector remains inertially fixed; both

the spin axis of the planet and the orbit normal precess about this vector.

This is most easily shown using a vector notation for the torque, T, exerted by the planet on the satellite;

T = T0sin i cos i x

= T0

s i n( ) s n

where s and n are unit vectors parallel to the planet's spin axis and the satellite's orbit normal, resp., and

T0=

3G C A( )m

2a3

.

Writing the spin and orbital angular momenta as

H = H s and h = h n

we have the equations of motion:

H = H s + H s = T = T0

s i n( ) s n

h = h n + h n = T = T0

s i n( ) s n

Now s and n are unit vectors, so s s and n n , and the right-hand sides of both equations are

n and s, so we must have

H = 0 and h = 0.

i.e., the magnitudes of H and h remain constant. Furthermore, the angle i between H and h is given by

H hcos i = H i h

or

cos i = s i n

so

d

dtcos i( ) = s i n + s i n = 0

since s ~ s n n and similarly for n is s. Thus the inclination remains constant also.

Finally, we look at the orientation of the plane defined by H and h and whose normal is given by s n

d

dts n( ) = s n + s n

= T0

s n( ) s n( )n

Hs

s n( )h

=T

0s n( )Hh

s n( ) hn + Hs( ){ }

=T

0cos i

Hhs n( ) H

T

where HT

is the (fixed) total angular momentum vector. Thus the vector s n( ) precesses around HT

at an angular rate

d

dt=

T0H

T

Hhcos i.

We can readily verify that this general expression reduces to our previous results in the limiting

cases h << H and H << h:

(i) small satellite, h << H :

HT

H , so d

dt

T0

hcos i

Now h GM a( )1

2 m for e << 1, m << M

so d

dtsatellite 3

2

G C A( )m cos i

GM( )1

2 a

7

2 m

= 32

GM( )1

2 J2

R2 cos i a7

2

= 32

n J2

Ra( )

2

cos i

as before, for e << 1.

(ii) large satellite, h >> H :

HT

~ h, so d

dt=

T0

Hcos i

H = C

d

dtplanet 3

2

G C A( )m cos i

C a3

= 32

GM m R2

C a3J

2cos i

as before.

*Almost all satellites fall in case (i), except for Earth's moon which satisfies case (ii), and possibly

Neptune's Triton, which may be an intermediate case. Case (ii) also applies to solar torques exerted on planetary

spin vectors, and to planetary torques exerted on satellite spin vectors.

Summary of precession rate formulae Orbital precession:

n0

32

J2

R

a( )2

154

J4

R

a( )4

+

n0

32

J2

R

a( )2

94

J2

2 154

J4

R

a( )4

+

due to planet

14

n0

ms

M

2b

32

(1) ( ) a

as

due to exterior satellite

14

n0

ms

Mb

32

(1) ( ) a

s

a

due to interior satellite

34

n2

n1 2sin2( )

34

n2

ncos

n = planet's mean motion = obliquity

due to the sun

Note : b3

2

(1) ( ) 3 +45

8

3+ 0 5( ) << 1

n0

GM

a3( )

1

2

n n0

1+ 34

J2

R

a( )2

+

Spin axis precession:

Free precession:

= C A

A

Forced precession, due to mass 'm' at distance 'a'

= 32

C A

C

Gm

a3

1 cos

=

32

C A

C

n2

cos due to Sun

32

C A

C

m

M

n2 cos i due to Satellite