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 precessionLet 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!ω

= Aω1 x1 + Bω2 x2 + Cω3x3 where !I is the inertia tensor and xi are the principal axes.

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

d!Hdt

=∂!H∂t

+!ω ×

!H = 0

=!!I i!"ω +!ω ×

!!I i!ω( )

Termduetorotatingco-

ordinatesystem

α

C !ω

In terms of Cartesian components,

A !ω1 + C − B( )ω2ω3 = 0 (1)

B !ω2 + A− C( )ω3ω1 = 0 (2)

C !ω3 + B − A( )ω1ω2 = 0 (3)

These equations are know as Euler's equations. In the case ofa planet flattened by rotation, so that A = B < C, they have avery simple solution: (3) ⇒ C !ω3 = 0⇒ω3 = constant

ddt

(1) : A !!ω1 + C − A( )ω3!ω2 = 0

∴ A !!ω1 + C − A( )2 ω 23ω1

A= 0 from (2)

i.e. !!ω1 +C − A

Aω3

⎡

⎣⎢

⎤

⎦⎥

2

ω1 = 0

Writing C − A

Aω3 = σ , the general solution is

ω1 = β cos σ t − t0( )⎡⎣ ⎤⎦

∴ω2 = +A

C − Aσβω3

sin σ t − t0( )⎡⎣ ⎤⎦

= β sin σ t − t0( )⎡⎣ ⎤⎦i.e., the instantaneous angular velocity vector,

"ω , precesses around the C axis at a rate σ , with a constant

angular displacement, α , given by

tan α = βω

(See diagram next slide)

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

amplitude of α = ʹ0 .2 = 1×10−6 rad. (The corresponding linear displacement of !ω from the C axis at the

Earth's poles is αR⊕∼ 6 meters.)

The observed period, however, is 2πσ obs

~ 343 days, whereas the period predicted by the above solution is

2πσ obs

=A

C − Ai

2πω3

# 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.

Observed variations in the Earth’s Length of Day (LOD) and pole position (x, y) since 1975.

Data from the Bureau of Time, Paris.

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!ω = spin vectori = inclination of orbit on planet's equator

!T = satellite/solar torque

!H = C

!ω = spin angular momentum

In time δt, the angular momentum changes by δ!H =

!T δt.

∴ the angular momentum vector rotates about n by an angle

δφ = −T δt

H sin i

⇒ precession rate =dφdt

= −T

H sin i

= −T

Cω 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/sunN = ascending node on Equator∠NP = φ

∠SP = π2−θ

C = planet's spin axis∠NS = uThe planet's gravitational field is

VG r,θ( ) ! −GMr

+GMR2

r3J2 P2 cosθ( )

which leads to an instantaneous torque on S of

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

= −m∂VG

∂θφ = 3

GMmR2

r3J2 sinθ cosθ φ

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 Ty , and average around one orbit:

φ = − sinφ x + cosφ y

∴ Tx = −T0 sinθ cosθ sinφ

Ty = T0 sinθ cosθ cosφ

⎫⎬⎪

⎭⎪ where T0 ≡

3GMmR2

r3J2

From spherical trigonometry we derive the relations cosθ = sin isinu tanφ = cos i tanu,while r is given by the equation of the orbit:

r −1 =1+ ecos u −ω( )⎡⎣ ⎤⎦

a 1− e2( ).

We can simplify the algebra by assuming that (i) e ! 0

(ii) i << π2

so that we can set r ! a,φ ! u, sinθ ! 1, and cosθ ! isinu.We then have the approximate results:

Tx ! −T0i sin2 u

Ty ! T0i sinu cosu

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

T = − 12T0i x

= − 32GMm

R2

r3J2i x

A slightly more complicated analysis valid for all i yield

T = − 34GMm

R2

a3J2 sin 2i x,

showing that T is zero for both i = 0 and i = π2

, and a maximum for i = π4.

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

dφdt

planet = − 32

GMmR2

Cωa3J2 cos i

This expression can be further simplified by substituting

J2 =C − A( )MR2

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

dφdt

= − 32

C − AC

⎛

⎝⎜⎞

⎠⎟m

M + m⎛

⎝⎜⎞

⎠⎟n2

ωcos i

The factor m

M + m is ~

mM

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

ω 2 /1d 2 /1d

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

π π

π π

m

M + mi

n2

ω

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

dφdt

= 5 ʹ0 .4 yr−1 = 6.69 ×10−7 d −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 =

1305

Terrestrial forced precession

The 18.6 yr nutation is caused by the Moon’s orbital precession.

Polaris

Vega

Forced Precession & Nutation of the Earth’s pole.

Source: Wikipedia.

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 − AMR2 = 0.001083

C

MR2 = 0.331

Examples of spin precession periods.

Object m ε TPREEarth Sun 23°.5 81,600yr.

Earth Sun+Moon 23°.5 25,700yr.

Mars Sun 25°.2 178,000yr.

Jupiter Sun 3°.1 500,000yr.*

Saturn Sun 26°.7 1,800,000yr.*

Neptune Sun 29°.4 23,000,000yr.*

Moon Earth 6°.7 78.5yr.

Callisto Jupiter ~0°. 200yr.

Titan Saturn ~0°. 200yr.

Triton Neptune ~0°. 65yr.

Iapetus Saturn ~9° 29,000yr?* Affected by solar torque on satellite orbits

Outer planet precession

Outer planet calculated precession periods and rates due to J2 alone, and to the combined effects of J2 and the principal equatorial satellites.

The bars show limits on the precession rate of Saturn’s ring plane, estimated at 0.5”/yr, or a period of ~2 Myr.

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 = −T0 sin 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 nwe 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 hor cos i = s i nso

ddt

cos 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

ddt

s × n( ) = !s × n + s × !n

= T0 s × n( ) s × n( ) × nH− s ×

s × n( )h

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

=T0 s × n( )

Hhs × n( ) × hn + Hs( ){ }

=T0 cos i

Hhs × n( ) × HT

where HT is the (fixed) total angular momentum vector. Thus the vector s × n( ) precesses around HT at an angular rate

dφdt

= −T0 HT

Hhcos i.

We can readily verify that this general expression reduces to our previous results in the limitingcases h << H and H << h:(i) small satellite, h << H :

HT ! H , so dφdt! −

T0

hcos i

Now h ! GM a( )12 m for e << 1, m << M

so dφdt

satellite ! − 32

G C − A( )m cos i

GM( )12 a

72 m

= − 32 GM( )

12 J2 R2 cos i a

−72

= − 32 n J2

Ra( )2

cos i

" as before, for e << 1.(ii) large satellite, h >> H :

HT ~ h, so dφdt

= −T0

Hcos i

H = Cω

∴dφdt

planet ! − 32

G C − A( )m cos i

Cω a3

= − 32

GM m R2

Cω a3 J2 cos i

" as before.*Almost all satellites fall in case (i), except for Earth's moon which satisfies case (ii), and possiblyNeptune's Triton, which may be an intermediate case. Case (ii) also applies to solar torques exerted on planetaryspin vectors, and to planetary torques exerted on satellite spin vectors.

Summary of precession rate formulaeOrbital precession:

!"ω # n032

J2Ra( )2

− 154

J4Ra( )4

+${ } "Ω # −n0

32

J2Ra( )2

− 94

J22 − 15

4J4

Ra( )4

+${ } $ due to planet

!"ω # 14

n0

ms

M( )α 2b32

(1) α( ) α ≡ aas

( ) "Ω # − !"ω $ due to exterior satellite

!"ω # 14

n0

ms

M( )αb32

(1) α( ) α ≡as

a( ) $ due to interior satellite

!"ω # 34

n⊙2

n 1− 2sin2 β( ) "Ω # − 3

4

n⊙2

n cosβ

⎫

⎬⎪

⎭⎪n⊙ = planet's mean motion β = obliquity

$ due to the sun

Note : b32

(1) α( ) # 3α + 458α 3 + 0 α 5( ) $α << 1

n0 ≡GMa3( )

12

n # n0 1+ 34

J2Ra( )2

+${ }

Spin axis precession:

Free precession: σ = C− A

C ω

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

Ω = − 32

C− AC( ) GM

a3 ω−1 cosβ

=− 3

2C− A

C

n⊙2

ωcosβ " due to Sun

− 32

C− AC

mM

n2

ωcos i " due to Satellite

⎧⎨⎪

⎩⎪

Note typo in expression for dW/dt above: M à m.