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/MR 2 ) can be derived from measurements of the oblateness and J 2 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 ω! α C 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 ω!!! = Aω 1 ˆx 1 + Bω 2 ˆx 2 + Cω 3 ˆx 3 where I is the inertia tensor and ˆx! i are the principal axes. If there are no external torques onthe planet, then we have d! H dt =! H t +! ω! H = 0 ( ) =! I i! "ω +! ω! I i! ω Term due to rotating coordinate system
In terms of Cartesian components, A!ω 1 + ( C B)ω 2 ω 3 = 0 (1) B!ω 2 + ( A C)ω 3 ω 1 = 0 (2) C!ω 3 + ( B A)ω 1 ω 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) C!ω 3 = 0 ω 3 = constant d (1) : A!! dt ω + C A 1 ( ) ω 3!ω 2 = 0 A!! ω 1 + ( C A) 2 ω 2 ω 3 1 A i.e.!! ω 1 + C A A ω 3 ω 1 = 0 2 = 0 from (2) Writing C A A ω = σ, the general solution is 3 ω 1 = β cos σ ( t t 0 ) A σβ ω 2 = + sin σ ( t t C A ω 0 ) 3 ( ) = β sin σ t t 0 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 = β ω (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 = 0 ʹʹ.2 = 1 10 6 rad. (The corresponding linear displacement of ω from the C axis at the! Earth's poles is α R 6 meters.) The observed period, however, is 2π σ obs ~ 343 days, whereas the period predicted by the above solution is 2π A = σ obs C A i 2π # 306 days. ω 3 (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: ˆn = orbit normal!ω = spin vector i = 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 ˆn by an angle δφ = T δt H sini precession rate = dφ dt = = T Cω sin i T H sin i
To calculate T, we consider the reaction torque exerted by the planet on the satellite, averaged around one orbit: S = satellite/sun N = ascending node on Equator NP = φ SP = π 2 θ Orbit Equatorial plane (xy plane) C = planet's spin axis NS = u The planet's gravitational field is V G ( r,θ)! GM + GMR2 J r r 3 2 P 2 ( cosθ) which leads to an instantaneous torque on S of T = r ( m V G ), where m is the satellite mass = m V G ˆφ = 3 GMmR2 J θ r 3 2 sinθ cosθ ˆφ Since T varies in both amplitude and direction as the satellite moves around its orbit, we resolve T into cartesian componnets T x (towards N ) and T y, and average around one orbit: ˆφ = sinφ ˆx + cosφ ŷ T x = T 0 sinθ cosθ sinφ where T T y = T 0 sinθ cosθ cosφ 0 3GMmR2 J r 3 2 From spherical trigonometry we derive the relations cosθ = sinisinu tanφ = cosi tanu, while r is given by the equation of the orbit: ( ) ( ) 1+ ecos u ω r 1 =. a 1 e 2
We can simplify the algebra by assuming that (i) e! 0 (ii) i << π 2 so that we can set r! a, φ! u, sinθ! 1, and cosθ! isinu. We then have the approximate results: T x! T 0 i sin 2 u T y! T 0 i sinu cosu Upon averaging arounnd one orbit ( 0 u 2π ), T y cancels and we have T = 1 2 T 0 i ˆx = 3 R2 GMm 2 r 3 J 2 i ˆx A slightly more complicated analysis valid for all i yield T = 3 R2 GMm 4 a J sin 2i ˆx, 3 2 showing that T is zero for both i = 0 and i = π, and a maximum for i = π. 2 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 dφ GMmR 2 planet = 3 J 2 dt Cωa 3 2 cosi This expression can be further simplified by substituting ( J 2 = C A ) MR 2 and using Kepler's 3rd law: n 2 a 3 = G( M + m) : The factor dφ dt = 3 2 C A C m M + m is ~ m M m M + m n 2 ω cosi for satellite-induced precession, but ~1 for solar-induced precession.
Terrestrial forced precession Let us evaluate the solar and lunar contributions to the Earth s precession rate: Sun Moon m M + m 1.0 1/81.3 n 2 π /365 d 2 π /27.3 d ω 2 π /1 d 2 π /1 d m M + m i n2 ω 4.72 x 10-5 + 10.37 x 10-5 = 15.1 x 10-5 d -1 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 dφ dt = 50 ʹʹ.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 = 1 305
Forced Precession & Nutation of the Earth s pole. Polaris Vega The 18.6 yr nutation is caused by the Moon s orbital precession. Source: Wikipedia.
This quantity may be combined with the measured value of To give the polar moment of inertia of the Earth: J 2 C A MR 2 = 0.001083 C MR 2 = 0.331 Note that this direct determination of C/MR 2 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.
Examples of spin precession periods. Object m ε T PRE Earth Sun 23.5 81,600 yr. Earth Sun + Moon 23.5 25,700 yr. Mars Sun 25.2 178,000 yr. Jupiter Sun 3.1 500,000 yr.* Saturn Sun 26.7 1,800,000 yr.* Neptune Sun 29.4 23,000,000 yr.* Moon Earth 6.7 78.5 yr. Callisto Jupiter ~0. 200 yr. Titan Saturn ~0. 200 yr. Triton Neptune ~0. 65 yr. Iapetus Saturn ~9 29,000 yr? * Affected by solar torque on satellite orbits
Outer planet precession
The bars show limits on the precession rate of Saturn s ring plane, estimated at 0.5 /yr, or a period of ~2 Myr. Outer planet calculated precession periods and rates due to J 2 alone, and to the combined effects of J 2 and the principal equatorial satellites.
Digression: mutual precession We have discussed (a) the nodal precession rate of satellite orbits due to the planetary J 2, 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 = T 0 sin i cosi ˆx = T 0 ŝ i ˆn ( ) ŝ ˆn where ŝ and ˆn are unit vectors parallel to the planet's spin axis and the satellite's orbit normal, resp., and T 0 = 3G ( C A )m. 2a 3 Writing the spin and orbital angular momenta as H = H ŝ and h = h ˆn we have the equations of motion:!h =!H ŝ + H ˆ!s = T = T 0 ŝ i ˆn!h =! h ˆn + h ˆ!n = T = T 0 ŝ i ˆn ( ) ŝ ˆn ( ) ŝ ˆn
Now ŝ and ˆn are unit vectors, so ˆ!s ŝ and ˆ!n ˆn, and the right-hand sides of both equations are ˆn and ŝ, 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 or so H hcosi = H i h cosi = ŝ i ˆn d ( cosi) = ˆ!s i ˆn + ŝ i ˆ!n = 0 dt since ˆ!s ~ ŝ ˆn ˆn and similarly for ˆ!n is ŝ. 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 ŝ ˆn d dt ( ŝ ˆn ) = ˆ!s ˆn + ŝ ˆ!n ( ) ( ŝ ˆn ) ˆn ( H ŝ ŝ ˆn ) = T 0 ŝ ˆn ( ) {( ) ( h ˆn + Hŝ) } = T ŝ ˆn 0 ŝ ˆn Hh = T cosi 0 ŝ ˆn Hh ( ) H T h where H T is the (fixed) total angular momentum vector. Thus the vector ( ŝ ˆn ) precesses around H T at an angular rate dφ dt = T 0 H T Hh cosi.
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 : H T! H, so dφ dt! T 0 h cosi Now so ( ) 1 2 m for e << 1, m << M h! GM a dφ dt satellite! G( C A)m cosi 3 2 7 GM 2 m ( ) 1 2 a " as before, for e << 1. ( ) 1 2 J 2 R 2 cosi a 7 2 ( ) 2 cosi = 3 2 GM = 3 2 n J 2 (ii) large satellite, h >> H : R a H T ~ h, so dφ dt = T 0 H cosi dφ dt " as before. H = Cω planet! 3 2 = 3 2 G( C A)m cosi Cω a 3 GM m R 2 Cω a 3 J 2 cosi *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:!"ω # n 0 3 "Ω # n 0 3!"ω # 1 4 n 0 J R 2 2 ( a ) 2 15 J R { 4 4 ( a ) 4 +$ } J R 2 2 ( a ) 2 9 J 2 15 J R { 4 2 4 4 ( a ) 4 +$ } m ( s M ) α 2 (1) b 32 $ due to planet ( α ) α a a s ( ) "Ω #!" ω $ due to exterior satellite (1) Note : b 32!"ω # 1 n m s (1) 4 0 ( M )αb 32!"ω # 3 4 "Ω # 3 4 n 2 n ( a ) ( α ) α a s $ due to interior satellite ( 1 2sin 2 β ) n cosβ = planet's mean motion n $ due to the sun n 2 ( α ) # 3α + 45α 3 + 0 α 5 8 ( ) $α << 1 β = obliquity ( ) 1 2 n 0 GM a 3 R { ( a ) 2 +$ } n # n 0 1+ 3 4 J 2
Spin axis precession: Free precession: σ = C A C ω Forced precession, due to mass 'm' at distance 'a' Ω = 3 2 = 3 2 3 2 ( ) GM C A C C A C C A C n 2 a 3 ω 1 cosβ ω cosβ m M n 2 ω " due to Sun cosi " due to Satellite Note typo in expression for dw/dt above: M à m.