Astronomy 6570: Physics of the Planets. Tidal Evolution

Transcription

1 Astronomy 6570: Physics of the Planets Tidal Evolution

2 Tidal Interactions x = r sinθ z = r cosθ Accel'n of point P due to m = ( aẑ r cosθẑ r sinθ ˆx ) a = Gm! d 3 " Gm d 3 d! = Gm d 3 a z ( ) ẑ xˆx And d = a z ( ) 2 + x 2 " a 1 z a 1 2 " a( 1 2z a ) 1 2 ( ) if r << a ( ) ( a z) ẑ xˆx 1+ 3 z a a " Gm! a 3 This accel'n is responsible for the centripetal accel'n of P in its orbit about the system's center of mass, at a distance of ma â c = n2 ma ẑ M + m = Gm ẑ (Kepler's 3rd law: G M + m a 2 The Tidal Accel'n, a! T, is the difference between a & a!! c : a! T = Gm 2 z a ẑ x ˆx 2 a a ( ) = n 2 a 3 ) ( M + m) from 0 :

3 If we write a! T in terms of a tidal potential, a! T = V T ( x, z), then V T = ( Gm z 2 x2 a 3 2 ) ( ) = Gm a 3 r cos2 θ 1 ( ) V T = Gm r 2 P a 3 2 cosθ Compare the centrifugal potential, V c = + 1 ω 2 r 2 P 3 2 ( cosθ) V c &V T have the same radial & angular dependence, but opposite signs. tidal bulge is a prolate spheroid. So we can apply the same theory as for centrifugal distortion. In particular, the tidal distortion of the planet's interior results in a P 2 component in the planet's gravitational potential, with an amplitude given by the Love number k 2 T : T Gm V G = k 2 i R5 P a 3 r 3 2 cosθ For the Earth, satellite measurements indicate that k 2 T " 0.29 ( ) (R = planet's radius) much less than the value derived previously for the rotational distortion: (Recall that k 2 = 3 2 k 2 = for a uniform density fluid planet.) (quadrupole poten.al)

4 The reduction of k T 2 in comparison with k 2 reflects the elastic component of the Earth's response to the time-varying tidal potential. Similarly, the tidal distortion of the Earth's solid surface is given by δr( θ) = h T R 2 i mr3 P 2 ( cosθ) Ma 3 where the observed value of the Love number is h 2 T = Compare this with the rotational Love number associated with the Earth's oblateness: Again, the reduction of h 2 T h 2 = is due to the Earth's elasticity. The oceans, of course, do not respond elastically. If their response were purely hydrostatic we would have h T T 2 ( ocean) = 1+ k 2! 1.29 so that the amplitude of the ocean tide would be approximately twice that of the solid body tide. In reality, the ocean's response is not hydrostatic, because the tidal time scale is comparable to the natural resonant frequencies of the large ocean basins: P = 2π ω res ~ basin dimension gd d = mean ocean depth ~ 4 km P ~ 10 9 cm cm s 1 ~ sec ~ 13 hrs.

5 Summary of Tide heights and Love numbers: At the surface of a body the total potential is: GM T + 1+ k r ( 2 )V T ( r,θ)! GM + GM T δr R ( 1+ k R 2 2 ) Gm R 2 P a 3 2 ( cosθ) If the surface is hydrostatic, the potential is a constant T δr( θ) H! ( 1+ k 2 ) mr4 P Ma 3 2 ( cosθ). In general, a solid surface does not deform hydrostatically, and we write for convenience T mr δr ( θ) S = h 4 2 P Ma 3 2 ( cosθ). For the Earth we find from measurements of solid earth tides h 2 T! 0.59, While the ocean responds closer to the hydrostatic limit: h 2 ( ocean)! 1+ k T 2 = 1.29 relative to the land surface, the ocean response is given by h 2 ( ocean) h T 2! 0.70 The tidal distortion is T = 3 2 h 2 mr 3 ( ) Ma 3! h 2 Lunar-induced solid body tide amplitude = T R = 41cm. Lunar-induced ocean tide = 58 cm (rel. to solid ). Solar induced tides are ~ 1 of these values. 3

6 Tidal Torques & Their Effects: If the tidal bulge were perfectly aligned with the tide-raising body, m, then symmetry requires that there be no net tidal torque between the two bodies. For real planets, however, the elastic response of the planet lags behind (in time) the periodic tidal potential, resulting in an angular offset of the maximum tide height from the direction to m. If the planet s spin rate, ω, is greater than the satellite s orbital motion, n, then the bulge will lead the satellite, as shown in the figure on the next slide. If ω < n, the bulge lags the satellite, and the torque changes sign. To calculate the torque exerted by the bulge on m, we return to the induced gravitational potential: V G = k 2 T GmR 5 ( ) a 3 r 3 P 2 cosθ Where θ is measured from the long axis of the bulge. The torque on m is T = m a 1 r V ( G θ ) evaluated at r = a, θ = ε = 3 k T GmR 5 sin 2ε 2 2 a 6 The torque acts to increase the orbital angular momentum of m, and the reaction torque of m on the bulge acts to decrease the spin angular momentum of the planet (if ε > 0).

7 ( ) Tidal potential, V T = GmR2 a 3 P 2 cosθ ( ) 1 2 3cos2 θ 1 P 2 µ ( ) Potential due to distorted planet, V D ( r,θ) = k 2 V T ( R,θ) ( R r ) 3! quadrupole field ~r 3 GmR V D ( a,ε) = k 5 2 P a 6 2 ( cosε)!including the lag angle Torque on satellite, T = m a i 1 a = 3 2 k 2 Gm 2 R 5 a 6 V D ε sin2ε Note that (a) the tidal torque scales as m 2, not m and (b) it decreases with distance as a -6. Both are because it is a torque exerted by the satellite on the bulge that it raises on the primary.

8 The lag angle, ε, is related to the energy dissipation rate in the planet due to the periodic tidal forcing. Approx., tan(2ε) = 1 Q where the quantity Q is given by Q 1 " E =! dt 2π E max!e is the rate of energy dissipation, E max is the peak potential energy stored in the bulge, & the integral extends over 1 tidal period. For the Earth, ε ~ 2 #.4, implying that Q $ 12. This value is NOT likely to be typical of other terrestrial planets, as the dissipation occurs largely in the Earth's oceans. The solid Earth probably has Q ~ 100, or ε ~ 0 #.3, based on observations of seismic oscillations. Despinning Timescales. The time for a satellite (or the sun) to despin a planet is simply where τ despin = Cω 0 T C = α MR 2 and ω 0 = initial spin rate. The same expression applies to tides raised by the planet on a satellite acting to despin the satellite, except that m % M R r, satellite radius and Q k 2 T now refers to tides in the satellite.

9 See the table of despinning times on the next slide. Note that the values of k 2 T used for the satellites are much less than unity this reflects the greater importance of mechanical strength (i.e., elasticity) vs. gravity in smaller bodies. For a uniform density body of rigidity µ, k T 2 Despite the reduced values of k 2 T, we see that satellite despinning times are quite short, except for very distant satellites such as Iapetus, Hyperion, and the outer jovian satellites. In fact, Iapetus is known to be despun to the synchronous state, but Hyperion is apparently not. Among the planets, only Mercury and Venus have despinning timescales less than the age of the solar system, yrs. Mercury spins with a period of 58 days, in a spin-orbit resonance [an alternative end-point to synchronism, for moderately eccentric orbits]. Venus spins retrograde, with a period of 243 days, for reasons which remain mysterious.

10 Tidal Despinning Timescales Planets by the Sun. k 2 T τ/q Mercury yrs. Venus yrs Earth (0.63) yrs. Mars yrs. Jupiter yrs. τ > age of solar system Satellites by their planets. Moon* Phobos* Callisto* Titan* Hyperion Iapetus* Rotates chaotically due to large e Oberon Triton* Charon* * Known to be synchronously rota.ng

11 ( ) 1 ( ) T 1 = 5T 1 4T 1 = d or ω n = 4 n n

12 Orbital Evolution Timescales The tidal torque on the satellite acts to increase the satellite/planet system's orbital angular momentum. For simplicity, we consider a zero-inclination, circular orbit: T = d dt ( h orb ) = d dt! m( GM ) 1 d 2 dt = 1 2 m GM mm M + m na2 a ( ) 1 2 a 1 1 "a = 3 k T 2 Q i G M 2 m R a "a The solution to this equation is straightforward ( ) a 2 = a β t t 2 0 where a 0 = initial semi-major axis. if we assume that m << M β a 11 2 since sin2ε! Q 1. At an given time, the timescalefor orbital expansion is given by t orbit ~ a "a = a 2 β 1 ~ a 2 m 1, where the very strong dependence on an almost guarantees that tidal orbital evolution is most important for the innermost satellite(s) of a planet. This dependence of a on m for a given value of t t 0 is illlustrated onthe following slide.

13 Note: recent measurements of da/dt for saturnian satellites show that this is wrong: in fact Q(S) ~ This suggests that the satellites are relatively young.

14 Satellite Tides Yoder (1979) Eccentricity tides Open ellipse = solid body, filled ellipse = tidal distortion. Low amplitude tide Large amplitude tide At the same time, the tidal amplitude varies as r -3. Both dissipate energy inside the satellite.

15 Satellite Tides vs Planet Tides: satellite tidal effects planet tidal effects! M p m s 2 5 R ( ) s k Q ( R p ) ( s k p )( p Q s ) R ( ) p k Q ( R s )( s k p )( p Q s ) ~ ρ p ρ s >>1 since R p > R s and Q p > Q s But, for synchronously rotating satellites the dominant semi-diurnal tide has zero frequency, since ω s = n, and thus the associated lag angle is zero. The most important effect is due to energy dissipation associated with orbital eccentricity, or spin axis obliquity, ε: "E = GM 2 p R 5 s n a 6 k ( s Q s ) 21 2 e sin2 ε = GM p m s 2a 2 "a Since no net torque is exerted a synchronous satellite, the orbital angular momentum is conserved e is damped: ( ) L z = GM p a 1 e 2 "e = 1 e2 2ea τ e = e "e = 2 21 "a if i! φ m s M p 1 2 cosi a ( R s ) 5 Q s n 1 k s Scales as m s and a 13/2 à affects small, close satellites! Q years for Triton, as illustrated in the numerical integation.

16 Neptune & Triton

17 Summary of the known and probable effects of tidal interactions: Tides raised on: By: Effect(s) Planets Sun Spin- down of Mercury & Venus Planets Satellites Spin- down of Earth, Pluto Orbit expansion of Moon and inner satellites of J, S, U Orbit decay of Phobos & Triton Establishment of orbital resonances (J, S) Satellites Planets De- spinning of most satellites to synch. rota.on Orbit decay & eccentricity damping due to radial and libra.on.des Tidal hea.ng of interiors of satellites (Io, Enceladus) è volcanism & plumes Other tides e.g., those raised on the sun by planets, can be shown to have negligible effects, even over the age of the solar system. Tides may also have played a role in the capture of satellites by planets or proto-planets (e.g., Triton).

18 Tidal Effects in the Earth-Moon System Since ~1750 it has been realized that ancient ( 0 AD) observations of solar eclipses were inconsistent with predictions based on modern observations and the theory of the moon s motion. In the 1930 s it was realized that the discrepancies in the predicted and actual eclipse locations, and in other ancient and historical observations (such as lunar occultations of stars and transits of Mercury across the sun) were attributable to a combination of 1. An orbital deceleration of the moon,!n and 2. A slowdown in the Earth s spin rate,!ω Early ( 1975) attempts to separate these 2 effects lead to a wide range of values for!n and!ω, depending on the particular date sets used. A commonly quoted value is from Spencer-Jones (1939) who obtained!n = 22 ".4 century 2. Note that solar or lunar eclipses really depend on their apparent longitude difference, relative to a co-ordinate system fixed on Earth, i.e., on of Δλ = λ m λ d and ( 2 Δλ) = "n n ( "ω dt 2 ω ) n ( ) "ω ω!! λ m!! λ

19 More recently, lunar-laser-ranging (LLR) experiments, and direct satellite determinations of the Earth s tidal lag angle, ε, as well as lunar occultations of stars referred to Atomic Time have all converged on a direct determination of!n alone:!n = ( 25 ± 1)" cy 2, corresponding to a rate of expansion of the Moon's orbit of!a = cm cy 1. The ancient eclipse observations (Chinese, Babylonian & Arab) may now be interpreted using this!n to derive an average value of!ω. Over the last 3000 years, we find!ω ω ( ) 10 9 cy 1, " 20.4 ± 1.2 corresponding to an increase in the length of day (LOD) of d (LOD) = (+1.8 ± 0.1) msec dt cy 1. [See diagram for data.] Direct modern determination of!ω is frustrated by decade and possibly century-scale random variations, in the Earth's spin due to changes in atmospheric circulation and/or interactions between the Earth's core and mantle. Curiously, the predicted value of!ω, based on tidal theory and the observed value of!n, is significantly different from that obtained above:!ω ω tidal = cy 1.

20 Measured length of day (LOD) and UT1 UTC, Leap seconds

21 The Earth therefore has a positive non-tidal acceleration in its spin rate of ~ cy 1. This appears to be due to a steady decrease in C, the polar moment of inertia, which has been independently observed from satellite tracking: ( )! 3 "C 2 MR 2 = ( 2.6 ± 0.6) 10 9 cy 1. d dt J 2 This decrease in C is presumed to be a result of post-glacial uplift of polar regions since the last Ice Age ~10,000 years ago. From "n, we can calculate the Earth's Q from Kepler's 3rd law and the equation for "a derived above: "n n = 3 2 = 9 2 Using k 2 T = 0.29, we get Q = "a a k 2 T Q 1 i G 2 mr 5 M a 2. This value of Q, if extrapolated into the past, implies a very close approach of the Moon to Earth < years ago, with unacceptable geological consequences to both bodies (e.g., melting the Moon!). Observations of growth lines on fossil corals & bivalve shellfish up to 400 My old support the current value of "n : ( )! 22 ± 1 fossils ( ) 10 9 cy 1, "ω ω so we are left with postulating a considerably higher value of Q at times before this epoch (perhaps as high as 30), if we are to avoid the above problem. It is likely that this variation in Q reflects the role of plate tectonics in rearrangaing the distribution of continents and ocean basins.

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24 Ancient Solar Eclipses & Tidal Evolution: Solar longitude, Lunar longitude, λ = λ 0 + n t λ = λ 0 + n t + 1 "nt 2 2 ( mod 2π ) Eclipse occurs when λ = λ i.e., λ 0 λ 0 + ( n n )t + 1 "nt 2 = 2π m 2 whereas the predicted eclipse time, T is given by λ 0 λ 0 + ( n n )T = 2π m t # T "nt 2 2( n n ) T "nt 2 2Δn The rotational position of the Earth at time t is and θ = θ 0 + ωt "ωt 2 θ pred = θ 0 + ωt ( ) "ωt 2 θ θ pred = ω t T = 1 "ω ω "n 2 ( Δn )T 2 The (east) longitude at which the eclipse is seen is φ = λ ( t) θ ( t) = λ ( T ) n "nt 2 θ 2Δn pred 1 "ω ω "n 2 ( Δn )T 2 = φ pred 1 2 ( ) "n "ω ω n n n T 2

25 Substituting appropriate numerical values, we get!n = 25" cy 1 = rad d 2!P = 1.8 ms cy 1!ω = rad d 2 n = rad d 1 n = rad d 1 ω = 6.28 rad d 1 t T # 27 T cy 2 sec, where T cy is the time before the present in centuries, and φ φ pred # T cy 2 = 26 0 over 3000 yr. Note that the rotational displacement of the Earth due to!ω alone is 1 ωt! 2 = T 2 2 cy 33 s cy 2 which coresponds to an error in sidereal clocks relative to Atomic time of 33 s cy 2 or almost 1 hour in 1000 yrs.

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27 Satellite Shapes Most planetary satellites have been tidally despun to a state of synchronous rotation (i.e., ω = n), with their spin axes approximately normal to their orbit planes. In this situation, satellites experience a permanent tidal potential in addition to the rotational potential V C. From the Rotational notes, the centrifugal potental is V C = 1 2 ω 2 r 2 sin 2 θ ( ) = 1 2 ω 2 x 2 + y 2 The tidal potential is symmetric about the X axis, and is given by V T = GM p 3 ( a 3 2 cos2 φ 1 2 )r 2 = GM p x 2 y 2 a ( z ) For a synchronous satellite, GM p a 3 = n 2 = ω 2 (Kepler III) so the combined potential is ( ) V TOT V C + V T = n x2 1 2 z 2

28 Like V C and V T, V TOT is a quadratic potential, so the distortion of the satellite leads to an additional gravitational potential = k 2 V TOT, so that the total potential at the surface of the satellite is (1 + k 2 ) V TOT = 1 2 ( n2 3sin 2 θ cos 2 ψ cos 2 θ)r 2 1+ k 2 ( )! where ψ is a longitude measured in the XY plane from the X (planet) direction. As before, the equilibrium shape is an equpotential surface: Gm r + Gm δr + (1 + k r 2 2 ) V TOT = const. δr( θ,ψ ) " n2 r k r 2Gm ( 2 ) 3sin 2 θ cos 2 ψ cos 2 θ k 2 ( 2 )q 3sin 2 θ cos 2 ψ cos 2 θ This describes a Triaxial Ellipsoid, with principal axes a(x), b(y), and c(z). The axis differences are: a c = 2( 1 + k 2 )qr a b = k 2 ( 2 )qr b c = k 2 ( 2 )qr since the value of! along each axis is given by +3( a), φ( b) and 1( c). If the satellite is uniform in its interior, then k 2 = 3 2 a c a b = 5q, = 15 q, b c = 5 q r r 4 r 4 and we have

29 Note that, for a synchronous satelllite, Kepler's 3 rd law implies q = n2 r3 Gm = M p m ( ) 3 = 3n2 r a 4πGρ. Note also that, regardless of the value of q, the equilibrium shape of such a satellite satisfies where ρ is the satellite's mean density. b c a c = 1 4 which can be used to test the hypothesis that a particular stellite has an equilibrium shape. EXAMPLE: Dermott & Thomas (1988) studied the shape of Saturn's satellite Mimas (r! 200 km) and found it to be triaxial, with a c = 17.7 ± 1.0 km b c = 4.8 ± 1.0 km ( ) = 0.27 ± 0.06 ( a c) b c in close agreement with an equilibrium figure. However, the predicted distortion is slightly greater: ( a c) pred = 5q r = 18.7 km ( n = radsec 1, ρ = g cm 3 q = ) A theory of the shape accurate to 0(q 2 ) predicts an even greater value of ( a c)! 20.3 km, so D & T concluded that Mimas is not homogeneous, but centrally condensed, with C Mr 2! ± This value was derived from a modified form of the Darwin-Radan relation for rotational oblateness.

30 The following Table and Figures give current values of the shapes of the Saturnian icy satellites, as derived by P.C. Thomas, et al., Icarus (2007) doi: /j.icarus ( Note that F = b c. a c )

31 P.C. Thomas, et al., Icarus (2007) doi: /j.icarus Figure 8: Roughness of limbs of saturnian satellites. Plotted quantities are the root mean square residuals to the limb fits (a) as a fraction of the object s mean radius and (b) as absolute amounts.

32 P.C. Thomas, et al., Icarus (2007) doi: /j.icarus Fig. 9. (a) Approach to equilibrium shapes. The observed factor F = (b c)(a c) plotted against object mean radius. Equilibrium values for F are 0.21 for Mimas, 0.23 for Enceladus, 0.24 for Tethys and 0.25 for Dione, and Rhea. Iapetus is not included in this plot. (b) Observed and predicted differences in long and short axes of saturian satellites. The predicted values assume that the object is homogeneous.