EART162: PLANETARY INTERIORS

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1 EART162: PLANETARY INTERIORS Francis Nimmo

2 Last Week Applications of fluid dynamics to geophysical problems Navier-Stokes equation describes fluid flow: Convection requires solving the coupled equations of heat transfer and fluid flow Behaviour of fluid during convection is determined by a single dimensionless number, the Rayleigh number Ra

play an important role in the orbital (and thermal) evolution of some bodies To

3 This Week Tides and Orbits Planetary tides are useful for two reasons: We can use observations of tides to constrain the internal structure of planetary bodies Tides play an important role in the orbital (and thermal) evolution of some bodies To understand the second point, we need to learn (a small amount) of orbital dynamics, if time permits

4 Tides (1) Body as a whole is attracted with an acceleration = Gm/a 2 But a point on the far side experiences an acceleration = Gm/(a+R) 2 The net acceleration is 2GmR/a 3 for R<<a On the near-side, the acceleration is positive, on the far side, it s negative For a deformable body, the result is a symmetrical tidal bulge: R a α m

5 Tides (2) It is often useful to think about tidal effects in the frame of reference of the tidally-deforming body E.g. tides raised on Earth by Moon (Earth rotates faster than Moon orbits, you feel the tidal bulge move past you) E.g. tides raised on the Moon by Earth (Moon rotates as fast as the Earth appears to orbit, the bulge is (almost) fixed ) Earth Moon Moon Earth If the Moon s orbit were circular, the Earth would appear fixed in space and the tidal bulge would be static

6 Tides (3) planet M R ϕ P a b m satellite Tidal potential at P (recall acceleration = - ) Cosine rule (R/a)<<1, so expand square root Constant Mean gravitational => No acceleration acceleration (Gm/a 2 ) Tide-raising part of the potential

7 Tides (4) We can rewrite the tide-raising part of the potential as Where P 2 (cos ϕ) is a Legendre polynomial, g is the surface gravity of the planet, and H is the equilibrium tide This is the tide raised on the Earth by the Moon m Does this make sense? (e.g. the Moon at 60R E, M/m=81) For a uniform fluid planet with no elastic strength, the amplitude of the tidal bulge is (5/2)H An ice shell decoupled from the interior by an ocean will have a tidal bulge similar to that of the ocean For a rigid body, the tide may be reduced due to the elasticity of the planet (see next slide)

8 Effect of Rigidity We can write a dimensionless number which tells us how important rigidity µ is compared with gravity: (g is acceleration, ρ is density) For Earth, µ~10 11 Pa, so ~3 (gravity and rigidity are comparable) For a small icy satellite, µ~10 10 Pa, so ~ 10 2 (rigidity dominates) We can describe the response of the tidal bulge and tidal potential of an elastic body by the Love numbers h 2 and k 2, respectively For a uniform solid body we have: E.g. the tidal bulge amplitude d is given by d= h 2 H (see last slide) The quantity k 2 is important in determining the magnitude of the tidal torque (see later)

2m. What is the effective rigidity of the Earth? For Earth, H=0.35m and d=h 2 H, so h 2 =0.

9 Why do Love numbers matter? Because we can measure the tidal bulge and deduce the rigidity of the planet! How do we measure the tidal bulge? A.E.H. Love Example: the solid part of the Earth has a fortnightly tidal amplitude d of about 0.2m. What is the effective rigidity of the Earth? For Earth, H=0.35m and d=h 2 H, so h 2 =0.6 So =3 So µ=100 GPa What do we conclude from this exercise? How do we reconcile this with mantle convection? Lord Kelvin

10 What can the Love number tell us about internal structure? Most planets are not uniform bodies If the planet has a dense core, then the Love number will be larger than that of a uniform body with equal rigidity If the planet has low-rigidity layers, the Love number will be larger than expected. Why is this useful? Moore & Schubert 2003

11 1) Tidal torques Synchronous distance Tidal bulge Effects of Tides In the presence of friction in the primary, the tidal bulge will be carried ahead of the satellite (if it s beyond the synchronous distance) This results in a torque on the satellite by the bulge, and vice versa. The torque on the bulge causes the planet s rotation to slow down The equal and opposite torque on the satellite causes its orbital speed to increase, and so the satellite moves outwards The effects are reversed if the satellite is within the synchronous distance (rare why?) Here we are neglecting friction in the satellite, which can change things. The same argument also applies to the satellite. From the satellite s point of view, the planet is in orbit and generates a tide which will act to slow the satellite s rotation. Because the tide raised by the planet on the satellite is large, so is the torque. This is why most satellites rotate synchronously with respect to the planet they are orbiting.

12 Tidal Torques Examples of tidal torques in action Almost all satellites are in synchronous rotation Phobos is spiralling in towards Mars (why?) So is Triton (towards Neptune) (why?) Pluto and Charon are doubly synchronous (why?) Mercury is in a 3:2 spin:orbit resonance (not known until radar observations became available) The Moon is currently receding from the Earth (at about 3.5 cm/yr), and the Earth s rotation is slowing down (in 150 million years, 1 day will equal 25 hours). What evidence do we have? How could we interpret this in terms of angular momentum conservation? Why did the recession rate cause problems?

13 Diurnal Tides (1) Consider a satellite which is in a synchronous, eccentric orbit Both the size and the orientation of the tidal bulge will change over the course of each orbit Empty focus Tidal bulge Fixed point on satellite s surface Planet a This tidal pattern consists of a static part plus an oscillation From a fixed point on the satellite, the resulting tidal pattern can be represented as a static tide (permanent) plus a much smaller component that oscillates (the diurnal tide) 2ae N.B. it s often helpful to think about tides from the satellite s viewpoint a

14 Diurnal tides (2) The amplitude of the diurnal tide d is 3e times the static tide (does this make sense?) Why are diurnal tides important? Stress the changing shape of the bulge at any point on the satellite generates time-varying stresses Heat time-varying stresses generate heat (assuming some kind of dissipative process, like viscosity or friction). NB the heating rate goes as e 2 we ll see why in a minute Dissipation has important consequences for the internal state of the satellite, and the orbital evolution of the system (the energy has to come from somewhere) Heating from diurnal tides dominate the behaviour of some of the Galilean and Saturnian satellites

15 Tidal Heating (1) Recall from Week 5 Strain depends on diurnal tidal amplitude d R d Strain rate depends on orbital period τ What controls the tidal amplitude d? Power per unit volume P is given by Here Q is a dimensionless factor telling us what fraction of the elastic energy is dissipated each cycle The tidal amplitude d is given by: α

16 Tidal Heating (2) Tidal heating is a strong function of R and a This is not exact, but good enough for our purposes The exact equation can be found at the bottom of the page Is Enceladus or Europa more strongly heated? Is Mercury strongly tidally heated? Tidal heating goes as 1/τ and e 2 orbital properties matter What happens to the tidal heating if e=0? Tidal heating depends on how rigid the satellite is (E and µ) What happens to E and µ as a satellite heats up, and what happens to the tidal heating as a result?

17 Important concepts for seminar Tidal heating Depends on R, e 2 and internal structure (h 2,k 2 ) Causes orbit to circularize (e 0) Orbital resonance Occurs when periods are in simple ratio Mutual perturbations can keep e > 0 Resonances prolong tidal heating

18 Important concepts for seminar Love numbers (h 2,k 2 ) Tell us how rigid a body is Larger Love numbers imply more tidal heating

20 Kepler s laws (1619) These were derived by observation (mainly thanks to Tycho Brahe pre-telescope) 1) Planets move in ellipses with the Sun at one focus 2) A radius vector from the Sun sweeps out equal areas in equal time 3) (Period) 2 is proportional to (semi-major axis a) 3 a ae b apocentre empty focus focus pericentre e is eccentricity a is semi-major axis

21 Newton (1687) Explained Kepler s observations by assuming an inverse square law for gravitation: Here F is the force acting in a straight line joining masses m 1 and m 2 separated by a distance r; G is a constant (6.67x10-11 m 3 kg -1 s -2 ) A circular orbit provides a simple example and is useful for back-of-the-envelope calculations: Period T Centripetal acceleration M r Mean motion (i.e. angular frequency) n=2 π/t Centripetal acceleration = rn 2 Gravitational acceleration = GM/r 2 So GM=r 3 n 2 (also true for elliptical orbits) So (period) 2 is proportional to r 3 (Kepler)

22 Orbital angular momentum For a circular orbit: Angular momentum = In For a point mass, I=ma 2 Angular momentum/mass = na 2 e is the eccentricity, a is the semi-major axis h is the angular momentum a ae b r focus m b 2 =a 2 (1-e 2 ) Angular momentum per unit mass. Compare with na 2 for a circular orbit An elliptical orbit has a smaller angular momentum than a circular orbit with the same value of a Orbital angular momentum is conserved unless an external torque is acting upon the body

23 Energy To avoid yet more algebra, we ll do this one for circular coordinates. The results are the same for ellipses. Gravitational energy per unit mass E g =-GM/r why the minus sign? Kinetic energy per unit mass E v =v 2 /2=r 2 n 2 /2=GM/2r Total sum E g +E v =-GM/2r (for elliptical orbits, -GM/2a) Energy gets exchanged between k.e. and g.e. during the orbit as the satellite speeds up and slows down But the total energy is constant, and independent of eccentricity Energy of rotation (spin) of a planet is E r =CΩ 2 /2 C is moment of inertia, Ω angular frequency Energy can be exchanged between orbit and spin, like momentum

24 Summary Mean motion of planet is independent of e, depends on GM and a: Angular momentum per unit mass of orbit is constant, depends on both e and a: Energy per unit mass of orbit is constant, depends only on a:

25 Angular Momentum Conservation Angular momentum per unit mass where the second term uses Say we have a primary with zero dissipation (this is not the case for the Earth-Moon system) and a satellite in an eccentric orbit. The satellite will still experience dissipation (because e is nonzero) where does the energy come from? So a must decrease, but the primary is not exerting a torque; to conserve angular momentum, e must decrease also- circularization For small e, a small change in a requires a big change in e Orbital energy is not conserved dissipation in satellite NB If dissipation in the primary dominates, the primary exerts a torque, resulting in angular momentum transfer from the primary s rotation to the satellite s orbit the satellite (generally) moves out (as is the case with the Moon).

26 Summary Tidal bulge amplitude depends on mass, position, rigidity of body, and whether it is in synchronous orbit Tidal Love number is a measure of the amplitude of the tidal bulge compared to that of a uniform fluid body Tidal torques are responsible for orbital evolution e.g. orbit circularization, Moon moving away from Earth etc. Tidal strains cause dissipation and heating Orbits are described by mean motion n, semi-major axis a and eccentricity e. Orbital angular momentum is conserved in the absence of external torques: if a decreases, so does e

27 How fast does it happen? The speed of orbital evolution is governed by the rate at which energy gets dissipated (in primary or satellite) Since we don t understand dissipation very well, we define a parameter Q which conceals our ignorance: Where ΔE is the energy dissipated over one cycle and E is the peak energy stored during the cycle. Note that low Q means high dissipation! It can be shown that Q is related to the phase lag arising in the tidal torque problem we studied earlier: ε

28 How fast does it happen(2)? The rate of outwards motion of a satellite is governed by the dissipation factor in the primary (Q p ) Here m p and m s are the planet and satellite masses, a is the semi-major axis, R p is the planet radius and k 2 is the Love number. Note that the mean motion n depends on a. Does this equation make sense? Recall Why is it useful? Mainly because it allows us to calculate Q p. E.g. since we can observe the rate of lunar recession now, we can calculate Q p. This is particularly useful for places like Jupiter. We can derive a similar equation for the time for circularization to occur. This depends on Q s (dissipation in the satellite).

29 Tidal Effects - Summary Tidal despinning of satellite generally rapid, results in synchronous rotation. This happens first. If dissipation in the synchronous satellite is negligible (e=0 or Q s >>Q p ) then If the satellite is outside the synchronous point, its orbit expands outwards (why?) and the planet spins down (e.g. the Moon) If the satellite is inside the synchronous point, its orbit contracts and the planet spins up (e.g. Phobos) If dissipation in the primary is negligible compared to the satellite (Q p >>Q s ), then the satellite s eccentricity decreases to zero and the orbit contracts a bit (why?) (e.g. Titan?)

30 Example results 1. Primary dissipation dominates satellite moves outwards and planet spins down 2. Satellite dissipation dominates orbit rapidly circularizes 2. Orbit also contracts, but amount is small because e is small

31 Tidal Heating (1) Recall diurnal tidal amplitude goes as in the limit when rigidity dominates ( ) So strain goes as Energy stored per unit volume = stress x strain In an elastic body, stress = strain x µ (rigidity) So total energy stored goes as µe 2 H 2 R s / For tide raised on satellite H=R s (m p /m s )(R s /a) 3 From the above, we expect the energy stored E to go as Note that here we have used

32 Tidal heating (2) From the definition of Q, we have We ve just calculated the energy stored E, so given Q s and n we can thus calculate the heating rate de/dt The actual answer (for uniform bodies) is But the main point is that you should now understand where this equation comes from Example: Io We get 80 mw/m 2, about the same as for Earth (!) This is actually an underestimate why?

33 Angular Momentum (1) The angular momentum vector of an orbit is defined by This vector is directed perpendicular to the orbit plane. By use of vector triangles (see handout), we have So we can combine these equations to obtain the constant magnitude of the angular momentum per unit mass This equation gives us Kepler s second law directly. Why? What does constant angular momentum mean physically? C.f. angular momentum per unit mass for a circular orbit = r 2 ω The angular momentum will be useful later on when we calculate orbital timescales and also exchange of angular momentum between spin and orbit

34 Elliptical Orbits & Two-Body Problem Newton s law gives us r m 1 r m 2 See Murray and Dermott p.23 where µ=g(m 1 +m 2 ) and is the unit vector (The m 1 +m 2 arises because both objects move) The tricky part is obtaining a useful expression for d 2 r/dt 2 (otherwise written as ). By starting with r=r and differentiating twice, you eventually arrive at (see the handout for details): Comparing terms in, we get something which turns out to describe any possible orbit

35 Elliptical Orbits Does this make sense? Think about an object moving in either a straight line or a circle The above equation can be satisfied by any conic section (i.e. a circle, ellipse, parabola or hyberbola) The general equation for a conic section is e is the eccentricity, a is the semi-major axis h is the angular momentum a ae r f focus For ellipses, we can rewrite this equation in a more convenient b form (see M&D p. 26) using b 2 =a 2 (1-e 2 ) θ=f+const.

Equation of orbital velocity: v 2 =GM(2/r 1/a) where: G is the gravitational constant (G=6.67x10 11 N/m 3 kg), M is the mass of the sun (or central

Equation of orbital velocity: v 2 =GM(2/r 1/a) where: G is the gravitational constant (G=6.67x10 11 N/m 3 kg), M is the mass of the sun (or central Everything in Orbit Orbital Velocity Orbital velocity is the speed at which a planetary body moves in its orbit around another body. If orbits were circular, this velocity would be constant. However, from