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ESS 298: OUTER SOLAR SYSTEM

ESS 298: OUTER SOLAR SYSTEM. Francis Nimmo. Io against Jupiter, Hubble image, July 1997. Course Outline. Week 1 – Introduction, solar system formation, exploration highlights, orbital dynamics Weeks 2-3 – Galilean satellites

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ESS 298: OUTER SOLAR SYSTEM

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  1. ESS 298: OUTER SOLAR SYSTEM Francis Nimmo Io against Jupiter, Hubble image, July 1997

  2. Course Outline • Week 1 – Introduction, solar system formation, exploration highlights, orbital dynamics • Weeks 2-3 – Galilean satellites • Week 4 – Titan and the other Saturnian satellites, Cassini results • Week 5 – Gas giants and ice giants – structure, atmospheres, rings, extra-solar planets • Weeks 6-7 – Computer project • Weeks 8-9 – Student presentations • Weeks 10 – The Outer Limits – Pluto/Charon, Kuiper Belt, Oort Cloud, future missions This schedule can be modified if someone is interested in a particular topic

  3. Logistics • Set texts – see website for suggestions http://www2.ess.ucla.edu/~nimmo/ess298 • Office hours – make appointments by email nimmo@ess.ucla.edu or drop by (4642 Geology) • Auditing? • Student presentations – ~30 min. talk on controversial research topic. Sign-up sheets week 4. • Grading – based on performance in student presentation (70%) and computer project writeup (30%). P/NP or letter grade. • Location/Timing –Tues/Thurs 11:15-12:30in room 4677 Geology • Questions? - Yes please!

  4. This Week • Where and what is the outer solar system? • What is it made of ? • How did it form? • How do we know? (spacecraft missions and ground-based observations) • Highlights • Orbital dynamics • Kepler’s laws • Moment of inertia and internal structure • Tidal deformation

  5. Where is it? • Everything beyond the asteroid belt (~ 3AU) • 1 AU=Earth-Sun distance = 150 million km • Jupiter, Saturn, Uranus, Neptune, Pluto, plus satellites • Kuiper Belt • Oort Cloud Inner solar system 1.5 AU 5 AU 30 AU Outer solar system

  6. Where is it? (cont’d) Distances on this figure are in AU. Areas of the planets are scaled by their masses. Percentages are the total mass of the solar system (excluding the Sun) contained by each planet. Note that Jupiter completely dominates. We conventionally divide the outer solar system bodies into gas giants, ice giants, and small bodies. This is a compositional distinction. How do we know the compositions?

  7. Basic Parameters Data from Lodders and Fegley, 1998. a is semi-major axis, e is eccentricity, R is radius, M is mass, r is relative density, Ts is temperature at 1 bar surface, m is magnetic dipole moment in Tesla x R3.

  8. Compositions (1) • We’ll discuss in more detail later, but briefly: • (Surface) compositions based mainly on spectroscopy • Interior composition relies on a combination of models and inferences of density structure from observations • We expect the basic starting materials to be similar to the composition of the original solar nebula (how do we know this?) • Surface atmospheres dominated by H2 or He: (Lodders and Fegley 1998)

  9. Compositions (2) • Jupiter and Saturn consist mainly of He/H with a rock-ice core of ~10 Earth masses • Uranus and Neptune are primarily ices covered with a thick He/H atmosphere • Pluto is probably an ice-rock mixture 90% H/He 75% H/He 10% H/He 10% H/He Figure from Guillot, Physics Today, (2004). Sizes are to scale. Yellow is molecular hydrogen, red is metallic hydrogen, ices are blue, rock is grey. Note that ices are not just water ice, but also frozen methane, ammonia etc.

  10. Temperatures • Obviously, the stability of planetary constituents (and thus planetary composition) depends on the temperature as the planets formed. We’ll discuss this in a second. • The present-day surface temperature may be calculated as follows: • Here F is the solar constant (1367 Wm-2), Ab is the Bond Albedo (how much energy is reflected), a is the distance to the Sun in AU, e is the emissivity (typically 0.9) and s is the Stefan-Boltzmann constant (5.67x10-8 in SI units). • Where does this equation come from?

  11. Temperatures (cont’d) • Temperatures drop rapidly with distance • Volatiles present will be determined by local temperatures • Volatiles available to condense during initial formation of planets will be controlled in a similar fashion (although the details will differ) Neptune Jupiter Uranus Saturn Plot of temperature as a function of distance, using the equation on the previous page with Ab=0.1 to 0.4

  12. Solar System Formation - Overview • 1. Nebular disk formation • 2. Initial coagulation (~10km, ~105 yrs) • 3. Orderly growth (to Moon size, ~106 yrs) • 4. Runaway growth (to Mars size, ~107 yrs), gas loss (?) • 5. Late-stage collisions (~107-8 yrs)

  13. Observations (1) • Early stages of solar system formation can be imaged directly – dust disks have large surface area, radiate effectively in the infra-red • Unfortunately, once planets form, the IR signal disappears, so until very recently we couldn’t detect planets (see later) • Timescale of clearing of nebula (~1-10 Myr) is known because young stellar ages are easy to determine from mass/luminosity relationship. This is a Hubble image of a young solar system. You can see the vertical green plasma jet which is guided by the star’s magnetic field. The white zones are gas and dust, being illuminated from inside by the young star. The dark central zone is where the dust is so optically thick that the light is not being transmitted. Thick disk

  14. Observations (2) • We can use the present-day observed planetary masses and compositions to reconstruct how much mass was there initially – the minimum mass solar nebula • This gives us a constraint on the initial nebula conditions e.g. how rapidly did its density fall off with distance? • The picture gets more complicated if the planets have moved . . . • The change in planetary compositions with distance gives us another clue – silicates and iron close to the Sun, volatile elements more common further out

  15. Disk cools by radiation Nebula disk (dust/gas) Polar jets Cold, low r Hot, high r Infalling material Dust grains Stellar magnetic field (sweeps innermost disk clear, reduces stellar spin rate) Cartoon of Nebular Processes • Scale height increases radially (why?) • Temperatures decrease radially – consequence of lower irradiation, and lower surface density and optical depth leading to more efficient cooling

  16. Temperature and Condensation Nebular conditions can be used to predict what components of the solar nebula will be present as gases or solids: Mid-plane Photosphere Earth Saturn Condensation behaviour of most abundant elements of solar nebula e.g. C is stable as CO above 1000K, CH4 above 60K, and then condenses to CH4.6H2O. From Lissauer and DePater, Planetary Sciences Temperature profiles in a young (T Tauri) stellar nebula, D’Alessio et al., A.J. 1998

  17. Accretion timescales (1) • Consider a protoplanet moving through a planetesimal swarm. We have where v is the relative velocity and f is a factor which arises because the gravitational cross-sectional area exceeds the real c.s.a. f is the Safronov number: Planet density r vorb Where does this come from? R fR where ve is the escape velocity, G is the gravitational constant, r is the planet density. So: Planetesimal Swarm, density rs

  18. Accretion timescales (2) f • Two end-members: • 8GrR2 << v2 so dM/dt ~ R2 which means all bodies increase in radius at same rate – orderly growth • 8GrR2 >> v2 so dM/dt ~ R4 which means largest bodies grow fastest – runaway growth • So beyond some critical size (~Moon-size), the largest bodies will grow fastest and accrete the bulk of the mass • If we assume that the relative velocity v is comparable to the orbital velocity vorb, we can show (how?) that Here f is the Safronov factor as before, n is the orbital mean motion (2p/period), ss is the surface density of the planetesimal swarm and r is the planet density

  19. Accretion Timescales (3) • Rate of growth decreases as surface density ss and orbital mean motion n decrease. Both these parameters decrease with distance from the Sun (as a-1.5 and a-1 to -2, respectively) • So rate of growth is a strong function (~a-3) of distance Approximate timescales t to form an Earth-like planet. Here we are using f=10, r=5.5 g/cc. In practice, f will increase as R increases. Note that forming Neptune is problematic!

  20. Runaway Growth • Recall that for large bodies, dM/dt~R4 so that the largest bodies grow at the expense of the others • But the bodies do not grow indefinitely because of the competing gravitational attraction of the Sun • The Hill Sphere defines the region in which the planet’s gravitational attraction overwhelms that of the Sun; the distance from which planetesimals can be accreted to a single body is a few times this distance rH, where Where does this come from? Here M and Ms are the planet and solar mass (2x1030 kg), and a is semi-major axis. Jupiter’s Hill Sphere is ~0.5 AU

  21. Late-Stage Accretion • Once each planet has swept up debris out to a few Hill radii, accretion slows down drastically • Size of planets at this point is determined by Hill radius and local nebular surface density, ~ Mars-size at 1 AU • Collisions now only occur because of mutual perturbations between planets, timescale ~107-8 yrs • This stage can be simulated numerically: Agnor et al. Icarus 1999

  22. Complications • 1) Timing of gas loss • Presence of gas tends to cause planets to spiral inwards, hence timing of gas loss is important • Since outer planets can accrete gas if they get large enough, the relative timescale of planetary growth and gas loss is also important • 2) Jupiter formation • Jupiter is so massive that it significantly perturbs the nearby area e.g. it scattered so much material from the asteroid belt that a planet never formed there • Jupiter scattering is the major source of the most distant bodies in the solar system (Oort cloud) • It must have formed early, while the nebular gas was still present. How?

  23. Giant planets? • Why did the gas giants grow so large, especially in the outer solar system where accretion timescales are slow?: • 1) original gaseous nebula develops gravitational instabilities and forms giant planets directly • 2)solid cores develop rapidly enough that they reach the critical size (~10-20 Me) to accrete local nebular gas (runaway) • Hypothesis 1) can’t explain why the gas/ice giants are so different to the original nebular composition, and require an enormous initial nebula mass (~1 solar mass) • Hypothesis 2) is reasonable, and can explain why Uranus and Neptune are smaller with less H/He – they must have been forming as the nebula gas was dissipating (~10 Myr) • In this scenario, the initial planet radius was ~rH, but the gas envelope subsequently contracted (causing heating)

  24. Summary • The Outer Solar System is Big and Cold • Cold - because disk density lower, radiative cooling more efficient. Means that volatiles can be accreted . . . • Big – planets are large because of runaway effect of accreting volatiles (while nebular gas is present) • Big – lengthscales separating planets set by Hill Sphere, which increases with planet mass and distance from the Sun

  25. Spacecraft Exploration • Three major problems (how do we solve them?): • Power • Communications • Transit time • Pioneers 10 & 11 were the first outer solar system probes, with fly-bys of Jupiter (1974) and Saturn (1979) Saturn with Rhea in the foreground

  26. Voyagers 1 and 2 • A brilliantly successful series of fly-bys spanning more than a decade • Close-up views of all four giant planets and their moons • Both are still operating, and collecting data on solar/galactic particles and magnetic fields Voyagers 1 and 2 are currently at 90 and 75 AU, and receding at 3.5 and 3.1 AU/yr; Pioneers 10 and 11 at 87 and 67 AU and receding at 2.6 and 2.5 AU/yr The Death Star (Mimas)

  27. Galileo antenna • More modern (launched 1989) but the high-gain antenna failed (!) leaving it crippled • Venus-Earth-Earth gravity assist • En route, it observed the SL9 comet impact into Jupiter • Arrived at Jupiter in 1995 and deployed probe into Jupiter’s atmosphere • Very complex series of fly-bys of all major Galilean satellites • Deliberately crashed into Jupiter Sept 2003(why?) • We’ll discuss results in a later lecture

  28. Cassini • Cassini is the “last of the Cadillacs”, a large (6 ton – why? ), very expensive and very sophisticated spacecraft. • Launched in 1997, it did gravity assists at Venus, Earth and Jupiter, and has now arrived in the Saturn system. • It carries a small European probe called Huygens, which will be dropped into the atmosphere of Titan, the largest moon • Cassini will do flybys of most of Saturn’s moons (particularly Titan), as well as investigating Saturn’s atmosphere and magnetosphere • We’ll discuss the new results later in the course False-colour Cassini image of Titan’s surface; greens are ice, yellows are hydrocarbons, white is methane clouds

  29. 2) Planetary accretion in action Outer Solar System Highlights (NB these reflect my biases!) • 1) The most volcanically active place in the solar system • 3) An ocean ~3 times larger than Earth’s

  30. Highlights (cont’d) • 4) Active nitrogen geysers • 5) “Hot Jupiters”

  31. Next time . . . • Orbital mechanics

  32. Orbital Mechanics • Why do we care? • Fundamental properties of solar system objects • Examples: synchronous rotation, tidal heating, orbital decay, eccentricity damping etc. etc. • What are we going to study? • Kepler’s laws / Newtonian analysis • Angular momentum and spin dynamics • Tidal torques and tidal dissipation • These will come back to haunt us later in the course • Good textbook – Murray and Dermott, Solar System Dynamics, C.U.P., 1999

  33. 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 ae a b apocentre pericentre focus e is eccentricity

  34. 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 m1 and m2separated by a distance r; G is a constant (6.67x10-11 m3kg-1s-2) • A circular orbit provides a simple example and is useful for back-of-the-envelope calculations: Period T Centripetal acceleration = rw2 Gravitational acceleration = GM/r2 So GM=r3w2 (this is a useful formula to be able to derive) So (period)2 is proportional to r3 (Kepler) Centripetal acceleration M r Angular frequency w=2 p/T

  35. 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? • C.f. angular momentum per unit mass for a circular orbit = r2w • The angular momentum will be useful later on when we calculate orbital timescales and also exchange of angular momentum between spin and orbit

  36. Elliptical Orbits & Two-Body Problem r Newton’s law gives us m1 r m2 where m=G(m1+m2) and is the unit vector (The m1+m2 arises because both objects move) See Murray and Dermott p.23 The tricky part is obtaining a useful expression for d 2r/dt2 (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

  37. 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 q=f+const. e is the eccentricity ae a r f For ellipses, we can rewrite this equation in a more convenient form using focus b b2=a2(1-e2)

  38. This is just Kepler’s third law again (Recall m=G(m1+m2)) Angular momentum per unit mass. Compare with wr2 for a circular orbit Timescale Where did that come from? • The area swept out over the course of one orbit is where T is the period • Let’s define the mean motion (angular velocity) n=2p/T • We will also use (see previous slide) • Putting all that together, we end up with two useful results: We can also derive expressions to calculate the position and velocity of the orbit as a function of time

  39. 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 Eg=-GM/r why the minus sign? • Kinetic energy per unit mass Ev=v2/2=r2w2/2=GM/2r • Total sum Eg+Ev=-GM/2r (for elliptical orbits, -m/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 Er=CW2/2 C is moment of inertia, W angular frequency • Energy can be exchanged between orbit and spin, like momentum

  40. Summary • Mean motion of planet is independent of e, depends on m (=G(m1+m2)) 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:

  41. Tides (1) • Body as a whole is attracted with an acceleration = Gm/a2 • But a point on the far side experiences an acceleration = Gm/(a+R)2 a R m • The net acceleration is 2GmR/a3 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:

  42. b R j m M a Tide-raising part of the potential Mean gravitational acceleration (Gm/a2) Constant => No acceleration Tides (2) P planet • Tidal potential at P • Cosine rule • (R/a)<<1, so expand square root satellite (recall acceleration = - )

  43. Tides (3) • We can rewrite the tide-raising part of the potential as • Where P2(cos j) is a Legendre polynomial, g is the surface gravity of the planet, and H is the equilibrium tide • Does this make sense? (e.g. the Moon at 60RE, 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) This is the tide raised on the Earth by the Moon

  44. Effect of Rigidity • We can write a dimensionless number which tells us how important rigidity m is compared with gravity: (g is acceleration, r is density) • For Earth, m~1011 Pa, so ~3 (gravity and rigidity are comparable) • For a small icy satellite, m~1010 Pa, so ~ 102 (rigidity dominates) • We can describe the response of the tidal bulge and tidal potential of an elastic body by the Love numbersh2 and k2, respectively • For a uniform solid body we have: • E.g. the tidal bulge amplitude is given by h2 H (see previous slide) • The quantity k2 is important in determining the magnitude of the tidal torque (see later)

  45. Synchronous distance Tidal bulge Effects of Tides In the presence of friction inthe 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 – see later. 1) Tidal torques 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.

  46. 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?

  47. 2ae Tidal bulge Fixed point on satellite’s surface a Empty focus Planet This tidal pattern consists of a static part plus an oscillation a 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 • 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) N.B. it’s often helpful to think about tides from the satellite’s viewpoint

  48. Diurnal tides (2) • The amplitude of the diurnal tide 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 e2 – why? • 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) • We will see that diurnal tides dominate the behaviour of some of the Galilean satellites

  49. 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 non-zero) – 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).

  50. 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 DE 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: e

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