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A A ∆ F A A A A ABA A AA A A AAAAAFFFEEDF EEDFEEDFED Supplementary Material

Formation of regular satellites from ancient massive rings in the .

Aur´elien Crida & S´ebastien Charnoz

1 The Roche radius 2 Torque, migration, and disk of the giant planets evolution 2.1 Satellite-disk interaction All the calculations and the model described in this paper In the case where ∆ = (r rR)rR 1, i.e., when a are independent of the expression of the Roche radius − ≪ rR. However, in order to make Figure 1b and to test our satellite is very close to the disk’s outer edge, the gravi- model against the actual satellites systems, we need to tational torque exerted by the disk onto the satellitesimal know the value of rR around every giant planet. is given at first order by (12) : 8 − The Roche radius around a planet of radius Rp and 2 7 2 3 Γ = q ΣrRΩR(r rR) (S2) ρp, for material of density ρ, is given by (7) : 27 − 3 where ΩR = GMprR is the angular velocity at the Roche radius. In response to this torque, the orbital 13 ρp 1 dr radius r changes following Γ = M 2 GMpr dt . Using rR = 2456Rp (S1) 2 ρ TR = 2πΩR, Mdisk = πΣr , D = MdiskM , and r R p ≈ rR, this gives:

dr 16 2 GMp −3 For , the average density(29) of the five largest M GMpr = qΣrR ∆ −3 dt 27 rR regular satellites is 1627 kg.m , so that Eq.(S1) gives 5 1 d(r rR) 2 GMp − rR = 57309 km. This is between the orbits of the second 3 − = 3 qD 32 ∆ √rR r dt 3 innermost satellite of Uranus (), and the third 2π rR one (). We adopt this value. 5 d∆ 2 −1 −3 = 3 q D TR ∆ (S3) For Jupiter, the average density(29) of the Galilean dt 3 −3 satellites is 2370 kg.m , leading to rR = 141 000 km. ∼ 2.2 Disk evolution In the case of , however, a similar calculation It is well-known that any viscous disk in Keplerian ro- gives a Roche radius that would be 66 000 km, out- tation spreads (e.g. (8) ). We note F the flow ∼ side the orbit of four out of the six main regular satel- through rR. Then, the life-time of the disk, normalized lites. We conclude that the density of these satellites to the at rR reads : must be ill determined. Thus, we take the Roche ra- Mdisk dius to be a little inside the orbit of (the inner- τdisk = (S4) most one), at 44000 km. This corresponds to a density FTR ρ = 4240 kg.m−3 in Eq. (S1). As we shall see below, this dimensionless parameter is central in the system. In the case of Saturn, the Roche radius is known to be From a hypothesis on the viscosity of the tidal disk, F between 138000 and 141000 km from the planet center, and τdisk are estimated as a function of the disk to planet constrained by the presence of the rings. So, we take mass ratio D in the next subsection. However, in what rR = 140000 km. This corresponds to a density ρ = follows, all expressions will be given both as a function of −3 720 kg.m in Eq.(S1). Note that the density of Janus τdisk independently of any assumption on the viscosity, is estimated to be 630 kg.m−3. and as a function of D under the hypothesis below.

6 Supplementary Material Crida & Charnoz 2012, Science, 338, 1196-

2.2.1 Estimation of F and τdisk 3.1 End of the continuous regime The self-gravity and mutual collisions inside a disk cause This regime applies only when the satellite is close a strong outward angular momentum flux. This can be enough to the disk that it can capture directly the ma- described as an effective disk’s viscosity. Using numerical terial flowing through the Roche radius. This puts a simulations of self-gravitating disks, a semi-empirical ex- constraint on the maximum distance of the satellite to pression for the effective viscosity has been derived (9) : the disk’s edge: it must be smaller than 2rH , where 13 ∗ r = r (q3) is the Hill radius of the satellite (13,14). ν 26 r 5 G2 Σ2 Ω3 (S5) H R ≃ H Thus the condition of validity of the continuous regime ∗ is ∆(t) < 2(q3)13. Using Eq.(S9), this translates into where rH is a particle’s Hill radius normalized by its ∗ an upper limit of the satellite’s mass or distance: diameter. Specifically, for r = rR, rH = 107275. The viscous spreading timescale, T , of a disk with ν q 13 effective viscosity ν and an outer radius rR can be defined ∆ < 2 as T r2 ν. Thus, 3 ν R 3 ≡ 3 2 3 2 √3 −12 2 rRΩR ∆ < τdisk ∆ τdisk = Tν TR = ⇔ 3 2 2πν r2 Ω 4 = R R ∗ 5 2 2 52πrH G Σ 3 2 2 ∆ < = ∆c 84 D (S10) 1 G Mp ⇔ τ ≈ = disk ∗ 5 2 2 4 52 52πrH G Σ rR 3 −32 3 q < 3 τdisk = qc 222 D (S11) π −2 2 ≈ τdisk = D (S6) 52r∗ 5 H So, the continuous regime applies close to the outer ∼00425 edge of the rings, for ∆ < ∆c given by Eq.(S10). At first, material that leaves the rings form one single satellite, With τdisk now expressed as a pure function of D, we can combine Eqs. (S4) and (S6) to obtain which grows continuously with M = F t and ∆ given by Eqs. (S8) and (S9). As long as q < qc, this satellitesimal 52r∗ 5 M migrates so slowly that all the material that leaves the F = H D3 p (S7) π TR ring through the Roche radius is within its influence, and immediately accreted. ∼235 It is remarkable that the limits of the continuous For massive planets with a small disk, the disk’s normal- regime, ∆c and qc, are simple expressions of one single ized lifetime is large and the flow is small (e.g. Saturn’s parameter τdisk (or D if one uses Eq.(S6) ). rings). The contrary stands for small planets surrounded Notice that qc is much larger than the mass of the by a massive tidal disk (e.g. the proto-lunar disk around largest fragments that would form spontaneously from the Earth). gravitational instabilities, which is 16πξ2D3 (30), where ξ < 1 (e.g. ξ2 = 01). So, the continuous regime corre- 3 Continuous regime sponds not only to the onset of the gravitational insta- bilities and the formation of clumps at rR, but leads to Let us consider that the material that crosses the Roche the formation of much larger bodies. radius is immediately captured by an existing satellite Since q = F tMp, the time needed to reach the end of outside. The mass of the latter grows continuously as the continuous regime is qcMpF , which, using Eq.(S7) dMdt = F , the flow of material coming through the is simply 95 orbits, independent of any other parameter. outer edge of the ring. We assume that F is constant. This is valid on a time scale small with respect to the evolution of the disk. It is the case of Saturn’s present rings, on the scale of the formation of the smaller up to Janus (10). The migration rate of the satellite is given by Eq.(S3); with M(t) = F t and ∆(t = 0) = 0, this gives :

232 M(t) 14 t 14 ∆(t) = D14 t12 (S8) 334 M T ∝ p R This can also be written as a mass-distance relation using Eq. (S4) : 3 √3 −12 2 q(t) = τdisk ∆(t) (S9) 2

7 Supplementary Material Crida & Charnoz 2012, Science, 338, 1196-

4 Discrete regime

After the critical mass qc and distance ∆c have been reached, the satellite is too far to accrete immediately the material that comes at rR from the tidal disk. So, it keeps migrating outwards at constant mass. A new satellitesimal forms, in the continuous regime. How- ever, Eq.(S8) shows that this new satellitesimal migrates faster than the first one. Thus, it enters the first satel- litesimal’s sphere of influence, and is accreted before it reaches ∆c. The mass of the first satellite then increases. And this process repeats itself. In the end, the first satel- lite still follows M = F t, but instead of accreting con- tinuously new material directly from the tidal disk, it accretes regularly small satellitesimals. We call this the discrete regime, which is a natural prolongation of the previous continuous regime. This is illustrated by a nu- merical simulation in SM 5. Figure S1: Numerical simulation of the continuous and 4.1 End of the discrete regime discrete regimes. Top panel : mass ratio of the satellite(s) to the planet The discrete regime is valid as long as as a function of time; the curve follows q = F tMp ; dia- monds correspond to short-lived satellitesimals, quickly r ∆ < 2 H + ∆ accreted by the larger one. r c R Bottom panel : ∆(t) (the curve matches Eq.(S8)), and 12 2 3 −16 23 the difference ∆ 2r r for the first satellite formed. ∆ < τ ∆ + ∆c H ⇔ 313 2 disk − ∆ ∆ 23 < + 1 5 An example ⇔ ∆ ∆ c c 3 13 0 < 1 + Z + Z with Z = (∆c∆) We have performed a numerical simulation, using a mod- ⇔ − Z0 < Z ified version of the code described in previous work (10). ⇔ In this simulation, we assume that the surface density of −2 where Z0 is given by (31) : the disk is constant Σ = 10 kg.m ; this value is arbi- trary (actually fairly low, of the order of that of Saturn’s 13 13 31 31 C ring). The disk is not evolved through viscosity, and its 1 + 27 1 27 Z = + − 06823278 profile remains flat all the time, so that the torque on the 0  2   2  ≈ satellite is exactly given by Eq.(S2). In addition, mass 5 −1     is input at the Roche radius at a rate F = 10 kg.s . So, the domain of the discrete regime is This flow is also arbitrary, and not given by Eq.(S7). Nonetheless, we can compute τdisk using Eq.(S4). The −3 545 ∆ < ∆d = Z0 ∆c 315∆c 264 D (S12) mass input at rR is first turned into a satellite that mi- ≈ ≈ √τdisk ≈ grates following Eq.(S3). Then, if this satellite is not −6 −32 3 further than 2 times its Hill radius from rR, the mass q < qd = Z0 qc 991qc 193 τdisk 2200 D (S13) ≈ ≈ ≈ input at rR is immediately acquired by it. The last satellitesimal accreted in the discrete regime The simulation is run around a Saturn mass planet, 8 4 has a mass qc, while the first satellite is almost ten times with rR = 14 10 m, so that TR = 535 10 s. We × 2 8 × more massive. So the growth of the mass of the first have τdisk = πΣr TR = 115 10 . Equations (S10) R × satellite proceeds through little steps. and (S11) give (using the expressions in τdisk, not in D): Similarly, as qd = 991 qc, the duration of the discrete −12 regime is qc = 16 10 × td = 937 TR (S14) −4 ∆c = 16 10 × The discrete regime lasts about a hundred orbits at rR, tc = qcMpF = 284 years whatever value the other parameters have. If D is large, then F is large and the migration is fast, The result is shown in figure S1, and perfectly matches so that in 10 (resp. 100) orbits, a satellite grows a these predictions: the continuous regime takes place un- ∼ ∼ lot and migrates far. If D is small, F is small and the til a mass qc is reached after a time tc. At this time, the migration is slow, so that in the end of these 10 or satellite is at ∆c, and ∆ 2r r becomes positive. − H 100 orbits, one gets a small satellite close to the∼ outer After, the discrete regime is entered, and many satel- edge∼ of the tidal disk. litesimals form before being accreted by the main satel-

8 Supplementary Material Crida & Charnoz 2012, Science, 338, 1196- lite. They appear in the top panel of the figure as dots. Two satellitesimals merge when their orbits are sep- As the first satellite migrates away, the satellitesimals arated by less than two mutual Hill radii rHm = 13 have more space and time to grow and reach higher 2 δm rR , that is when . 3 Mp The transition between the continuous and discrete 13 regime is very smooth, as the steps for the growth of the 2 ∆1 ∆2 < 2 q main satellite are so small that the curve still appears − 3 continuous. In fact, Eq.(S13) shows that the largest ac- creted satellitesimals are one order of magnitude smaller Then the merger of the two satellites occurs when t = than the big one. tmerge such that

t + t 34 1 112 δt merge 0 = D14 (S17) 6 Pyramidal regime T 219 35q T R R In this section, we consider many satellites of constant The merger takes place at a distance: mass, migrating outwards. Satellite-satellite interactions are here neglected, and we assume that the orbits are ∆1(tmerge) ∆2(tmerge) ∆merge circular. First, we study migration under the influence ≈ ≡ of the tidal disk, then under the influence of the plane- 119 13 2 13 δt 29 tary tides. In both cases, the migration speed decreases ∆merge = 89 D q (S18) 3 TR with the distance. Therefore, the distance between two satellites decreases with time, as the faster inner one Such a merger leads to the formation of a satellite of catches up with the slower outer one. Consequently, two mass 2δm at distance ∆merge. It occurs every 2δt, as satellites eventually have an encounter, and merge. The it needs 2 satellites. Then, the same process repeats newly formed satellite is assumed to have the total mass itself with the products of the merging events. The of its two progenitors, and keeps migrating outwards. above equations apply, replacing δm by 2δm and δt by Assuming that satellitesimals of mass δm are produced 2δt. The distance of the merging, ∆merge, is propor- 13 29 59 every time interval δt at a fixed radius r, then mergers tional to δt δm , so it is multiplied by 2 . In fact, 59 59 ∆merge δm δt . In this pyramidal regime, the between two of these first generation satellitesimals take ∝ ∝ place every 2 δt at the same place rm1. So, satellitesimals many satellitesimals have their masses roughly propor- of mass 2 δm are produced every 2 δt at a fixed radius. tional the distance to the power 95 : Then, the process gets repeated, and these second gen- δm ∆95 (S19) eration satellitesimals will merge a little further, every ∝ 4 δt, leading to satellitesimals of mass 4 δm. And so on. Also, the number density of satellites should decrease The regular production of satellitesimals of given mass, with ∆. The position of the nth satellite is ∆ 25n9 migrating outwards at a decreasing speed with distance, n 147n. So n log(∆) and dnd∆ 1∆. ∝ ≈ leads to a hierarchical or pyramidal merging history, like ∝ ∝ a genealogical tree. We call this the pyramidal regime. It is analytically described below. Recall that we saw in 6.2 Migration due to planetary tides SM 4 that satellites of mass Md = qdMp are produced Assuming the satellite is beyond the synchronous or- at ∆d every δt = MdF . So, the pyramidal regime is the natural continuation of the continuous and discrete bit, the planetary tides lead to the following migration regimes. rate (32) :

12 5 dr 3k2p(GMp) R − 6.1 Migration due to the tidal disk = p q r 112 (S20) dt Qp For constant mass satellites, the solution of Eq.(S3) is: with k2p, Mp, Rp, and Qp are standing for the planet’s 74 14 tidal Love number, mass, equatorial radius, and dissipa- 2 t + t0 ∆(t) = q14D14 (S15) tion factor. With K = 39k (GM )12R 52Q assumed 334 T 2p p p p R constant, the solution of this equation is: where t0 is an integration constant, determined by ∆(t = 213 213 r(t) = (Kq) (t + t0) (S21) 0) ∆0 . Take≡ two satellitesimals of mass δm, numbered 1 and Again, we take two satellitesimals of mass δm, num- 2, with number 1 born at time t = 0 and number 2 bered 1 and 2, with number 1 born at time t = 0 and at t = δt, at the location ∆0. Under the assumption number 2 at t = δt, at the location rt0. Under the as- (t + t0) δt, Eq.(S15) gives: ≫ sumption (t + t0) δt, Eq. (S21) gives: ≫ 14 −34 q D δt t + t0 2 213 −1113 ∆1 ∆2(t) = 3 (S16) r1 r2(t) = (Kq) (t + t0) δt (S22) − 2 3 TR TR × − 13

9 Supplementary Material Crida & Charnoz 2012, Science, 338, 1196-

So, a merger takes place when System planets in SM 7: D > 10−5 except for Neptune

13 and Mars). 2 Note in passing that the torque from the planetary r1 r2(t) = 2 q r(t) − 3 tides is always negligible for ∆ < ∆d, which validates a 213 13 posteriori the use of the disk torque in the calculations 2 (Kq) 2 213 213 δt = 2 q (Kq) (t + t0) relative to the continuous and discrete regimes. 13 (t + t )1113 3 0 Satellites of same mass should be regularly delivered at 13 δt 2 ∆ by the pyramidal regime dominated by the torque = q 2:1 13(t + t ) 3 from the tidal disk. Then, the pyramidal regime dom- 0 313 δt inated by the planetary tides takes it over. In the t + t0 = end, satellites formed in the pyramidal regime have their 26 q13 masses such that: That is at 95 (∆∆2:1) ∆ < ∆2:1 239 39 213 3 213 −239 213 439 q (∆) = ∆+1 (S25) rmerge = (Kq) δt q δt q ∝ Q ∆ > ∆2:1 26213 ∝ ∆2:1+1 (S23) As δt and q are doubled every merging event, the many This relation is plot on Figure 1b, and fits quite well satellitesimals have their orbital radius roughly propor- the satellites systems of Saturn, Uranus, and Neptune. tional the mass to the power 1039 : For very low massive disks, the transition between the two power laws could take place at smaller ∆, but this 3910 δm r (S24) would not change much the shape of the curve in the ∝ figure, as the transition is relatively smooth. Note this power-law is independent of the numerical coefficient, and in particular of the efficiency of the dissi- pation inside the planet k2pQp, provided the latter is in- 7 Application to Solar System dependent of r and constant with time. All the satellites that form in the pyramidal regime with migration domi- bodies: Estimation of the tidal nated by the planetary tides will align on a straight line disks’ initial masses and τdisk of slope 39 in a log(mass) – log(orbital radius) diagram. The orbital radius of the outer most satellite, though, is How can we estimate the mass of the tidal disk, that may determined by K and by the total time of evolution of have given birth to the satellite population of the differ- the system. This sets the domain of application of our ent planets? A good starting point is to consider the model. satellites systems as seen today. Indeed, several works have shown that when a tidal disk spreads, the mass 6.3 Conclusion implanted in the satellite(s) is comparable to the mass of the full tidal disk, and that some mass is lost onto Two different mass-distance relations are found. When the planet (10,11,18,19). So, if today’s satellites have the distance of a satellite to the tidal disk is such that formed from a tidal disk, the mass of the latter, Mdisk no more inner Lindblad resonances with the satellite fall must have been comparable to the present mass of the within rR, then the torque from the disk vanishes. This satellites. Thus, we apply the following simple rule: we happens when the 2:1 inner Lindblad resonance falls at assume that Mdisk 15 Ms, where Ms is the total mass rR, and corresponds to ∆ = ∆2:1 0587. Beyond that, of all satellites that≃ we assume to be born from the tidal Eq.(S24) applies. For smaller ∆, one≈ should compare the disk. Ms must be estimated for each planet. From this two migration rates given by Eqs. (S3) and (S20). With and Eq.(S6), we can derive a value of D and of τdisk for a¯ = rrR and rR given by Eq.(S1), one finds: these planets. 4 −53 3 (dadt¯ )tides 3 − ρ k2 1 ∆ 22 = π 2456 5 p p Earth : Ms = 73 10 kg, D = 00183, τdisk = (d∆dt) 2 ρ Q D a¯4 • × disk p 125. ∼ × −5 4 10 Mars : Although most researchers favor a capture • where we have taken ρpρ 1 and k2pQp = 23 origin for both and Deimos, it has been sug- 10−4 (15). ≈ × gested (33) that Mars had a debris disk with mass 18 −6 Finally : Mdisk 10 kg, so that D 23 10 , and τ ≈77 109. ≈ × −6 disk 10 D ∆ < ∆2:1 ≈ × −7 ∀ 23 (dadt¯ )tides 10 D ∆ < 017 Jupiter : Ms = 2 10 kg, mass of all Galilean <  −9 ∀ • × −4 6 (d∆dt)disk  10 D ∆ < 003 satellites, D = 16 10 , τdisk = 17 10 .  2 ∀ × × 07 D ∆ < ∆d 21 ∀ Saturn : Ms = 5 10 kg mass of all satellites  • × So, the torque from the planetary tides is negligible in below (as most researches favor the theory most cases for ∆ < ∆2:1 (see estimates of D around Solar that Titan and formed in Saturn’s nebula),

10 Supplementary Material Crida & Charnoz 2012, Science, 338, 1196-

−5 8 D = 13 10 , and τdisk = 24 10 . in these simulations, only 10-15 hours after the impact, × 23 × Taking Ms = 14 10 kg (mass of all satellites up the disk is emptied or the proto- is formed (20). to and including Titan,× which fits well in our model) Assuming that the proto-Pluto was about Pluto’s mass, −4 5 gives D = 37 10 , and τdisk = 31 10 . this yield an orbital period at the disk’s edge about 12 × × 22 hours, which means that τdisk is about 1. Our computed Uranus : Ms = 10 kg, mass of all satellites up value ( 14) is in the right order of magnitude, and is • to (and including) , D = 17 10−4, and ∼ 6 × compatible with an emptying timescale comparable with τdisk = 14 10 . × the orbital period. 19 Our qualitative agreement with SPH simulations in Neptune : Ms = 5 10 kg, mass of all satellites • apart from (which× is thought to be captured), these extreme conditions shows that we may still catch −7 10 the important physics at play. D = 73 10 , and τdisk = 79 10 . × × 21 Pluto : Ms = 15 10 kg, Charon’s mass, D = 7.1.2 Rapidly evolving disks: first insight into • × 017, and τdisk = 14. thermodynamical effects

The horizontal lines on Figure 3 correspond to these In the case of massive disks and small τdisk, Eq. (S13) numbers. for qd doesn’t apply because Σ and D can’t be consid- ered as constant when qd D. In addition, for massive ≈ 7.1 Discussion about massive disks: disks, τdisk is not given by Eq.(S6) because the viscosity is probably not given by Eq.(S5) in the -forming cases of the Earth and Pluto disk: the viscous heating would lead to its melting or 7.1.1 Comparison with numerical simulation vaporization. A model of viscosity limited by the black-body cooling Here, our model is compared with previous work, in of the photosphere of the Moon-forming disk (34) makes which N-body simulations of the impact were made. No the dissipation constant, hence the product Σν and the thermodynamics is at play in this case, so that the vis- flow F . In this case, we still have M(t) = F t. Thus, cosity is by construction given by Eq.(S5), and τdisk by Eq.(S3) can still be integrated, with D = D0 q, D0 Eq.(S6). This may not be physical (see below), but al- being the initial disk to planet mass ratio. It gives:− lows an easy comparison. −3 3 For the Earth, with ρ = ρMoon = 3346 kg.m , we have 2 q √3 −12 2 r 185 107m, and T = 7h. So, we expect the disk q 1 = τ ∆ (S26) R R − 3 D 2 disk to disappear≈ × in about 365 days, in correct agreement 0 with N-body simulations suggesting about a month (19). with τdisk = D0MpF TR. The condition for the continu- In this frame, the Moon should have ended at ∆ = 7 ous regime turns into an implicit condition for the mass ∆2:1 = 0587, which is at r = 158rR = 29 10 m. of the satellite: At this distance, the orbital period around the× Earth is 32 52 13h50min. Thus, one could think that the Moon was be- 2 q 3 −32 q 1 < 3 τdisk = qc (S27) low the synchronous orbit. However, for the Earth-Moon − 3 D0 2 system to have the present total angular momentum, the of the Earth at this time needs to be The continuous regime can now be maintained with q 6h6min. So, in our model, the Moon forms well beyond slightly larger than qc (but note that here, qc is not any 3 the , and the Earth-Moon system has more 222 D ). This is because in this model, the flow evolved through tidal effects to the present state. (accretion) is constant, while Σ and D (which govern mi- Also, τdisk = 125 gives qd = 00140, which is just a gration) decrease with time. In fact, the left hand side 72 little bit more than the mass ratio of the Moon to the of Eq.(S27) is always smaller than 3D02 , so that the 13 −23 Earth (00122), and just a little bit less than the puta- continuous regime is never left if τdisk < 2 3 D0 . tive D = 00183; so, the Moon should have finished its This corresponds to a spreading time of one month for accretion at the end of the discrete regime. This allows the Earth, and 6 days for Pluto. So, with a constant flow for the formation of satellitesimals of mass a less than F (which is justified by thermodynamical constraints), a tenth of the Moon, which should be accreted by the taking into account the fact that D decreases with time, Moon. However, it is possible that one of these satel- we find that Charon grew entirely in the continuous litesimals is captured in a horseshoe orbit (it happened regime, while the Moon probably started the discrete to Epimetheus with Janus). This supports the idea that regime. But the thermal and dynamical evolution of the a small companion to the Moon was accreted later, form- Moon-forming disk is a very complex problem, far be- ing the lunar highlands (21). yond the scope of this paper. Concerning Pluto, we find that τdisk 14, which Of course, in such a high disk/planet mass ratio and on means a spreading time of the order of the≈ orbital period such short timescale, other approximations of our model at the disk’s edge. Whereas shocking at first sight, this are violated. More precisely: (a) the gravitational po- result is in agreement with SPH simulations of Charon tential may be far from Keplerian; (b) the computation formation, after an impact on the proto-Pluto. Indeed, of disk’s torque assumes that all resonances have time to

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respond to the satellite’s , which is not pos- 0.001 sible in one orbital period; (c) for such a value of τdisk, 0.0001 ∆c is not small, and the approximation ∆ 1 is not ≪ valid (see SM 9 for the large ∆ case). 1e-05 planet 1e-06 / M 7.2 About Mars satellite 1e-07

18 q = M Assuming a disk of mass 10 kg, we find that the spread- 1e-08 ing of this disk should have been dominated by the pyra- Amalthea 1e-09 midal regime. Indeed, τdisk is extremely high, so that These −14 9 Q(∆) qd 10 , which correspond to objects of 5 10 kg 1e-10 ≈ ∼ × 1 10 only. If Phobos and Deimos were to be formed this way, ∆ = (r-rR)/rR it means that a chain of satellites formed, of masses in- creasing with distance as described in SM 6. Phobos and Figure S2: The satellites of Jupiter in the q ∆ plane, Deimos would have been the two last and most massive 8 − with rR = 141 10 m (see SM 1 and 7.3). ones. × Subsequent orbital evolution would have made all The dashed line follows Eq.(S25). satellites inside the orbit of Phobos crash on Mars, be- cause they were below the synchronous orbit. This sce- age of the Jovian moons, or have slightly migrated inward nario has been recently proposed (33), in which a giant since they formed. Anyway, the fit with (∆) defined by impact creates the disk around Mars from which Pho- Eq.(S25) is not good. So, it seems that theQ regular satel- bos and Deimos form. Our model shows that in this lites of Jupiter didn’t form from a tidal disk. Note that case, the pyramidal regime should prevail, explaining the there is so far no explanation for the strong differences many past Martian moons and the two presently surviv- between these two groups of satellites. In particular, the ing ones. Note that only a few moons would have had mass-distance distribution and the composition of the masses of the order of Phobos, the other ones being much smaller ones is still puzzling. In addition, the Laplace smaller. As a consequence, we don’t expect too many resonance couples the orbital evolution of Io, Europa and large oval craters on the surface of Mars created by this Ganymede, which erases the initial conditions. Thus, it population. is difficult to reach a conclusion about the origin of this Nonetheless, the origin of Phobos and Deimos is still system. Note that models indicate that the conditions a matter of debate. We tried here to apply our model to inside Jupiter’s circum-planetary disk were significantly Mars, but it is difficult to reach a conclusion, and this different from that in Saturn’s one. The four Galilean issue is beyond the scope of this paper. In particular, it moons could have formed in such a circum-planetary should be noted that Deimos may not have formed in the disk (in terms of orbital architecture and composition), pyramidal regime because it can’t be transported beyond in contrast to that of Saturn, where only one Titan is Mars’ synchronous orbit. Then, only Phobos may be the expected to form (1-4). Because it formed earlier, and last member of a population of satellites formed from quickly opened a gap in the circum-solar nebula (16), the spreading of a tidal disk, and Deimos may have been Jupiter could be the only giant planet whose satellite captured. system formed entirely inside its circum-planetary disk. Also, it could well be that Jupiter simply never had a 7.3 About Jupiter massive .

We find that if Jupiter had been surrounded by a tidal 7.4 About Uranus disk of mass equivalent to its satellite system, it should have been dominated by the Pyramidal regime. This is Uranus has many small satellites close to its Roche ra- consistent with the fact that Jupiter has 8 regular satel- dius. This is in agreement with the beginning of the pyra- lites, and not one. However, the system of the regular midal regime. Their masses globally follow q ∆95, but satellites of Jupiter is divided in 2 groups of 4 bodies: some scatter is observed. Actually, it has∝ been shown the Galilean satellites which orbit beyond 4 108 m and that this system is chaotically unstable (35), but the dy- × have masses above 1022 kg; Metis, Adrastea, Amalthea namical instabilities occur on time scales much longer and These, which are inside 2 108 m and are lighter than that of the formation of this system (τ corre- × disk than 2 1018 kg. Among these two groups, the mass is sponds to 1600 years). Our assumption that the satel- × not an increasing function of the radius. This is not in lites don’t∼ perturb each other’s orbit is therefore valid for qualitative agreement with the Pyramidal regime. the duration of the formation of the satellites. The Ura- The system is represented in Figure S2, which is sim- nian satellites probably formed with q ∆95, and this ilar to Figure 1b in the main text for the other giant distribution has been altered afterwards∝ by chaotic mo- planets. Note that Metis and Adrastea do not appear in tion, leading to the observed scatter. In fact, the poten- this figure because they orbit inside the Roche radius as tially destructive chaos in the smallest calculated in SM1; they could be denser than the aver- within times of the order of 4 100 million years argues −

12 Supplementary Material Crida & Charnoz 2012, Science, 338, 1196- in favor of a younger formation for these bodies than for the planet itself (35). At least the small moons of Uranus probably didn’t form in a circum-planetary gaseous disc. It should be noted that the tiny rings shepherded by satellites around Uranus shouldn’t be considered as rem- nants of the original tidal disk. They are more likely remnants of the accretion process and of collisions be- tween satellites.

7.5 About Neptune and Triton Triton is the largest satellite of Neptune, but is an ir- regular satellite with a retrograde orbit. Therefore, is is thought to have been captured. In the case of a capture through an impact, this event 8 Numerical exploration of all likely perturbed the orbits of the pre-existing satellites outside of 5 Neptune radii, leading to their possible ejec- regimes using exact resonant tion or merging with Triton (36). So, the 6 regular satel- torque lites observed today could have escaped destruction, if they were already formed. It could also be that Nep- Since τdisk was found to be the only free parameter, we tune had more satellites in the past, before the capture investigate numerically how the mass of the first moon of Triton, which prolonged the q ∆ relation observed − increases as a function of ∆ and τdisk. For a given value in Fig. 1b. In this case, the disk mass estimated here is 8 of τdisk (ranging from 01 to 10 ) we solve simultaneously a lower bound. the following equations: One could also imagine that the formation of the regu- (a) the evolution of r, the distance of the satellite to lar satellites took place after Triton’s capture. After all, the planet’s due to the disk’s torque (11) : the impact that led to the capture of Triton could also +∞ have produced a tidal disk; this idea is appealing, but dr 2r12 considering Triton’s angular momentum, this tidal disk = Γm (S28) dt M(GM )12 p m=2 would have been retrograde, which is not in agreement with the present prograde orbits. However, if Triton im- (b) the evolution of the mass of the satellite pacted Neptune, other bodies of similar size may well have done so at a later time, giving eventually birth to dM = F (S29) the regular satellites through a tidal disk. dt A more recent model considers the capture of Triton from a binary object (37). This event may have been pri- Γm is the torque applied by the disk onto the satellite mordial, and may have not destroyed the putative satel- by the m 1 : m resonance inside the disk (38). We used − lite system. In that case, the formation of the 6 regular an exact torque expression (39), valid either for low and satellites of Neptune could be primordial as well and not high m. At each time step, as r increased, we checked much perturbed, or easily posterior to Triton’s capture. whether the satellite was in the continuous, discrete, or In conclusion, the presence of Triton and its necessary pyramidal regime. The results are plotted in Figure 3. capture doesn’t contradict the idea that our model ap- The discontinuous aspect of the frontiers between the plies to Neptune’s regular satellites. different regimes, for high ∆ (or low τdisk), is due to the discontinuous leaving of the resonances from the disk as the big satellite goes away: the 4:3 resonance, then the 3:2 and finally, the last one, the 2:1. When the 2:1 leaves the disk, then the satellite stops. This corresponds to ∆ = ∆2:1 = 0587.

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9 Large ∆ So, the condition Eq.(S34) reads:

When the first moon is very far from the Roche radius, 26 ∆4 then the hypothesis ∆ 1 does not hold anymore. To τ < 1 36∆ + O(∆2) disk 3f ′ 4 explore this regime, we look≪ for a more precise expression − −6 2 of ∆c. When ∆ is not small, the expression of the torque ∆ (1 + 6∆ + O(∆ )) ′ × exerted by the disk on the outer satellite should be taken 3f τdisk − < ∆ 2 1 + 24∆ + O(∆2) from (39), their Eq. (47b). It gives: 24 4 2 d∆ 2 ′ > ∆ 1 12∆ + O(∆ ) (S35) ′ 92 −1 −3 √ − = f q D a¯ TR ∆ (S30) 3f τdisk dt 3 ′ where f 25. Not only the numerical coefficient is To zeroth order, one recognizes the condition ∆ < ≈ −12 slightly different, but more importantly, a dependence in ∆c τ where only the numerical coefficient has ∝ disk a¯ = rrR = ∆ + 1 appears, which makes the integration slightly changed with respect to the previous case less easy. Using q = F tMp, da¯ = d∆, and separating Eq.(S10). The right hand member of Eq.(S35) is an the variables, one finds: approximation of (4 h(¯a))−12 ; while the zeroth order ′ differs from this exact value by more than 10% for − 2f F t (¯a 1)3a¯ 92 da¯ = D dt (S31) ∆ > 01, the first order is accurate to less than 10% − 3TR Mp up to ∆ < 025. Eq.(S35) is a more accurate definition The right hand term integrates into: for the boundary of the continuous regime.

f ′τ disk q2 3 9.1 Condition for a never-ending contin- With the condition that when t = 0, q = 0 anda ¯ = 1, uous regime at small τdisk the left hand term of Eq.(S31) integrates into: Interestingly, expanding Eq.(S35) to first order gives: − − 6 − 2 − 32 2¯a 12+2¯a 32 a¯ 52+ a¯ 72+ f(¯a) (S32) − − 5 7 35 ≡ 2 4 12∆ ∆ + ′ > 0 (S36) So, now, the mass-distance relation reads: − √3f τdisk

3 −12 ′ q = τ f(¯a) (S33) which is always true if 1 192√3f τdisk < 0. So, the f ′ disk − continuous regime never ends if Input into the condition of validity of the continuous regime, it gives: 1922 τdisk < 50 (S37) 3f ′ ≈ ∆ < 2rH rR q 13 a¯ 1 < 2¯a The largest possible ∆c from Eq.(S36) is 124 042. − 3 At this point, the first order differs from the exact≈ value 6 6 1 2 (4 h(¯a))−12 by 30%. Thus, the value given above 1 < ′ f(¯a) ∼ − a¯ 3f τdisk should be considered with care. However, it is true that − 26 1 6 26 h(¯a) admits a minimum, and that for low enough values τdisk < f(¯a) 1 h(¯a) (S34) of τ , only one satellite forms. The numerical integra- 3f ′ − a¯ ≡ 3f ′ disk tion with accurate torque (SM 8) confirms that τdisk 50 ≈ Instead of an expression for ∆c, we get an implicit is the limit (see Figure 3). equation fora ¯, that can be solved numerically. But this turns also into an explicit condition on τdisk. It is possible to expand f and h to first order in ∆. Actually, f is the integral of

− 9 ∆3(1 + ∆) 92d∆ = ∆3 ∆4 + O(∆5) d∆ − 2 ∆4 So, f(∆) = 1 36∆ + O(∆2) . 4 − −6 1 − − 1 = (∆(1 + ∆)) 6 = ∆ 6(1 + 6∆ + O(∆2)) − a¯

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10 On the influence of the moon on the disk in the continuous and discrete regimes

One can wonder if the torque exerted by the Moon on the disk (Γ, given by Eq. (S2), SM 2) will not at some point prevent the disk from spreading, and stop the growth of the Moon. To check this, this torque should be compared to the viscous torque exerted by the tidal disk on its outer edge, given by :

2 Γν = 3πνΣrRΩR (S38)

10.1 Constant D, and small moons 10.2 Varying D, and large moons Below, we assume that D and Σ are constant, that is we In this section, we consider the case where the assump- consider the case where the mass of the growing moon tion q D doesn’t hold anymore, so that D can’t be is small compared to that of the disk: q D. Using considered≪ as constant. Assuming that at least F is con- ≪ the relation between the mass and the distance in the stant (see SM 7.1.2), the mass-distance relation is given continuous and discrete regime Eq.(S9), one finds by Eq.(S26), which input into Eq.(S2) gives

−32 −34 12 4 2 −34 −32 Γ = 2 3 q ΣrRΩRτdisk − − 2q − Γ = 2 323 34q12 1 Σr4 Ω2 τ 34 − 3D R R disk Then, using Eq.(S5) and Eq.(S38), one finds: 0 The viscosity is not any more given by Eq.(S5), but 32 74 ∗5 2 2 Γν 2 3 π 26 r G Σ 2 h considering that τdisk = Tν TR, one gets ν = r TRτdisk, = − R Γ 12 2 4 34 q rR ΩR τdisk and thus : 52 74 ∗5 2 2 3 13 r D 32 h 32 74 2q = −3 4 2 3 1 12 Γν 3D0 π q τdisk − = 14 (S40) Γ 967 Γ q12 τ ν = (S39) disk Γ 12 14 q τdisk The numerator is always larger than 37. So, the vis- cous torque always wins, and the continuous or discrete The ratio between the two torques decreases as q in- regimes are not interrupted as long as: creases. Using the expression of qc Eq.(S11), the ratio between the two torques at the end of the continuous 192 12 regime is 69√τdisk, generally much larger than 1. Us- q < (S41) ∼ τdisk ing the expression of qd Eq.(S13), the viscous torque is an order of magnitude larger than the gravitational torque For the Earth’s Moon, this makes τ < 13 106, or a during all the discrete regime, provided τ > 20. disk disk spreading time shorter than 103 years for the× spreading Using Eq.(S6), the ratio of the two torques is ∼ never to be blocked by the Moon’s torque on the disk. 21 Dq 400 MdiskM, so the viscous torque in the disk always≈ dominates the satellite’s torque if the satel- lite isn’t more massive than the disk. In conclusion, the influence of the moon(s) on the disk can be neglected in the continuous and discrete regimes as long as q < D.

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