11 The Gravity Field and Gravitational Response to Rotation: Moments of Inertia

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1 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 11.1 The Gravity Field External to a planet (in a region where there is no or negligible mass), the gravitational potential V satisfies Laplace s equation 2 V = 0 (11.1) It makes sense to use a planet-centered spherical coordinate system, and in that case the general solution to Laplace s equation can be written in the form: V = 1 a =0 m=0 a r +1 (C m cos mϕ + S m sin mϕ)p m (cosθ) (11.2) Figure 11.1 Angle ϕ is longitude. The reference radius a is conventionally taken to be the equatorial radius. The solution assumes no external sources of mass (i.e., all terms decay as r goes to infinity). The P s are associated Legendre functions, and along with the sines and cosines of longitude they define spherical harmonics usually written Y m. The C s and S s are called spherical harmonic coefficients. [Warning: There are various ways of writing this and defining the coefficients in this expansion. Some people use a minus sign in front of the whole thing. Some people use complex coefficients, with exp(imφ) instead of sines and cosines. You can always check your sign convention by making sure that the gravitational acceleration is pointing in the right direction, but to get the rest right you have to find out the author s normalization convention, etc.] Obviously, C 00 is nothing other than GM. (By the way, GM can be measured to twelve figure accuracy for Earth, but that doesn t mean we know M that well! G is the least wellknown fundamental constant and very hard to measure). Precise tracking of an orbiting

2 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 114 spacecraft can give you these coefficients. As in any mathematical representation, you always truncate the representation and thus fail to characterize very high harmonics. This is a potential problem even in quite low planetary orbits, where the geometric attenuation (the high powers of a/r) is not large. In the special case where the planet is a rotating hydrostatic fluid, symmetry arguments alone dictate that the potential is axisymmetric (only m=0 terms allowed) provided we choose our polar axis to be coincident with the rotation axis. Moreover, there is no physical distinction between northern and southern hemispheres, so odd values are excluded. (This assumes we ve chosen the origin of coordinates to be the center of mass). We can then write the potential in the form: V = GM r [1 J a 2 2 r P 2 (cosθ)] (11.3) =1 In this simple case, the P s are now the simple Legendre polynomials, and the J s are called gravitational moments. In rapidly rotating planets, J 2 is generally far larger than any of the other harmonics (except of course C 00 ). The fundamental definition of the gravitational potential is of course V( r ) = G ρ( r )d 3 r r r (11.4) all space obtained by adding up the contributions of all masses and appealing to the superposition principle (the linearity of Newtonian gravity). Outside the planet, we can appeal to the fundamental theorem (also known in mathematics as the generating function): 1 r r r r +1 =0 = P (cosγ ) (11.5) where γ is the angle between vectors r and r, and r is outside the planet (r is inside the planet). Moreover, ( m)! P (cosγ ) = P m (θ)p m (θ ')cos[m(φ φ ')] (11.6) (l + m)! m=0 But we only need to keep track of m=0, for evaluating J 2 (because these are the only terms that contribute to the integral): P (cosγ ) = P (cosθ)p (cos θ ) + m 0 terms (11.7) so it follows immediately that J 2 = 1 Ma r 2 P 2 2 (cos θ )ρ( r )d 3 r (11.8) One can see already why the J s are called gravitational moments since they are integrals of the internal density distribution, weighted by progressively higher powers of the radius.

3 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 115 Recalling that P 2 (cosθ) = (3cos 2 θ- 1)/2, one can see immediately that Ma 2 J 2 = [ 3 z (x2 + y 2 + z 2 )]ρ(r)d 3 r (11.9) where x,y,z is a Cartesian coordinate system in which z is along the rotation axis. Now if we define the axial moment of inertia as C and the other two principle moments of inertia as A and B then we have: C (x 2 + y 2 )ρ(r)d 3 r A (z 2 + y 2 )ρ(r)d 3 r; B (z 2 + x 2 )ρ(r)d 3 r and it is easy to see that (11.10) Ma 2 J 2 = C 1 [A + B] (11.11) 2 In the special (and highly relevant case) where A=B (i.e., the planet is a body of rotation rather than triaxial), we have Ma 2 J 2 = C A (11.12) So this moment is related to the difference between axial and equatorial moments of inertia. In a similar manner you can show that C 22 =(B-A)/4Ma 2. (With appropriate choice of zero longitude, S 22 will be zero.) But you cannot get the actual values of A, B and C from the gravity field; there is insufficient information. You can only get their differences (e.g., C-A) The Response of a Planet to its Own Rotation We saw that J 2 can be related to the difference in the moments of inertia about equatorial and polar axes. An independent piece of information is often available from the precession rate of a planet. Since the torque acting on a planet depends on C-A and the rate of change of angular momentum of a planet is proportional to C, precession rate gives you (C-A)/C. Together with eqn 11.12, one can then solve for the individual moments of inertia C and A. This is how we know Earth, Mars and lunar moments of inertia. (Determination of the Mars moment of inertia using Mars Pathfinder tracking was one of the major accomplishments of that mission). A somewhat more complicated version of this approach can work for Mercury. But we ought to be able to figure out something more from J 2 alone because it s value depends on how the planet responds to its own rotation, and this response depends on its density distribution. It should be obvious, for example, (just by looking at the definition as an integral over the interior weighted by radius squared) that J 2 will be small for a body that is centrally concentrated.

4 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 116 This body has a dense core and low density envelope so it has a small J 2 This body is uniform density but has the same mean density, mass and rotation as the one on the left. It therefore has a similar oblateness but larger J 2. (See also problem 11.1) Figure 11.2 Although this is intuitively reasonable, the appropriate theory is quite nasty, and rather little insight emerges from wallowing in the nastiness. (The full theory involves integrodifferential equations that must be solved on a computer and even then converge slowly). I will only give a feeling for the theory. This will only work if J 2 is dominated by hydrostatic effects. The theory explicitly assumes hydrostaticity The Constant Density Limit (Maclaurin Spheroid) Consider, first, the constant density body. The external potential is obviously completely determined by the shape of the free surface. Approximate the free surface of this body by the lowest order non-spherical shape permitted, i.e., r s = r 0 (1+ ε.p 2 ) (11.13) where r 0 is some mean radius and ε is a dimensionless constant. Inserting in the fundamental equation for the gravitational potential and using the generating function, one immediately finds that the only part that depends on P 2 is of the form 1 V 2 = GP 2 r P 3 2 (cos θ )d(cos θ ) x 2.2π x 2.ρ 0.dx (11.14) 1 r 0 (1+εP 2 ) 0 (V 2 is the amplitude of the coefficient of the P 2 term in the external potential. Here and below, the explicit angular dependence is sometimes omitted since it should be obvious and it makes the equations less cluttered). Since the P s are orthogonal to each other (and remember that P 0 is unity), the only part of the integral over x that contributes is the part proportional to P 2. Therefore:

5 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia V 2 = GP 2 2 P r 3 2 (cos θ )d(cos θ ).ε.2πρ 0 a 5 1 = 3G 5r 3 Ma2.ε [where we ve used the fact that the normalization integral for Legendre functions is 2/(2 +1).] We ve set r o =a (equatorial radius), which is correct to this order of approximation. We must compare with the expression that defines J 2 in terms of the external expansion of the field: V 2 P 2 = GMa2 r 3 J 2 P 2 (11.15) J 2 = 3 5 ε (11.16) Consider, now, the gravitational potential evaluated at the actual surface of the body. We must of course include the effect of rotation (the centrifugal effect). Recall that the acceleration is ω 2 s where s is the perpendicular distance from the rotation axis. The potential that yields this acceleration is (by integration) obviously ω 2 s 2 /2 = ω 2 r 2 sin 2 θ/2 = ω 2 r 2 (1-P 2 )/3. Now the total potential must be constant at the surface. This is where the assumption of hydrostaticity enters. To lowest non-vanishing order in P 2 this implies that: GM r 0 (1 + εp 2 ) GM a J 2 P ω 2 a 2 [1 P 2 ] (11.17) must have no dependence on P 2. If it had a dependence on P 2 then it wouldn t be a constant on that surface! Notice that we can ignore the differences between r and a, etc., in terms that are already small (J 2 and ε are small parameters). Also, ε<<1 means that 1/(1+εP 2 ) = 1-εP 2 to an excellent approximation. In other words, { GM a [ ε J 2 ] 1 3 ω 2 a 2 }P 2 0 ε J 2 q 3 = 0 (11.18) where q is a dimensionless measure of planetary rotation: q ω 2 a 3 GM But we already have J 2 =-3ε/5. Substituting, we get J 2 = q 5q ; ε=- 2 6 (11.19) (11.20) This is what we mean by the response of the planet to rotation... the gravitational moment is related to a dimensionless measure of the strength of rotation. In general, we expect that

6 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 118 J 2 = Λ 2,n q n+ (11.21) n=0 with the n=0 term dominating. For example, Λ 2,0 = 0.5 for a Maclaurin spheroid (the technical name for the uniform density case we studied here) and Λ 4,0 = Of course, all that we ve done here is find these coefficients in the special case of a uniform density body (a case where we already know the moment of inertia). But it should be plausible to you, and turns out to be actually true that these coefficients are diagnostic of the density structure. For example, 5 Λ 2,0 = π = (11.22) for the case of the model we studied for Jupiter where P=Kρ 2. This is very different from The Radau-Darwin Approximation There is an approach, conceptually the same as described above, that works for bodies that are close to constant density. This is not a bad approximation for terrestrial planets and icy satellites but is rather poor for giant planets. In this approach, you write the total gravitational potential in the form ρ U = U s + (dp / dρ')(dρ' / ρ') (11.23) ρ s where this form comes from using hydrostatic equilibrium: P =ρ U. The subscript s refers to some (arbitrary) reference surface. This form of the potential must also be identical to that derived from the fundamental formula for gravity, i.e. U W + G ρ(r')d 3 r ' (11.24) r r' where W is the external potential (tides or rotation). One then assumes that the equipotential surfaces inside the planet can be written in the form of r=r 0 (1+ε 2 (r)p ) and derives a integro-differential equation for the function ε 2 (r). To make a long story short (look at Hubbard s book) it turns out that in place of the simple previous result (Λ 2,0 =0.5 for C/Ma 2 =0.4), we obtain a more general result C Ma = {1 2 5 [ 5 3Λ 2, ]1/2 } (11.25) (This is called the Radau-Darwin approximation. It is not merely an approximate theory; it is also an approximation to that theory.) Recall that at this level of approximation, Λ 2,0 =J 2 /q. Here are some values that this formula predicts:

7 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 119 Table 11.1 Λ 2,0 C/Ma Note: The formula fails badly for small values of either parameter, as you might expect, given the approximations. It is, however, possible to generalize the approach to a layered body, and this works moderately well if each individual layer is of nearly constant density What are the Data? Bodies fall into two classes: those for which hydrostatic effects dominate (i.e., rapid rotators ) and those for which the rotational bulge is no bigger than the other effects on topography and gravity. The rapid rotators further subdivide into those for which Radau- Darwin is roughly valid (terrestrial bodies and icy satellites) and the gaseous bodies (in which the density variations are too large, and a more complex and detailed theory has been devised -one example is the solution given above for P ρ 2.) In addition, for satellites we must correct for the tidal effect on J 2. For synchronously rotating bodies (all large satellites), the tidal distortion is a factor of three larger than the rotational distortion. From this one can readily show that J 2,r the rotational part of J 2 is only 2/5 of the total. This correction is made below for those cases where the body is a synchronously rotating satellite, excluding the Moon. (The Moon and Mercury are so far from hydrostatic equilibrium that it makes on difference whether the correction is made).

8 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 120 Table 11.2 Body Measured J 2 or Inferred Measured q J J 2,r /q 2,r C/Ma 2 Mercury (8±6)x10-5 1x10-6 ~1 * Venus (6±3)x x10-8 ~10 2 * Earth x x Moon 2.024x X10-6 ~30 * Mars 1.959x x Jupiter x ~0.25 Saturn 1.646x ~0.23 Uranus 3.352x ~0.20 Neptune 3.538x ~0.22 Io x x Europa x x Ganymede x x Callisto 1, x x Titan 1, x x Enceladus 1,2 2.09x x *Non-hydrostatic (at this low level of rotation) so method does not work. (other methods exist for Mercury, Moon). Hydrostatic but Radau-Darwin doesn t work. More complex theory can provide a highly precise relationship between density structure and measured J 2 but the results of this theory are not expressible in the form of a solution for C/Ma 2. As a consequence, I ve inserted a ~ even though the more general theory is highly accurate. (See next chapter for more results). Note that the result for Jupiter does agree quite well with the exact prediction of Λ 2,0 =0.173 for P ρ 2. J 4 and even J 6 are used for these planets to improve the estimates of internal structure. 1 The inversion of the Galileo spacecraft data assumes that the rotational and tidal responses are related hydrostatically. The values quoted above for J 2 are not what you will find in the published papers (see references in Showman and Malhotra, Science, Oct.1, 1999). The value in the table is for J 2,r, the rotational effect on J 2 only. 2 In the case of Callisto and perhaps Titan, there must now be some question as to whether this interpretation is correct, since significant non-hydrostatic gravity may exist and does exist on Titan. As of January 2008, the Titan gravity data have resisted simple interpretation because the non-hydrostatic terms are important. See Gao, P. and Stevenson, D.J. (2013) Nonhydrostatic effects and the determination of Icy satellites Moment of Inertia. Icarus 226,

9 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia Simple Models for the Moment of Inertia We saw how one can get an estimate of the axial moment of inertia from just the gravity field; specifically from the response of the planet to its own rotation. In a few special cases where precession has been measured (Earth, Moon, Mars) we can do even better. In the case of very slowly rotating bodies (Mercury, Venus) we can do very little (we don t know their moments of inertia). It may be possible to get Mercury s I/MR 2 from it s obliquity and libration (discussed later in the text). If we approximate the moment of inertia by assuming spherical symmetry, then of course I=(A+B+C)/3, or I = 1 3 { [(x2 + y 2 ) + (x 2 + z 2 ) + (y 2 + z 2 )]ρ( r )d 3 r} R = 2 3 4πr 4 ρ(r)dr 0 (11.26) and we can proceed to evaluate for interior models and compare with observations. There are two particularly interesting cases to consider: The Simple Two-Layer Model Assume ρ(r) = Aρ 0 for 0 < r < xr and ρ(r) = ρ 0 for xr < r < R. Then I MR = 2 + (1 x 5 )] 2 5.[Ax5 [Ax 3 + (1 x 3 )] (11.27) Of course, this should be combined with our knowledge of the observed mean density ρ av = ρ o [Ax 3 +(1-x 3 )] (11.28) Now A~2 roughly corresponds to iron core and silicate mantle and A~3 roughly corresponds to rock core and ice mantle. (Further subdividing rock into silicates and iron will not change things much.) This model cannot be applied for Earth because the density of the silicates changes too much across the mantle. The model is plotted below in parametric form (mean density vs. moment of inertia factor) with the larger curve being A=3 (rock/ice) and the smaller curve being A=2 (iron/rock).

10 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 122 Figure 11.3 Ganymede is about the lowest I/MR 2 conceivable for any solid object in the solar system. Europa fits with a large rock core model. Io and Mars fit with modest cores, Moon with a small core at most. Callisto does not fit the model at all. It is shown in brackets at the correct mean density and I/MR 2 but well off the A=3 curve. The likely interpretation is that Callisto is only partially differentiated. Problem 11.2 asks you to consider whether it might instead be explained by a non-hydrostatic structure. Mercury is shown where it might be based on other information. Titan and Enceladus are explainable by low density cores (A~2.5). It is also useful to see (below) how these bodies fit on the I/MR 2 vs. core size model. Provided they fit the previous curve you can use this to read off the core size. All of this is approximate of course. Figure 11.4

11 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 123 Notice that there are two values of x for a given I/MR 2, so the correct choice of x is of course dictated by the need to get the right mean density. There is not enough information in the gravity data alone to decide whether the large icy bodies have separated iron from silicates to form an iron core The P=Kρ 2 Model (for Giant Planets) For this we have ρ(r) = ρ c sin(kr)/kr from which one gets: I MR 2 = x 3 sin(π x)dx xsin(π x)dx = 2 3 [1 6 ] 0.26 (11.29) 2 π Notice that this result is not changed much if you add heavy elements uniformly since (as discussed previously) that will not change the form of the density profile (even though the radius of the body may change substantially!) However, if you add a dense core then I/MR 2 ~0.26(1-M core /M total ). This comes from including the cos(kr)/kr term in the density profile (see earlier chapter). Since I/MR 2 is somewhat smaller for Saturn than for Jupiter, it is likely that Saturn has a core. Table 11.3 Observational Moments of Inertia Body Observed I/MR 2 Comments Mercury Large iron core, thin mantle Earth Large core, but also action of pressure on constituents Moon Does not demand an iron core (but there probably is one; maybe 500km in radius) Mars Requires iron core, slightly less mass fraction than for Earth Jupiter 0.25 Need for dense core is marginal Io Substantial Fe core Europa H 2 O layer on rock Ganymede Ice mantle on rock Callisto 2, ±0.005 Partially differentiated Saturn 0.24 Smaller value than Jupiter suggests presence of dense core Titan 2, Either partially differentiated or fully differentiated but a low density core (`2.5 g/cc)

12 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 124 Enceladus 2, Fully differentiated with a low density core Uranus 4 ~0.20 Interior much denser, but interpretation non-unique Neptune 4 ~0.22 Denser body but less densely concentrated than Uranus? 1 Determined by precession combined with gravity. Mars value now better known (to 0.2%) using Pathfinder tracking (Folkner et al Science ,1997). 2 Uses both the rotational and permanent tidal deformation effect on gravity field, so one can (in principle) test for hydrostatic equilibrium In practice, some ambiguities arise in flybys (i.e. C 22 and J 2 are not always independently determined). 3 Recall previous comments about the possibility that Callisto is non-hydrostatic so that the quoted value may be wrong. 4 In these bodies, and to some extent Saturn, the constraints provided by J 2 and J 4 are quite strong, but they cannot be inverted into a sharply defined value for the moment of inertia. The reason is that the response to the rotation is non-linear (the rotational parameter q is not sufficiently small). This is a technical difficulty, not a fundamental shortcoming! 5 For Mercury, a combination of gravity & spin data are used. see Margot et al (2013), J. Gepohys. Rev, 117 DOI: /2012JE Other Ways of Getting Moment of Inertia We have focused here on the use of gravity, sometimes in combination with the precession constant (for Earth, Moon and Mars). In some cases (Mercury, Jupiter, Saturn) we can also use the theory for the obliquity evolution developed by Peale and others to get information on the moment of inertia. This is not a measurement of I/MR 2 in the same sense as what we have discussed above, because it makes a heavier reliance on the correctness of theoretical ideas about spin-orbit coupling, etc. But it has in principal the promise of accurate values. It may also be possible (eventually) to directly measure the precession rates of giant planets. Juno should do this for Jupiter. Remarkably, Juno may also measure the angular momentum of Jupiter using the dragging of inertial frames (the Lense-thirring effect in Einstein s theory of gravity). Ch. 11 Problems 11.1) On p120, the claim is made that the two bodies depicted in the cartoon have similar oblateness. How similar? How different are the values of J 2? Solution: If the mass is overwhelmingly in a core of radius a c then (modifying the derivation of p114) we get V 2 Ma c 2.ε c where ε c is the distortion of the core only. Then from the definition of J 2 we have J 2 = - 3(a c /a) 2 ε c /5. From consideration of the equipotential at the core radius (ignoring the negligible mass external to it), we get - ε c -J 2.(a/a c ) 2 (a c /a) 3 q/3 =0, where q is defined for the entire planet. This implies that ε c = -5(a c /a) 3 q/6. (This result is independent of a, as it should be). From

13 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 125 consideration of the constancy of the potential (including rotational effect) on the outer radius, we get -ε -J 2 q/3=0. Putting this together, we get ε = - [ 1 / / 2 (a c /a) 5 ]q. This varies from q/3 to -5q/6 as the core radius varies from zero to the full radius of the body. In this sense, the oblateness is similar (though of course it does nonetheless vary by a factor of two or so). By contrast the value of J 2 can vary enormously. For example, if we choose a c /a =0.7 then the magnitude of ε is reduced by a factor of two (relative to the value at a c /a =1) but J 2 is reduced by almost a factor of six! 11.2) The gravity field of Callisto is assumed to be hydrostatic in the analysis of Galileo data. But suppose instead that we interpret the measured* J 2 to consist of two parts, i.e. J 2 =J 2h +J 2nh, respectively the hydrostatic and non-hydrostatic parts. [*Strictly speaking, they don t measure J 2 (they actually measure a combination of J 2 and C 22 ) but I want you to assume that the value given in the notes is in fact a correct measurement.] (a) Assuming Callisto has the same structure (i.e., same C/Ma 2 ) as Ganymede (and assuming Ganymede is fully hydrostatic), estimate J 2nh. (b) What size mass anomaly on the icy surface of Callisto would be required to explain J 2nh? What size structure does this imply for the anomaly? (There s no unique answer to this, so quote plausible things, e.g. hockey puck -like structures or holes~1000km in radius, that would be needed.) Don t do anything that requires fooling around with Legendre functions; I m only looking for rough answers. But take care to think through the consequences of moving the center of mass. (That is, you can t just put a mesa on one side of the body and then think of that as the only effect on evaluating the gravity field. The center of mass has moved!) (c) What size anomaly would be required on the rocky core, if we suppose (instead of (b) above) that this is the sole source of J 2nh? (The statement of this problem implicitly assumes that you think of Callisto as having a fully differentiated structure like that we think is appropriate for Ganymede). (d) Which is more likely (c or d)? Commentary: The data for Ganymede are better than for Callisto and it has been possible to determine the non-hydrostatic part. It is much smaller than that required to justify the alternative explanation for Callisto that is examined in this problem. This supports (but does not prove) the simpler, incompletely differentiated interpretation for Callisto. However, the latest data for Titan suggest a substantial non-hydrostatic effect, and Titan has a similar q value to Callisto. 11.3) As the Moon evolved tidally outward from Earth, its spin rate decreased, always maintaining synchroneity (i.e., spin period=orbital period). Assuming that the observed J 2 for the moon is a frozen bulge that arose in the early history of lunar orbital history, at what orbital distance was this bulge acquired?

14 11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia 126 Solution: From p117, J 2 /q = 26.6 for the current Moon. The equilibrium expected value for this almost constant density body is J 2 /q = 1 / 2. Accordingly, we must go to an epoch when the square of the angular velocity was larger by (26.6)/( 1 / 2 )= By Kepler s laws, this happened when the orbital distance was smaller by (53.2) 2/3 ~ 14. Since the current orbit is at 60 Earth radii, this would be have been at ~4.3 Earth radii! (This is unlikely to be the explanation because the orbital evolution at that location would have been very much faster than the cooling time of a rotational bulge). 11.4) Using Mathematica (or any numerical method you prefer), explore the parameter space for simple three layer models of Uranus in which the rocky core has density 8g/cm 3, the intermediate ice layer has density 2.5 g/cm 3 and the outer gas envelope has density that declines linearly with radius from ρ gas,0 at the bottom of the envelope to zero at the outer radius. (Obviously, this is a crude way of representing the outer region.) Your models must satisfy a mean density of 1.32 g/cm 3 and moment of inertia I/MR 2 =0.20. Since the actual radius and mass don t matter when you fit these values, you have three adjustable parameters: the mass fractions of rock and ice and the value of ρ gas,0 (which must be positive but less than 2.5). This leaves you one degree of freedom. Your results could be a plot of core mass fraction, ice mass fraction and gas mass fraction as a function of ρ gas,0. (Part of the purpose of this is to find the range of parameters for which a physical solution exists.) 11.5) Enceladus was once closer to Saturn. Suppose it froze up at 90% of its current orbital radius (that is, its shape at that time became rigid). Conseqeuntly the gravity should then be interpreted using the value of q appropriate to that epoch. How would this change the moment if inertia estimate for Enceladus? What model might you them construct (assume a rocky core and an ice shell of density 1 g/cc and derive the rocky core density. The mean density must be 1.62 g/cc). Commentary: This possibility was discounted in the recent paper (Iess et al, Science, 344, 78, (2014)) because the observed shape is not consistent with this model.