Haumea's Shape and Composition
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HAUMEA’S SHAPE, COMPOSITION, AND INTERNAL STRUCTURE
Dunham, E. T, Desch, S. J., and Probst, L. School of Earth and Space Exploration, Arizona State University San Francisco University High School
ABSTRACTWe have calculated the figure of equilibrium of a rapidly rotating, differentiated body to determine the shape,structure, and composition of the dwarf planet Haumea. Previous studies of Haumea’s light curve have suggestedHaumea is a uniform triaxial ellipsoid consistent with a Jacobi ellipsoid with axes ≈ × ×
513 km, and bulkdensity ≈ − . In contrast, observations of a recent stellar occultation by Haumea indicate its axes are ≈ × ×
523 km and its bulk density ≈ − ; these results suggest that Haumea cannot be a fluid inhydrostatic equilibrium and must be partially supported by interparticle forces. We have written a code to reconcilethese contradictory results and to determine if Haumea is in fact a fluid in hydrostatic equilibrium. The code calculatesthe equilibrium shape, density, and ice crust thickness of a differentiated Haumea after imposing (semi-) axes lengths a and b . We find Haumea is consistent with a differentiated triaxial ellipsoid fluid in hydrostatic equilibrium withaxes of best fit a = 1050 km, b = 840 km, and c = 537 km. This solution for Haumea has ρ avg = 2018 kg m − , ρ core = 2680 kg m − , and core axes a c = 883 km, b c = 723 km, and c c = 470 km, which equates to an ice mantlecomprising ∼
17% of Haumea’s volume and ranging from 67 to 167 km in thickness. The thick ice crust we infer allowsfor Haumea’s collisional family to represent only a small fraction of Haumea’s pre-collisional ice crust. For a widerange of parameters, the core density we calculate for Haumea suggests that today the core is composed of hydratedsilicates and likely underwent serpentinization in the past.
Keywords:
Kuiper belt objects: individual (Haumea); planets and satellites: interiors; planets andsatellites: composition; planets and satellites: formation a r X i v : . [ a s t r o - ph . E P ] A p r Dunham, E. T. et al. INTRODUCTIONThe Kuiper Belt Object (KBO) and dwarf planetHaumea is one of the most intriguing and puzzling ob-jects in the outer Solar System. Haumea orbits be-yond Pluto, with a semi-major axis of 43.2 AU, andis currently near its aphelion distance ≈ . V = 17 .
3, due to itslarge size and icy surface. Haumea’s mean radius isestimated to be ≈
720 km (Lockwood et al. 2014) to ≈
795 km (Ortiz et al. 2017), and its reflectance spec-tra indicate that Haumea’s surface is uniformly coveredby close to 100% water ice (Trujillo et al. 2007; Pinilla-Alonso et al. 2009). Haumea is the third-brightest KBO,after the dwarf planets Pluto and Makemake (Brownet al. 2006). Haumea has two small satellites, Hi’iakaand Namaka, which enable a determination of its mass, M H = 4 . × kg (Ragozzine & Brown 2009); itis the third or fourth most massive known KBO (afterPluto, Eris, and possibly Makemake). Despite its largesize, it is a rapid rotator; from its light curve Haumea’srotation rate is found to be 3 . ± . >
100 km) object in the SolarSystem (Rabinowitz et al. 2006). Haumea is also asso-ciated with a collisional family (Brown et al. 2007) andis known to have a ring (Ortiz et al. 2017). Based onits rapid rotation and its collisional family, Haumea isinferred to have suffered a large collision (Brown et al.2007), > m ≈ .
28 in 2005 (Rabinowitz et al. 2006), ∆ m = 0 . m = 0 .
32 in 2009(Lockwood et al. 2014). Since Haumea’s surface is spec-trally uniform, such an extreme change in brightnesscan only be attributed to a difference in the area pre-sented to the observer. Haumea has been modeled as atriaxial ellipsoid with axes a > b > c , the c -axis beingaligned with the rotation axis. In that case, the changein brightness from peak to trough would be given by∆ m = 2 . (cid:20) log (cid:16) ab (cid:17) − log (cid:18) r r (cid:19)(cid:21) , (1)where r = (cid:0) a cos φ + c sin φ (cid:1) / (2)and r = (cid:0) b cos φ + c sin φ (cid:1) / , (3) φ being the angle between the rotation axis and the lineof sight (Binzel et al. 1989). If φ = 0 ◦ , then ∆ m = 0,because the same a × b ellipse would be presented tothe observer. Instead, ∆ m is maximized when φ = 90 ◦ ,because then the ellipse presented to the observer wouldvary between a × c and b × c . In that case, ∆ m =2 . ( a/b ) and the axis ratio is related directly to∆ m . Assuming φ = 90 ◦ in 2009, when ∆ m = 0 . b/a = 0 .
75. Taking into account thescattering properties of an icy surface, Lockwood et al.(2014) refined this to b/a = 0 . ± .
01. Thus, Haumea isdistinctly non-spherical, and is not even axisymmetric.Haumea appears to be unique among large So-lar System objects in having such a distinctly non-axisymmetric, triaxial ellipsoid shape. Based on itsrapid rotation (angular velocity ω = 4 . × − s − ),Haumea is inferred to have assumed a particular shapeknown as a Jacobi ellipsoid. This is a class of equilibriumshapes assumed by (shearless) fluids in hydrostatic equi-librium when they rotate faster than a certain threshold(Chandrasekhar 1969, 1987). For a body with angularvelocity ω and uniform density ρ , the axis ratios b/a and c/a of the ellipsoid are completely determined, andthe axis ratios are single-valued functions of ω / ( πGρ ).For a Jacobi ellipsoid with b/a = 0 .
806 and Haumea’srotation rate, the density must be ρ = 2580 kg m − ,and c/a = 0 . a = 960km then yields b = 774 km and then c = 499 km,(4 π/ abc ρ exactly matches Haumea’s mass. The meanradius of Haumea would be 718 km. Moreover, thecross-sectional area of Haumea would then imply a sur-face albedo p V ≈ . − .
84 (Rabinowitz et al. 2006;Lacerda & Jewitt 2006; Lellouch et al. 2010; Lockwoodet al. 2014), consistent with the albedo of a water icesurface.To explain Haumea’s icy surface and ρ = 2580 kg m − ,one would have to assume that the interior of Haumeawas close to 2600 kg m − in density (an interior of hy-drated silicates (Desch & Turner 2015)) while its surfacewas a very thin ice layer ( § th mag-nitude star by Haumea in January 2017. The shadowof Haumea traced out an ellipse, as expected for theshadow of a triaxial ellipsoid; but the (semi-)axes ofthe shadow ellipse were much larger than expected: aumea’s Shape and Composition b (cid:48) = 569 ±
13 km by a (cid:48) = 852 ± a = 1161 ± b = 852 ± c = 513 ±
16 km (6).This new shape causes Haumea to look significantlydifferent than previous models: the mean radius ofHaumea is larger at 798 km, the albedo is a smaller p V ≈ .
51 (and would require a darkening agent inaddition to water ice), and the bulk density a lower1885 ±
80 kg m − . Moreover, the axis ratio c/a ≈ .
44 issignificantly lower than previous estimates ≈ .
52, andis inconsistent with a Jacobi ellipsoid or a fluid in hydro-static equilibrium. Ortiz et al. (2017) point out the pos-sibility that shear stresses may be supported on Haumeaby granular interparticle forces (Holsapple 2001).In either case, Haumea is likely to have a rocky coresurrounded by ice, but no analytical solution exists forthe figure of equilibrium of a rapidly rotating, differen-tiated body. Therefore it is not known whether or notHaumea is a fluid in hydrostatic equilibrium. In thispaper we attempt to reconcile the existing data fromHaumea’s light curve and occultation shadow, with thegoal of deriving its true shape and internal structure.Besides its axes, important quantities to constrain arethe ice fraction on Haumea today, and the size, shape,and density of its core. A central question we can solveusing these quantities is whether Haumea is a fluid inhydrostatic equilibrium or demands granular physics tosupport it against shear stresses. In addition, we can usethe core density and ice fraction to constrain the geo-chemical evolution of Haumea and, by extension, otherKBOs, as well as models of the origin of Haumea’s col-lisional family.In §
2, we examine whether it is possible for Haumea tobe a differentiated (rocky core, icy mantle) body with aJacobi ellipsoid shape. We show that only homogeneousbodies are consistent with a Jacobi ellipsoid shape. In § § ≈ × ×
537 km withbulk density ≈ − . In § CAN A DIFFERENTIATED HAUMEA HAVEJACOBI ELLIPSOID AXES? Observations have suggested that Haumea has axesconsistent with a Jacobi ellipsoid: for example, Lock-wood et al. (2014) inferred axes of a = 960 km, b = 770km, and c = 495 km (yielding axis ratios b/a = 0 . c/a = 0 . ≈ − .These axes are within 1% of the Jacobi ellipsoid solu-tion: a Jacobi ellipsoid with Haumea’s mass, rotationrate, and a = 960 km, has axes b = 774 . c = 498 . b/a = 0 .
806 and c/a = 0 . − . Uni-form density is a central assumption of the Jacobi ellip-soid solution, and yet Haumea is manifestly not uniformin density. Its reflectance spectra robustly show the exis-tence of a uniform water-ice surface (Trujillo et al. 2007;Pinilla-Alonso et al. 2009). The density of this ice, whichhas structure Ih at 40K, is about ≈
921 kg m − (Deschet al. 2009), much lower than Haumea’s mean density.Haumea, therefore, is certainly differentiated. A basicquestion, then, is whether a differentiated Haumea canbe consistent with axes that match a Jacobi ellipsoid.To answer this question, we have calculated the grav-itational potential of a differentiated Haumea, as de-scribed in Probst (2015). We model Haumea as twonested, aligned triaxial ellipsoids. We assume Haumea’souter surface is an ellipsoid with axes a = 960 km, b = 770 km, and c = 495 km, and we allow its coreto have arbitrary density ρ core , and arbitrary axis ratios p c = b c /a c and q c = c c /a c . For a given density ρ core and axis ratios p c and q c , the core axis a c is chosen sothat the mass of the core plus the mass of the ice man-tle, with density ρ ice = 921 kg m − , equal the mass ofHaumea. We then calculate whether the surface andthe core-mantle boundary (CMB) are equipotential sur-faces.An equilibrium solution must have the equipotentialsurfaces coincident with the surface and CMB, or elsevortical flows will be generated. In the absence of ex-ternal and viscous forces and large internal flows, thevorticity −→ ω follows the equation D −→ ωDt = 1 ρ −→∇ ρ × −→∇ P (4)where P is the pressure and ρ is the density. If thegradients of ρ and P are misaligned by an angle Θ, thenvorticity will be generated at a rate ∼ ( P/ρ ) /R (sin Θ),where R is comparable to the mean radius. In a time τ ,the vortical flows will circulate at rates comparable tothe rotation rate, ω H , where τ ∼ ω H ρR /P (sin Θ) − .Assuming ω H = 4 . × − s − , ρ = 921 kg m − , P =18 MPa, and R ∼
725 km, the timescale τ ∼ / sin Θyears. Even a small mismatch, with Θ ∼ ◦ , would leadto significant vortical flows within hundreds of years. Dunham, E. T. et al. a b
Surface fit angle ( ∘ )Core axis ratios minimize surface fit angleCore axis ratios minimize CMB fit angle S u r f a c e f i t ang l e ( ∘ ) C M B f i t ang l e ( ∘ ) Figure 1.
The “fit angle” Θ on the surface ( a ) and the core-mantle boundary ( b ), as functions of the core axis ratios p c = b c /a c and q c = c c /a c (note that p c ≥ q c ). A core density 2700 kg m − has been assumed. The core axis ratios that minimize thesurface fit angle and are most consistent with equilibrium are p c ≈ .
80 and q c ≈ .
51 (denoted by the white star in b ). Thecore axis ratios that minimize the fit angle on the CMB are p c ≈ .
82 and q c ≈ .
55 (denoted by the yellow star in b ). On eithersurface the fit angle is at least a few degrees, and it is not possible to minimize Θ on both surfaces with the same core ratios. Equilibrium solutions demand Θ = 0 ◦ , i.e., that −→∇ ρ and −→∇ P are parallel. In hydrostatic equilibrium, thenet force is −→ F = −−→∇ P + ρ −→ g eff , (5)where −→ g eff measures the acceleration due to gravity aswell as centrifugal support due to Haumea’s rotation. Ifthe net force is zero, then it must be the case that −→∇ ρ is parallel to −→ g eff . In other words, the gradient in theeffective gravitational potential must be parallel to thegradient in density, and surfaces of density discontinuitymust be equipotential surfaces.We solve for the effective gravitational potential bydiscretizing an octant of Haumea on a Cartesian gridwith 60 evenly spaced zones along each axis. If a rectan-gular zone is entirely inside the triaxial ellipsoid definedby the core, its density is set to ρ core ; if it is entirely out-side the triaxial ellipsoid defined by the surface, its den-sity is set to zero; and if it is entirely between these twoellipsoids, its density is set to ρ ice . For zones straddlingthe core and ice mantle, or straddling the ice mantle andthe exterior, the density is found using a Monte Carlomethod. An array of about 100 points on the surface isthen generated, by generating a grid of N θ ≈
10 angles θ from 0 to π radians, and of N φ ≈
10 angles φ from0 to 2 π radians. The points on the surface are definedby x ∗ = a sin θ cos φ , y ∗ = b sin θ sin φ , z ∗ = c cos θ . At each point we find the vector normal to the surface, −→ n = (2 x ∗ /a )ˆ e x + (2 y ∗ /b )ˆ e y + (2 z ∗ /c )ˆ e z , as well asthe gravitational acceleration −→ g , found by summing thegravitational acceleration vectors from each zone’s con-tribution. To this we add an additional contribution dueto centrifugal support: −→ g eff = −→ g + ω x ∗ ˆ e x + ω y ∗ ˆ e y . (6)Once these vectors are found at each of the surfacepoints defined by θ and φ , we find the following quantityby summing over all points: M = 1 N θ N φ (cid:88) N θ (cid:88) N φ −→ n · −→ g eff |−→ n | |−→ g eff | . (7)We also define “fit angle,” the mean angular deviationbetween the surface normal and the effective gravita-tional field: Θ = cos − M . (8)If the equipotential surfaces are coincident with the sur-face, then M = 1 and Θ = 0 ◦ . In a similar fashion wedefine an identical metric M on the core-mantle bound-ary as well.In Figure 1, we plot the fit angle on the surface andcore-mantle boundary of Haumea, as a function of theassumed core axis ratios p c and q c . A core density aumea’s Shape and Composition − is assumed. The core axis ratios that min-imize Θ on the outer surface are p c = b c /a c ≈ . q c = c c /a c ≈ .
51. For this combination, Θ ≈ ◦ .Meanwhile, the core axis ratios that minimize Θ on thecore-mantle boundary are p c ≈ .
82 and q c ≈ .
55. Forthis combination, Θ ≈ ◦ . Significantly, the parametersthat minimize Θ on the outer surface are not those thatminimize Θ on the core-mantle boundary. Given thechange in Θ with changes in p c and q c , either Θ on thesurface or core-mantle boundary must be at least a fewdegrees. Figure 1 shows numerous patches of higher-than-expected angles. These patches are caused by therandom nature in which the grid cells straddling theCMB are populated. The coarseness of the grid leadsto the code creating a bumpiness in the CMB surfacewhich then upwardly skews the calculation of the aver-age fit angle in places. This is because where the surfaceis bumpy, the surface normal vector can be significantlydifferent from the gravitational acceleration vector. Thepatches become more numerous, but smaller in magni-tude, with increasing numerical resolution.We have repeated the analysis for other core densi-ties of 3000 kg m − and 3300 kg m − . In those cases thediscrepancy between what parameters p c and q c mini-mize Θ on the surface vs. what parameters minimizeit on the CMB grows even larger. If Haumea has axes a = 960 km, b = 770 km, and c = 495 km, and is di-vided into a rocky core and icy mantle, the only wayto maintain hydrostatic equilibrium on the surface andcore-mantle boundary is for the core density to be asclose as possible to the inferred bulk density of Haumea, ≈ − , and for the core and surface axis ratiosto converge. The solution is naturally driven to one ofuniform density. In this case, the core comprises over96% of the mass of Haumea, and the ice thickness is <
10 km on the a and b axes, and < c axis.Even for this case, though, the effective equipotentialsurfaces fail to coincide with the surface or core-mantleboundary, by several degrees. This suggests that theonly way for Haumea to have axes consistent with a Ja-cobi ellipsoid is for it to have essentially no ice mantle,less than a few km thick.These investigations reveal two facts. First, a Haumeadivided into a rocky core and icy mantle cannot haveaxes equal to those of a Jacobi ellipsoid. This lendssome support to the finding by Ortiz et al. (2017) thatHaumea’s axes deviate significantly from a Jacobi ellip-soid’s. Second, if Haumea cannot conform to a Jacobiellipsoid, then it is not possible to use analytical formu-las to describe its shape, and a more powerful techniquemust be used to derive its internal structure. METHODSTo calculate the shape of a rapidly rotating Haumeawith a rocky core and icy mantle, we have written a code,named kyushu , that calculates the internal structureand figure of equilibrium of a differentiated body under-going rapid, uniform rotation. Our algorithm is adaptedfrom that of Hachisu (1986a,b; hereafter H86a,H86b),for calculating the structure of stars orbiting each otherin binary systems, as follows.The Hachisu (1986a,b) algorithm relies on using a gov-erning equation derived from the Bernoulli equation, ateach location on a three-dimensional grid: (cid:90) ρ − dP + Φ − (cid:90) Ω r ⊥ dr ⊥ = C, (9)where the first term is the enthalpy, H , the second termis the gravitational potential energy, the third term isthe rotational energy, and C is a constant. Here r ⊥ isthe distance from the rotation axis.The grid is defined in spherical polar coordinates,the variables being distance from the origin, r , the co-sine of the polar angle, µ , and the azimuthal angle, φ . A discretized grid of r i , with i = 1 , , ...N r is de-fined, with r uniformly spaced between r = 0 and r Nr = R . Likewise, a discretized grid of µ j , with j = 1 , , ...N µ is defined, with µ uniformly spaced be-tween µ = 0 and µ Nφ = 1, and a discretized grid of φ k ,with k = 1 , , ...N φ , is defined, with φ uniformly spacedbetween φ = 0 and φ Nφ = π/
2. Symmetries across theequatorial plane and n = 2 symmetry about the polaraxis are assumed. Quantities in the above equation, in-cluding density ρ ijk , gravitational potential Φ ijk , etc.,are defined on the intersections of grid lines. Typicalvalues in our calculation are R = 1300 km, N r = 391, N µ = 33, and N φ = 33, meaning that quantities arecalculated at 33 × ×
391 = 425 ,
799 locations.The enthalpy term can be calculated if the densitystructure and equation of state are provided. For ex-ample, if P = Kρ γ (as for an adiabatic gas), then H = ( γ ) / ( γ − P/ρ , and is immediately known as afunction of the local pressure and density. For planetarymaterials such as olivine, clays, or water ice, it would bemore appropriate to use a Vinet (Vinet et al. 1987) orBirch-Murnaghan (Birch 1947) equation of state, withthe bulk modulus and the pressure derivative of thebulk modulus specified. This equation of state can thenbe integrated to yield the enthalpy. The recent paperby Price & Rogers (2019) provides formulas for this.For our purposes, we neglect the self-compression of theplanetary materials inside Haumea. The bulk moduli ofplanetary materials are typically 10s of GPa, while themaximum pressure inside Haumea is < Dunham, E. T. et al. compression can be ignored. For ease of calculation wetherefore assume uniform densities in the ice mantle andin the rocky core, and compute the enthalpy accordingly.The gravitational potential term is found by numeri-cal integration of an expansion of the gravitational po-tential in spherical harmonics, using equations 2, 3, 33,34, 35 and 36 of H86b, using n = 2 symmetry, andtypically N l = 16 terms in the expansion. The rota-tional term is defined to be Ω Ψ, where Ψ = − r ⊥ / r ⊥ again being the distance from the axis. The terms H , Φ and Ω Ψ all spatially vary, but their sum is aconstant C at all locations. The Hachisu algorithm ex-ploits this fact by fixing two spatial points “A” and “B”to be on the boundary of the body. Point “A” lies at r = r A , µ = 0 (in the equatorial plane), φ = 0, or x = r (1 − µ ) / cos φ = r A , y = r (1 − µ ) / sin φ = 0, z = r µ = 0. Point “B” lies at r = r B , µ = 0 (in theequatorial plane), φ = π/
2, or x = 0, y = r B , z = 0. Fora triaxial ellipsoid ( x/a ) + ( y/b ) + ( z/c ) = 1, points A and B refer to the long and intermediate axes of thebody in the equatorial plane, with r A = a and r B = b .At these two locations, H = 0, and the Hachisu algo-rithm then solves for the only two values of C and Ωthat allow H = 0 at both these boundary points. With C and Ω defined, H is found at all locations, and theenthalpy integral H = (cid:82) ρ − dP is inverted to find thedensity ρ at each location. Locations with H < z axis can be equated with c of a tri-axial ellipsoid, although of course the shape need notnecessarily be a triaxial ellipsoid. After adjusting thedensity everywhere, the code then recalculates the grav-itational potential and performs the same integrations,solving iteratively until the values of Ω and the densi-ties ρ at all locations converge. Because the densitiesand the volume of the body are changed with each iter-ation, the mass of the object is a varying output of themodel.We apply the H86a,b algorithms as part of a largeriterative procedure that introduces two new variables toan equation of state: P cmb and ρ core . We assume thatat locations within the body with pressures P < P cmb , ρ = ρ ice ≡
921 kg m − . At higher pressures P > P cmb ,we assume ρ = ρ core . This divides the body into acore and an icy mantle, each with distinct densities,with the pressure equal to a uniform value P cmb every-where on the core-mantle boundary. For P > P cmb , H = P cmb /ρ ice + ( P − P cmb ) /ρ core . This form of theequation of state ignores self-compression, as is appro-priate in bodies of Haumea’s size made of materials likewater ice, olivine, or hydrated silicates. The bulk mod-ulus of water ice is 9.2 GPa (Shaw 1986), far higher than the likely pressures in the ice shell, <
20 MPa(section 4). Likewise, the bulk moduli of olivine is 126GPa (N´u˜nez-Valdez et al. 2013), and that of the hy-drated silicate clay antigorite is 65 GPa (Capitani &Stixrude 2012), far higher than the maximum pressureinside Haumea ( <
300 MPa). We therefore expect self-compression to change the densities by < H < P cmb /ρ ice , the density is simply ρ ice , and for H > P cmb /ρ ice , ρ = ρ core .With this definition, we iterate as follows to find P cmb and ρ core . In each application of the Hachisu algorithms,we initialize with a density distribution with ρ = 0 for( x/a ) + ( y/b ) + ( z/c ) > ρ = ρ ice for ( x/a ) +( y/b ) + ( z/c ) <
1, but ρ = ρ core for ( x/a ) + ( y/b ) +( z/c ) < ξ . That is, we define Haumea’s surface to bea triaxial ellipsoid, and its core to be a similar triaxialellipsoid with aligned axes, smaller in size by a factor of ξ . The value of ξ is chosen so that the total mass of theconfiguration equals M H , the mass of Haumea: ξ = (cid:20) M H / (4 πabc ) − ρ ice ρ rock − ρ ice (cid:21) / . (10)The value of c is unknown and an output of the code,so we initialize our configuration with c = b . We applythe Hachisu algorithms several times. First we define P cmb ≈ ρ core = 2700 kg m − , applythe Hachisu algorithms, and calculate the mass of thebody, M . Because this will not match Haumea’s mass, M H , we multiply ρ core by a factor M H /M , and reapplythe Hachisu algorithms. We do this until we have foundan equilibrium configuration with Haumea’s mass. Out-puts of the code include ρ core and Ω, and in general Ωwill be smaller than Haumea’s true angular frequencyΩ H = 2 π/P rot for P cmb = 0 MPa ( P rot is the rota-tional period). We then repeat the procedure, findingan equilibrium configuration with Haumea’s mass, hav-ing P cmb = 40 MPa. An output of the code will be adifferent ρ core and a different Ω, which in general willbe > Ω H . If the solution is bracketed, we use standardbisection techniques to find P cmb that yields an equi-librium configuration consistent not just with Haumea’smass, but with its period as well.Thus, inputs of the code include r A = a and r B = b , and Haumea’s mass M H and angular velocity Ω H .The outputs of the code include the density ρ at alllocations, the calculated mass M (which should complywith M = M H ), the angular frequency Ω (which shouldequal 2 π/P rot ), and the values of c and P cmb and ρ core .To benchmark the kyushu code, we ensure that it re-produces a Jacobi ellipsoid when a homogeneous den- aumea’s Shape and Composition a = 960 km, wouldhave axes b = 774 . c = 498 . . − : this is similar to, but does notexactly equal, the solution favored by Lockwood et al.(2014), who fit Haumea’s light curve assuming a = 960km, b = 770 km, c = 495 km, and uniform densityof 2600 kg m − . Running kyushu and assuming axesof a = 960 km and b = 774 . b/a = 0 . .
03% and the rotation period to within 0 . P cmb = 0 .
42 MPa, so that the body is essentially uni-form in density, with an ice layer < . − , and the short axis haslength c = 499 . c/a = 0 . c -axis match a uniform-density Jacobi ellipsoid to within0 .
03% and 0 . a and b axes areconsistent with such a solution. RESULTSWe have run the kyushu code for 30 different combina-tions of a and b axes, with a varying from 950 km to 1075km in increments of ∼
25 km, and b varying from 800 kmto 900 km in increments of ∼
25 km. In comparing a sub-set of these runs with runs performed with 10 km incre-ments, we found that convergence of key outputs (axes,densities) is 0.3% or less, so we consider 25 km to be nu-merically converged. We find families of solutions thatcan conform to Haumea’s mass ( M = 4 . × kg)and rotation period ( P rot = 3 . c axis, the shape of the core-mantle boundary(CMB), and the thickness of the ice layer above the core.In Figure 2 we plot the following quantities as func-tions of imposed a and b axes: the average (bulk) den-sity, ρ avg ; the (semi-)axis c ; and the thickness of theice layer along the a , b and c axes. The density of theice layer was imposed to be 921 kg m − . In Figure 3we plot as functions of a and b the following: the coredensity, ρ core ; the pressure at the core-mantle boundary, P cmb ; and the semi-axes a c , b c and c c of the core. En-tire families of solutions are found across the range of a and b that we explored, with the exception of simul-taneous combinations of large a and large b . The input a × b combinations of 1025 × , × , × ×
800 did not yield solutions because wheninitializing with these parameters, kyushu was not able converge at Haumea’s mass and rotation rate. Simulta-neously imposing large a and b yields a low bulk density ρ avg and large parameter ω / ( πGρ avg ) that cannot yielda c axis consistent with Haumea’s mass.Across the explored parameter space that yielded so-lutions, the average density of Haumea ranges from 1905to 2495 kg m − . As might be expected, the average(bulk) density of Haumea is equally sensitive to both a and b , being inversely proportional to the volume andtherefore to the product ab . As an example solution,we take a = 1050 km and b = 840 km, for which theaverage density is 2018 kg m − .For allowed solutions, the shortest (semi-)axis c rangesfrom 504 to 546 km. The c axis is equally sensitive toboth a and b , being large when the product ab is large.For the case with a = 1050 km and b = 840 km, we find c = 537 km.Likewise, the thickness of the ice layer increases withincreasing a and b (and therefore c ). The ice is alwaysthickest along the a axis, ranging from 15 to 210 kmacross the explored range; it is intermediate in thicknessalong the b axis, ranging from 15 to 150 km; and it isthinnest along the c axis, ranging from 5 to 80 km. Forthe case with a = 1050 km and b = 840 km, we find icethicknesses of 167 km, 117 km, and 67 km along the a , b , and c axes.Across the explored parameter space that yielded so-lutions, the density of Haumea’s core ranges from 2560to 2740 kg m − . Core density is much more sensitive tothe a axis than the b axis, and tends to be greater when a is greater. For the case with a = 1050 km and b = 840km, we find a core density 2680 kg m − .Across the explored range, the pressure at the coremantle boundary ranges from 3.2 to 36.6 MPa, with thelowest values corresponding to the thinnest ice layersand smallest values of a and b . For the case with a =1050 km and b = 840 km, P cmb = 30 . a axis. For the extreme case with a = 950 km, b = 800 km, and c = 504 km (a smallHaumea), we find core axes of a c = 935 km, b c = 785km, and c c = 499 km. For this configuration, b c /a c =0 .
840 and c c /a c = 0 . b/a = 0 .
842 and c/a = 0 . ξ = 0 . ∼
1% of the radius of Haumea, and comprises1 .
5% of Haumea’s volume. For the opposite extremecase of a large Haumea, with a = 1050 km, b = 875km, and c = 546 km, we find core axes of a c = 840 km, b c = 725 km, and c c = 466 km. For this configuration, Dunham, E. T. et al.
Figure 2.
Contour plots of quantities in the equilibrium shape models of a differentiated Haumea, as functions of imposed axes a and b . The panels depict: the average (bulk) density of Haumea, ρ avg ; the c (semi-)axis; the thickness of the ice layer alongthe a axis; the thickness of the ice layer along the b axis; and the thickness of the ice layer along the c axis. Solutions werenot found for simultaneous combinations of large a and large b , but otherwise entire families of fluid hydrostatic equilibriumsolutions exist. b c /a c = 0 .
86 and c c /a c = 0 .
55, close to the axis ratiosfor the surface, b/a = 0 .
83 and c/a = 0 .
52. The meansize of the core, relative to the mean size of the surface,is ξ = 0 . a = 1050 km and b = 840 km, the core semi-axes are ≈ × × b c /a c = 0 .
82 and c c /a c = 0 .
53. Themean size of the core, relative to the mean size of thesurface, is ξ = 0 . . x/a c ) + ( y/b c ) +( z/c c ) , where x , y , and z are computed for each anglecombination θ and φ and radius defined by r = (r(ir core − core )) /
2. Here ir core is the index of the first radialzone outside the core at that θ and φ . We find thata triaxial ellipsoid shape is consistent with the core towithin 1.5% and the surface to within 0.5%. Both arewithin the code’s resolution error. It is remarkable thatthe core mantle boundary solution is driven to the shapeof a triaxial ellipsoid. This justifies the assumption in § DISCUSSION aumea’s Shape and Composition Figure 3.
Contour plots of quantities in the equilibrium shape models of a differentiated Haumea, as functions of imposedaxes a and b . The panels depict: the density of the core, ρ core ; the pressure at the core-mantle boundary, P cmb , in MPa; the a (semi-)axis of the core; the b (semi-)axis of the core; and the c (semi-)axis of the core. As in Figure 2, solutions were not foundfor simultaneous combinations of large a and large b , but otherwise entire families of fluid hydrostatic equilibrium solutionsexist. Reconciling the light curve and occultationdatasets
Under the assumption of uniform density, there is onlyone Jacobi ellipsoid solution that can match Haumea’smass and rotation rate, and its inferred b/a axis ratio.Once M , ω and b/a are specified, the average density ρ avg is fixed, which determines c/a and all the axes. Incontrast, our modeling demonstrates that once the as-sumption of uniform density is dropped, a wide rangeof solutions exists with different semi-axes a and b , andeven with the same b/a axis ratios. These would makelinear cuts from the lower left to the upper right throughthe contour plots of Figures 2 and 3, and would yield arange of ρ avg and other properties. This additional free-dom suggests it may be possible to have a differentiatedHaumea be a fluid in hydrostatic equilibrium, and simul- taneously fit the shadow observed by Ortiz et al. (2017)during the occultation.We find one solution, the example case consideredabove, to be quite favorable. This solution has outersemi-axes a = 1050 km, b = 840 km, and c = 537km. The core-mantle boundary is defined to lie at P c = 30 . a c = 883 km, b c = 723 km, and c c = 470 km. The core density is ρ core = 2680 kg m − , and the average density of Haumeain this case is ρ avg = 2018 kg m − . The ice mantlein this case comprises 17.2% of Haumea’s volume andranges in thickness from 170 km on the a axis, to 120km on the b axis, to 71 km on the c axis. The albedois p V ≈ .
66, slightly lower than the range 0 . − . ≈ .
51 calculated by Ortiz et al. (2017). Likewise, theaxes and average density we favor are intermediate be-0
Dunham, E. T. et al. tween the previous solutions assuming a Jacobi ellipsoidwith ρ avg = 2600 kg m − , and the Ortiz et al. (2017)case with ρ avg = 1885 kg m − .The projection of this triaxial ellipsoid onto the Earth(its shadow) is a complicated function that depends onits orientation relative to the line of sight. It is moredifficult to invert the shadow axes to find the axes ofHaumea’s surface, but it is possible. We outline ourmethods in Appendix A (6).If the tilt of Haumea’s rotation axis out of the planeof the sky is ι = 13 . ◦ and the rotational phase were ψ = 0 ◦ , we concur with Ortiz et al. (2017) that only atriaxial ellipsoid with axes 1161 × ×
513 km wouldbe consistent with shadow axes 852 ×
569 km. However,small changes in Haumea’s rotational phase have a largeimpact on the shadow size. We find that if Haumea’srotational phase during the occultation were ψ = 13 . ◦ ,then the shadow axes would be a (cid:48) = 853 . b (cid:48) = 576 . a (cid:48) = 852 ± b (cid:48) = 569 ±
13 km.Ortiz et al. (2017) favored ψ = 0 ◦ , but inspection oftheir light curve (their Extended Data Figure 6) showsthat the rotational phase at minimum brightness was atleast 0.04 (14 . ◦ ), and would not be inconsistent witha value of 0.06 (21 . ◦ ), relative to the phase of 0.00 atthe time of the occultation. Finally we note that theaxis ratio b/a = 0 .
80 for this case is the same as previ-ous estimates ( b/a ≈ .
80; Lockwood et al. (2014)), andyields a light curve with ∆ m ≈ .
23 during the epoch ofthe occultation (6). This approximates the actual lightcurve amplitude of ∆ m = 0 .
26 observed by Lockwoodet al. (2014).A much more extensive parameter study must beundertaken to simultaneously fit all the data. As-trometry of Haumea’s moons can better constrain themoons’ orbital poles and, if Hi’iaka’s orbit is alignedwith Haumea’s equator, Haumea’s rotational pole and ι . More and consistent analyses of the now 15 yearsof light curve data, especially considering different re-flectance functions, can better constrain the rotationalphase ψ during the occultation. Further exploration ofparameter space may yield a shape for Haumea that isexactly consistent with the light curve data and the oc-cultation shadow. As shown here, though, Haumea canbe a fluid in hydrostatic equilibrium and can conform tothe occultation shadow.5.2. Aqueous alteration of Haumea’s core and itsastrobiological potential
A large range of axes a (from 950 to 1075 km) and b (from 800 to 900 km) are consistent with bodies withHaumea’s mass and rotation rate, and are fluid configu- rations in hydrostatic equilibrium. Our favored solutionwith a = 1050 km and b = 840 km has a mass frac-tion of ice of 17.2%, but this value could range from1% to 22% across the range we explored. Across thisrange, however, the allowable core density varies onlyslightly, from ρ core = 2560 kg m − to 2740 kg m − , devi-ating by only a few percent from our favored density of ρ core = 2680 kg m − . This is very close to the averagedensity previously inferred for Haumea, but this appearsto be coincidental. A robust result of our analysis is thatHaumea’s core has a density ≈ − , overlain byan ice mantle.Comparison of the density of Haumea’s core toother planetary materials provides strong clues toHaumea’s history. Grain densities of ordinary andenstatite chondrites are typically > − , andtheir bulk densities typically ≈ − becauseof ∼
10% porosity (Consolmagno et al. 2008; Wilkisonet al. 2003). Carbonaceous chondrites are marked bylower grain densities, average 3400 kg m − (range from2400 − − depending on the type of chon-drite), higher porosities ≈ − − (Macke et al. 2011; Consol-magno et al. 2008). The difference is that carbonaceouschondrites are largely composed of products of aqueousalteration. In fact, the more oxidized groups of car-bonaceous chondrites have higher porosity (Macke et al.2011). Hydrated silicates typically have densities in thisrange. Clays such as montmorillonite, kaolinite, illite,and mica typically have densities between 2600 kg m − and 2940 kg m − (Osipov 2012). This strongly suggeststhat Haumea’s core is composed of hydrated silicates,and that Haumea’s core was aqueously altered in itspast.Serpentinization is the process by which silicates typ-ical of the dust in the solar nebula react with water onan asteroid or planet, producing new phyllosilicate min-erals. A archetypal reaction would be:1 .
000 (Mg . Fe . ) SiO + 1 .
142 H O → .
474 Mg Si O (OH) + 0 .
193 Fe O +0 .
052 SiO + 0 .
194 H . In this reaction, 1 kg of olivine with fayalite contenttypical of carbonaceous chondrites may react with 0.129kg of water to produce 0.826 kg of chrysotile, 0.281kg of magnetite, 0.020 kg of silica, and 0.002 kg ofhydrogen gas, which escapes the system. The totaldensity of olivine (density 3589 kg m − ) plus ice (den-sity 921 kg m − ) before the reaction is 2697 kg m − .After the reaction, the mixture of chrysotile (den-sity 2503 kg m − ) plus magnetite (density 5150 kg m − ) aumea’s Shape and Composition − ) has a total density of2874 kg m − (Coleman 1971). Including a 10% poros-ity typical of carbonaceous chondrites, the density ofthe aqueously altered system would be 2612 kg m − ,remarkably close to the inferred density of Haumea’score.If Haumea’s core underwent significant aqueous alter-ation, some of this material may have dissolved in thewater and ultimately found its way into the ice mantleof Haumea. In fits to Haumea’s reflectance spectrum,Pinilla-Alonso et al. (2009) found the most probable sur-face composition was an intimate mixture of half crys-talline and half amorphous water ice, with other com-ponents comprising <
10% of the surface; but similarmodeling by Trujillo et al. (2007) found that Haumea’ssurface is best fit by a mixture of roughly 81% crys-talline water ice and 19% kaolinite. Kaolinite was addedto the fit to provide a spectrally neutral but blueish ab-sorber; few other planetary materials contribute to thereflectance spectrum in this way. Kaolinite is a commonclay mineral [Al Si O (OH) ] very similar in structureto chrysotile, produced by weathering of aluminum sili-cate minerals like feldspars.A variety of phyllosilicates have been observed by theDawn mission on the surface of Ceres (Ammannito et al.2016), strongly suggesting aqueous alteration of silicateswithin a porous, permeable core or a convecting mud-ball (Bland et al. 2006; Travis 2017). If kaolinite canbe confirmed in Haumea’s mantle, this would providestrong support for the aqueous alteration of Haumea’score.Preliminary modeling by Desch & Neveu (2015) sug-gests that aqueous alteration of Haumea’s core is a verylikely outcome. Haumea, or its pre-collision progeni-tor, could have been differentiated into a rocky olivinecore and icy mantle. Desch & Neveu (2015) show thatmany factors can lead to cracking of a rocky core onsmall bodies. Microcracking by thermal expansion mis-match of mineral grains or by thermal expansion of porewater during heating (as the core heats by radioactivedecay over the first < ∼ yr, although further geochemical modeling isneeded to test more proposed scenarios for Haumea’sstructure and evolution.A long ( ∼ yr) duration of aqueous alteration sug-gests a period of habitability within Haumea. To de-velop and survive, life as we know it requires waterand a long-lasting environment with little temperaturevariability (Davis & McKay 1996). With central tem-peratures approaching 750 K, and surface temperaturesnear 40 K, a large fraction of Haumea’s interior wouldhave had intermediate temperatures consistent with liq-uid water (Castillo-Rogez & Lunine 2012). The originof life is also thought to require a substrate to protectand localize biochemical reactions. Clays such as mont-morillonite can act as this substrate because they canbind substantial water, and are soft and delaminate eas-ily. Clays can also promote the assembly of RNA fromnucleosides, and can stimulate micelles to form vesicles(Travis 2017). The interior of Haumea may have at onepoint resembled regions beneath the seafloor experienc-ing hydrothermal circulation. These regions are con-ducive to life: Czaja et al. (2016) discovered archaeaanaerobically metabolizing H S in such environments,and other studies have confirmed that microbes existdeep in fractures of hot environments (Jannasch & Mottl1985).5.3.
Implications for the mass of the collisional family
An ongoing mystery is why Haumea’s collisional fam-ily contains so little ice. The total masses of Hi’iakaand Namaka, plus 2002 TX and the other collisionalfamily members, amount to about 2.4% of Haumea’smass (Vilenius et al. 2018). This is much smaller thanthe amount of ice that has been presumed to have beenejected. As described in §
2, if Haumea really were aJacobi ellipsoid with uniform density ≈ − , itwould have to have a very thin ice layer comprising per-haps only ≈
4% of Haumea’s mass. This is much lowerthan the mass fraction of ice in typical KBOs. If theKBO has bulk density ρ , the mass fraction of ice wouldbe f ice = ( ρ ice ) / ( ρ ) × ( ρ rock − ρ ) / ( ρ rock − ρ ice ). Atypical KBO may form from a mixture of pure olivinewith 10% porosity and density ρ rock = 3300 kg m − , andnon-porous ice with density ρ ice = 921 kg m − . The ρ in such a KBO could range from 1500 kg m − to2500 kg m − which equates to f ice ranging from 46% to12%. If Haumea was comparable to these end membercases, it would need to lose 91% and 67% of its ice re-2 Dunham, E. T. et al. spectively to end up with a post-collisional ice fractionof 4%. It is difficult to explain why Haumea would lose91% of its ice instead of 100%. Also, neither of these sce-narios match with the 2.4% of ice thought to be ejected,which is also difficult to reconcile.This discrepancy is ameliorated by our results. Ourmodeling of Haumea’s structure shows that it may re-tained a significant fraction of ice. Across the parameterspace we explored, Haumea’s present-day bulk densityvaries from 1900 kg m − to 2500 kg m − (core density2550 kg m − to 2750 kg m − ), which corresponds to f ice ranging from 1% to 22%. The lower end of this rangeis unlikely from the standpoint of the occultation data.We favor that today, Haumea has a high ice fraction: f ice =17% is our favored case.In addition to this argument, our model suggests thatHaumea underwent serpentinization, meaning the coreexperienced pervasive aqueous alteration. This pro-cess would reduce the fraction of ice below that whichHaumea started. As an example, if Haumea initiallyhad a density ρ = 2500 kg m − , like that of Eris (Brown& Schaller 2007), and original ρ rock = 3300 kg m − , itstarted with f ice =34%. Serpentinization would havethen consumed ice into the rocky core to lower thecore density to ρ rock = 2612 kg m − , which would alterHaumea’s ice fraction to 23%. So, if the collision ejected2.4% of the ice, Haumea’s ice fraction today would be f ice ∼ CONCLUSIONSThis paper presents numerical modeling designed totest three questions about the KBO Haumea: 1) IsHaumea a Jacobi ellipsoid? If it is differentiated, whatis Haumea’s shape? 2) Is Haumea a fluid in hydrostaticequilibrium? 3) Can Haumea’s occultation and lightcurve data be reconciled? We aimed to address thesequestions with the goal of understanding the composi-tion and structure of Haumea to learn about its colli-sional history and evolution.We have written a code kyushu based on the algo-rithms of Hachisu (H86a,b) to calculate the internalstructure of a rapidly rotating differentiated body basedon input parameters such as the semi-axes a and b . Al- though we did not explore all parameter space, Haumeaappears to be best approximated as a differentiated tri-axial ellipsoid body in hydrostatic equilibrium with axes a = 1050 km, b = 840 km, and c = 537 km. This shapefits the Ortiz et al. (2017) occultation shadow and isclose to light curve data. As this shape, Haumea hascore axes a c = 883 km, b c = 723 km, c c = 470 km, ρ avg = 2018 kg m − , ρ core = 2680 kg m − which equatesto an ice mantle comprising ∼
17% of Haumea’s massand ranging from 71 to 170 km in thickness. Haumea’salbedo is p v ∼ .
66 in this case.In contrast to previous studies (Lockwood et al.2014; Rabinowitz et al. 2006), our results suggest thatHaumea’s ice crust amounts to a significant portion ofthe body. Due to the thickness of the ice, Haumea’score has a relatively high density indicating the com-position of the core is a hydrated silicate (the closestmatch is kaolinite). For the core to be hydrated, a longperiod ( ∼ yr) of serpentinization must have occuredduring which regions of the core were potentially hab-itable. The thick ice crust also suggests that Haumea’scollisional family (icy objects a few percent the massof Haumea) was produced from only a small portion ofthe ice Haumea started with, before Haumea sufferedthe collision. Insights into this type of mantle strippingcollision could be applicable to modeling metal-rich,fast-rotating triaxial ellipsoid 16 Psyche, the focus ofthe upcoming NASA Psyche mission (Elkins-Tantonet al. 2016).As this study continues, we would like to expandparameter space to obtain more precise results. Wecan explore how Haumea would change shape or com-position if we use different ice densities, porosity, an-gles/orientations to better match the shadow in addi-tion to matching the light curve amplitude/phase moreprecisely and using an appropriate equation of stateto include the compressibility of materials. Haumeais a unique and interesting body worthy of study forits own sake, but understanding Haumea can provideinsights into fundamental processes such as subsurfaceoceans/aqueous alteration on small bodies and dynam-ics of mantle-stripping collisions, acting across the SolarSystem.We thank Darin Ragozzine and Sarah Sonnett forhelpful discussions about the collisional family andHaumea’s light curve. We thank Steve Schwartz and Vi-ranga Perera for useful discussions about how to modelHaumea using smoothed particle hydrodynamics codes.We thank Leslie Rogers and Ellen Price for introducingus to the Hachisu (H86a,b) algorithm and for generaldiscussions about how they implemented the Hachisu aumea’s Shape and Composition Ammannito, E., Desanctis, M. C., Ciarniello, M., et al.2016, Science, 353Binzel, R. P., Gehrels, T., & Matthews, M. S. 1989,Asteroids II (University of Arizona Press), 1258Birch, F. 1947, Physical Review, 71, 809Bland, P., Zolensky, M., Benedix, G., & Sephton, M. 2006,Meteorites and the Early Solar System II, 853Brown, M. E., Barkume, K. M., Blake, G. A., et al. 2006,The Astronomical Journal, 133, 284Brown, M. E., Barkume, K. M., Ragozzine, D., & Schaller,E. L. 2007, Nature, 446, 294Brown, M. E., & Schaller, E. L. 2007, Science, 316, 1585Capitani, G. C., & Stixrude, L. 2012, AmericanMineralogist, 97, 1177Castillo-Rogez, J. C., & Lunine, J. 2012, in Frontiers ofAstrobiology, ed. C. Impey, J. Lunine, & J. Funes(Cambridge University Press), 201Coleman, R. G. 1971, Bulletin of the Geological Society ofAmerica, 82, 897Consolmagno, G. J., Britt, D. T., & Macke, R. J. 2008,Chemie der Erde, 68, 1Czaja, A. D., Beukes, N. J., & Osterhout, J. T. 2016, 44,983Davis, W. L., & McKay, C. P. 1996, Origins of life andevolution of the biosphere, 26, 61Desch, S., & Neveu, M. 2015, Lunar and Planetary ScienceConference, Dunham, E. T. et al.
APPENDIXHere we derive the formulas needed to calculate the axes of Haumea’s shadow as it occults a star. We assumeHaumea’s surface is a triaxial ellipsoid with long axis along the x direction, with axes a > b > c , defined by thosepoints that satisfy f ( x, y, z ) = x a + y b + z c = 1 . We assume the star lies in a directionˆ e LOS = cos ψ sin φ ˆ e x + sin ψ sin φ ˆ e y + cos φ ˆ e z Here φ is the angle between the line of sight (from us through Haumea to the star) and Haumea’s pole (along the z axis). We can define two unit vectors in the plane of the sky:ˆ e = − sin ψ ˆ e x + cos ψ ˆ e y , and ˆ e = ˆ e × ˆ e LOS = + cos ψ cos φ ˆ e x + sin ψ cos φ ˆ e y − sin φ ˆ e z . Haumea’s limb is the locus of those points, defined by r , such that the line of sight is tangent to the surface, orperpendicular to the normal, so that ∇ f · ˆ e LOS = 0 . All of these points satisfy z = − c tan φ (cid:18) x cos ψa + y sin ψb (cid:19) , which define a plane inclined to the sky. The intersection of the plane with the ellipsoid defines an ellipse, and theprojection of this ellipse onto the plane of the sky—Haumea’s shadow—also is an ellipse.We project the points on Haumea’s limb onto the plane of the sky by recasting r in the coordinate system using ˆ e ,ˆ e , and ˆ e LOS : r = ( r · ˆ e ) ˆ e + ( r · ˆ e ) ˆ e + ( r · ˆ e LOS ) ˆ e LOS = s ˆ e + t ˆ e + u ˆ e LOS , with s = r · ˆ e = − x sin ψ + y cos ψ and t = r · ˆ e = x cos ψ cos φ + y sin ψ cos π − z sin φ. All the points on the limb have z related to x and y as above, so the boundary of the shadow, which equals theprojection of the limb onto the plane of the sky, is defined by s = − x sin ψ + y cos ψ and t cos φ = x cos ψ (cid:18) c a tan φ (cid:19) + y sin ψ (cid:18) c b tan φ (cid:19) . Inverting, we find x , y and z in terms of s and t for points along the limb: x = 1∆ (cid:20) − sin ψ (cid:18) c b tan φ (cid:19) s + cos ψ t cos φ (cid:21) ,y = 1∆ (cid:20) + cos ψ (cid:18) c a tan φ (cid:19) s + sin ψ t cos φ (cid:21) , and z = − tan φ ∆ (cid:20) sin ψ cos ψ (cid:18) c b − c a (cid:19) + (cid:18) c a cos ψ + c b sin ψ (cid:19) (cid:18) t cos φ (cid:19)(cid:21) , aumea’s Shape and Composition c tan φ (cid:18) cos ψa + sin ψb (cid:19) . Subsituting these expressions for x , y and z into the equation for the ellipsoidal surface, we find the projection ofHaumea’s limb onto the plane of the sky satisfies P s + Qst + Rt = 1 , where P = 1∆ a sin ψ (cid:18) c b tan φ (cid:19) + 1∆ b cos ψ (cid:18) c a tan φ (cid:19) + tan φ ∆ c sin ψ cos ψ (cid:18) c b − c a (cid:19) ,R = 1∆ cos φ (cid:18) cos ψa + sin ψb (cid:19) + c tan φ ∆ cos φ (cid:18) cos ψa + sin ψb (cid:19) , and Q = 2 sin ψ cos ψ ∆ c cos φ (cid:18) c b − c a (cid:19) (cid:20) c tan φ (cid:18) cos ψa + sin ψb (cid:19)(cid:21) . These also define an ellipse, rotated in the s - t plane.After rotating the ellipse in the plane of the sky by an angle θ , defined bytan 2 θ = Q/ ( R − P ) , we find it has axes a (cid:48) and b (cid:48) defined by1( a (cid:48) ) = (cid:2) P cos θ + R sin θ − Q sin θ cos θ (cid:3) and 1( b (cid:48) ) = (cid:2) P sin θ + R cos θ + Q sin θ cos θ (cid:3) . We have written a simple code that takes a , b , and c , and ψ and φ as inputs, and solves for θ and then the semi-axes a (cid:48) and b (cid:48) of Haumea’s shadow.One end-member case includes φ = 0 ◦ , in which Haumea’s pole is pointed toward the star; we derive θ = − ψ andregardless of ψ , Haumea’s shadow has axes a (cid:48) = b and b (cid:48) = a . Another end-member case is φ = 90 ◦ , in which case theline of sight to the star is parallel to Haumea’s equator. The shadow will have b (cid:48) = c regardless of ψ , and the otheraxis will be a (cid:48) = ab (cid:20) cos ψa + sin ψb (cid:21) / , in which case a (cid:48) = b if ψ = 0 ◦ (looking along the long a axis), or a (cid:48) = a if ψ = 90 ◦ (looking along the b axis). Onemore end-member case is ψ = 0 ◦ but arbitrary φ , in which case a (cid:48) = b and b (cid:48) = a cos φ (cid:20) c a tan φ (cid:21) +1 / . This is the case considered by Ortiz et al. (2017). Assuming a = 1161 km, b = 852 km, c = 513 km, ψ = 0 ◦ and φ = 76 . ◦ (a tilt of Haumea’s pole with respect to the plane of the sky of 13 . ◦ ), we find a (cid:48) = 852 km and b (cid:48)(cid:48)