11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
113
11 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
(11.1)
∇ 2V = 0
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:
1 ∞  ⎡a⎤
V = ∑∑⎢ ⎥
a = 0 m = 0 ⎣ 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, C00 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
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:
2
∞
GM
⎡a⎤
V=
[1 − ∑ J 2 ⎢ ⎥ P2 (cosθ )]
r
⎣r ⎦
=1
(11.3)
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, J2 is generally far larger than
any of the other harmonics (except of course C00).
The fundamental definition of the gravitational potential is of course

ρ(r ′ )d 3r ′

V (r ) = G ∫
 
r − r′
all space
(11.4)
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′
  = ∑ +1 P (cos γ )
r − r ′ = 0 r
(11.5)
where γ is the angle between vectors r and r′, and r is outside the planet (r′ is inside the
planet). Moreover,
P (cos γ ) =

( − m)!
∑ (l + m)!P
m=0
m

(θ )Pm (θ ')cos[m(φ − φ ')]
(11.6)
But we only need to keep track of m=0, for evaluating J2 (because these are the only
terms that contribute to the integral):
P (cos γ ) = P (cosθ )P (cosθ ′ ) + m ≠ 0 terms
(11.7)
so it follows immediately that
J 2 = −
1
r ′ 2 P2 (cosθ ′ )ρ(r ′ )d 3r ′
2 ∫
Ma
(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.
11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
115
Recalling that P2 (cosθ) = (3cos2θ- 1)/2, one can see immediately that
3
1
Ma 2 J 2 = − ∫ [ z 2 − (x 2 + y 2 + z 2 )]ρ(r)d 3r
2
2
(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 3r
A ≡ ∫ (z 2 + y 2 )ρ(r)d 3r; B ≡ ∫ (z 2 + x 2 )ρ(r)d 3r
(11.10)
and it is easy to see that
1
Ma 2 J 2 = C − [A + B]
2
(11.11)
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 C22=(B-A)/4Ma2. (With appropriate choice
of zero longitude, S22 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).
11.2 The Response of a Planet to its Own Rotation
We saw that J2 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 J2 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 J2 will be small for a body that is centrally concentrated.
11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
This body has a dense core and low density
envelope so it has a small J2
116
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 J2. (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 J2 is dominated by
hydrostatic effects. The theory explicitly assumes hydrostaticity.
11.2.1 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.,
rs = r0 (1 + ε .P2 )
(11.13)
where r0 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 P2 is of the form
GP
V2 = 3 2
r
r0 (1+ ε P2 )
1
∫ P (cosθ ′)d(cosθ ′) ∫
2
−1
x 2 .2π x 2 .ρ0 .dx
(11.14)
0
(V2 is the amplitude of the coefficient of the P2 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 P0 is unity), the only part of the integral over x that contributes is the part
proportional to P2. Therefore:
11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
GP
V2 = 3 2
r
117
1
∫P
2
2
(cosθ ′ )d(cosθ ′ ).ε .2πρ0 a 5
−1
3G
= 3 Ma 2 .ε
5r
(11.15)
[where we’ve used the fact that the normalization integral for Legendre functions is
2/(2+1).] We’ve set ro=a (equatorial radius), which is correct to this order of
approximation. We must compare with the expression that defines J2 in terms of the
external expansion of the field:
GMa 2
J 2 P2
r3
3
⇒ J2 = − ε
5
V2 P2 = −
(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 ω2s2/2 = ω2r2sin2θ/2 =
ω2r2(1-P2)/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 P2 this implies that:
GM
GM
1
−
J 2 P2 + ω 2 a 2 [1 − P2 ]
r0 (1 + ε P2 )
a
3
(11.17)
must have no dependence on P2. If it had a dependence on P2 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 (J2 and ε are small parameters). Also, ε<<1 means that
1/(1+εP2) = 1-εP2 to an excellent approximation. In other words,
GM
1
[−ε − J 2 ] − ω 2 a 2 }P2 ≡ 0
a
3
q
⇒ −ε − J 2 − = 0
3
{
(11.18)
where q is a dimensionless measure of planetary rotation:
q≡
ω 2a3
GM
(11.19)
But we already have J2=-3ε/5. Substituting, we get
q
5q
J 2 = ; ε =2
6
(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
11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
∞
J 2 = ∑ Λ 2,n q n + 
118
(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 = −0.536 .
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 1⎞
Λ 2,0 = ⎜ 2 − ⎟ = 0.173
⎝π
3⎠
(11.22)
for the case of the model we studied for Jupiter where P=Kρ2. This is very different from
0.5.
11.2.2 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 / d ρ ')(d ρ ' / ρ ')
(11.23)
ρs
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.
ρ(r')d 3r '
U ≡ W + G∫
r − r'
(11.24)
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=r0(1+ε2(r)P2+...)
and derives a integro-differential equation for the function ε2(r). To make a long story
short (look at Hubbard’s book) it turns out that in place of the simple previous result
(Λ2,0=0.5 for C/Ma2 =0.4) , we obtain a more general result
C
2
2
5
= {1 − [
− 1]1/2 }
2
Ma
3
5 3Λ 2,0 + 1
(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,
Λ2,0=J2/q. Here are some values that this formula predicts:
11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
119
Table 11.1
Λ 2,0
0.50
0.45
0.40
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0.0
C/Ma2
0.4
0.383
0.366
0.347
0.326
0.303
0.278
0.249
0.217
0.179
0.133
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.
11.3 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 RadauDarwin 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∝ρ2.)
In addition, for satellites we must correct for the tidal effect on J2. 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 J2,r the rotational part of J2 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).
11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
120
Table 11.2
Body
Mercury
Venus
Earth
Moon
Mars
Jupiter
Saturn
Uranus
Neptune
Io1
Europa1
Ganymede1
Callisto1,2
Titan1,2
Enceladus1,2
Measured J2 or
J2,r
(8±6)x10-5
(6±3)x10-6
1.0826x10-3
2.024x10-4
1.959x10-3
1.4733x10-2
1.646x10-2
3.352x10-3
3.538x10-3
0.745x10-3
0.175x10-3
0.051x10-3
0.013x10-3
0.013x10-3
2.09x10-3
Measured q
J2,r/q
1x10-6
6.1x10-8
3.5x10-3
7.6X10-6
4.6x10-3
0.089
0.153
0.035
0.028
1.712x10-3
0.505x10-3
0.190x10-3
0.037x10-3
0.040x10-3
6.26x10-3
~1
~102
0.31
~30
0.43
0.166
0.107
0.096
0.125
0.435
0.346
0.268
0.35
0.335
0.333
Inferred
C/Ma2
*
*
0.33
*
0.375
~0.25†
~0.23†
~0.20†
~0.22†
0.375
0.346
0.311
0.35
0.34
0.335
*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 J2 but the results of this
theory are not expressible in the form of a solution for C/Ma2. 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 Λ2,0=0.173 for P∝ρ2. J4 and even J6 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 J2 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 J2,r , the rotational effect on J2 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 ,1185-1191
11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
121
11.4 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/MR2 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
1
 
I = { ∫ [(x 2 + y 2 ) + (x 2 + z 2 ) + (y 2 + z 2 )]ρ(r )d 3r}
3
R
2
= ∫ 4π r 4 ρ(r)dr
30
(11.26)
and we can proceed to evaluate for interior models and compare with observations. There
are two particularly interesting cases to consider:
11.4.1 The Simple Two-Layer Model
Assume ρ(r) = Aρ0 for 0 < r < xR and ρ(r) = ρ0 for xR < r < R. Then
I
2 [Ax 5 + (1 − x 5 )]
=
.
MR 2 5 [Ax 3 + (1 − x 3 )]
(11.27)
Of course, this should be combined with our knowledge of the observed mean density
ρav = ρo[Ax3 +(1-x3)]
(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).
11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
122
Figure 11.3
Ganymede is about the lowest I/MR2 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/MR2 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/MR2 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. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
123
Notice that there are two values of x for a given I/MR2, 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.
11.4.2 The P=Kρ 2 Model (for Giant Planets)
For this we have ρ(r) = ρcsin(kr)/kr from which one gets:
1
∫x
I
2
= . 01
2
MR
3
3
sin(π x)dx
∫ x sin(π x)dx
2
6
= [1 − 2 ] ≈ 0.26
3
π
(11.29)
0
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/MR2 ~0.26(1-Mcore/Mtotal). This comes from including the cos(kr)/kr term in the density
profile (see earlier chapter). Since I/MR2 is somewhat smaller for Saturn than for Jupiter,
it is likely that Saturn has a core.
Table 11.3 Observational Moments of Inertia
Mercury5
Observed
I/MR2
0.346
Earth1
0.3308
Moon1
0.391
Mars1
0.366
Jupiter
Io2
Europa2
0.25
0.378
0.346
Large iron core, thin mantle
Large core, but also action of pressure on
constituents
Does not demand an iron core (but there
probably is one; maybe 500km in radius)
Requires iron core, slightly less mass
fraction than for Earth
Need for dense core is marginal
Substantial Fe core
H2O layer on rock
Ganymede2
Callisto2,3
0.3105
Ice mantle on rock
0.359±0.005
Saturn
0.24
Titan2,3
0.33-0.34
Partially differentiated
Smaller value than Jupiter suggests
presence of dense core
Either partially differentiated or fully
differentiated but a low density core (`2.5
g/cc)
Body
Comments
11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
Enceladus2,3
0.33-0.34
Uranus4
~0.20
Neptune4
~0.22
124
Fully differentiated with a low density core
Interior much denser, but interpretation
non-unique
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 278 1749,1997).
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. C22 and J2 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 J2 and J4 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: 10.1029/2012JE004161
11.4.3 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/MR2 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 J2?
Solution: If the mass is overwhelmingly in a core of radius ac then
(modifying the derivation of p114) we get V2 ∝ Mac2.εc where εc is the
distortion of the core only. Then from the definition of J2 we have J2 = 3(ac/a)2εc/5. From consideration of the equipotential at the core radius
(ignoring the negligible mass external to it), we get - εc -J2.(a/ac)2 –
(ac/a)3q/3 =0, where q is defined for the entire planet. This implies that εc =
-5(ac/a)3q/6. (This result is independent of a, as it should be). From
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 -J2 –q/3=0. Putting this together, we get ε = [1/3 +1/2(ac/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 J2 can vary enormously. For example, if we
choose ac/a =0.7 then the magnitude of ε is reduced by a factor of two
(relative to the value at ac/a =1) but J2 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* J2 to consist
of two parts, i.e. J2=J2h+J2nh, respectively the hydrostatic and non-hydrostatic
parts.
[*Strictly speaking, they don’t measure J2 (they actually measure a combination
of J2 and C22) 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/Ma2) as Ganymede
(and assuming Ganymede is fully hydrostatic), estimate J2nh.
(b) What size mass anomaly on the icy surface of Callisto would be required to
explain J2nh? 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 J2nh? (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 J2 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?
11. The Gravity Field and Gravitational Response to Rotation: Moments of Inertia
126
Solution: From p117, J2/q = 26.6 for the current Moon. The equilibrium
expected value for this almost constant density body is J2/q = 1/2.
Accordingly, we must go to an epoch when the square of the angular
velocity was larger by (26.6)/(1/2)= 53.2. 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/cm3, the intermediate ice layer has density 2.5 g/cm3 and the
outer “gas” envelope has density that declines linearly with radius from ρgas,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/cm3 and moment of inertia I/MR2 =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 ρgas,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 ρgas,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.