Recent observations of a stellar occultation by Neptune give an oblateness of 0.022 + or - 0.004 for Neptune's atmosphere at the 1-microbar pressure level. This results is consistent with hydrostatic equilibrium at a uniform atmospheric rotation period of 15 hours, although the error bars on quantities used in the calculation are such that an 18-hour period is not excluded. The oblateness of a planetary atmosphere is determined from stellar occultations by measuring the times at which a specified point on immersion or emersion occultation profiles is reached. Whether this standard procedure for deriving the shape of the atmosphere is consistent with what is known about vertical and horizontal temperature gradients in Neptune's atmosphere is evaluated. The nature of the constraint placed on the interior mass distribution by an oblateness determined in this manner is consided, as is the effects of possible differential rotation. A 15-hour Neptune internal mass distribution is approximately homologous to Uranus', but an 18-hour period is not. The implications for Neptune's interior structure if its body rotation period is actually 18 hours are discussed

Hubbard, W. B.

English

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N87-17644
ON THE OBLATENESS AND ROTATION RATE OF NEPTUNE'S ATMOSPHERE
W. B. Hubbard
Lunar and Planetary Laboratory, University of Arizona
Recent observations of a stellar occultation by Neptune (Hubbard et
al., 1985, Astron. J. 90, 655-667) give an oblateness of 0.022 ± 0.004
for Neptune's atmosphere at the l-microbar pressure level. This
result is consistent with hydrostatic equilibrium at a uniform
atmospheric rotation period of 15 hours, although the error bars on
quantities used in the calculation are such that an 18-hour period is
not excluded. The oblateness of a planetary atmosphere is determined
from stellar occultations by measuring the times at which a specified
point on immersion or emersion occultation profiles is reached. We
critically evaluate whether this standard procedure for deriving the
shape of the atmosphere is consistent with what we know about vertical
and horizontal temperature gradients in Neptune's atmosphere. We then
consider the nature of the constraint placed on the interior mass
distribution by an oblateness determined in this manner, considering
the effects of possible differential rotation. A 15-hour Neptune
internal mass distribution is approximately homologous to Uranus', but
an 18-hour period is not. We discuss the remarkable implications for
Neptune's interior structure if its body rotation period is actually
18 hours.
This morning, you heard Ellis Miner talking about two different values for the
rotation period of Neptune. Some questions came up about which one was cor-
rect. The purpose of this talk is to try to elucidate that matter and to
attempt to convince you that there is some possibility that maybe both are
correct.
The first point I want to make is the reason that I, at least, would like to
know the rotation period of Neptune is because of what it tells us about the
interior. If we expand the external gravity potential V e of the planet in the
usual form where 8 is the colatitude and J2 is the second-degree zonal
harmonic:
V e = (GM/r) [I - J2 (a/r) 2 P2(cos 8)] , (1)
where a is the equatorial radius; then for a rotating body in hydrostatic
equilibrium, J2 is going to be proportional to a small parameter given by
q = _2a3/GM , (2)
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https://ntrs.nasa.gov/search.jsp?R=19870008211 2020-03-20T12:36:16+00:00Z
where _ is the angular rotation rate, G is the gravitational constant, and M
is the mass. We express the proportionality in the form
J2 = A2 q (3)
to lowest order in q. A 2 is the response coefficient. For the Jovian plan-
ets, we can look at these values which are known for Jupiter and Saturn,
compare them with the homogeneous case, and we see that these planets are
centrally condensed. The more centrally condensed they are, the smaller A2.
For a slxteen-hour period of Uranus, which is sort of becoming a consensus
value (although there may still be some holdouts), we obtain for Uranus,
A2 = 0.09.
Now the 18-hour period for Neptune, which was discussed this morning, and
which comes from looking at cloud features going around on Neptune_ gives us
A2 = 0.20, derived from the fact that the J2 of Neptune is 4 x 10-0 • Now
A2 = 0.20 means that Neptune is less centrally condensed than any of the
Jovian planets, and that leads to very weird interior models. For example,
one may have to consider models that are composed entirely of water. Morris
Podolak recently completed a study with Ray Reynolds and Rich Young, and he
might want to say a word or two about this.
There is another way to measure the rotation period, and this is where the
atmosphere comes in. One can look at the shape of the planet's surface, and
the way one does this is to assume that the atmosphere defines an equipoten-
tial surface. Then we calculate the potential at the pole of the planet where
b is the polar radius, and we equate that to the full corotating potential at
the equator, where we add the centrifugal potential to expression (I). We
then equate these potentials, use that to calculate the relative difference in
the equatorial and polar radius, and that then can be expressed in terms of J2
and this rotation parameter q, which can also be expressed in terms of these
other variables as previously described:
e = (a-b)/a = (3/2)J 2 + (I/2)q
= q(3A 2 + 1)/2
= J2(3 + A2-I)/2 . (4)
In 1983, we had a very good stellar occultation by Neptune, which was observed
by our group, and we obtained data from eight stations over a rather large
baseline (Hubbard et al., 1985). We were able to observe from extremely
southern locations as well as extremely northern locations. The timings were
used to define the equlpotential surface. There was a very measurable differ-
ence between a spherical and oblate object. Thus, we were able to deduce the
oblateness. One of the difficulties that comes in when one is trying to
measure oblateness is that the planets with atmospheres have fuzzy edges.
265
Weget a lot of scattering of light due to refractive focusing and defocusing.
It's difficult to say where a particular pressure level occurs, but there is a
standard procedure for doing this; namely, we look for the half intensity point
as obtained by fitting an isothermal light curve (i.e., a theoretical light
curve computedfor a strictly isothermal atmosphere) to the observed data.
That seems to work reasonably well, although I should point out that you see
variations from one station to another which don't reproduce. So, in detail,
the planet is not globally layered, at least beyond somelimit. The immersion
and emersion profiles are grossly similar: we have non-correlation in detail
although the overall structure is similar.
If we considerably idealize the atmosphere, in terms of a perfectly isothermal
quiet atmosphere, then the rays of light coming from the star to the observer
will fall in a pattern such that the angle through which Neptune bends each
ray will increase exponentially with depth of penetration of the ray into
Neptune's atmosphere. At the ideal half intensity point, the linear deflec-
tion of the ray at the Earth is precisely equal to the scale height H, which
in the case of Neptune is about 50 km. Nowthe argument in trying to say how
accurately wecan define the edge of the planet, goes as follows: Wehave
density fluctuations (with respect to the ideal isothermal atmosphere) which
are perturbing the rays so that they actually don't fall in the idealized
path, but they, in fact, sometimes converge, producing a spike. At other
times they maydiverge by an extra amount. But the argument is that any extra
bending that they will suffer due to imperfections in the isothermal atmos-
phere will be small comparedwith the value of the overall bending angle. The
reason for that is that we assumethat the density fluctuations are small com-
pared to the background density, which is certainly quite reasonable. If that
argument is true, then any residuals that we get in our fit to the overall
profile of the atmosphere should be considerably less than the scale height.
That is the type of accuracy for which we aimed in fitting the solution. With
suitable discarding of anomalousdata points, we succeeded.
Someof the data were affected by knownsystematic problems, and indeed it was
those stations that turned out to have the largest residuals. On the ingress
side, we got residuals less than a kilometer at our Taiwan station, and at our
Hobart station it was again less than a kilometer (a very nice fit to the best
fit solution). A portable station gave a 37 km residual and was thrown out in
the final solution, though I could mention that we did include it in one solu-
tion, which gave a slightly shorter rotation period. On the emersion side,
there was a measurableseparation between the spherical profile and the oblate
profile. The residual was -I km at ChungLi, Taiwan. The Guamstation was
thrown out. The overall fit had all retained stations with residuals of sub-
stantially less than 50 km. The results of this solution I'll give to you in a
moment. Let mealso mention that as a by-product of this analysis, we obtained
the temperature at the occultation level, which was computedby fitting an
isothermal light curve to the data, and obtaining the scale height. The scale
height was then corrected for the variable gravity, due to the oblateness of
the planet. Finally, the temperature at the one microbar level was deduced as
a function of the latitude. What we find is in general not really any indica-
tion of a large temperature gradient as a function of latitude. In fact, at
this level in the atmosphere it looks like the temperature is reasonably
constant, both with latitude and with time. The average value turns out to be
266
about 155 K. So if that is the case, then that is further support for the
model that we are assuming, that we are dealing with a surface that is at a
constant potential and a constant density.
Wehad to fit not only to the oblateness of course, but also solve simultane-
ously for the equatorial radius as well as the center of figure. Though we
have a very nicely defined minimumin the rms residuals, because of the large
number of the parameters that we had to vary, the probability that the solu-
tion actually lay within a narrow interval in the oblateness e is still rather
small. But the deducedvalue is e = 0.022 ± 0.004, which is an improved re-
suit over the one that I reported at the Uranus/Neptune meeting (which led to
a 13.6 hour period, which was mentioned by Ellis Miner this morning). What we
did in this analysis was to use Harris' (1984) improved pole position for
Neptune. That leads to a reasonable value for A2: 0.12, and a period P = 15h
(+3h, -2h). However, notice that becauseof the large error bars we still
can't rule out an 18-hour period. Nevertheless, as I mentioned earlier, I
don't like the 18-hour period.
Here is the model for the atmospheric dynamicist to figure out. Suppose that
we had differential rotation on cylinders, and suppose that the bulk of the
planet is actually rotating with a period consistent with the models and the
J2 value (say, P = 15h). Supposethat we have an equatorial zone which is in
differential rotation, but unlike Jupiter and Saturn, it's going backwards.
In other words, it's rotating at a slower rate of, say, 18 hours. Wehave
spots on it, so that's what we see going around. Wewant the q of the planet
as a whole to be 0.033, of course, the value corresponding to a 15-hour
rotation period. The oblateness is nowgoing to be given by this formula:
e = (3/2)J 2 + (I/2)q 0 + (1/2) Aq , (5)
where qo = q(15h) is the q corresponding to the deep interior (about 0.033),
and Aq is a correction due to the differentially rotating outer layer. But
we want this oblateness to remain almost the same as the one we just got [eq.
(4)]. Thus we want Aq to be much smaller than q0" It turns out that if you
do the calculation, assuming the planet is rotating on cylinders so that you
can derive the centrifugal force from a potential, then the correction Aq/q0
is just given by the formula:
Aq/q0 = -0.31 cos2e0 , (6)
where e0 is the colatitude where the bounding cylinder pierces the surface, the
bounding cylinder being the one which divides the inner region which rotates
with a period of 15h from the outer region which rotates with a period of 18h.
Thus, by adjusting the parameters an appropriate amount (for example, taking e0
to be 60 deg which gives you a big band in latitude of ± 30 deg to have spots
in), you can make Aq = -0.076 q0, less than a tenth of q0" Thus one can con-
struct a model which would do the job, i.e., reconcile a 15 h deep-interlor
period with an 18h equatorial-atmospheric period. But as to whether that seems
plausible or not, I would have to refer you to a dynamicist.
267
REFERENCES
Harris, A. W. (1984). Physical properties of Neptune and Triton inferred from
the orbit of Triton. In Uranus and Neptune (Ed. J. T. Bergstralh),
pp. 357-373. NASA CP 2330.
Hubbard, W. B., H. P. Avey, B. Carter, J. Frecker, H. H. Fu, J.-A. Gehrels,
T. Gehrels, D. M. Hunten, H. D. Kennedy, L. A. Lebofsky, K. Mottram, T.
Murphy, A. Nielsen, A. A. Page, H. J. Reitsema, B. A. Smith, D. J. Tholen,
B. Varnes, F. Vilas, M. D. Waterworth, H. H. Wu, B. Zellner (1985).
Results from observations of the 15 June 1983 Occultation by the Neptune
System. Astron. J. 90, 655-667.
DR. PODOLAK: l'd like to pick up on the speaker's first statements. In order
to match models to Uranus with a 16-hour period you need very much different
models from those required to match an 18-hour period for Neptune. The amount
of ice necessary is not so unreasonable, but the distribution differences
between Uranus and Neptune are great.
Briefly, there are three types of material you have to worry about: things
that are always solid in the solar nebula (like things we call rock), things
that are always gaseous like hydrogen and helium, things that could be solid
or gaseous depending on the temperature that we call ice (it's just a generic
name not meaning the stuff is solid and frozen) and that would be water, meth-
ane, ammonia or similar substances. Then if you try to construct models made
up of those three materials that match Uranus with a 16-hour period and
Neptune with an 18-hour period, what you find is that the amount of rock that
you put in the core is similar for both planets. It's a little bit more for
Neptune because Neptune has a slightly higher density. The amount of ice that
you put in the planets is similar, and in fact the ice-to-rock ratio is about
three, which is exactly what you'd expect for solar compositions. All that is
very nice except that for Uranus, most of the ice sits in a shell around the
core (it's very centrally condensed like Bill said). For Neptune, most of the
ice has to spread out throughout the envelope because it's rotating so much
more slowly that you still want to have a high moment of inertia. The problem
is, why do these things have such different structure? Just to give you a
feeling for how strange this is, if you were to say "O.K., suppose the stuff
on Uranus just fell down and the stuff on Neptune is going to fall down in
another couple of years." It turns out that the amount of gravitational ener-
gy you release is so large that even with an effective temperature of about
I00 K, it would still take two billion years to radiate away. You're talking
about a really substantial amount of energy and a really big difference in
structure. It is hard to see how that came about. That is the reason that I
would like a 15-hour period for Neptune too.
DR. ORTON: Does anyone else want to reconstruct the solar system?
DR. ALLISON: Bill, do you think that there is a realistic prospect of even-
tually refining these occulatation measurements to the point where they could
be used as a discriminant between models of rotation on cylinders versus thin-
layer models?
268
DR. HUBBARD:I think that it could be done if we can find a way to pin down
the location of the center of Neptune accurately. In the case of Uranus, we
have the advantage of the rings. They don't have atmospheres, so you can
determine excatly where they are. Nowif we can get enough observations of a
Neptunian ring, or portions of a Neptunlan ring to do the same, then yes, in
the long run it should be possible to do the samething. In that case, I
believe that enough precision would be possible to check this point out.
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W. BOOR QUALm 
270 


On the oblateness and rotation rate of Neptune's atmosphere

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