Determination of the average density of Jupiter from the
characteristics of the orbits of its moons indicates that Jupiter must
be largely composed of gas rather than of rock, like the terrestrial
planets.
We can use Newton’s law of gravitation and our knowledge of the
behavior of materials under pressure to get a good idea of the
interior structure of Jupiter, even though we are unable to observe
the interior directly.
If we know the “equation of state” – the pressure a gas exerts at
any given density and temperature – then we can build a model of
a planet like Jupiter as a spherical object supported against its own
gravity by the pressure its gaseous components exert.
Assuming a cold interior (cold in that we believe there are no
nuclear reactions going on, as there are in the sun) allows us to
build an approximate spherical model of Jupiter, because the
behavior of most constituents at zero temperature is believed to be
known from theory.
Even if we did not have this theory available, we could use
information about the density and temperature of various materials
at high pressures that we can obtain on the earth. We can obtain
these measurements, for example, by driving the materials to high
pressures by subjecting them to shock waves from high explosives.
Our knowledge of material properties is therefore gained from a
combination of theoretical calculation and from laboratory
experiments.
This knowledge tells us how much pressure a given mix of
materials will exert when compressed to a particular density and
maintained at a particular temperature.
At relatively low temperatures, the pressure is determined mainly
by the density of the material and does not depend much at all
upon the particular value of the temperature.
From Newton’s law of gravitation, we can determine the
gravitational force at any particular distance from the center of a
spherical planet, even at a distance that is inside the planet.
This gravitational force must be balanced by the pressure of the
material making up the planet at that radius (i.e. at that distance
from the center of the planet).
Balancing the gravitational and pressure forces at each distance
from the planet’s center, and using our knowledge of the material
density we require to generate the pressure, we can build a model
to tell us how the density of the material must vary with distance
from the center of the planet.
We do this on a computer, and it takes about one second.
For each material composition we choose for our model of the
planet, and for each planet mass we might obtain from
“weighing” the planet by observing the orbital motion of its
moons, we get a run of density with radius.
The density is largest in the planet’s core, and we find that at a
large enough distance from the core the density becomes zero.
This is the surface of our model planet.
Therefore:
1. given a material composition,
2. given a mass for the planet, and, of course,
3. given the assumption that the temperature is everywhere in the
range of cold values for which the pressure is insensitive to
temperature,
we can compute a radius for the model of the planet.
The result of such model building is the set of mass-radius curves
on the next slide.
Points J, S, U, and N on this plot indicate the observed data for the
planets Jupiter, Saturn, Uranus, and Neptune.
The point J, representing Jupiter, lies very close to the mass-radius
relations for cold planets constructed from pure hydrogen (the
top curve) and for hydrogen-helium mixture (next curve down,
labeled x=0.25) with 25% helium by mass.
We may therefore conclude that Jupiter is largely hydrogen and
helium, not only near its surface but in the interior as well.
We also note that the point S, representing Saturn, lies right on the
curve for cold planets consisting of hydrogen-helium mixtures
with 25% helium by mass.
Mass-radius curves for
objects of various compositions at zero temperature.
Curve labeled X = 0.25 is
for an approximately solar
mixture of hydrogen and
helium. Points J, S, U, and
N represent the outer
planets. Radius is in units
of hundredths of a solar
radius (1 R= 6.96×105 km)
and mass is in units of solar
masses (1 M = 1.99×1011 g).
Jupiter and Saturn are
clearly composed
predominantly of hydrogen
and helium; Uranus and
Neptune must have a large
complement of heavier
elements.
(from Encyclopedia of the Solar System)
It is interesting to note that the figure indicates that if Jupiter were
much more massive, it would actually be smaller!
This strange result can be true because the gases that make up
planets like Jupiter are compressible, just like the pillows in the
illustration from your text book that appears on the next slide.
Jupiter has enormous pressures in its interior which compress
hydrogen into a liquid metallic state.
No one has observed liquid metallic hydrogen in the laboratory.
Instead, we believe it exists under the conditions in Jupiter’s
interior as a result of theoretical calculations done on
computers.
Liquid metals should not be unfamiliar to you. Mercury is a metal
that is a liquid at room temperature.
Ingredients for an interior structure model of a giant planet:
1. Total mass of the planet.
2. Chemical make-up of the planet.
3. Equation of state of the planet’s material.
4. Sources of heat within the planet.
5. Rate of heat radiation from surface into space.
We will simplify the problem by the following assumptions,
which are not too far off the mark for Jupiter and Saturn:
1. Planet is all hydrogen.
2. No heat sources inside the planet.
3. No heat loss from the surface.
The method we will use to determine the proper locations of the
mass shells in our model for Jupiter’s interior is a balance at the
edges where the mass shells meet between
1. The force of gravity pulling inward and
2. Pressure forces pushing outward
On the next slide a sequence of mass shells are indicated, going
outward from the center of Jupiter, located far to the left of the
region shown in the diagram, toward the surface, located well
beyond the far right in the diagram.
The average pressure in each shell, which is determined from the
density alone, is indicated by a horizontal dashed line.
We can estimate the strength of the outward pressure force at the
interface between any 2 mass shells by seeing how much
stronger the pressure is in the inner shell relative to that in the
outer of the 2 shells.
The method we will use to build a model of a planet like Jupiter is
to assume that it is perfectly spherical, and then to divide it up
into a series of concentric, spherical shells.
If we have lots of these shells in our model, then we can
approximate the conditions within each one as essentially
uniform.
With uniform conditions in each shell, uniform density, pressure,
and temperature, we can easily calculate the mass of each shell
from its volume and density.
The net pressure force acting outward on each shell will be the
difference between the pressure of the shell just inside it and the
pressure of the shell just outside it.
The gravitational force pulling inward on each shell will be the
same as the gravitational force we would get if all the mass of
all the shells inside it were concentrated at the center of the
planet.
This last point about gravity is a very big deal.
The famous mathematician Gauss proved that with Newton’s law of
gravity a spherical shell of material of constant density
everywhere within the shell exerts a gravitational force on
masses outside the shell that is identical to the force it would
exert if all the mass in the shell were instead concentrated at the
point at its center.
Gauss also proved that the gravitational force from this shell is
precisely zero on any mass located inside the shell’s inner
radius.
These two astounding facts make it very easy for us to compute the
gravitational force at the interface between any two of our
concentric shells making up our model of Jupiter.
There is no force from shells outside this interface, and the force
from all the shells inside this radius is obtained by concentrating
all that mass at the center of the model planet.
The diagram on the next slide shows a few of the mass shells that
make up our model Jupiter.
At the inner and outer edges of each shell, we can compute the
gravitational force generated by all the shells inside this radius.
That force is indicated by the blue arrows pointing inward (to the left
in the diagram).
By comparing the pressures in the shells on either side of each
interface between shells, we can get the net pressure force outward
there.
These pressure forces are indicated by the red arrows in the diagram.
What the computer does for us in building this model is to find a set of
radii for the shell interfaces that make the two opposing forces,
gravity and pressure, balance precisely, so that our shells will not
move either in or out. The result is a model of Jupiter that gets
better and better as we use more and more shells, each with less
and less mass inside it but all adding up to the mass we have
measured for the planet by observing the orbits of its moons.
How the computer program works:
The following slide sequence on how the computer program works
in detail are for the intensely curious student.
Less curious students need only understand that we build our
model by balancing the outward force of pressure with the
inward force of gravity at the interfaces between
homogeneous, spherical, concentric shells of material.
We must know the material composition, and assume that it is
relatively cold, so that its pressure depends mainly upon its
density alone. For each composition, given the total mass of
the planet, we get a radius for the planet and an internal
structure.
We can establish force balance at the two interfaces we have been
discussing by moving the inner one of them inward and moving
the outer one outward. This is shown on the following slide.
The new interface positions are indicated by the red vertical lines.
When moving the 2 interfaces shown to their new positions (the red
vertical lines), we must keep the mass unchanged in each of the
mass shells affected by this motion.
This means that the first mass shell, which is now thinner, has a
smaller volume holding the same, original mass.
Therefore its density, and hence its pressure, is now higher.
The second mass shell, which has become thicker, now has a larger
volume to hold its same, original mass.
Therefore its density, and hence its pressure, is now lower.
Similar changes must occur in the third and fourth mass shells in the
diagram.
The outward motion of the interface between these shells causes the
pressure in the third mass shell to decrease and that in the fourth
one to increase.
These changes are indicated in the following diagram.
The result of the motion of the 2 interfaces between the mass shells
is shown in the following diagram.
At the left, where the interface was displaced inward, the density
and pressure difference between the inner shell and the one just
outside it has increased. Thus the slope of the blue line joining
the pressures at the mass shell centers has increased, indicating a
stronger outward pressure force at the interface.
The inward motion of this interface at the left has caused the
gravitational force there to increase a bit, since the distance from
the interface to the center of Jupiter has been reduced slightly.
At the right, where the interface was displaced outward, the density
and pressure difference between the inner and outer shells
adjacent to this interface has been diminished. Thus the slope of
the blue line joining the pressures at the centers of these mass
shells is much smaller. The pressure force at this interface has
therefore been greatly diminished, and now actually acts very
weakly inward.
The results of moving the mass shell interfaces, discussed and
illustrated above, suggest a way in which we can refine any
initial guess that we might make for our model of the interior of
Jupiter.
This initial model is determined by dividing up the total mass of
Jupiter among a series of concentric, spherical mass shells.
The initial densities, and hence the pressures, in these shells is
simply guessed in some reasonable way.
The result could be the plot of density versus radius that appears on
the following slide.
First we plot our initial guess.
From this initial data, we can compute the estimated strength of
gravity pulling inward at each mass shell interface and also the
strength of the pressure force pushing outward there.
In general, these will not be in balance.
If we move one interface while keeping all the others fixed, it is a
simple matter to find the new location at which the gravity and
pressure forces are in balance.
Finding this location of force balance would be much harder if we
allowed all the interfaces to move at once.
If we fix the locations of every second mass shell interface, then it
is simple to find the locations of the other half of the interfaces
at which the two opposing forces are in balance.
This is illustrated on the next 2 slides.
Now we compute the gravitational and
pressure forces on the inner surfaces
of the even numbered mass shells,
at the points indicated by the
arrows.
We now allow these surfaces to move until the gravitational
and pressure forces on them balance.
We mark the original positions in this
and the next slide by the
white circles.
Surfaces allowed to
move are indicated
here by the white
arrows.
In this case, balance is achieved by reducing the pressure
gradient at each shell surface that is allowed
to move (except for the first one).
This is evident from the flatter
slopes of the lines
joining the new
shell centers
Slopes of the lines joining
for these
the mass shell centers
interfaces.
indicate the strength of the
outward pressure force.
This is accomplished by moving each shell surface slightly
outward. (Only shell surface #2 moves inward.)
This can be seen by comparing the new
locations of the shell centers with
When the density
the white circles marking
of one mass shell is only
their original
slightly larger than that
locations.
of the shell just outside it,
then the outward pressure
from the inner shell exceeds
only slightly the inward
pressure from the outer shell.
This procedure has fixed our force imbalances at only half of the
mass shell interfaces.
At the other interfaces, it might have even made force imbalances
worse than before.
If we now hold fixed the positions of the inner edges of the even
numbered mass shells, which we previously allowed to move,
then we can readjust the location of the other half of the
interfaces to achieve force balance there.
This process is illustrated in the next sequence of slides.
We now reset the positions of the circles
on our diagram, and then we allow
the odd numbered shell surfaces
to move.
Force balance is once again accomplished
by a slight movement of the shell
surfaces outward. Again,
the pressure
gradients are
reduced by
this motion.
This procedure has fixed our force imbalances once again at half of
the mass shell interfaces.
However, this has happened at the cost of creating new force
imbalances at the other half of the interfaces.
One could imagine a back-and-forth process, fixing first one half
and then the other of these interfaces and never improving the
global picture of force balance.
However, our first set of adjustments moved half the interfaces
outward, and our second set did the same thing to the other half.
What we have accomplished by creating force balance at one set of
interfaces and then the other is to move all the interfaces
outward.
This has reduced the densities in all the mass shells and has
increased the radius of our model Jupiter.
Doing this 10 more times has a very visible effect, shown on the
next slide.
After 10 iterations of this type, where
in each iteration alternate shell
surfaces move outward,
the shell radii and
densities are
as shown
here.
After 100 more such iterations,
the shell radii and
densities are
as shown
here.
After 20,000 more such iterations,
the shell radii and densities are
as shown here. Clearly,
the model interior
structure is
settling.
After 50,000 more such iterations,
the shell radii and densities are
as shown here. There has
been hardly any further
adjustment.
Here we have rescaled our diagram,
so that all 10 shells fit onto
the display.
Now, to get a more accurate representation
of the interior structure, we have
divided each of our shells
into 2.
After 100,000 iterations on this refined
set of mass shells, we obtain
the interior structure
shown here.
We now move the circles on our diagram,
so that we can see what improvements
we obtain from dividing each of
these 20 mass shells into 2.
Just as with our previous refinement, the largest
adjustments are near the center.
And of course we see a
bit more of the outermost structure.
We readjust the circles on the diagram one last
time, in order to see what changes one
final refinement of our set of
mass shells might
bring.
We are still refining our representation of the
deep interior, but the rest of our model
is no longer changing as we use
more and more mass shells
of less and less mass
each.
We readjust our first 2 circles, and refine once more.
After 200,000 iterations with these 160 mass shells,
we obtain the interior structure shown by the
smooth curve. We can see from this
curve that our representation
with only 80 mass shells
was already quite
sufficient.
Now we double the mass of the planet and go through
the entire iteration and refinement process
to arrive at the smooth curve shown here.
The central density has increased by
a factor of 4, and the radius has
actually decreased.
We double the mass of the planet again & go through
the entire iteration and refinement process
to arrive at the smooth curve shown here.
The central density has increased by
another factor of 4, and the radius
has decreased still further.
When we double the mass of the planet
again, something very different
happens. The dense
interior condenses
into a more densely
packed form of
hydrogen, a
liquid metal.
Actually, we have made up our pressuredensity relationship, and therefore
this is not real hydrogen.
But the behavior is
similar. At a certain
high density and
pressure, the gas
condenses into a
liquid metal.
The liquid metal phase is much denser
(at the same pressure) and is
much stiffer (that is, it
resists further
compression
more strongly
than does
the gas).
The “phase diagram” on the next slide divides a pressuretemperature plane into regions in which hydrogen is either
liquid or solid, molecular or metallic.
The interiors of models of the outer planets trace lines in this phase
diagram, where progressing along one of these lines
corresponds to moving in or out in radius within the planet.
Uranus and Neptune do not reach the metallic region of this phase
diagram, but both Jupiter and Saturn do.
The shaded area, which is crossed by the line representing the
Saturn interior model, indicates a region of pressure and
temperature in which it is believed that any helium mixed in
with the hydrogen would naturally separate out.
It is believed that such separation of helium from hydrogen is
happening inside Saturn, with the helium falling toward the
center of the planet, and releasing gravitational potential
energy in the form of heat as it does so.
(from Encyclopedia of the Solar System)
(from Encyclopedia of the Solar System)
To get a better idea of the interior structure of giant planets like
Jupiter, we can take advantage of their rapid rotation.
If Jupiter were not rotating, its gravity would force it to assume a
spherical shape.
A spherical planet exerts a gravitational force on a satellite just as
if all its mass were concentrated at a single point at its center
(don’t worry why this is true, but it is).
The orbit of a satellite can then tell us the mass of the planet, but it
cannot tell us anything about the distribution of the mass with
radius within the planet.
Through the centrifugal force, rapid rotation causes a fluid planet
like Jupiter or Saturn to become distorted.
Then the gravitational forces felt by a spacecraft orbiting the planet
can reveal details of the internal distribution of mass in the planet.
Figure 11.3 (a) Gravity alone makes a planet spherical, but rapid
rotation flattens out the spherical shape by flinging material near
the equator outward. (b) Saturn is clearly not spherical.
Jupiter is not actually a perfectly spherical object, as we have
assumed in building our model of its interior.
It is not spherical because it is rotating rapidly, and the centrifugal
force acts to fling the material near Jupiter’s equator outward,
giving the planet a slightly oblate aspect.
This phenomenon is more noticeable for Saturn, which is roughly
the same radius as Jupiter and spinning at about the same rate,
but which has only about a third of Jupiter’s mass.
Thus the centrifugal acceleration outward of the material near
Saturn’s equator is about the same as for Jupiter, but the
gravitational acceleration inward is only about a third as large.
The result is that Saturn is significantly more visibly oblate than
Jupiter.