Solar System Formation & Dynamics

Nothing on the exam is closely related to the problems in homework #2. However, it may be useful to review some basic points touched on in questions (2) and (3), as described below…

Question 2: you were asked to estimate the rotation period of the Sun if the angular momentum in Jupiter’s orbit were suddenly given to the Sun. To do this, we first note that the angular momentum,

where v is the tangential (i.e. orbital) velocity. For a Keplerian orbit,

so for the angular momentum content in the Jovian orbit we have,

.

Let’s suppose that the Sun is not spinning at all (not true, but a reasonable approximation that we can justify after the fact once we show that adding Jupiter’s angular momentum spins up the Sun tremendously). How fast will it spin if we put of angular momentum into it? There are various ways to estimate this. One of the simplest is to note that for a rotating body, the angular momentum is,

where is the rotation frequency and the moment of inertia. A uniform density sphere (which is not a good approximation for the Sun, but suffices for a quick estimate) has,

.

Equating the angular momentum of Jupiter’s orbit with the expression for the angular momentum of a spinning solid sphere gives us an estimate for the spin frequency, which we can then convert into a spin period if we want via,

.

Substituting numbers you’ll get a spin period of less than a day.

Question 3: the most important parts of this question were parts (a) and (b), where you were asked to calculate how much energy would need to be added to the protoplanetary disk to unbind it. One first needs to calculate the binding energy per unit mass for gas in a circular orbit at radius r from the star. To do this, note that gas in orbit around a star has both gravitational potential energy and kinetic energy. Per unit mass,

.

For a circular orbit, , so we can simplify this to read,

.

This is the binding energy (or just the total energy) per unit mass… i.e. it’s the energy of 1 gram (or kg) of gas at this specific radius. To work out the total binding energy, we consider an annulus in the disk between radii r and (r + dr). The mass in that annulus is,

and hence that gas has binding energy (mass x specific binding energy) that is equal to,

.

The total binding energy is then just the integral of the above over all radii,

.

For the specified surface density profile, the radial dependence of the integrand is , so the integral gives you a natural log (hence you do need limits on both inner and outer radii to avoid divergence).

The gas and dust in the protoplanetary disk can be heated by two distinct sources of energy,

• Stellar irradiation: some fraction of the starlight will hit the disk (at the top and bottom), be absorbed by the dust, and heat it. A completely flat disk of negligible thickness absorbs 1/4 of the total emitted stellar radiation, and a thicker or “flared” disk (one with a bowl-like shape) absorbs even more.
• Accretion energy: if gas is flowing through the disk toward the star, a fraction (half, by the virial theorem) of the potential energy goes into heating the gas. This is sometimes described as “viscous” heating, since the transport of angular momentum within the disk that leads to accretion is somewhat akin to the viscosity in a fluid.

If the disk extends inward to radius , the accretion energy if the accretion rate is (units grams per second) is just,

,

where the factor of two comes from the fact that only half of the liberated potential energy goes into heat – the other half goes into kinetic energy which is higher for the faster orbital speeds close to the star. Plugging in typical numbers (e.g. an accretion rate of , which is a fairly typical number for a young Solar mass star with a disk) one can derive the accretion luminosity. The result is that, for most disks, stellar irradiation is the dominant energy source. The temperature scales with orbital radius as something like,

.

Knowing the temperature within the disk (and also the pressure, which we can obtain following the arguments given in the previous lecture), we can try and work out what types of solid material ought to be present at different locations within the disk. This is the concept known as the condensation sequence. First, we measure the abundances of the different elements within the Sun, using spectroscopic observations of the Solar surface (these measurements may be supplemented by lab measurements of the abundances within primitive meteorites). If we assume that these abundances reflect what the initial composition of the protoplanetary disk was, they tell us (e.g.) how much carbon was present relative to the amount of oxygen, hydrogen, magnesium etc. They do not, however, specify the form that carbon was in. Was it elemental carbon (e.g. graphite), a molecule like carbon dioxide or methane, or some mineral such as calcium carbonate? To determine this, we calculate the thermodynamically preferred mix of chemical compounds that would be present at given temperature and pressure for the known set of elemental abundances (for the physicists, we minimize the Gibbs free energy of the system). This mix defines the condensation sequence. To give some examples, methane is predicted to be present for temperature T < 40 K, important minerals such as perovskite below about 1400 K, and the hardiest materials such as aluminum oxide below about 1700 K. At higher temperatures, all elements are predicted to be in the gas phase.

Water deserves special attention, both because of its critical role in planetary habitability and because ice is a major component of the disk by mass in regions where it’s cold enough for ice to be present (recall that both oxygen and hydrogen are abundant elements in the Sun). The phase diagram for water is described in the Wikipedia article “Properties of Water”. It looks like,

At atmospheric pressure on Earth (1 bar, 100,000 Pa) water can exist in any of its three phases: vapor, liquid and solid. The protoplanetary disk, however, has much lower pressures that are far below the triple point (“TP” in the above diagram). At low P, the only stable phases are water ice and water vapor. Under disk conditions, we thus expect water to be in the form of ice for temperatures below about 150-170 K, and to be in the form of vapor for higher temperatures. The radial location where the transition from vapor to ice occurs is known as the snow line. In the Solar System, meteorites that originated from asteroids interior to about 2.7 AU are found to be water-poor, while those that came from further out are quite water-rich. We thus estimate that the Solar system snow line was at about 2.7 AU (this is a potentially misleading statement, as presumably the location of the snow line moved around as the Solar Nebula evolved and dissipated, but it’s reasonable as an empirical estimate).

The most important thing to note here is the distinction between the location of the snow line (in the disk), and the “habitable zone” defined as the range of distances from the star where a planet could sustain liquid water on its surface. A glance at the water phase diagram suffices to convince one that the snow line is invariably further from the star than the outer edge of the habitable zone, because at the low pressures in the disk water remains a vapor down to much lower temperatures than on the surface of a planet. This reasoning leads one to think that, when the Earth formed, the bulk of material at its location would have been “dry” minerals with very little water content. If so, then the water that is so critical to life on Earth must have been delivered later, from bodies that were initially further out… such as asteroids or potentially comets.

The radial distribution of surface density in the protoplanetary disk is not predictable from any simple physical consideration (it depends on how efficiently angular momentum is transported or lost at different radii, which is a largely open research problem). The vertical distribution of gas, however, is easier. It is a consequence of hydrostatic equilibrium.

Before considering the disk case, let’s consider (hopefully review!) how this works for a planetary atmosphere. If the planet’s gravity is , a constant, the equation of hydrostatic equilibrium gives,

.

This just says that the vertical pressure gradient has to balance the gravitational force in equilibrium. The pressure is related to the density (and potentially other things) via the equation of state. The simplest equation of state is an isothermal equation of state,

,

where $c_s$ is the sound speed. Substituting, and assuming the sound speed is a constant,

.

This is a simple differential equation, which can immediately be integrated,

.

The solution is,

.

The density in an isothermal planetary atmosphere thus falls off as an exponential with height above the surface.

Let’s now try the same calculation for gas in the protoplanetary disk. The geometry looks like,

Most of the gravitational force from the star acts inward, but for there is a component of the stellar gravity, , that acts “downward” (i.e. toward the disk mid-plane). Note that we will neglect any gravitational force from the disk gas itself, under the assumption that the disk is low mass (roughly, that ). From the geometry we have,

,

where is the radial distance. If the disk is thin, in the sense that the only heights above and below the mid-plane that we have to concern ourselves with are small compared to the distance from the star, we can make a simplifying approximation. The spherical radius is almost the same as the cylindrical radius , and then,

,

where is the Keplerian angular velocity for an orbit around the star.

Having worked out the effective gravity (which in this case is not a constant as in the planetary atmosphere example), we can solve for hydrostatic equilibrium assuming isothermality just as before,

.

We have a differential equation,

.

The right-hand side now integrates to rather than $z$, so we get a Gaussian fall off of density with height rather than an exponential. The vertical thickness, , is related to the angular velocity and the sound speed via,

.

Equivalently, remembering that , the Keplerian orbital velocity, we can divide both sides by to get,

.

What this means is that the thickness of the protoplanetary disk depends on how how the gas is (hotter gas has a higher sound speed). For reasonable values, in protoplanetary disks, i.e. the disk is moderately “thin”. Knowing how thick the disk is in the vertical direction allows us to estimate the volume density (say at the mid-plane) from the surface density. For example, for a Minimum Mass Solar Nebula model where the surface density at 1 AU is , then if we have at 1 AU that . The volume density is then, roughly,

.

This is evidently a low value… even the “dense” inner regions of protoplanetary disks are very low density by terrestrial standards.