Solar System Formation & Dynamics

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,

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

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The solution is,

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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,

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We have a differential equation,

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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,

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

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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,

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