Solar System Formation & Dynamics

ASTR 3710 Fall 2013

Lecture #8: Particle interactions with gas disks

Leave a comment Posted by philarmitage on November 10, 2013

Although there is vastly more gas than solid material in protoplanetary disks (typically by a factor of the order of 100), it is the evolution of the solid component that is critical to the formation of both terrestrial planets and the cores of gas giants. As long as the solid particles are relatively small (anywhere between microns in size up to tens or hundreds of meters) their dynamics is closely coupled to that of the gas via aerodynamic forces. These forces comes in two flavors, depending upon the particle size,

Epstein drag

mean free path

Stokes drag

Protoplanetary disks are extremely tenuous gases. The mean free path at the mid plane of the disk may be of the order of tens of cm or meters in size, depending upon the distance from the star and the surface density of the disk. As a result, the earliest phases of solid growth are always securely in the Epstein regime of drag. Consider a solid particle modeled (probably not very well) as a sphere of radius $s$ . The particle has a velocity relative to the local gas of ${\bf v}$ (note that this is not the orbital velocity of either the gas or the particle, but generally a much smaller number). The gas has density $\rho$ , and a mean thermal speed $v_{\rm th}$ . Under these conditions, the Epstein drag is given by,

${\bf F}_D = - \frac{4 \pi}{3} \rho s^2 v_{\rm th} {\bf v}$ .

As you might guess, the drag depends upon the frontal area of the particle (i.e. it scales with the square of the particle radius), but as you might not guess the force is only linear in the relative velocity. It is this feature of Epstein drag which is distinct from the Stokes drag that we’re more familiar with in terrestrial situations.

The aerodynamic interactions between solid particles and the gas are particularly important for the vertical and radial mobility of solids with the disk. Let us first consider a dust particle suspended in the disk above (or equivalently below) the mid plane. The gas within the disk is maintained in hydrostatic equilibrium by a balance of forces: the vertical component of stellar gravity, acting downward, is balanced by an upward-acting vertical pressure gradient. A particle, however, does not feel the pressure force within the gas. With the only force acting on the particle being gravity, there is a tendency for dust to settle toward the mid plane, potentially forming a dense particle layer.

We can calculate how quickly dust settling occurs in a laminar (non-turbulent) disk quite easily. We begin by defining a quantity known as the friction time (or stopping time),

$t_{\rm fric} \equiv m v / | F_D |$ .

Since $mv$ is the linear momentum associated with the relative motion of the particle against the gas, this time scale is the time on which aerodynamic drag tries to slow down the particle to match the local gas speed. Often, it is useful to use a dimensionless version of the same quantity, which we accomplish by multiplying by the local Keplerian orbital frequency,

$\tau_{\rm fric} \equiv t_{\rm fric} \Omega_K$ .

Substituting for the drag in the Epstein regime (and remembering that $m = (4/3) \pi s^3 \rho_m$ , with $\rho_m$ the material density of the particle) we find,

$t_{\rm fric} = \frac{\rho_m}{\rho} \frac{s}{v_{\rm th}}$ .

For small particles this time scale is very small indeed. For micron sized particles at 1 AU, for example, a few seconds is a representative estimate. As a consequence, it is generally reasonable to assume that small dust particles attain terminal velocity as they settle toward the mid plane, in much the same way as a skydiver jumping from a plane accelerates downward only up to the point where air resistance matches gravity. In our situation, we have the same two forces acting, gravity, acting vertically downward toward the mid plane,

$F_{\rm grav} = m \Omega_K^2 z$ ,

and aerodynamic drag, given by the Epstein formula as before,

${\bf F}_D = - \frac{4 \pi}{3} \rho s^2 v_{\rm th} {\bf v}$ .

Equating these forces defines the settling speed,

$v_{\rm settle} = \frac{\rho_m}{\rho} \frac{s}{v_{\rm th}} \Omega_K^2 z$ ,

from which we could also compute a rough estimate of the settling time scale,

$t_{\rm settle} = z / |v_{\rm settle}|$ .

Even for quite small particles (micron sized dust grains, which are coupled the gas very strongly) this time scale is generally shorter than the lifetime of the disk. At 1 AU, the settling time scale, ignoring both collisions with other particles (which may lead to coagulation and faster settling) and turbulence (which by stirring up particles opposes settling), typically may be of the order of 100,000 yr.

What about the effects of aerodynamic drag on the radial motion of solid particles? It is not immediately obvious that there are any effects, as a solid body such as a planet orbits a star in a balance between a radial gravitational force and an outward centrifugal force. There are thus no “unbalanced” forces of the kind that we appealed to when discussing settling. This is too simple, however, The gas in a protoplanetary disk is subject to additional non-gravitational forces, which cause it to orbit the star at a very slightly different speed from the Keplerian value, $v_K = \sqrt{GM_*/r}$ . This difference means that solid bodies do have a velocity differential with respect to the gas, and suffer aerodynamic effects as a consequence.

Quantitatively, we consider a gas disk that is circular and static (for the curious, the terms we are about to neglect that arise from radial gas flows are, indeed, generally negligible). Working in the disk mid plane, the orbital velocity is determined by the sum of the radial gravitational force and the pressure gradient,

$\frac{v_\phi^2}{r} = \frac{GM_*}{r} + \frac{1}{\rho} \frac{{\rm d}P}{{\rm d}r}$ .

If we neglect pressure, then this simplifies to the usual formula for the Keplerian orbital velocity. The pressure term, however, contributes a small but important correction. If – as is normally the case – the disk mid plane pressure is high at small radius and lower further out, then the pressure gradient term ${\rm d}P / {\rm d}r$ is negative. The pressure gradient provides an outward “push” that partially opposes gravity, and as a result the gas does not have to rotate as fast to maintain radial force balance.

The key result is thus: gas in protoplanetary disks typically rotates at less than the local Keplerian velocity.

For many purposes, this effect is small and can be safely ignored. If we write the pressure as a power law in radius,

$P \propto r^{-n}$ ,

then substituting we find,

$v_\phi = v_K \left[ 1 - n \left( \frac{h}{r} \right)^2 \right]^{1/2}$ .

Since $h / r \ll 1$ , the gas velocity differs from the Keplerian value by a small amount, normally less than 1%. This is still, however, quite a large absolute velocity – at 1 AU of the order of 100 meters per second – and it is has a profound effect on the dynamics of solid bodies. Since the gas rotates slower than Keplerian velocity, while the solid particles want to orbit at the Keplerian velocity, the solids feel a headwind as they orbit faster than the local gas. The aerodynamic effect of this headwind causes the solids to lose angular momentum, and spiral in toward the star.

The striking fact is how fast this radial drift is. Weidenschilling (1977) showed that the rate of aerodynamic radial drift (not to be confused with the gravitational migration we will discuss later) is a function of the particle size. The effects of drift slow down for very massive bodies (obviously, as those have a lot of inertia relative to their surface area), but they are also small for extremely small particles that are effectively frozen into the gas by tight aerodynamic coupling. The strongest effects occur for a dimensionless friction time $\tau_{\rm fric} = 1$ . Plotted against the friction time, the radial velocity has the following form,

The values on the y-axis depend upon the disk properties (especially on $h/r$ , which sets the velocity differential that drives radial drift) – here I have assumed $h/r = 0.05$ . The peak drift speeds in this case exceed 0.1% of the Keplerian orbital velocity. This implies very short decay times,

For the same disk model (the middle line in the plot), particles with $\tau_{\rm fric} \simeq 1$ are predicted to be lost into the star on a time scale that is only of the order of 100 or 1000 years in the planet forming region! What size these most vulnerable particles are again depends upon the disk model, but in the terrestrial planet forming region values of tens of centimeters to around a meter are typical.

Radial drift, due to aerodynamic forces against a sub-Keplerian gas disk, is the cause of the radial drift problem or meter-sized barrier in planet formation. Clearly, if meter-sized bodies only survive for a brief period within the disk, either growth through the adjacent size scales must proceed very quickly, or the disk must possess “traps” within which radial drift is halted or slowed. In class we discussed a number of possible ways to get around the radial drift problem, but this is still an open research problem. Indeed, how to form planetesimals (the km-sized bodies that are the smallest objects immune to radial drift) out of smaller solid particles is often considered to be the most serious unsolved problem of planet formation.

lecture

(Partial) solutions to HW#2

Leave a comment Posted by philarmitage on October 7, 2013

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,

$L = m v r$

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

$v = \sqrt{GM_* / r}$

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

$L_J = M_J \sqrt{G M_\odot r}$ .

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 $L_J$ 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,

$L = I \omega$

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

$I = \frac{2}{5} M_\odot R_\odot^2$ .

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,

$P = \frac{2 \pi}{\omega}$ .

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,

$E_B = -\frac{GM_*}{r} + \frac{1}{2} v^2$ .

For a circular orbit, $v = \sqrt{GM_*/r}$ , so we can simplify this to read,

$E_B = -\frac{GM_*}{2r}$ .

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,

$2 \pi r \Sigma dr$

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

$-\frac{GM_*}{2r} \times 2 \pi r \Sigma dr$ .

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

$\int_{r_{in}}^{r_{out}} -\frac{GM_*}{2r} \times 2 \pi r \Sigma dr$ .

For the specified surface density profile, the radial dependence of the integrand is $1/r$ , so the integral gives you a natural log (hence you do need limits on both inner and outer radii to avoid divergence).

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Lecture #7: The condensation sequence

Leave a comment Posted by philarmitage on October 2, 2013

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 $R_{in}$ , the accretion energy if the accretion rate is $\dot{M}$ (units grams per second) is just,

$L_{acc} = \frac{G M_* \dot{M}}{2 R_{in}}$ ,

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 $10^{-8} M_\odot yr^{-1}$ , 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,

$T(r) \propto r^{-1/2}$ .

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.

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Lecture #6: Disk structure

Leave a comment Posted by philarmitage on October 2, 2013

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 $g$ , a constant, the equation of hydrostatic equilibrium gives,

$\frac{dP}{dz} = - \rho g$ .

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,

$P = \rho c_s^2$ ,

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

$\frac{d \rho}{d z} = - \frac{g}{c_s^2} \rho$ .

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

$\int \frac{d \rho}{\rho} = - \frac{g}{c_s^2} \int dz$ .

The solution is,

$\rho = \rho(z=0) \exp [ - g z / c_s^2 ]$ .

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 $|z| > 0$ there is a component of the stellar gravity, $g_z$ , 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 $M_{disk} << M_*$ ). From the geometry we have,

$g_z = \frac{GM_*}{d^2} \sin \theta = \frac{GM_*}{d^2} \frac{z}{d}$ ,

where $d$ 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 $d$ is almost the same as the cylindrical radius $r$ , and then,

$g_z \simeq \frac{GM_*}{r^3} z = \Omega_K^2 z$ ,

where $\Omega_K$ 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,

$\frac{dP}{dz} = - \rho g_z$

$P = \rho c_s^2$ .

We have a differential equation,

$\frac{d \rho}{\rho} = -\frac{\Omega_K^2}{c_s^2} z dz$ .

The right-hand side now integrates to $z^2 / 2$ rather than $z$, so we get a Gaussian fall off of density with height rather than an exponential. The vertical thickness, $h$ , is related to the angular velocity and the sound speed via,

$h = \frac{c_s}{\Omega_K}$ .

Equivalently, remembering that $r \Omega_K = v_K$ , the Keplerian orbital velocity, we can divide both sides by $r$ to get,

$\frac{h}{r} = \frac{c_s}{v_K}$ .

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, $h / r \sim 0.05$ 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 $1700 g cm^{-2}$ , then if $h = 0.05 r$ we have at 1 AU that $h = 7.5 \times 10^{11} cm$ . The volume density is then, roughly,

$\rho \sim \frac{\Sigma}{2 h} \sim 10^{-9} g cm^{-3}$ .

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

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Problem set #2 (due Thursday Oct 3 in class)

Leave a comment Posted by philarmitage on September 27, 2013

Problem set #2

Several people have had trouble exporting the data into Excel. Here is the full data set (no selections on mass or radius made yet) in Excel (xls) and Excel (xlsx) formats.

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Writing assignment #2 (due Tuesday October 1st)

Leave a comment Posted by philarmitage on September 27, 2013

Read the article about dusty disks, linked below:

Space Dust Bunnies Could Unravel a Planetary Mystery

When you are finished, write a paragraph (between 100 and 200 words) summarizing it at a level that a member of the general, non-specialist public will be able to understand. You can submit your paragraph by following this link

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Reading assignment (for class discussion Tuesday 24th)

2 Comments Posted by philarmitage on September 17, 2013

Please read the short article on hot exoplanets, linked below. Besides
reading for content, also pay attention to things that the author does
and doesn’t do well in writing for the general public.

Atmospheric circulation on hot exoplanets: What about magnets?

Remember, some things to consider when writing for the general public
include:

Jargon: Is the writer being careful to avoid or explain words that the
public might not know?
Readability: Does the writer avoid run-on sentences and
make his/her message clear?
Correctness: Is the writer presenting correct information?
Relevance: Has the writer told you why you should care about this?
Organization: Does the writer’s paragraph build upon itself in a clear,
logical way?
Connection: Does the writer use analogies or make connections to
things his/her audience might have experienced in everyday life?
Appeal: Does the writer somehow hook the reader, use humor
in his/her writing, or tell a story to interest his/her audience?

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Lecture #5: Introduction to protoplanetary disks

Leave a comment Posted by philarmitage on September 17, 2013

Protoplanetary disks – like those seen above in HST images of the Orion Nebula – are a natural consequence of the broader star formation process. Stars within the Galaxy form within molecular cloud cores, relatively dense regions of gas that are embedded within much larger structures known as giant molecular clouds. The Orion Nebula, which can readily be seen through a small telescope, is part of one of the nearest GMCs.

Observationally, molecular cloud cores have scales of around 0.1 pc, and masses of perhaps 1 to a few Solar masses (these are obviously very rough numbers). It is thus plausible that one such core forms one star, or perhaps a binary or small multiple system. We can estimate how long collapse of a core to much higher densities ought to take, under the assumption that there is no persistent source of support against gravity (e.g. by magnetic fields). By equating kinetic energy to gravitational potential energy,

$\frac{1}{2} v^2 = \frac{GM}{r}$ ,

we obtain the escape velocity,

$v_{esc} = \sqrt{ \frac{2 GM}{r} }$ .

The collapse time scale can then be estimated as,

$t \sim \frac{r}{v_{esc}} \sim \sqrt{ \frac{r^3}{2 GM} }$ .

By noting that the mass of the cloud, $M = (4/3) \pi r^3 \rho$ , with $\rho$ the mean density, this can be rewritten in the more transparent form,

$t \sim 1 / \sqrt{G \rho}$ ,

i.e. the time to collapse under gravity depends solely on the density of the collapsing cloud. Numerically, for a Solar mass cloud that is 0.1 pc in radius, we estimate that the collapse time scale is about 300,000 yr. This is the characteristic time scale on which the formation of low mass stars like the Sun proceeds.

One tenth of a parsec is about 20,000 AU, or 4 million times the current radius of the Sun. Since the specific angular momentum,

$l = r v$ ,

the very large disparity between molecular cloud core scales and those of stars means that even vanishingly small amounts of core rotation translate into too much angular momentum to form a star directly. Instead, we expect a disk to form. To be a bit more quantitative, we can equate the specific angular momentum of a core, with scale $r_{core}$ and rotation speed $\delta v$ ,

$l_{core} \sim r_{core} \delta v$ ,

to the specific angular momentum at distance $r$ from a protostar of mass $M_*$ ,

$l_K = \sqrt{GM_* r}$ .

Doing so allows us to estimate a characteristic disk size, which works out to be,

$r_{disk} = \frac{ \delta v^2 r_{core}^2}{GM_*}$ .

If we take $\delta v = 0.01 km/s$ for a core of radius 0.1 pc and mass one Solar mass, we get,

$r_{disk} \simeq 50 AU$ .

In other words, even a very small rotational component on cloud core scales implies that infalling gas will form a disk much larger than stellar scales. There are a couple of caveats to this conclusion. If a binary forms, then much of the angular momentum of the initial core could end up in the orbital angular momentum of the binary, with correspondingly less ending up in disks. If angular momentum is lost on the collapse time scale, for example because magnetic fields act to brake the rotation as the cloud collapses, then again disks may be smaller or in principle even non-existent.

Once a star and a disk form, the specific angular momentum of gas orbiting within the disk,

$l_K \propto r^{1/2}$ ,

is an increasing function of orbital radius. This has important consequences. It implies that absent either,

Angular momentum transport, i.e. redistribution of angular momentum within the disk, or
Angular momentum loss from the disk as a whole

the disk ought to be stable. Since both angular momentum transport and loss processes are thought to be rather slow, protoplanetary disks are able to persist for a very large number of orbital times, and in particular to outlast the dynamical collapse phase of the cores that form them.

These theoretical considerations are closely related to the scheme used to observationally classify Young Stellar Objects. Four basic stages are envisaged:

Class 0 – this is the dynamical collapse phase, characterized by a deeply embedded protostar that is so shrouded by surrounding dust that it is only visible in the far-IR and sub-mm wavebands.
Class I – at this stage a star and disk (perhaps accompanied by an outflow or jet) have formed, but the remnants of the cloud core are still being accreted. The system becomes visible in the near-IR.
Class II – a star and disk only… by now the envelope has all been accreted. The stellar photosphere is now visible even in the optical.
Class III – the disk is dispersed, leaving a pre-main-sequence star only

A cartoon of these stages looks like this,

How long the disk-bearing phases of YSO evolution last is an observational question that is not easy to answer. Young stars do not display clocks, and estimating their ages is notoriously tricky. The standard approach is to measure the disk fraction in clusters, by looking for the excess near-IR emission indicating warm dust in a surrounding disk, and then estimate the age of the cluster by fitting the position of the stars in a Hertzprung-Russell diagram to evolutionary tracks for young stars. Following this procedure, one obtains a plot of disk fraction vs age,

…from which one can read off the median disk lifetime. The standard estimate is about 3 Myr, though very recent work suggests that this may be a modest underestimate… 5 Myr may be a better number. In any event, the estimated lifetimes of protoplanetary gas disks have two important implications for planet formation,

We have less than 10 Myr to form gas giants, before the gas is gone.
If the Moon formation is correctly dated at 50-100 Myr after the formation of the earliest solids in the Solar Nebula, the final assembly of the terrestrial planets in the Solar System must have occurred in a gas-free environment.

lecture

Lecture #4

Leave a comment Posted by philarmitage on September 16, 2013

Here are the slides Rosalba showed last week on observations of extrasolar planetary systems:

Observations of exoplanets

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Campus closed Thursday: all deadlines extended to Tuesday

Leave a comment Posted by philarmitage on September 12, 2013

Deadlines for the first problem set and the first writing assignment are extended to Tuesday (please note the writing assignment is required work, even though you’ll get automatic credit once it’s turned in satisfactorily). Stay safe, and of course we’ll be flexible for those of you who’ve been flooded.

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Solar System Formation & Dynamics

Lecture #8: Particle interactions with gas disks

(Partial) solutions to HW#2

Lecture #7: The condensation sequence

Lecture #6: Disk structure

Problem set #2 (due Thursday Oct 3 in class)

Writing assignment #2 (due Tuesday October 1st)

Reading assignment (for class discussion Tuesday 24th)

Lecture #5: Introduction to protoplanetary disks

Lecture #4

Campus closed Thursday: all deadlines extended to Tuesday

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