4 Quiz

Answer

The scale height is given by . The Keplerian angular speed at a given radius will be the same for all four discs since it depends only on the stellar mass and the radius, which are the same in all four cases.

The sound speed is inversely proportional to the mean molecular mass of the gas in the disc. For the disc composed entirely of molecular hydrogen, and for the disc composed entirely of helium, . The disc composed of a mixture of molecular hydrogen and helium will have a mean molecular mass that is somewhere between 2u and 4u, and as noted in the question, the disc composed of molecular hydrogen, helium and other heavier elements has a mean molecular mass of 2.3u.

The largest scale height will be for the disc with the largest sound speed, and this will be for the disc with the smallest mean molecular mass. Therefore the protoplanetary disc composed entirely of molecular hydrogen will have the largest scale height at a given radius. (Conversely, the disc composed entirely of helium will have the smallest scale height.)

Answer

The difference between the orbital and Keplerian speeds at a given radius is given by

As noted in the answer to Question 1, the Keplerian angular speed at a given radius will be the same for all four discs since it depends only on the stellar mass and the radius, which are the same in all four cases. So the difference in speeds will be smallest when the term under the square root is largest. This term will be largest when H/r is smallest. From the information in Question 1, this will be for the disc composed entirely of helium.

Answer

For particles with a Stokes parameter of τ_S = 1, the radial drift speed from Equation 16 is v_rad = -v_Kη/2. As noted in the answer to Question 1, the Keplerian angular speed at a given radius will be the same for all four discs since it depends only on the stellar mass and the radius, which are the same in all four cases. So the radial drift speed will be largest for the disc with the largest value of η. Since this is given by η = n(H/r)², and the disc aspect ratio is largest for the disc with the largest scale height, this corresponds to the disc composed entirely of hydrogen, as revealed in Question 1.

Answer

The isolation mass is given by Equation 21:

Therefore the first three statements are true, since M_iso ∝ Σ, M_iso∝ a³ and M_iso∝ 1/ M_*^1/2. Furthermore, since the isolation mass in Activity 5 at 1.0 au is 3.94 × 10²³ kg, the corresponding masses at distances 10× smaller and 10× larger are 1000× smaller and 1000× larger respectively, therefore the next two statements are also true. Hence all statements are true except the last one.

Answer

See Figure 6 for details.

Figure 6 (repeated) Schematic view of possible outcomes of the core-accretion model.

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The figure has three parts. Each part is comprised of two stages: formation and output. In part (a), in the formation stage, two hemispheres with common centres are shown on two sides of a vertical line. The left side of the line is labelled ‘core formation by solid accretion’. Two identical smaller spheres are drawn on the left side of the left hemisphere. An arrow from each of the smaller spheres points towards the centre of the left hemisphere. The right side of the line is labelled ‘gas accretion beyond critical mass’. The hemisphere on the right is bigger in size. A blue ring is shown around the right hemisphere. A set of radial arrows points towards the ring around the right hemisphere. In the output stage, a photo of Jupiter (a gas giant) is shown. In part (b), in the formation stage, a similar diagram to that in part (a) is shown. The left side of the line is labelled ‘core formation by solid accretion’. The right side of the line is labelled ‘slow core accretion’. Here, the blue ring is thinner compared with part (a) and there are fewer radial arrows. In the output stage, a photo of Neptune (an ice giant) is shown. In part (c), in the formation stage, another similar diagram is shown. The left side of the line is labelled ‘core formation by solid accretion’. The right side of the line is labelled ‘growth in the region of the disc with little solids’. Here, both hemispheres are of the same size. The blue ring is far thinner compared with parts (a) and (b), and surrounds both hemispheres. In the output stage, a photo of Earth (a terrestrial planet) is shown.

Figure 6 (repeated) Schematic view of possible outcomes of the core-accretion model.

Answer

The Toomre criterion for fragmentation is

The sound speed may be written c_s = H ω_K, where H is the scale height, so this becomes

Then we note that the Keplerian angular speed is so this now becomes

So, calculating in this case

Since this is greater than 1, the Toomre condition is not satisfied and the disc will not fragment.

Answer

The Jeans mass is given by Equation 25 as

So, if the Jeans mass exceeds the mass of Jupiter, we have

In this case

So the disc aspect ratio must be greater than 0.055 (2 s.f.).

Answer

The core-accretion scenario can explain the formation of most types of exoplanets. However, it struggles to explain the formation of giant planets at orbital distances larger than a few astronomical units (corresponding to orbital periods longer than a few years), because of the extended time needed to form big enough cores at these distances. These planets probably formed by the disc-instability scenario.