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# 4 Quiz

Answer the following questions in order to test your understanding of the key ideas that you have been learning about.

## Question 1

Consider four protoplanetary discs, with the same temperature, around stars of similar mass. One is composed entirely of molecular hydrogen, one is a mixture of molecular hydrogen and helium, one is composed entirely of helium, and one is a mixture of hydrogen, helium and other heavier elements giving a mean molecular mass of 2.3u (where u is the atomic mass unit, 1.66 × 10-27 kg). Which disc will have the largest scale height at a given radius?

The disc composed of molecular hydrogen only.
The disc composed of helium only.
The disc composed of molecular hydrogen and helium.
The disc composed of molecular hydrogen, helium and heavier elements.
All four discs will have the same scale height.

### 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.)

## Question 2

Consider the same four protoplanetary discs as in Question 1. Which disc will have the smallest difference between the orbital and Keplerian speeds at a given radius?

The disc composed of molecular hydrogen only.
The disc composed of helium only.
The disc composed of molecular hydrogen and helium.
The disc composed of molecular hydrogen, helium and heavier elements.
All four discs will have the same difference in speeds.

### 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.

## Question 3

Consider again the same four protoplanetary discs as in Question 1. Which disc will have the largest radial drift speed at a given radius for particles with a Stokes parameter of τS = 1?

The disc composed of molecular hydrogen only.
The disc composed of helium only.
The disc composed of molecular hydrogen and helium.
The disc composed of molecular hydrogen, helium and heavier elements.
All four discs will have the same radial drift speed.

### Answer

For particles with a Stokes parameter of τS = 1, the radial drift speed from Equation 16 is vrad = -vKη/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.

## Question 4

Which of the following statements about the isolation mass involved in the growth of planetesimals are true?

The isolation mass increases as the surface density of the protoplanetary disc increases, for a given star at a given orbital radius.
The isolation mass increases with orbital distance from the central star, for a given star and a given disc surface density.
The isolation mass decreases as the mass of the star increases, for a given orbital distance and a given disc surface density.
For the situation described in Activity 5, the isolation mass at 0.1 au is 3.94 × 1020 kg.
For the situation described in Activity 5, an isolation mass of 3.94 × 1026 kg corresponds to a distance of 10 au.
The isolation mass can never be larger than the mass of the Earth.

### Answer

The isolation mass is given by Equation 21:

Therefore the first three statements are true, since Miso ∝ Σ, Miso∝ a3 and Miso∝ 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.

## Question 5

Match the following core-accretion scenarios to the type of planet that results.

Core formation by solid accretion followed by gas accretion beyond the critical mass.

Gas giant planet

Core formation by solid accretion followed by slow core accretion.

Ice giant planet

Core formation by solid accretion followed by growth in the region of the disc with little solids.

Terrestrial planet

Using the following two lists, match each numbered item with the correct letter.

1. Core formation by solid accretion followed by gas accretion beyond the critical mass.

2. Core formation by solid accretion followed by slow core accretion.

3. Core formation by solid accretion followed by growth in the region of the disc with little solids.

1. 1

   Gas giant planet

2. 2

   Ice giant planet

3. 3

   Terrestrial planet

The correct answers are:

- 1 =
- 2 =
- 3 =

### Answer

See Figure 6 for details.

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

## Question 6

A protoplanetary disc around a star of mass M* = 0.75 M☉ has a surface density of Σ = 4700 kg m-2 at a radius of r = 3.3 au. If the disc aspect ratio is H/r = 0.065, determine whether the disc satisfies the Toomre criterion for fragmentation.

The Toomre parameter is less than 1 and so the disc does meet the Toomre criterion.
The Toomre parameter is greater than 1 and so the disc does not meet the Toomre criterion.

### Answer

The Toomre criterion for fragmentation is

The sound speed may be written cs = H ωK, 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.

## Question 7

In a protoplanetary disc around a star of mass M* = 0.45 M☉, what is the minimum value of the disc aspect ratio H/r to ensure that the Jeans mass exceeds the mass of Jupiter?

H/r > 0.0055
H/r > 0.055
H/r > 0.55
H/r > 5.5
H/r > 55

### 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.).

## Question 8

The formation of which of the following types of planets cannot be explained by the core-accretion scenario?

Hot Jupiter planets at very small orbital distances.
Giant planets at orbital distances larger than a few au.
Terrestrial planets in Earth-like orbits around their stars.
Mini-Neptune planets at orbital periods of a few months.
Super-Earth sized planets at orbital periods of less than a year.

### 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.

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