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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?

Answer

The scale height is given by cap h equals c sub s solidus omega sub cap k . 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, m macron equals two times u and for the disc composed entirely of helium, m macron equals four times u . 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?

Answer

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

normal cap delta times v of r equals left square bracket one minus Square root of one minus n times left parenthesis cap h divided by r right parenthesis squared right square bracket times v sub cap k of r

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 τ_S = 1?

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.

Question 4

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

Answer

The isolation mass is given by Equation 21:

cap m sub normal i times normal s times normal o equals eight divided by Square root of three times left parenthesis pi times cap sigma times cap c right parenthesis super three solidus two times a cubed divided by cap m sub asterisk operator super one solidus two full stop

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.

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.

Core formation by solid accretion followed by gas accretion beyond the critical mass.
Core formation by solid accretion followed by slow core accretion.
Core formation by solid accretion followed by growth in the region of the disc with little solids.

1
Gas giant planet
2
Ice giant planet
3
Terrestrial planet

The correct answers are:

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.

Answer

The Toomre criterion for fragmentation is

multirelation cap q equals omega sub normal cap k times c sub normal s divided by pi times cap g times cap sigma less than one full stop

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

multirelation cap q equals omega sub normal cap k squared times cap h divided by pi times cap g times cap sigma less than one full stop

Then we note that the Keplerian angular speed is omega sub normal cap k equals left parenthesis cap g times cap m sub asterisk operator solidus r cubed right parenthesis super one solidus two so this now becomes

multirelation cap q equals cap g times cap m sub asterisk operator divided by r cubed times cap h divided by pi times cap g times cap sigma equals cap m sub asterisk operator divided by pi times r squared times normal cap sigma times cap h divided by r less than one full stop

So, calculating in this case

cap q equals 0.75 multiplication 1.99 multiplication 10 super 30 kg divided by pi multiplication left parenthesis 3.3 multiplication 1.496 multiplication 10 super 11 times normal m right parenthesis squared multiplication 4700 kg m super negative two multiplication 0.065

cap q equals 27 left parenthesis two s full stop f full stop right parenthesis

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?

Answer

The Jeans mass is given by Equation 25 as

cap m sub Jeans equals four times pi times cap m sub asterisk operator times left parenthesis cap h divided by r right parenthesis cubed

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

four times pi times cap m sub asterisk operator times left parenthesis cap h divided by r right parenthesis cubed greater than cap m sub Jup

cap h solidus r greater than left parenthesis cap m sub Jup divided by four times pi times cap m sub asterisk operator right parenthesis super one solidus three

In this case

cap h solidus r greater than left parenthesis 1.90 multiplication 10 super 27 kg divided by four times pi multiplication 0.45 multiplication 1.99 multiplication 10 super 30 kg right parenthesis super one solidus three

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?

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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OpenLearn - The formation of exoplanets
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