# Astronomy college course/Why planets lose their atmospheres/Quiz answers explained

## Atmospheric loss_Study

### __--1. It is important to distinguish between molecules (collectively) in a gas and one individual molecule. This question is about an individual molecule. For a planet with a given mass, size, and density, which has the greater escape velocity?

- a) the heavier molecule has the greater escape velocity
- b) the lighter molecule has the greater escape velocity
- c) no molecules have escape velocity
- d) all molecules move at the escape velocity
+ e) all molecules have the same escape velocity

#### Explanation

The phrase "for a given planet" implies that we are comparing atoms on the same planet. But the escape velocity of a planet is a property of the entire planet: Massive, dense planets tend to have the larger escape velocity. --Guy vandegriftSock1 (discusscontribs) 22:10, 22 December 2015 (UTC)

### __2. --It is important to distinguish between molecules (collectively) in a gas and one individual molecule. This question is about a typical molecule in the gas. For a planet with a given mass, size, and density, which type of gas is more likely to escape?

- a) atoms in a denser gas are more likely to escape
- b) atoms in a colder gas are more likely to escape
- c) all types of gas are equally likely to escape
+ d) atoms in a hotter gas is more likely to escape
- e) atoms in a gas with more atomic mass are more likely to escape

#### Explanation

For a given planet, all escape velocities are equal. A gas is more likely to escape if its atoms are moving quickly. Obviously hotter gasses are associated with more internal energy, and therefore faster atoms. But it also turns out that the low mass atoms tend to have greater speed. This is due to the formula kT = βmv2, where m is atomic mass and v is the average speed. In other words, the speed, v, can be solved to yield v=κ(T/m)1/2

This is a difficult problem for a course like this. To make matters even more difficult, it would not be difficult to create several closely related variations of this question.--Guy vandegriftSock1 (discusscontribs) 22:10, 22 December 2015 (UTC)

### __3. Which type of gas is likely to have the faster particles?

- a) a hot gas with high mass atoms
+ b) a hot gas with low mass atoms
- c) all gasses on a given planet have the same speed
- d) a cold gas with high mass atoms
- e) a cold gas with low mass atoms

#### Explanation

Remember that kinetic energy is directly proportional to temperature for the ideal gas. Kinetic energy is the product of mass and velocity squared. Hence velocity squared is poroportional to the ratio of temperature/mass. A fast average speed implies either a high temperature or a low atomic mass.--Guy vandegrift (discusscontribs) 01:16, 23 December 2015 (UTC)

### __4. What is it about the isotopes of Argon-36 and Argon-38 that causes their relative abundance to be so unusual on Mars?

- a) different half-life
- b) identical abundance
- c) identical mass
+ d) different speed
- e) different chemical properties

#### Explanation

Because of their different atomic masses, the two isotopes have slightly different speeds. Over time one of them is more likely to escape, especially from a planet as small as Mars.--Guy vandegrift (discusscontribs) 00:55, 23 December 2015 (UTC)

### __5. In the formula, ${\displaystyle {\frac {1}{2}}m_{\mathrm {atom} }v_{\mathrm {escape} }^{2}=G_{\mathrm {Newton} }{\frac {M_{\mathrm {planet} }m_{\mathrm {atom} }}{r_{\mathrm {planet} }}}}$, which of the following is FALSE?

- a) vescape is independent of matom
- b) the formula is valid only if the particle is launched from the surface of planet of radius rplanet
- c) the formula can be used to estimate how fast an atom must move before exiting the planet
- d) the particle is assumed to have been launched vertically
+ e) the formula is valid for all launch angles

#### Explanation

The escape velocity depends on the angle at which an object is directed. It takes less energy to put a satellite into low orbit around the Earth than to send it into deep interplanetary space. Any formula relating the escape speed to the mass and radius of the planet must reflect this fact.--Guy vandegrift (discusscontribs) 01:17, 23 December 2015 (UTC)

### __6. What statement is FALSE about ${\displaystyle {\frac {1}{2}}m_{\mathrm {atom} }\langle v_{\mathrm {atom} }^{2}\rangle _{ave}={\frac {1}{2}}k_{\mathrm {B} }T}$?

+ a) Temperature is measured in Centigrades
- b) The kinetic energy is directly proportional to temperature.
- c) Temperature is measured in Kelvins
- d) The average speed of a low mass particle is higher than the average speed of a high mass particle
- e) This equation does not involve the size or mass of the planet.

#### Explanation

This famous equation relates temperature to the average speed of atoms in a gas. Experiment in the nineteenth century showed that there was a linear relationship, and that the graph extrapolated to zero speed at what was then the unimaginable temperature of negative 273 Celsius. An ideal gas in which none of the atoms are moving is obviously a zero energy state. It was later understood that this is the coldest any substance can get.--Guy vandegrift (discusscontribs) 01:18, 23 December 2015 (UTC)

### __7. ${\displaystyle {\frac {1}{2}}m_{\mathrm {atom} }\langle v_{\mathrm {atom} }^{2}\rangle _{ave}={\frac {1}{2}}k_{\mathrm {B} }T}$, where T is temperature on the Kelvin scale. This formula describes:

- a) The the speed an atom needs to escape the planet, where m is the mass planet.
- b) The speed an atom needs to escape the planet, where m is the mass of the atom.
- c) The speed of a typical atom, where m is the mass of the planet.
+ d) The speed of a typical atom, where m is the mass of the atom.
- e) The speed an atom needs to orbit the planet, where m is the mass of the atom.

#### Explanation

Though not 100% accurate, this equation allows us to relate the the temperature of a gas to the propoerties of its atoms. It has nothing to do with the planet's gravitational field. It says that temperature is directly proportional to the mass of the atom times the square of its average velocity. This, in turn is to with a factor of roughly 2 of this particle's kinetic energy.--Guy vandegrift (discusscontribs) 01:23, 23 December 2015 (UTC)