# Astronomy college course/Why planets lose their atmospheres

${\displaystyle {\mathcal {TIP}}}$   This is a very difficult section. Most students should skip it and attempt to learn the answers in the subpage /Quiz answers explained

This discussion combines ideas taken from two Wikipedia pages[1] [2]and one Wikiversity page[3]. The purpose of this learning resource is to show how calculations can be used to gain insight into complex phenomena such as the ability of a planet to retain an atmosphere. From the numbers alone, it is possible to see evidence of recent volcanism on Venus and Triton. And it is possible to understand why we are confident that a meteorite found on Earth actually came from Mars. All by looking at numbers and equations!

Atmospheric escape is the loss of planetary atmospheric gases to outer space. We shall focus first on atmospheric escape due to thermal particles exceeding the escape velocity. In order to do so, we shall borrow from the Wikipedia article Effective temperature to construct a table of nominal temperatures of the planets based solely on the distance from the planet or moon from the Sun. This table facilitates an understanding of why the Argon isotope ratio on Mars differs from that of the Sun, Earth, and Jupiter. A few of the other mechanisms by which planets and moons lose or retain their atmosphere are briefly outlined.

## Energy

Energy is stored work. It has the same units as work, the Joule (J).

There are many forms of energy:

Spring energy: Work has been done on a spring to compress or stretch it; the spring has the ability to push or pull on another object and do work on it. The force required to stretch a spring is proportional to the distance it is stretched: F = kx where x is the stretch distance and k is a constant characteristic of the spring (big heavy springs have larger k values). The work done in stretching a spring from 0 to x is the integral of dW = Fdx. Since the force function is linear, we can just take the average force of kx/2 and avoid using calculus:

   W = average F x distance = (kx/2)(x) = ½kx²


Assuming 100% efficiency, the energy stored in a stretched spring is the same as the work done in stretching it, so Spring E = ½kx²

Example: How much energy is stored in a spring with k = 2000 N/m that has been stretched 1 cm away from its equilibrium length?

E = ½kx² = ½(2000)(0.01)²  = 0.1 J


Gravitational potential energy: a mass has been lifted to a height; when released it will be pulled down by gravity and can do work on another object as it falls.

Example: Find the energy stored in a tonne of water at the top of a 20 m high hydroelectric dam.

The long way is to use F = mg and then W = Fd to find the work needed to lift the water up.

The short way is to combine the formulas, replacing F with mg and using h (height) in place of d:

Gravitational energy = W = Fd = mgh

Egravity = mgh = (1000 kg)(9.81 m/s²)(20 m) = 196200 kg m²/s² = 1.96 x 105 J

Kinetic energy: A mass is moving and can do work when it hits another object. Ekinetic = ½mΔV2 = ½m(Vf2-Vi2)

Example: A 8kg ball is moving at 5m/s. EK = ½(8 kg)(5 m/s)2 = 200 J.

Electrical energy: Electrons can flow out of a battery or capacitor and do work on another electrical component such as a light bulb.

## Escape velocity

What goes up does not necessarily go down. If an object is thrown upward it never returns to the planet from which it was thrown. A good rule of thumb is

${\displaystyle {\frac {1}{2}}m_{\mathrm {atom} }v_{\mathrm {escape} }^{2}=G_{\mathrm {Newton} }{\frac {M_{\mathrm {planet} }m_{\mathrm {atom} }}{r_{\mathrm {planet} }}}}$,

where

• GNewton = 6.67×10−11
m3·kg−1·s−2
is Newton's universal constant of gravitation
• rplanet is the radius of the planet (or the height from which the projectile is directed)
• Mplanet is the mass of the planet
• matom is the mass of the atom
• vescape is the critical speed of an upward directed atom called the escape velocity. If the speed is less than vescape, the atom returns. If the speed is greater than vescape, the atom does not return.

This formula only applies to objects thrown directly upward. Objects thrown at other angles might go into an orbit and require a slightly different formula, but for our purposes this is good enough. Note that the mass of the atom cancels, which means that the critical velocity for escape does not depend on the mass: A massive rocket launched into space has the same escape velocity as an individual atom. This extra factor (matom) could have been omitted from the formula, but its inclusion makes it easier for physics majors to understand that this is an energy conservation argument. The left hand side of the above equation is the initial kinetic energy, and the right hand side the gain in potential energy as the particle "rises" to an infinite height.

### Thermal energy

Atoms in a gas or atmosphere do not all move at the same speed, but instead follow the so-called "Bell curve" (properly called a normal or Gaussian distribution). The formula for the average component of a squared speed (see root mean square, ${\displaystyle v_{ave}}$, in any direction is also easy to remember if you think about energy:

${\displaystyle {\frac {1}{2}}m_{\mathrm {atom} }\langle v_{\mathrm {atom} }^{2}\rangle _{ave}={\frac {3}{2}}k_{\mathrm {B} }T}$,

where Boltzmann's constant is kB1.38 × 10-23 Joules/degree. (Like G, kB is a universal constant of nature that we assume to be constant everywhere and for all time.) This equation is a well known formula from a theoretical analysis of the ideal gas developed at by Maxwell and Boltzmann in 1860-1877.

### Defining an "escape" temperature

It is convenient to define a "escape temperature" for any planet as that temperature for which an average thermal molecule has the escape velocity. While the actual temperature of a planet is typically much less than this "escape temperature", the ratio of "escape temperature" to actual temperature is an indicator of how well the planet can retain its atmosphere. The "escape temperature" is obtained by eliminating speed as a variable in the above two equations. Before we combine the equations, it is convenient to think about radius-and-density instead of radius-and-mass. The density, ${\displaystyle \rho }$, (or mass density) is mass divided by volume. Using a well-known formula for the radius of a sphere, the mass, M, and radius, r, of a planet are related by,

${\displaystyle M_{\mathrm {planet} }={\frac {4\pi }{3}}\rho _{\mathrm {planet} }\cdot r_{\mathrm {planet} }^{3}}$.

Setting ${\displaystyle T_{\mathrm {Kelvins} }}$ equal to ${\displaystyle T_{\mathrm {esc} }}$ and solving yeilds:

${\displaystyle T_{\mathrm {esc} }=\left({\frac {8\pi G}{9k_{B}}}\right)m_{\mathrm {atom} }\cdot \rho _{\mathrm {planet} }\cdot r_{\mathrm {planet} }^{2}}$

The physical significance of ${\displaystyle T_{\mathrm {esc} }}$ is that if a planet has this temperature, the average molecule in the atmosphere will has enough speed to escape from the planet. It is constructive to compare this critical temperature with the actual temperature of the planet.

### Handy units

This is a wonderful formula if you happen to be a human calculating machine. The rest of us need to convert this formula into more "handy" units. This is accomplished by taking convenient values for as many variables as we can. Using values taken from wikipedia,

• Rplanet = 6.37×106 m (meters) = the radius of Earth
• Mplanet = 5.97×1024 kg (kilograms)= the mass of Earth
• matom = 1.66×10-27 kg (kilograms) = 1 AMU ≈ mass of proton or neutron
• ρwater = 1 gm/cm3 = 1000 kg/m3.

Using these values, we calculate the critical temperature to be 2736 Kelvins. This is astonishingly high, but a planet with this critical temperature would lose all it's atmosphere immediately. Remember that any planet that loses 0.1% of its atmosphere each year due to gravitational escape would lose its atmosphere in a few thousand years.

The use of convenient numbers to calculate this effective temperature is useful because it can be used to generate the following "handy" formula:

${\displaystyle T_{\mathrm {esc} }=2736\,m\rho r^{2}}$

where the critical temperature, T, is in Kelvins, the atomic mass, m, is in amu, the density, ρ is specific density (i.e. normalized to water), and the planet's radius, r, is measured in earth radii. Although the physical meaning of this formula remains unexplained, it allows us to quantify the statement that atmospheres tend to be associated with large cold planets, and that it is easier for a planet to retain an atmosphere if the atmosphere consists of molecules with large atomic mass. (See atmospheric escape)

## Estimate of temperature as function of distance from the Sun

The question of whether a planet retains an atmosphere depends on the temperature of the planet, and the single most important factor that influences temperature is the distance from the Sun. Using arguments analogous to the critical temperature for escape from a planet's gravity, the Wikipedia article Effective temperature) discusses how a planet's distance from the Sun helps determine the planet's temperature. An important result of that discussion is the following calculation estimate of what a planet's temperature would be if distance from the Sun were the only factor that influences a planet's temperature:

${\displaystyle T_{\rm {nom}}=\left({\frac {4A_{\rm {abs}}}{A_{\rm {rad}}}}{\frac {1-a}{\varepsilon }}\right)^{1/4}\left({\frac {R_{\odot }}{2D}}\right)^{1/2}T_{\odot }\approx (1-a)^{1/4}\left({\frac {R_{\odot }}{2D}}\right)^{1/2}T_{\odot }\;}$,

where ${\displaystyle D}$ is the distance from the Sun, ${\displaystyle T_{\odot }}$≈5778K is the Sun's temperature, and ${\displaystyle R_{\odot }}$≈ 6.955×108is the Sun's radius. In science and engineering, nominal often represents an accepted approximation as opposed to an exact, typical, or average measurement. Here we use it to keep the discussion simple, as well as to obtain a calculation that will never require updating as new measurements are obtained.

This formula for a planets temperature, ${\displaystyle T_{\rm {nom}}}$, makes the simplifying assumption that temperature is determined by an energy balance: The planet releases all the energy it gains from the Sun by warming to the point where the the energy released due to thermal radiation equals the energy absorbed due to proximity to the Sun. When the actual temperature is not close to the nominal temperature, we seek a reason why. In the case of Venus, the mechanism that causes a much higher actual temperature is the w:Greenhouse effect.

The first term (raised to the ${\displaystyle 1/4}$ power) is essentially a fudge factor that includes complications beyond the scope of this article. These complications include cloud cover and how a planet transfers heat from the hot and cold regions. (See: atmospheric escape for more information on these terms.) The most important term is a, which is the planet's albedo. [4] Taking a=1, we can obtain the following formula for the nominal temperature of a planet at a given distance from the Sun:

${\displaystyle T_{\rm {nom}}={\frac {278.6}{D^{1/2}}}}$ ,

where T is in Kelvins and D is the distance to the moon or planet in AU.

## Defining the alpha factor

It is convenient to define alpha (α) as a dimensionless ratio that tells us whether the planet's proximity to the sun, its size, density, and atmospheric composition, all are conducive to retention of an atmosphere:

${\displaystyle {\rm {alpha=\alpha ={\sqrt {\frac {T_{esc}}{T_{nom}}}}}}}$.

When α is calculated for objects in the solar system, we expect the following:

• If α is large, one would expect a dense atmosphere.
• If α is small, one would expect a tenuous atmosphere, or virtually no atmosphere.

The decision to take the square root of the temperature ratio was arbitrary, and made so that α would represent the ratio of the escape velocity to thermal speed.

## Table of parameters relevant to atmospheric retention

The following table includes the terrestrial planets and two moons most likely to contain atmospheres:

Object Dist.
(AU)
(earth
density
g/cm3
Tave
(K)
Tnom
(K)
gas amu Tesc
(K)
α Psurf
(atm)
Earth 1 1 5.515 288 279 N2 28 422493 39 1
Venus 0.72 0.95 5.24 737 328 CO2 44 569308 42 91
Mars 1.5 0.53 3.94 210 227 CO2 44 134242 24 0.006
Saturn's Titan 9.5 0.4 1.88 98 90 N2 28 23507 16 1.45
Jupiter's Ganymede 5.2 0.41 1.936 ?* 122 O2 16 14456 11 0
Jupiter's Io 5.2 0.29 3.528 ?* 122 S02 64 50531 20 0

In this chart, Tave is the actual (measured) average surface temperature, Tnom is a 'nominal' temperature based solely on the planet's distance from the sun, and Tesc is a measure of how much gravity is present. Tesc is the temperature the planet would need to have in order for the atmosphere to almost instantly disappear. Inspection of the following (hidden) table establishes that, except for the gas planets, the objects in the above list are most likely to retain an atmosphere against gravitational escape by thermal molecules. Psurf is the pressure at the surface of the planet, in atmospheres.

Click to view or hide full table
Object Dist.
(AU)
(earth
density
g/cm3
Tave
(K)
Tnom
(K)
gas amu Tesc
(K)
α Psurf
(atm)
Jupiter 5.2 11 1.33 N/A 122 H2 2 875812 85 N/A
Saturn 9.5 9.1 0.69 N/A 90 H2 2 315419 59 N/A
Neptune 30.1 3.9 1.64 N/A 51 H2 2 133710 51 N/A
Uranus 19.2 4 1.27 N/A 64 H2 2 110082 41 N/A
Earth 1 1 5.515 288 279 N2 28 422493 39 1
Venus 0.72 0.95 5.24 737 328 CO2 44 569308 42 91
Mars 1.5 0.53 3.94 210 227 CO2 44 134242 24 0.006
Saturn's Titan 9.5 0.4 1.88 98 90 N2 28 23507 16 1.45
Jupiter's Ganymede 5.2 0.41 1.936 ?* 122 O2 16 14456 11 0
Jupiter's Io 5.2 0.29 3.528 ?* 122 S02 64 50531 20 0
Jupiter's Callisto 5.2 0.38 1.83 ?* 122 CO2 44 31478 16 0
Neptune's Triton 30.1 0.21 2.061 ?* 51 N2 28 7096 12 0
Mercury 0.39 0.38 5.43 ?* 446 O2 16 34869 9 0
Jupiter's Europa 5.2 0.25 3.01 ?* 122 O2 16 7909 8 0
Earth's Moon 1 0.27 3.346 ?* 279 Ar* 40 27295 9 0

## discussion of the table

A few observations can be made:

Earth and Venus have nearly the same parameters. Moreover, Venus is hotter than one would calculate based on its proximity to the sun, and a hotter Venus would tend to favor more thermal escape from its gravitational field. The factor of 100 between the surface pressure of Venus and Earth needs to be explained.

Saturn's Titan has a dense atmosphere, while Jupiter's Io has a tenuous atmosphere, something that is not reflected by their values. This needs to be explained.

Of all terrestrial planets with significant atmosphere (Venus, Earth, Mars), Mars has the lowest value of . It should be no surprise that Mars also has the most tenuous atmosphere among these three planets. In a later section we shall discuss how this marginal ability of Mars to hold an atmosphere has modified the relative isotopic abundances of Argon in the Martian atmosphere.

## Argon isotope ratio on Mars

Argon Isotope Ratios are a signature of atmospheric escape on Mars (Curiosity rover) (April, 2013)[5][6]

The table of beta values shown above suggests that Mars would have considerably more difficulty maintaining its atmosphere than Earth or Venus. This can be seen in the abundance ratio of the two isotopes of Argon (36Ar and 38Ar). The lighter isotope has a slightly higher chance of escaping from the planet, and as reflected by the figure, the result is a significant difference in the abundance ratio of these isotopes. This figure raises a two interesting points:

1. The fact that this abundance exists and can be explained lends credibility to the assertion that we have samples of Martian rocks that have landed on Earth as meteorites from Mars.
2. Isotopes of the same element have virtually the same chemistry, which renders isotope separation technically difficult. It is fortunate that only a few nations on Earth have the ability to separate (uranium-238 and uranium-235). So how does Mars manage to accomplish this separation with Argon?

#### Plausibility argument that isotope separation can occur on Mars

The difference in mass between the two isotopes of Argon is about 5.5%, which implies that at a given temperature, the average speed of the two isotopes differs less than 3%. Yet the difference in the isotope ratios on Mars, as compared with Earth is quite large. On Earth it is about 5.5, and on Mars it is closer to 3.5. The ratio is 5.5/3.5 ≈ 1.6, How does a difference in atomic mass less than 6% lead to a 60% change in isotope abundance?

To answer this question we must review the normal probability distribution. Many college students take a statistics course where the area under the tail of a normal distribution is used to determine probability. Physicists know this area as the erfc function. When z is sufficiently large, so that a sufficient small area is inside the tail, the following approximation is useful:

${\displaystyle \mathrm {erfc} (z)\approx {\frac {e^{-z^{2}}}{z{\sqrt {\pi }}}}\quad {\rm {if\;z>>1}}}$
Tail of a normal distribution.

Inspection of this expression verifies that for large z, a small percent change in z will cause a large difference in the area under the tail. Compare for example the ratio of probabilities if z is increased slightly from z0 to z1=z0+ε.

${\displaystyle {\frac {P_{1}}{P_{2}}}=\exp \left((z_{0}+\varepsilon )^{2}-z_{0}^{2}\right)\approx \exp 2z_{0}\varepsilon }$

For example, if z0 = 10.0 and z1 = 10.5 is 5% larger, the ratio of the tails is:

${\displaystyle {\frac {P_{1}}{P_{2}}}=\exp(2\times 10\times .05)=e\approx 2.7}$.

To understand why only gas molecules at the extreme tail of the normal distribution, we use our "handy" formula to calculate the critical temperature for Mars. The specific density of Mars is 3.94 g/cm3, the atomic mass of CO2 is 44, and Mars has a radius equal to .532 Earth radii. The critical temperature at which Mars immediately loses its atmosphere is

${\displaystyle T_{\mathrm {esc} }=2736\,m\rho r^{2}=(2736)(44)(3.94)(.532)^{2}\approx 134000\;{\rm {Kelvins}}}$

It is clear that molecules that escape the gravitational attraction of Mars are at the extreme tail of the distribution. And at this tail, even the smallest percent change in atomic mass will greatly affect the fraction of molecules that escape.

## Other mechanisms that influence atmospheric retention

The previous discussion is not intended to model atmospheric escape, but to illustrate the nature of the methods used. A large number of complications have been ignored. Correct modelling of such a complex process requires extensive use of computer modelling, as well as a bit of luck if a correct model is to emerge.

#### Hydrodynamic escape

While it has not been observed, it is theorized that an atmosphere with a high enough pressure and temperature can undergo a "hydrodynamic escape." In this situation atmosphere simply flows off into space, driven by thermal energy.[7] Here it is possible to lose heavier molecules that would not normally be lost.

#### Significance of solar winds

The relative importance of each loss process is a function of planet mass, its atmosphere composition, and its distance from its sun. A common erroneous belief is that the primary non-thermal escape mechanism is atmospheric stripping by a solar wind in the absence of a magnetosphere. Excess kinetic energy from solar winds can impart sufficient energy to the atmospheric particles to allow them to reach escape velocity, causing atmospheric escape. The solar wind, composed of ions, is deflected by magnetic fields because the charged particles within the wind flow along magnetic field lines. The presence of a magnetic field thus deflects solar winds, preventing the loss of atmosphere.

Depending on planet size and atmospheric composition, however, a lack of magnetic field does not determine the fate of a planet's atmosphere. Venus, for instance, has no powerful magnetic field. Its close proximity to the Sun also increases the speed and number of particles, and would presumably cause the atmosphere to be stripped almost entirely, much like that of Mars. Despite this, the atmosphere of Venus is two orders of magnitudes denser than Earth's.

While Venus and Mars have no magnetosphere to protect the atmosphere from solar winds, photoionizing radiation (sunlight) and the interaction of the solar wind with the atmosphere of the planets causes ionization of the uppermost part of the atmosphere. This ionized region, in turn induces magnetic moments that deflect solar winds much like a magnetic field. This limits solar-wind effects to the uppermost altitudes of atmosphere, roughly 1.2–1.5 planetary radii away from the planet. Beyond this region, called a bow shock, the solar wind is slowed to subsonic velocities.[8][source?] Nearer to the surface, solar-wind dynamic pressure achieves a balance with the pressure from the ionosphere, in a region called the ionopause. This interaction typically prevents solar wind stripping from being the dominant loss process of the atmosphere.

#### Phenomena of non-thermal loss processes on moons with atmospheres

Several natural satellites in the Solar System have atmospheres and are subject to atmospheric loss processes. They typically have no magnetic fields of their own, but orbit planets with powerful magnetic fields.

#### Sequestration

This is a loss, not an escape; it is when molecules solidify out of the atmosphere onto the surface. This happens on Earth, when water vapor forms glacial ice or when carbon dioxide is sequestered in sediments. The dry ice caps on Mars are also an example of this process.

One mechanism for sequestration is chemical; for example, most of the carbon dioxide of the Earth's original atmosphere has been chemically sequestered into carbonate rock. Very likely a similar process has occurred on Mars. Oxygen can be sequestered by oxidation of rocks; for example, by increasing the oxidation states of ferric rocks from Fe2+ to Fe3+. Gases can also be sequestered by adsorption, where fine particles in the regolith capture gas which adheres to the surface particles.

8. Cite error: Invalid <ref> tag; no text was provided for refs named Shizgal, 1996