Escape Velocity

How did the gas giants sweep up and hold onto their gaseous atmospheres?
How does the Earth hold onto it's atmosphere?
Why doesn't the Moon have an atmosphere?
Can atmosphere's change over time?
What is a black hole?

Whereas solids are bound to a planet (or a tiny rock or asteroid) by molecular bonds, gases can only be held to a planet by gravity.
In order to know how strongly they are held, we must understand what energy is required to escape from a planet (or any object) -- and we will measure this energy in terms of velocity -- and we then must understand how fast a gas is moving under specified conditions.

Escape velocity: Measuring the gravitational strength of an object

The escape velocity is the exact amount of energy you would need to escape the gravitational clutches of an object with mass.  Since all objects have mass, they all have a measureable gravitational strength.  A good way to think about escape velocity is to think about a deep well (physicists like to think of this as an energy well).  If you are at the bottom of the well and want to get out (to escape), you need enough energy to climb out.  The deeper the well, the more energy you will have to expend in order to climb to
the top.  If you have only enough energy to get half way out, you will eventually fall back to the bottom.  The escape velocity is a way of measuring the exact amount of energy needed to reach the lip of the well -- and have no energy left over for walking away.

When a ball is thrown up into the air from the surface of the Earth, it does not have enough energy to escape.  So it falls back down. How might we enable the ball to escape?  Throw it harder, give it more energy.  How hard must we throw it?  Just hard enough to get over the top, over the edge of the well.

We can find this energy directly by saying that the kinetic energy of the thrown ball must exactly equal the 'potential energy' of the well.  From basic physics we know that the potential energy for an object at a height above a surface is:

Epotential= GMm/R

where
G = Newton's universal constant of gravity = 6.67 x 10-11 N-m2/kg3
M = the mass of the 'attracting object' [the planet] [in units of kg]
m = the mass of the object trying to escape [e.g., me or a ball or a rocket or a molecule] [in kg]
R = the distance between the centers of objects M and m [in units of m]
note: provided we do everything in the same units, we don't have to worry about units
while the kinetic energy we know from above:
Ekinetic=0.5 m v2
where
m = mass of the moving object [in kg]
v = the velocity of object m [in m/sec]
If we set these two energies equal to each other, and solve for v, we find the exact velocity needed to escape from the energy well:
0.5 m v2= GMm/R
v= (2GM/R)0.5
and since this velocity is exactly what is needed to 'escape,' it is called the escape velocity:
vescape= (2GM/R)0.5

Note what extremely important parameter is not in the escape velocity equation: the mass of the moving object.  The escape velocity depends only on the mass and size of the object from which something is trying to escape.  The escape velocity from the Earth is the same for a pebble as it would be for the Space Shuttle.

example #1: What is the escape velocity from the Earth?

MEarth=5.97 x 1024 kg
REarth=6378 km = 6.378 x 106 m
vescape= (2 x 6.67 x 10-11 N-m2/kg3 x 5.97 x 1024 kg / 6.378 x 106 m)0.5
vescape=1.12 x 104m/sec=11.2 km/sec
(this is equivalent to about 7 miles/sec or 25,200 miles per hour)
example #2: What is the escape velocity from the Sun?
MSun= 1.99 x 1030 kg
RSun= 700,000 km = 7 x 108 m
vescape= (2 x 6.67 x 10-11 N-m2/kg3 x 1.99 x 1030 kg/ 7 x 108 m)0.5
vescape=6.15 x 105m/sec=615 km/sec
example #3: What would be the escape velocity from Sun if it were the size of the Earth?
vescape= (2 x 6.67 x 10-11 N-m2/kg3 x 1.99 x 1030 kg/ 6.378 x 106 m)0.5
vescape=6.45 x 106m/sec=6450 km/sec
example #4: What would be the escape velocity from the Earth if it were the size of a pea? Take the diameter the pea to be one centimeter (radius = 0.005 m).
vescape= (2 x 6.67 x 10-11 N-m2/kg3 x 5.97 x 1024 kg / 0.005 m)0.5
vescape=3.99 x 108m/sec=399,000 km/sec
note: 399,000 km/sec is faster than the speed of light (300,000 km/sec).  Thus, we have created (in our imaginary experiment) an object from which light cannot escape.  According to the laws of modern physics, in this case according to a postulate of Albert Einstein, nothing, including light, can travel faster than light.  Thus, if light cannot escape, then neither can anything else. This is a black hole.  Clearly, for any object, we can find a size such that the escape velocity would be greater than the speed of light.  Similarly, for any size object (or region of space), we can find a mass such that a mass crammed into that volume of space would be a black hole.  The 'size' of the black hole is called the 'Schwarzchild Radius."
Question: To what size would I have to shrink the Sun in order to make it a black hole?

Question: Imagine that, just like now, the Earth orbits the Sun. Suddenly, I snap my fingers and turn the Sun into a black hole.  What would happen to the Earth?

The Kinetic Theory of Gases: What is the speed of a molecule in a gas?

Now we know the velocity required to escape from a planet.  This is equivalent to knowing the maximum velocity an object can have and still be gravitationally bound to that object.  We next need to know how fast gas molecules move.

Gases are characterized by their temperature, with the temperature determing the average velocity of a gas molecule.  According to the kinetic theory of gases developed in the 19th century, the amount of thermal energy per atom or molecule in a gas is given by the formula

Ethermal= 3kT/2

where
T is the temperature (in Kelvins),
k is the Boltzman constant = 1.38 x 10-16 gm cm2/(sec2 degrees).
For a moving object, the thermal energy is equal to the kinetic energy

Ekinetic=1/2  m vaverage2

where
m = mass of the moving atom or molecule in the gas (in gm)
vaverage = the average velocity of an atom or molecule in the gas (in cm/sec)
Since Ethermal= Ekinetic, we can set these two formulae equal to each other
1/2  m vaverage2 =3kT/2
and then we can solve for vaverage to find
vaverage = (3kT/m)0.5
Note what the average velocity depends on: it is directly proporational to the temperature and inversely proportional to the molecular/atomic mass.  Thus, at a single temperature, the average velocity of light elements (hydrogen, for example) will be faster than the average velocity of heavy elements (e.g., radon) at the same temperature.  And, for a given element/molecule, the average velocity increases as the temperature increases.

Note that in a gas some molecules are moving faster and some slower than average. The distribution of velocities is known as the Boltzman distribution.  Fully 0.5% of all the molecules (1 out of 200) are moving faster than twice the average and only 0.5% are moving slower than 20% of the average.  But, as we shall see, it is these 0.5% fast movers that are critically important when we talk about atmospheres.

As an illustration of a Boltzman-like distribution, think of cars on the interstate.  Most are moving at speeds close to the speed limit of 65 mph.  In fact, probably 70-90% of the cars are moving between 60-70 mph.  However, a few are moving faster and a few are moving slower. And perhaps 1 out of ever 200 cars is moving 100 mph while another 1 out of 200 is puttering along at 45 mph. Now supposed that a highway patrol officer pulls over and tickets the 100 mph driver.  What happens?  Someone else speeds up and replaces the fastest car so that there is always one bozo driving much too fast.

example #1: We can calculate the average velocity of a hydrogen atom at the surface of the sun .  We need to know the mass of a hydrogen atom (mhydrogen = 1.67 x 10-24 gm) and the temperature of the surface of the sun (T = 5800 K):

vaverage (H) = [ 3 x (1.38 x 10-16 gm cm2 sec-2 deg-1) x (5800 deg) / (1.67 x 10-24 gm) ]0.5
vaverage (H) = [1.44 x 1012]0.5 = 1.20 x 106 cm/sec x [1 km/ 105 cm]
vaverage (H) = 12.0 km/sec
example #2: We can calculate the average velocity of an aluminum atom (note that even aluminum will be a gas in atmospheres of stars) at the surface of the sun. Note that the aluminum atom is 27 times more massive than a hydrogen atom:
vaverage (Al) = [3 x (1.38 x 10-16 gm cm2 sec-2 deg-1)x(5800 deg)/(27 x 1.67 x 10-24 gm)]0.5
vaverage (Al) = [5.3 x 1011]0.5 = 2.3 x 105 cm/sec x [1 km/ 105 cm]
vaverage (Al) = 2.3 km/sec
Note that the escape velocity of the Sun (calculated above) is 615 km/sec, which is more than 50 times greater than the average velocity of H at the surface of the Sun.  And since all other elements are heavier than H, they will have lower average velocities, it is clear that these materials cannot escape from the Sun.

example #3: The Sun has an outer layer called the corona, in which the temperature is well above a million degrees. Can hydrogen atoms escape from the corona, if T = 4,000,000?

vaverage (H) = [3 x (1.38 x 10-16 gm cm2 sec-2 deg-1)x(4,000,000 deg)/(1.67 x 10-24 gm)]0.5
vaverage (H) = [9.92 x 1014]0.5 = 3.15 x 107 cm/sec x [1 km/ 105 cm]
vaverage (H) = 315 km/sec
??? It appears that the average H atom cannot escape, even at 4 million K.  However, remember that this is an average velocity, and that 1 out of every 200 atoms will be moving twice as fast. So, 0.5% of the atoms will have velocities in excess of 630 km/sec, which is above the Sun's escape velocity.  These atoms can stream outwards and escape from the Sun.

2 + 2: Escape velocity and the kinetic theory of gases and the snow line

At the distance of the snow line (let's call this 5 AU), the temperature of material illuminated and heated by the Sun is about 150 K (-120 C).  Thus, the averge velocity of a hydrogen atom

vaverage (H2) = [ 3 x (1.38 x 10-16 gm cm2 sec-2 deg-1) x (150 deg) / (1.67 x 10-24 gm) ]0.5
vaverage (H2) = 1.92 km/sec
Theory and modeling work suggest that a planet can hold onto it's atmosphere if the escape velocity is about 6 times greater than the average kinetic velocity.  Thus, a planet with vescape greater than 6vaverage can hold on to H2 at 150 K.

First, the easy situation: imagine an object with vescape = 1.9 km/sec (e.g., the Moon) and a gas temperature of 150 K.  Clearly, even our average H2 molecule (1.92 km/sec) can escape.  So quickly, all the hydrogen would escape from the Moon to space.

Now for the intermediate case: what about a planet with vescape = 8 km/sec  and a  gas with vaverage = 2 km/sec? In this case, vescape is less than 6vaverage  (12 km/sec).  This means that the fastest few molecules will escape.  Once they escape, a few other molecules will find there way to higher speeds and then they will escape.  And bit by bit, molecule by molecule, all the gas will trickle out this small 'opening' until the atmosphere is completely depleted in this gas.

Finally, the other limit: to hold onto  H2 at 150 K we only need an object with vescape greater than 11.5 km/sec and we're all set.  Since the Earth has vescape = 11.2 km/sec, it would appear that an Earth-mass sized object at the distance of Jupiter could hold onto hydrogen gas.

However, a planet can only hold on to gas already captured in its atmosphere.  So we must ask whether the planet can capture an atmosphere.
Note that Jupiter's orbital velocity around the Sun (Jupiter's orbital period divided by the circular path traveled by Jupiter in its orbit) is about 13 km/sec.  Thus, the gas orbiting the Sun at 5 AU is moving at about this speed and our protoJupiter must be able to capture and hold onto this gas, which is harder than holding on to gas already captured.

A good rule of thumb is that we need an escape velocity twice as large as the orbital velocity in order to capture orbiting gas. So, what mass object do we need.  If we doubled the radius of the Earth, the volume of the Earth would increase by a factor of two cubed (8). Thus, an object with a radius of 2 x 6378 km, and made of Earthy material, would have a mass about eight times larger than that of Earth. The escape velocity from such an object would be:

vescape= (2 x 6.67 x 10-11 N-m2/kg3 x 8 x 5.97 x 1024 kg) / (2 x 6.378 x 106 m)0.5
vescape= 22.3 km/sec
which isn't quite as large as 2 x 13 km/sec = 26 km/sec.  This is how we get our rough estimate that we need an object with ten (a bit more than 8) Earth masses in order to capture H gas from the solar nebula.  Clearly, once the gas is caught, the protoplanet can easily hold these gases.