1. ### davidjonsson

4
$$\gamma = \frac{g}{c_{p}}$$

where $$\gamma$$ is the rate that temperature falls when rising in an atmosphere. g is gravitational acceleration and $$c_{p}$$ is the heat apacity. On Earth it is 9.8 Kelvin per kilometer close to the surface of the Earth.

Remember that g has a negative centrifugal acceleration term like this
$$g = \frac{ G m}{r^{2}} - \frac{v^{2}}{r}$$
where v is the speed due to rotation of the planet.

My question is if v should be determined only from Earths rotation around its axis or both from Earths rotation and the molecular motion in the gas.

If molecular motion is considered the value of g for air on Earth sinks by 0.95 % and the gravitational acceleration becomes
$$g = \frac{ G m}{r^{2}} - \frac{v^{2}}{r} - \frac{5 v_{rms}^{2}}{3 r}$$

Where $$v_{rms}$$ is the root mean square speed of thermal motion on Earth approximately 500 m/s.

Do you understand how I derived the last term?

David

2. ### Andrew Mason

6,805
You cannot take the molecular motion into account. The centripetal acceleration is required in order to keep a mass moving at the earth's rotational speed. For a given volume of air, the centre of mass of that air, presumably, rotates with the earth. At the molecular level, the molecules move randomly but the centre of mass does not. So for each molecule that may be moving in one direction (say in the direction of the earth's rotation), there is another moving in the opposite direction relative to the centre of mass of that volume of air. If you were to factor in the molecular motion as you have done, you would be assuming all the molecules in the air were moving all in the same direction (in the direction of the earth's rotation) which is not the case.

AM

3. ### davidjonsson

4
My assumption is based on a even distribution of molecular motions in all three dimensions.

Remeber that the effect is based on the square of the speed and not the mean velocity you refer to which is linear. Do calculations on pairs of oppositely moving particles and you will find that they do not average up to the mean flow of the gas. I began this investigation by doing exactly that.

For particles moving parallell to the planet the molecule speed will for the forward moving molecule be
$$v + v_{rms}$$
and the speed for the oppositely backwards moving particle
$$v - v_{rms}$$

Leading to the net mean centrifugal acceleration
$$((v + v_{rms})^{2}/r + (v - v_{rms})^2/r)/2 = (v^{2} + v_{rms}^2)/r$$

This is not equal to the acceleration for the mean gas flow $$v^{2}/r$$

Likewise for the sideways north-south moving molecule pair where the north-south component adds perpendicular to the planet motion component v like this
Speed north $$\sqrt{v^{2} + v_{rms}^{2}}$$
Speed south $$\sqrt{v^{2} + (-v_{rms})^{2}}$$

The mean value of this molecule pair of centrifugal acceleration becomes

$$((\sqrt{v^{2} + v_{rms}^{2}})^{2}/r + (\sqrt{v^{2} + (-v_{rms})^{2}})^2/r)/2 = (v^{2} + v_{rms}^2)/r$$

Again the same size of the effect. For the up and down motion of the molecule there is no tangential component and thus no effect for centrifugal acceleration.
So for these three motions in the three dimensions the mean value is
$$((v^{2} + v_{rms}^2)/r + (v^{2} + v_{rms}^2)/r + v^{2}/r)/3 = (v^{2} + 2 v_{rms}^2/3)r$$

This is a bit forward but not identical to what I initially wrote. I have also added the rotation of the molecule in my first post. Can anyone imagine how I derived that?

David

4. ### Andrew Mason

6,805
You appear to be assuming that individual molecules moving randomly about the atmosphere are prescribing circular motion about the earth at that speed. What is the basis for that assumption?

Suppose I place 4 walls on the earth's surface to make a column of air. Since the air in that column is definitely not moving relative to the earth in circular motion at a speed greater than the speed of the earth's surface, if your analysis is correct, the weight of the air in that column, and therefore the pressure of that air on the earth's surface, would be greater than the surrounding air weight and pressure. Is that, in fact, observed?

AM

5. ### davidjonsson

4
The gas molecules collide on average 9 billion times a second but since collision is perfectly elastic that doesn't change the total momentum of involved particles. It doesn't matter if the molecules collide or not. Despite frequent collisions most of the time the molecules are freely floating.

Make Cartesian decomposition of the molecules' speed and you will find that motion in x and y, the tangential directions, has the effect. The vertical z component has no effect.

Analyzing the case in a small cube can be meaningful. That would allow to determine the entire stress tensor. On arXiv I have derived the shear stress that the difference in molecular acceleration results in. That stress is up to 20% of the total shear stress in Venus' atmosphere. At first I didn't realize there would be a change to the normal stress but then on January 11 I also found that.

In your assumed situation more than half of the molecules will have a net speed higher than the surface speed. They will mostly be in free fall in regard to gravity. They don't move in circular orbits but in small parts of highly eccentric elliptic orbits between collisions. If you stop that motion, by cooling the gas, the weight of air in that column will increase by almost 1%.

Your question is not irrelevant because on many planets and stars there is a strong wind or current flowing around the latitudes. So there is a net mean circular motion of gas and plasma in and around astronomical bodies.

And if temperature is hot enough the orbits can be circular and weight would drop to a minimum. At even higher temperatures most orbits would be hyperbolic and the gas would leave the planet. This problem is already considered by astronomers. They say that gas will leave the planet when the thermal speed exceeds the escape velocity.

David

6. ### davidjonsson

4
I suppose I should use the speed

$$\overline{v^{2}_{x}} =\overline{v^{2}_{y}} =\overline{v^{2}_{z}} = \overline{v^{2}_{rms}}/3$$

instead of $$v_{rms}$$ in my calculation above. Right?

Seems like the effect will shrink to one third then.

David