by davidjonsson
 P: 4 The adiabatic heat gradient is determined as $$\gamma = \frac{g}{c_{p}}$$ 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
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 Quote by davidjonsson;3091803 My question is if [whether 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.
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
P: 4
 Quote by Andrew Mason 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
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

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P: 6,352

 P: 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