Register to reply 
The adiabatic heat gradient is temperature dependent 
Share this thread: 
#1
Jan1911, 03:58 PM

P: 4

The adiabatic heat gradient is determined as
[tex]\gamma = \frac{g}{c_{p}} [/tex] where [tex]\gamma[/tex] is the rate that temperature falls when rising in an atmosphere. g is gravitational acceleration and [tex] c_{p} [/tex] 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 [tex]g = \frac{ G m}{r^{2}}  \frac{v^{2}}{r}[/tex] 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 [tex]g = \frac{ G m}{r^{2}}  \frac{v^{2}}{r}  \frac{5 v_{rms}^{2}}{3 r}[/tex] Where [tex] v_{rms} [/tex] 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
Jan2011, 07:11 AM

Sci Advisor
HW Helper
P: 6,654

AM 


#3
Jan2011, 05:21 PM

P: 4

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 [tex]v + v_{rms}[/tex] and the speed for the oppositely backwards moving particle [tex]v  v_{rms}[/tex] Leading to the net mean centrifugal acceleration [tex]((v + v_{rms})^{2}/r + (v  v_{rms})^2/r)/2 = (v^{2} + v_{rms}^2)/r[/tex] This is not equal to the acceleration for the mean gas flow [tex]v^{2}/r[/tex] Likewise for the sideways northsouth moving molecule pair where the northsouth component adds perpendicular to the planet motion component v like this Speed north [tex]\sqrt{v^{2} + v_{rms}^{2}}[/tex] Speed south [tex]\sqrt{v^{2} + (v_{rms})^{2}}[/tex] The mean value of this molecule pair of centrifugal acceleration becomes [tex]((\sqrt{v^{2} + v_{rms}^{2}})^{2}/r + (\sqrt{v^{2} + (v_{rms})^{2}})^2/r)/2 = (v^{2} + v_{rms}^2)/r[/tex] 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 [tex]((v^{2} + v_{rms}^2)/r + (v^{2} + v_{rms}^2)/r + v^{2}/r)/3 = (v^{2} + 2 v_{rms}^2/3)r[/tex] 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
Jan2111, 09:34 AM

Sci Advisor
HW Helper
P: 6,654

The adiabatic heat gradient is temperature dependent
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
Jan2111, 07:33 PM

P: 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
Jan2411, 07:20 PM

P: 4

I suppose I should use the speed
[tex]\overline{v^{2}_{x}} =\overline{v^{2}_{y}} =\overline{v^{2}_{z}} = \overline{v^{2}_{rms}}/3[/tex] instead of [tex]v_{rms}[/tex] in my calculation above. Right? Seems like the effect will shrink to one third then. David 


Register to reply 
Related Discussions  
Heat and Internal Energy and Heat and Temperature Change: Specific Heat Capacity  Advanced Physics Homework  4  
Heat Transfer::Assigning direction to the temperature gradient  Engineering Systems & Design  7  
: Heat transfer  temperature gradient value at a certain point  Introductory Physics Homework  15  
Temperature Gradient  Introductory Physics Homework  3 