sarkus, you need to do the last , the final step, before you will advice to NASA your own good designs of measurer of the staticoressure: to search Internet and find out how the static preassure is meassured ... in science. When you find it out, come back and tell us what you have discovered. Then we will be able to conclude this thread with ... right solution. OK?
Hum, Just a small off-topic note to those that haven’t heard but it was recently found that the air is slightly heavier than we previously thought. The composition of air is mostly nitrogen (around 72%) and oxygen (21%). The rest is a mixture of carbon dioxide, water vapour and argon. All straight forward and well understood, however, it is the amount of argon that is the critical factor here, because the more argon present, the denser the air. The density of air under standard conditions is only 1.239 milligrams per cubic centimetres. In 1969, it was calculated that the amount of argon in the air was around 0.917%. However, recently, a team from the Korea Research Institute of Standards and Science (KRISS) and the International Bureau of Weights and Measures (BIPM) in France, have determined that the actual amount of argon is 0.9332%, ± 0.0006. Air density is increased. So this in turn will affect buoyancy, or static pressure… Anyway the presence of argon 40 was also discovered in the atmosphere of titan, though I haven’t seen the data; but it would be a similar process to work out the density of Titan’s air. but i think they used the data from the decent time (allowing for the size of the parachutes) they could narrow down the column height of the air and work out its density.
I expressed the math in terms of how much the mass of the atmosphere weighs in newtons, which is a unit of weight. Read it again while you're thinking again. Incorrect. Weight is the resistance to acceleration.
If the whole atmosphere gets heated uniformly, the isometric pressure on the surface too increases, that is weight of the atmosphere changes with temperature? The main reason for wind is the difference in atmosphere temperature hence the difference in pressure. If you take PA as the weight of the isometric atmosphere you have to bring in the temperature also into the picture.
everneo, Wonderful! Please, explain why you think so? (I guess, the answer will be interested to everybody here...)
Yuiry, I was expecting an answer from you! It was an ideal situation, ( but in reality we know that the atmosphere has different layers like troposhpere and lower & upper stratospheres with generally decreasing temperature from the earth surface to the upper spheres. Pressure reduces with decreasing temperature and increasing altitude. .. with all sort of thermodynamic, fluiddynamic properties ) where i assumed the tempeature is uniform throughout the atmosphere that has isometric pressure. In that case, with uniform increase in temperature, the atmosphere tends to increase in thickness because of pressure, but it is bounded by the earth surface at lowest level and bounded by gravity at upper level. So, the uniform increase in temperature throughout the atmosphere increases the pressure at surface (lower boundary) significantly.
>“ Originally Posted by 1100f >You are talking about the mass of the atmosphere. ” >I expressed the math in terms of how much the mass of the atmosphere weighs in >newtons, which is a unit of weight. Read it again while you're thinking again. Absolutely! "weight" is a force, not mass. >“ Originally Posted by 1100f >The weight of the atmosphere is the gravitational force acted on the atmosphere by >earth. ” >Incorrect. Weight is the resistance to acceleration. Now, THAT'S incorrect. MASS is the resistance to acceleration, which can take place even in deep space with no gravitational masses nearby. WEIGHT, is by definition, the gravitational force of the earth on a body (on the earth- weight on another planet, moon, etc. would be that body's gravitational force).
Guys, One more time: weight is not a gravitational force! It is very important! The gravitational force exists and acts on body all the time (till it is into the gravitational field) but weight depends on motion of body and even can vanish at some type of motions (at free falling on center of gravity). Therefore, weight is not a gravitational force! In some conditions weight has the same value and direction as the gravitational force, in other conditions - it is not the same. The state of weightless is the best proof of that. The direct and correct definition of weight in Physics is the following: Weight of body is the force with which this body is pressing its support (bearing) or is pulling its suspension bracket/clip. The force of weight is applied not to the body, but to its support or suspension!Therefore, weight always is caused by a combination of the gravitational force and all other forces acting on the body (usually – inertial ones, but it might be the Coulomb forces, EM forces and any other). So, forget useless dispute based upon wrong definition of weight, especially in regard to a gas atmosphere, which in any gravitational field will remain to be … an atmosphere with a density distribution according to the Boltzmann’s formula.
Yuriy, Wouldn't the mass of the heated atmosphere increase for the same reason the mass of an accelerated particle seems to increase when v -> c? Geistkiesel
Everneo, It is impossible to heat the atmosphere uniformly. Half the atmosphere is opposite to the sun, and is therefore cooln; further for the atmosphere exposed to the sun the heat absorption can never be uniform due, for one reason, to the tilt of the earth's axis.The total weight of the atmosphere must always vary, if indeed the temperature increases the weight. Geistkiesel
everneo posted: Actually in this post we have mentioned everything that is needed to be considered to come to the right conclusion in this thread. all what we need now - a little connection with Science... 1. The first of all we should recognize how the contemporary Physics describes the atmospheres of planets. In the equilibrium state (the rested atmosphere, no winds, no gradients of temperature, etc) the distribution of gaseous atmosphere (in Physics term “atmosphere” is used in regard of many other situations: for instance, when vortex moves in liquid it caries on some amount of liquid, which is called as “the vortex’s atmosphere”; this atmosphere is streamlined by rest liquid) is described by famous Boltzmann’s formula: D(h |T) = D(0|T)exp(-mgh/kT)…………………(1) D(x|T) is the gas’ density on the level x (above ground x = 0), m is gas’ molecules’ mass, g =981 cm/s^2, k is the Boltzmann’s constant and T is absolute temperature. So, the atmosphere has no “upper boundary”! But D essentially depends on T. 2. If we will change the temperature of atmosphere, T, the distribution of gases will change, but the pressure, Po, on the planet’s surface will remain the same – equal to the weight of atmosphere per unit of square of the planet’s surface. So, measuring this (the static or taken at total rest of atmosphere!) pressure on the ground of planet, we actually are measuring the weight of the planet’s atmosphere (it assumes that we know the area of the planet’s surface!). this pressure of a totally rested gas is called “the stagnation pressure” or “the static pressure”. 3. We can not apply the ideal gas laws of thermodynamics to the atmosphere as a whole – the atmosphere as a whole is not a closed thermodynamic system! But we can apply those laws to some parts of atmosphere if we are considering some fast enough processes with these parts (for instance, to kinematics of wind). Such processes usually are adiabatic thermodynamic processes and are very well ‘describable” by Poisson’s formulas and usual hydrodynamics of an ideal gas. 4. As we know, the major consequence of the stationary motion of an ideal gas is the Bernulli’s law: 5. Po = P + D(0|T)v^2 /2 ……………………..(2) So, the measured pressure P will be differ (will be less) from the pressure Po on the value of so called “dynamical pressure” D(0|T)v^2 /2. Of course, this pressure has nothing to do with weight of atmosphere! Therefore, to measure Po, we should use such a tool that … measures namely pressure Po, not P. This tool was invented by Pitot and is called “the Pitot’s tube”. Pitot’s tube measures the dynamical pressure in a gas’ flow. If we we can measure only P, not Po, we have to accompany it with measuring of temperature T and velocity of gas, v. Then knowing what gas we deal with, we will be able to find D(0|T) and calculate Po.
Thanks, superL, for good reference. I fixed some errors in my post after this material. Now it looks better.
