The front page of our Forum again is fulfilled with crank.net museum’s exhibitors. So it is time to return to pleasant Physics… There is my first question: The probe grounded on Titan easily measured the weight of the entire Titan’s atmosphere. How it did that? How heavy the atmosphere of Earth is?
Is its surface area known? If so, F = PA so the force of the entire atmosphere is mean atmospheric pressure at the surface times the surface area of Titan. I guess that would only be an approximation though, as the pressure thing would be basically impossible to measure and well, surface area is difficult to calculate on an irregular surface. It would probably be a very useful approximation though if you were needing a good ball park of the weight of the atmosphere.
At that point, I edit my post to account for the fact that those questions came to mind after I'd already posted. Please Register or Log in to view the hidden image! I'm still spoiled from all those "ideal" shapes etc. from college. I'm still trying to think of another way to do it.
It will be wonderful if you will find another way, NASA will appreciate it very much... But you was on very good way already; all what you need now to think about why wind changes the pressure you need to calculate weight of atmosphere by your first manner? P.S. And irregularities of surface do not spoil your approach: in your formula A is indeed "a regular surface area": it is a magic feature of pressure to squeeze planet with the same tension no meter what shape its surface is (recall Pascal’s law!)
OK, 1100F, Now suck it all up and squeeze it all into a bag and put it onto some scales. Now how much does it weigh? Has the atmosphere got a weight. I think it has.
Well, it would be more appropriate to can ask for the atmosphere mass instead of weight. you can use for example the method proposed by wesmorris to find that it is about 5.1 10<sup>18</sup>kg
http://n93.cs.fiu.edu/measures/ The atmosphere weighs 15 pounds for every square inch. Converted to metric so I can think, that's a mass of 6,900 kilograms per square meter. The surface area of the Earth ( http://www.vendian.org/envelope/dir1/earth_jupiter_sun.html ) is 510 trillion square meters. That makes the mass of the atmosphere 3.5 quintillion kilograms, and at 9.8 newtons per kilogram it weighs 35 quintillion (35 x 10^18) newtons. That's a tad more than zero.
You are talking about the mass of the atmosphere. The weight of the atmosphere is the gravitational force acted on the atmosphere by earth. The total force that is acted by gravitation on the atmosphere will still be zero
Only if you're talking about the sum of the gravitational force vectors. If you purely talking numbers and ignoring vectors then the atmosphere DOES have weight - otherwise there would be no pressure. Weight = Mass x Acceleration (in this case gravity). If there was no acceleration (i.e. no gravity) then you could put a 1,000,000kg block of concrete above my head and it would have no weight and I would feel no pressure from it. However, on Earth there is gravity, and luckily I don't have a 1,000,000kg concrete block over my head. It is the WEIGHT of the air-molecules that create the pressure - which is why pressure drops the higher you get into the atmosphere. It is not the MASS of the air-molecules that does that.
Weight is defined as the gravitational force vector. Well, what you show is that each part of the atmosphere has a weight that is not zero. If you take a column of atmosphere over som surface, it will have a weight. But again, since the weight is defined as the gravitational force, the weight of the whole atmosphere will be zero.
Well, given F=MA, weight's not a tough calculation given mass and our standard 9.8 m/s^2 convention for accelleration. Oh I see your point. Since the accelleration of air toward the surface is net zero... the atmosphere is weightless, but has a whole lotta mass.
Guys, The discussion in this thread went in the wrong directions, leaving the really interesting issues aside… 1100f, you are mistaken: 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! 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. So, the main question stays the same: What if Titan had a very windy weather? P.S. wesmorris, why you change your way of thoughts so easy, instead of challenge the opponent's way first?
Because I was confused at the difference between weight and gravitational force. I was thinking of them as the same. I was thinking F=ma, so weight = MA. The term is very specific as you've pointed out, and highlights one of the reasons I'm no physicist. Please Register or Log in to view the hidden image! Hehe.
Is it not the net force of weight of the atomosphere applied to the whole sphere (support) is 0, as 1100f said ?
But you were absolutely right in the case of our problem: weight of any athmosphere indeed is equal to the product of its static preassure and area of support! So, the problem now is the only one: how probe should measure the static pressure if there is a strong wind?
If that were true, atmospheric pressure at the surface would be zero. Okay.. I'm going to think about static pressure with a wind. The rotation of the body is known, so if you subtract the pressure added by the drag of the atmosphere on the body, you should be able to find static pressure. So you could get the mean wind, which could be looked at as the pressure vector tangential to the radius of the body. So you stick a fan up, see how fast it rotates, calculate pressure from that... substract it from the atmospheric pressure measurement and shazaam, static pressure.
everneo, No, in case of pressure you can not add elementary forces acting on support as they are the free vectors: they are not, they are a coupled vectors, that are tied with their points of action. Isotropic pressure is squeezing body very much. Zero force would not effect on body at all. So, one can understand that doing like 1100f suggests we will come to simple ... nonsense.
Maybe I'm being a bit simplistic, but wouldn't you just shield the measuring device from the wind - i.e. build it inside the probe, assuming the probe can equalise pressures (is open at some place to the environment). Given that the weather will affect the readings of pressure (on Earth it can range from 970mb to 1040mb or more (probably far more) regardless of actual exposure to the physical effects of the weather, you would just need to read measurements over a long period of time. Assuming the wind never dies down, though, I wouldn't know how you would be able to measure anything other than the localised atmospheric pressure.
