8
$\begingroup$

Surface pressure given is p = 1000 hPa. (the temperature along the atmosphere is constant (T = −10 °C; hydrostatic equilibrium assumed). I tried using the hypsometric equation to find the height of the atmosphere's top and then going through that result, but it's not giving me the results I expect. I expect 500 hPa because I've seen the solution, but I obtained 700-ish hPa.

$\endgroup$
  • 4
    $\begingroup$ If your homework is about the mass, then try an equation that involves the mass. Two hints: What quantity would you need to know to calculate the total mass of the atmosphere? And what law governs this quantity that also involves pressure? $\endgroup$ – AtmosphericPrisonEscape Feb 4 '18 at 23:21
  • $\begingroup$ What results do you expect? What makes you suspect the different answer? $\endgroup$ – JeopardyTempest Feb 5 '18 at 5:28
  • 1
    $\begingroup$ Show your logic. $\endgroup$ – Communisty Feb 5 '18 at 8:32
  • 1
    $\begingroup$ casey and Communisty, I'll try to put my thought process, in detail, as soon as possible, give me three days. Trully busy right now. $\endgroup$ – n.mont Feb 5 '18 at 23:37
  • 1
    $\begingroup$ I posted an answer, I figured it out today. I was confused about the idea that exactly half pressure divided exactly half of the atmospheric mass. A pressure level isn't the same as height level. As the atmosphere, here is assumed isothermic, using the hypsometric equation we can conclude that the height obtained for this pressure level is really low comparing to the atmosphere's top height, which proves this result to be successful. $\endgroup$ – n.mont Feb 12 '18 at 17:31
4
$\begingroup$

Simple?

Only assumption: hydrostatic relation $dp = -g \rho dz$

Expanding the RHS yields

$dp = -g \frac{dM}{dV} dz \frac{dA}{dA} = -\frac{g}{dA} dM$

($dA dz \equiv dV$)

So any $\Delta p = c \cdot \Delta M$, which means that

  • a pressure difference is proportional to a difference in mass

($c$ is relatively constant as long as the gravitational acceleration constant, otherwise you'd have to integrate the hydrostatic eqn.)

So if the total mass is proportional to a $\Delta p$ of x hPa then the pressure level which separates equal parts is 1/2 of that value.

$\endgroup$
4
$\begingroup$

I'll answer my own question. If there is hydrostatic equilibrium, then we have enter image description here

So, assuming that the surface pressure is p (in this case 1000hPa) and that the atmosphere's top has zero (0) pressure and infinite height, then enter image description here

In this case:

enter image description here

So, as the atmospheric mass above a surface area is rho times integral of "dz", and assuming that this area is 1 m^2 to simplify (dx.dy=1):

enter image description here

enter image description here Pa

If we want half of that mass M, then

enter image description here

enter image description here Pa , which is indeed 500 hPa.

$\endgroup$
  • $\begingroup$ Why do you need all the equations? It seems obvious that the answer is 500 hPa, just from knowing the behavior of gasses under compression. That is, the pressure at any level is equal to the weight of atmosphere above it. Perhaps more obvious when you're used to thinking of pressure as pounds per square inch (or the equivalent kilograms per square meter for metric), instead of using a named unit. $\endgroup$ – jamesqf Feb 13 '18 at 3:39
  • 1
    $\begingroup$ While the answer is correct, @Lukas in his answer was able to get there with fewer steps. $\endgroup$ – Communisty Feb 14 '18 at 10:50
  • 2
    $\begingroup$ Not that getting a result in a different way is necessarily a bad thing. In fact it shows creativity and resilience :-) Hopefully you can learn and improve your methods and insight given the other answer(s), but much appreciated coming back to provide your answer, wish more would do so :-) $\endgroup$ – JeopardyTempest Feb 16 '18 at 7:53
  • $\begingroup$ You can accept your own answer (or any other) by clicking on the tick mark on the left side near the top. $\endgroup$ – Pont Feb 20 '18 at 7:29
4
$\begingroup$

I always found this style question interesting as a meteorology graduate teaching assistant, as there often seemed to be a variety of ways to attempt to answer it.

It's also often one of the earlier abstract type of questions in lower-level meteorology courses, and thus one I really encouraged my students to avoid resorting to help with the answer too quickly on, as exercises like these prove vital to developing persistence and creativity in trying to derive meteorological equations - and if you can't learn to fight through questions like these, you'll be lost in dynamics/thermodynamics/etc and beyond...

So if you're in a course with this type of question assigned, as much as it's painful, if haven't gotten your own good answer yet, please keep retrying and rereading on your own until you come up with something, even if only partial or uncertain. You'll learn more that way than by reading answers in the long run. And then come back after the assignment is over and learn some more here!


But, call me crazy or handwavy, but I believe you can get there even easier by recalling back to earlier equations:

$${\text{Pressure} \ ≡ \ \frac{\text{Force}}{\text{Area}}}$$

Assuming we're talking atmospheric pressure, we should be talking only about the the force due to the weight of air above, so $${\text{Force}} \ = \ {m_{above}}\cdot g$$

And setting up the two levels:

$${\text{Pressure}_{(Level\,=\,Ground)}} \ = \ {m_{(Full\,Atmospheric\,Column)}}\cdot g$$ $${\text{Pressure}_{(Value\,Where\,1/2\,Mass)}} \ = \ 0.5 \cdot m_{(Full\,Atmospheric\,Column)}\cdot g$$

I'll leave the one or two fairly basic high school algebra substitution steps for future readers needing proper completeness, but it quickly becomes the simple:

$${\text{Pressure}_{(Value\,Where\,1/2\,Mass)}} \ = \ 0.5 \cdot {\text{Pressure}_{(Level\,=\,Ground)}}$$

$$\text{So} \ 0.5 \cdot 1000 \ {\text{mb}} = \fbox{500 mb} $$

Naively the only place I can really see anyone even trying to take issue is in defining atmospheric pressure as simply the weight of air above it, overlooking things like vertical perturbation pressures. But of course those are just that, perturbations, and thus not a part of the mean atmospheric pressure. But perhaps I'm wrong somewhere along the way???

$\endgroup$
3
$\begingroup$

Pressure in a fluid is, literally, the weight of the fluid above the point of measure, per unit area, so if you have half the amount of air (or water, for the matter) above you, then you will measure exactly half the amount of pressure.

So the answer is 500 hPa.

That is valid under the standard hydrostatic approach (no turbulence forces; no dynamic forces).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.