<p>I'll answer my own question.
If there is hydrostatic equilibrium, then we have <a href="https://i.sstatic.net/AujXp.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/AujXp.gif" alt="enter image description here"></a> </p>

<p>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 <a href="https://i.sstatic.net/YZalw.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YZalw.gif" alt="enter image description here"></a></p>

<p>In this case: </p>

<p><a href="https://i.sstatic.net/6qRO9.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/6qRO9.gif" alt="enter image description here"></a></p>

<p>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):</p>

<p><a href="https://i.sstatic.net/swXFR.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/swXFR.gif" alt="enter image description here"></a></p>

<p><a href="https://i.sstatic.net/qqrUy.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/qqrUy.gif" alt="enter image description here"></a> Pa</p>

<p>If we want half of that mass M, then</p>

<p><a href="https://i.sstatic.net/YWMhl.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YWMhl.gif" alt="enter image description here"></a></p>

<p><a href="https://i.sstatic.net/gkC8W.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/gkC8W.gif" alt="enter image description here"></a> Pa , which is indeed 500 hPa.</p>

CC BY-SA 3.0
2018-02-13T03:39:14.840Z
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.
jamesqf
CC BY-SA 3.0
2018-02-14T10:50:49.280Z
While the answer is correct, @Lukas in his answer was able to get there with fewer steps.
Communisty
CC BY-SA 3.0
2018-02-16T07:53:02.127Z
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 :-)
JeopardyTempest
CC BY-SA 3.0
2018-02-20T07:29:19.477Z
You can accept your own answer (or any other) by clicking on the tick mark on the left side near the top.
Pont
CC BY-SA 3.0
2018-02-12T17:23:30.027Z
2018-02-14T10:02:27.470Z
2018-02-14T10:02:27.470Z
n.mont
<p>Simple?</p>

<p>Only assumption: 
hydrostatic relation
$dp = -g \rho dz$ </p>

<p>Expanding the RHS yields </p>

<p>$dp = -g \frac{dM}{dV} dz \frac{dA}{dA} = -\frac{g}{dA} dM$</p>

<p>($dA dz \equiv dV$)</p>

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

<ul>
<li>a pressure difference is proportional to a difference in mass</li>
</ul>

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

<p>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.</p>

CC BY-SA 3.0
2018-02-13T02:22:09.630Z
2018-02-13T02:22:09.630Z
Lukas
<p>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.</p>

<p>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...</p>

<p><strong>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!</strong> 
<hr>
But, call me crazy or handwavy, but I believe you can get there even easier by recalling back to earlier equations:</p>

<p>$${\text{Pressure} \ ≡ \ \frac{\text{Force}}{\text{Area}}}$$</p>

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

<p>And setting up the two levels:</p>

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

<p>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:</p>

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

<p>$$\text{So} \ 0.5 \cdot 1000 \ {\text{mb}} = \fbox{500 mb} $$</p>

<p>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, <strong>perturbations</strong>, and thus not a part of the <strong>mean</strong> atmospheric pressure.  But perhaps I'm wrong somewhere along the way???</p>

CC BY-SA 3.0
2018-02-16T08:42:46.047Z
2018-02-20T04:55:44.383Z
2018-02-20T04:55:44.383Z
JeopardyTempest
<p>Pressure in a fluid is, literally, the <strong>weight of the fluid above</strong> 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. </p>

<p>So the answer is 500 hPa. </p>

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

CC BY-SA 3.0
2018-02-18T01:31:52.470Z
2018-03-14T07:57:09.543Z
2018-03-14T07:57:09.543Z
DrGC
<p>Surface pressure given is <em>p</em> = 1000 hPa. (the temperature along the atmosphere is constant (<em>T</em> = −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.</p>

CC BY-SA 3.0
2018-02-04T23:21:23.687Z
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?
AtmosphericPrisonEscape
CC BY-SA 3.0
2018-02-05T05:28:12.800Z
What results do you expect?  What makes you suspect the different answer?
JeopardyTempest
CC BY-SA 3.0
2018-02-05T06:40:58.503Z
I expect 500 hPa because I've seen the solution, but I obtained 700-ish hPa.
n.mont
CC BY-SA 3.0
2018-02-05T08:32:58.873Z
Show your logic.
Communisty
CC BY-SA 3.0
2018-02-05T14:54:39.480Z
@n.mont please edit your question to show us how you arrived at 700hPa. It will be much more beneficial for you if we are able to see your thought process and help you see where it went wrong.
casey
CC BY-SA 3.0
2018-02-05T23:37:27.277Z
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.
n.mont
CC BY-SA 3.0
2018-02-12T17:31:54.637Z
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.
n.mont
<meteorology><atmosphere><homework>
How do I find the horizontal pressure level that divides the atmosphere into 2 layers of equal mass?
n.mont
# How do I find the horizontal pressure level that divides the atmosphere into 2 layers of equal mass?

(Asked by: username_0 on 2018-02-04)

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.

**Tags:** meteorology, atmosphere, homework

> **Comments:**
> **username_1**: 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?  
> **username_2**: What results do you expect?  What makes you suspect the different answer?  
> **username_0**: I expect 500 hPa because I've seen the solution, but I obtained 700-ish hPa.  
> **username_3**: Show your logic.  
> **username_4**: @username_0 please edit your question to show us how you arrived at 700hPa. It will be much more beneficial for you if we are able to see your thought process and help you see where it went wrong.  
> **username_0**: username_4 and username_3, I'll try to put my thought process, in detail, as soon as possible, give me three days. Trully busy right now.  
> **username_0**: 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.  

---

## 4 answers

### Answer by username_0 on 2018-02-12. Score: 4

I'll answer my own question.
If there is hydrostatic equilibrium, then we have [![enter image description here](https://i.sstatic.net/AujXp.gif)](https://i.sstatic.net/AujXp.gif)

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](https://i.sstatic.net/YZalw.gif)](https://i.sstatic.net/YZalw.gif)

In this case:

[![enter image description here](https://i.sstatic.net/6qRO9.gif)](https://i.sstatic.net/6qRO9.gif)

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](https://i.sstatic.net/swXFR.gif)](https://i.sstatic.net/swXFR.gif)

[![enter image description here](https://i.sstatic.net/qqrUy.gif)](https://i.sstatic.net/qqrUy.gif) Pa

If we want half of that mass M, then

[![enter image description here](https://i.sstatic.net/YWMhl.gif)](https://i.sstatic.net/YWMhl.gif)

[![enter image description here](https://i.sstatic.net/gkC8W.gif)](https://i.sstatic.net/gkC8W.gif) Pa , which is indeed 500 hPa.

> **Comments:**
> **username_5**: 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.  
> **username_3**: While the answer is correct, @username_7 in his answer was able to get there with fewer steps.  
> **username_2**: 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 :-)  
> **username_6**: You can accept your own answer (or any other) by clicking on the tick mark on the left side near the top.  

---

### Answer by username_7 on 2018-02-13. Score: 4

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.


---

### Answer by username_2 on 2018-02-16. Score: 4

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???


---

### Answer by username_8 on 2018-02-18. Score: 3

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).


---

