Calculating the Mass of the Atmosphere (using a triple integral)












4












$begingroup$


I have been given the task of calulating the mass of the atmosphere given the following (not quite correct) information. It is intended to be a thought experiment for an undergraduate course in atmospheric science. I recognize that I have chosen a difficult way of doing this but I believe it should still work, and I am confused as to where I may be going wrong in my reasoning.



Radius of Earth: 6000 km.



Density of Atmosphere at sea level: $3times10^{19}$ $frac{molecules}{m^3}$.



Now, taking the composition of the atmosphere to be approximately 78.1% N2 (28.014 g/mol), 20.9% 02 (31.998 g/mol), and 0.9% Ar (39.948 g/mol), we get an average molecular weight of 28.926 g/mol.



Converting the density (Avogadro's and average molecular weight from above), we get a density of 0.00144 $frac{g}{m^3}$.



Now I am assuming that the density of the atmosphere decreases exponentially proportionate to height. That is, $d(h)=e^{lambda h}$ and calculated lambda using my density and the radius of the earth, $d(6000000)=0.00144=e^{6000000lambda}$. Solving for the proportionality constant we get $lambda = -1.09times10^{-6}$. Note, prior to writing this question I had been using my original given density to calculate this proportionality constant and that was not valid. However, this is also wrong.



Taking the 'top' of the atmosphere to be 500 km above the Earth, and plugging this into a triple integral we get:



$int_0^{2pi} int_0^pi int_{6000000}^{6500000} r^2e^{(-1.09times10^{-6}r)}sinphi, dr, dphi, dtheta$



I have been using Wolfram Alpha to compute, but have not been able to finagle an answer that is even close to correct.



Is there anything glaringly wrong with what I have done thus far? I am deeply suspicious of my treatment of the density function, and also of my units. Intuitively, this feels like a valid (albeit complicated) approach to the problem. I am not very interested in a simpler approach, but rather want to know where I have gone wrong. My calculus is also quite rusty.



This is my first StackExchange post, so thank you beforehand for any help.










share|cite|improve this question









$endgroup$












  • $begingroup$
    I wanted to come back and point out that the number density of air at sea level (which was given to me) is 6 orders of magnitude off. This is apparent in the 0.00144 g/m^3 calulated mass density (again, 6 orders of magnitude off).
    $endgroup$
    – N Lowe
    Jan 19 at 19:49
















4












$begingroup$


I have been given the task of calulating the mass of the atmosphere given the following (not quite correct) information. It is intended to be a thought experiment for an undergraduate course in atmospheric science. I recognize that I have chosen a difficult way of doing this but I believe it should still work, and I am confused as to where I may be going wrong in my reasoning.



Radius of Earth: 6000 km.



Density of Atmosphere at sea level: $3times10^{19}$ $frac{molecules}{m^3}$.



Now, taking the composition of the atmosphere to be approximately 78.1% N2 (28.014 g/mol), 20.9% 02 (31.998 g/mol), and 0.9% Ar (39.948 g/mol), we get an average molecular weight of 28.926 g/mol.



Converting the density (Avogadro's and average molecular weight from above), we get a density of 0.00144 $frac{g}{m^3}$.



Now I am assuming that the density of the atmosphere decreases exponentially proportionate to height. That is, $d(h)=e^{lambda h}$ and calculated lambda using my density and the radius of the earth, $d(6000000)=0.00144=e^{6000000lambda}$. Solving for the proportionality constant we get $lambda = -1.09times10^{-6}$. Note, prior to writing this question I had been using my original given density to calculate this proportionality constant and that was not valid. However, this is also wrong.



Taking the 'top' of the atmosphere to be 500 km above the Earth, and plugging this into a triple integral we get:



$int_0^{2pi} int_0^pi int_{6000000}^{6500000} r^2e^{(-1.09times10^{-6}r)}sinphi, dr, dphi, dtheta$



I have been using Wolfram Alpha to compute, but have not been able to finagle an answer that is even close to correct.



Is there anything glaringly wrong with what I have done thus far? I am deeply suspicious of my treatment of the density function, and also of my units. Intuitively, this feels like a valid (albeit complicated) approach to the problem. I am not very interested in a simpler approach, but rather want to know where I have gone wrong. My calculus is also quite rusty.



This is my first StackExchange post, so thank you beforehand for any help.










share|cite|improve this question









$endgroup$












  • $begingroup$
    I wanted to come back and point out that the number density of air at sea level (which was given to me) is 6 orders of magnitude off. This is apparent in the 0.00144 g/m^3 calulated mass density (again, 6 orders of magnitude off).
    $endgroup$
    – N Lowe
    Jan 19 at 19:49














4












4








4





$begingroup$


I have been given the task of calulating the mass of the atmosphere given the following (not quite correct) information. It is intended to be a thought experiment for an undergraduate course in atmospheric science. I recognize that I have chosen a difficult way of doing this but I believe it should still work, and I am confused as to where I may be going wrong in my reasoning.



Radius of Earth: 6000 km.



Density of Atmosphere at sea level: $3times10^{19}$ $frac{molecules}{m^3}$.



Now, taking the composition of the atmosphere to be approximately 78.1% N2 (28.014 g/mol), 20.9% 02 (31.998 g/mol), and 0.9% Ar (39.948 g/mol), we get an average molecular weight of 28.926 g/mol.



Converting the density (Avogadro's and average molecular weight from above), we get a density of 0.00144 $frac{g}{m^3}$.



Now I am assuming that the density of the atmosphere decreases exponentially proportionate to height. That is, $d(h)=e^{lambda h}$ and calculated lambda using my density and the radius of the earth, $d(6000000)=0.00144=e^{6000000lambda}$. Solving for the proportionality constant we get $lambda = -1.09times10^{-6}$. Note, prior to writing this question I had been using my original given density to calculate this proportionality constant and that was not valid. However, this is also wrong.



Taking the 'top' of the atmosphere to be 500 km above the Earth, and plugging this into a triple integral we get:



$int_0^{2pi} int_0^pi int_{6000000}^{6500000} r^2e^{(-1.09times10^{-6}r)}sinphi, dr, dphi, dtheta$



I have been using Wolfram Alpha to compute, but have not been able to finagle an answer that is even close to correct.



Is there anything glaringly wrong with what I have done thus far? I am deeply suspicious of my treatment of the density function, and also of my units. Intuitively, this feels like a valid (albeit complicated) approach to the problem. I am not very interested in a simpler approach, but rather want to know where I have gone wrong. My calculus is also quite rusty.



This is my first StackExchange post, so thank you beforehand for any help.










share|cite|improve this question









$endgroup$




I have been given the task of calulating the mass of the atmosphere given the following (not quite correct) information. It is intended to be a thought experiment for an undergraduate course in atmospheric science. I recognize that I have chosen a difficult way of doing this but I believe it should still work, and I am confused as to where I may be going wrong in my reasoning.



Radius of Earth: 6000 km.



Density of Atmosphere at sea level: $3times10^{19}$ $frac{molecules}{m^3}$.



Now, taking the composition of the atmosphere to be approximately 78.1% N2 (28.014 g/mol), 20.9% 02 (31.998 g/mol), and 0.9% Ar (39.948 g/mol), we get an average molecular weight of 28.926 g/mol.



Converting the density (Avogadro's and average molecular weight from above), we get a density of 0.00144 $frac{g}{m^3}$.



Now I am assuming that the density of the atmosphere decreases exponentially proportionate to height. That is, $d(h)=e^{lambda h}$ and calculated lambda using my density and the radius of the earth, $d(6000000)=0.00144=e^{6000000lambda}$. Solving for the proportionality constant we get $lambda = -1.09times10^{-6}$. Note, prior to writing this question I had been using my original given density to calculate this proportionality constant and that was not valid. However, this is also wrong.



Taking the 'top' of the atmosphere to be 500 km above the Earth, and plugging this into a triple integral we get:



$int_0^{2pi} int_0^pi int_{6000000}^{6500000} r^2e^{(-1.09times10^{-6}r)}sinphi, dr, dphi, dtheta$



I have been using Wolfram Alpha to compute, but have not been able to finagle an answer that is even close to correct.



Is there anything glaringly wrong with what I have done thus far? I am deeply suspicious of my treatment of the density function, and also of my units. Intuitively, this feels like a valid (albeit complicated) approach to the problem. I am not very interested in a simpler approach, but rather want to know where I have gone wrong. My calculus is also quite rusty.



This is my first StackExchange post, so thank you beforehand for any help.







calculus integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 18 at 2:28









N LoweN Lowe

234




234












  • $begingroup$
    I wanted to come back and point out that the number density of air at sea level (which was given to me) is 6 orders of magnitude off. This is apparent in the 0.00144 g/m^3 calulated mass density (again, 6 orders of magnitude off).
    $endgroup$
    – N Lowe
    Jan 19 at 19:49


















  • $begingroup$
    I wanted to come back and point out that the number density of air at sea level (which was given to me) is 6 orders of magnitude off. This is apparent in the 0.00144 g/m^3 calulated mass density (again, 6 orders of magnitude off).
    $endgroup$
    – N Lowe
    Jan 19 at 19:49
















$begingroup$
I wanted to come back and point out that the number density of air at sea level (which was given to me) is 6 orders of magnitude off. This is apparent in the 0.00144 g/m^3 calulated mass density (again, 6 orders of magnitude off).
$endgroup$
– N Lowe
Jan 19 at 19:49




$begingroup$
I wanted to come back and point out that the number density of air at sea level (which was given to me) is 6 orders of magnitude off. This is apparent in the 0.00144 g/m^3 calulated mass density (again, 6 orders of magnitude off).
$endgroup$
– N Lowe
Jan 19 at 19:49










1 Answer
1






active

oldest

votes


















1












$begingroup$

Two issues that I see:




  1. You calculated $lambda$ incorrectly

  2. You can simplify the integral


First, the Earth is not made up of gas. The exponential drop in density applies to distances above the Earth. If the density of the atmosphere is $0.00144 frac{g}{m^3}$ at sea level, this implies:



$$Ce^{lambda 0} = 0.00144$$



not



$$Ce^{lambda 6,000,000} = 0.0014$$



Unfortunately, this means that you need one more data point (at some distance above sea level) to calculate $lambda$ (since below sea level we have water/rock, not gas).



Second, there is a symmetry about $r$ if we treat the Earth as a sphere. You can take advantage of this to integrate the mass of the atmosphere using "shells" of radius $6000000 + h$ ($h$ is height above sea level). This reduces a triple integral to a single integral:



$$M_{atm} = int_{6,000,000}^{6,500,000} Ce^{lambda h}4pi (6000000+h)^2dh$$



This is a realtively easy one for WolframAlpha...just need more data to get $lambda$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! I knew it would be something silly. You are correct, the Earth is not made up of gas. Do you have any suggestions for how to go about calculating a correct value for lambda? I have been sitting here scratching my head over it. Does that invalidate my entire approach? I cannot think of a way to cause the exponential function to 'decay starting at 6,000,000. You mention a second data point but I do not see how that would help the situation. This is intended to be an extremely rough approximation, I can invent an approximate data point.
    $endgroup$
    – N Lowe
    Jan 18 at 3:35










  • $begingroup$
    @NLowe yes, it invalidates your approach. Why not set the density at 1000km above sea level to that of interplanetary space (5 atoms Hydrogen per cubic cm)? The equation to solve will be $0.00144e^{lambda 1000000}=d_{space}$
    $endgroup$
    – Bey
    Jan 18 at 4:02














Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077767%2fcalculating-the-mass-of-the-atmosphere-using-a-triple-integral%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Two issues that I see:




  1. You calculated $lambda$ incorrectly

  2. You can simplify the integral


First, the Earth is not made up of gas. The exponential drop in density applies to distances above the Earth. If the density of the atmosphere is $0.00144 frac{g}{m^3}$ at sea level, this implies:



$$Ce^{lambda 0} = 0.00144$$



not



$$Ce^{lambda 6,000,000} = 0.0014$$



Unfortunately, this means that you need one more data point (at some distance above sea level) to calculate $lambda$ (since below sea level we have water/rock, not gas).



Second, there is a symmetry about $r$ if we treat the Earth as a sphere. You can take advantage of this to integrate the mass of the atmosphere using "shells" of radius $6000000 + h$ ($h$ is height above sea level). This reduces a triple integral to a single integral:



$$M_{atm} = int_{6,000,000}^{6,500,000} Ce^{lambda h}4pi (6000000+h)^2dh$$



This is a realtively easy one for WolframAlpha...just need more data to get $lambda$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! I knew it would be something silly. You are correct, the Earth is not made up of gas. Do you have any suggestions for how to go about calculating a correct value for lambda? I have been sitting here scratching my head over it. Does that invalidate my entire approach? I cannot think of a way to cause the exponential function to 'decay starting at 6,000,000. You mention a second data point but I do not see how that would help the situation. This is intended to be an extremely rough approximation, I can invent an approximate data point.
    $endgroup$
    – N Lowe
    Jan 18 at 3:35










  • $begingroup$
    @NLowe yes, it invalidates your approach. Why not set the density at 1000km above sea level to that of interplanetary space (5 atoms Hydrogen per cubic cm)? The equation to solve will be $0.00144e^{lambda 1000000}=d_{space}$
    $endgroup$
    – Bey
    Jan 18 at 4:02


















1












$begingroup$

Two issues that I see:




  1. You calculated $lambda$ incorrectly

  2. You can simplify the integral


First, the Earth is not made up of gas. The exponential drop in density applies to distances above the Earth. If the density of the atmosphere is $0.00144 frac{g}{m^3}$ at sea level, this implies:



$$Ce^{lambda 0} = 0.00144$$



not



$$Ce^{lambda 6,000,000} = 0.0014$$



Unfortunately, this means that you need one more data point (at some distance above sea level) to calculate $lambda$ (since below sea level we have water/rock, not gas).



Second, there is a symmetry about $r$ if we treat the Earth as a sphere. You can take advantage of this to integrate the mass of the atmosphere using "shells" of radius $6000000 + h$ ($h$ is height above sea level). This reduces a triple integral to a single integral:



$$M_{atm} = int_{6,000,000}^{6,500,000} Ce^{lambda h}4pi (6000000+h)^2dh$$



This is a realtively easy one for WolframAlpha...just need more data to get $lambda$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks! I knew it would be something silly. You are correct, the Earth is not made up of gas. Do you have any suggestions for how to go about calculating a correct value for lambda? I have been sitting here scratching my head over it. Does that invalidate my entire approach? I cannot think of a way to cause the exponential function to 'decay starting at 6,000,000. You mention a second data point but I do not see how that would help the situation. This is intended to be an extremely rough approximation, I can invent an approximate data point.
    $endgroup$
    – N Lowe
    Jan 18 at 3:35










  • $begingroup$
    @NLowe yes, it invalidates your approach. Why not set the density at 1000km above sea level to that of interplanetary space (5 atoms Hydrogen per cubic cm)? The equation to solve will be $0.00144e^{lambda 1000000}=d_{space}$
    $endgroup$
    – Bey
    Jan 18 at 4:02
















1












1








1





$begingroup$

Two issues that I see:




  1. You calculated $lambda$ incorrectly

  2. You can simplify the integral


First, the Earth is not made up of gas. The exponential drop in density applies to distances above the Earth. If the density of the atmosphere is $0.00144 frac{g}{m^3}$ at sea level, this implies:



$$Ce^{lambda 0} = 0.00144$$



not



$$Ce^{lambda 6,000,000} = 0.0014$$



Unfortunately, this means that you need one more data point (at some distance above sea level) to calculate $lambda$ (since below sea level we have water/rock, not gas).



Second, there is a symmetry about $r$ if we treat the Earth as a sphere. You can take advantage of this to integrate the mass of the atmosphere using "shells" of radius $6000000 + h$ ($h$ is height above sea level). This reduces a triple integral to a single integral:



$$M_{atm} = int_{6,000,000}^{6,500,000} Ce^{lambda h}4pi (6000000+h)^2dh$$



This is a realtively easy one for WolframAlpha...just need more data to get $lambda$.






share|cite|improve this answer











$endgroup$



Two issues that I see:




  1. You calculated $lambda$ incorrectly

  2. You can simplify the integral


First, the Earth is not made up of gas. The exponential drop in density applies to distances above the Earth. If the density of the atmosphere is $0.00144 frac{g}{m^3}$ at sea level, this implies:



$$Ce^{lambda 0} = 0.00144$$



not



$$Ce^{lambda 6,000,000} = 0.0014$$



Unfortunately, this means that you need one more data point (at some distance above sea level) to calculate $lambda$ (since below sea level we have water/rock, not gas).



Second, there is a symmetry about $r$ if we treat the Earth as a sphere. You can take advantage of this to integrate the mass of the atmosphere using "shells" of radius $6000000 + h$ ($h$ is height above sea level). This reduces a triple integral to a single integral:



$$M_{atm} = int_{6,000,000}^{6,500,000} Ce^{lambda h}4pi (6000000+h)^2dh$$



This is a realtively easy one for WolframAlpha...just need more data to get $lambda$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 18 at 12:11

























answered Jan 18 at 2:59









BeyBey

1464




1464












  • $begingroup$
    Thanks! I knew it would be something silly. You are correct, the Earth is not made up of gas. Do you have any suggestions for how to go about calculating a correct value for lambda? I have been sitting here scratching my head over it. Does that invalidate my entire approach? I cannot think of a way to cause the exponential function to 'decay starting at 6,000,000. You mention a second data point but I do not see how that would help the situation. This is intended to be an extremely rough approximation, I can invent an approximate data point.
    $endgroup$
    – N Lowe
    Jan 18 at 3:35










  • $begingroup$
    @NLowe yes, it invalidates your approach. Why not set the density at 1000km above sea level to that of interplanetary space (5 atoms Hydrogen per cubic cm)? The equation to solve will be $0.00144e^{lambda 1000000}=d_{space}$
    $endgroup$
    – Bey
    Jan 18 at 4:02




















  • $begingroup$
    Thanks! I knew it would be something silly. You are correct, the Earth is not made up of gas. Do you have any suggestions for how to go about calculating a correct value for lambda? I have been sitting here scratching my head over it. Does that invalidate my entire approach? I cannot think of a way to cause the exponential function to 'decay starting at 6,000,000. You mention a second data point but I do not see how that would help the situation. This is intended to be an extremely rough approximation, I can invent an approximate data point.
    $endgroup$
    – N Lowe
    Jan 18 at 3:35










  • $begingroup$
    @NLowe yes, it invalidates your approach. Why not set the density at 1000km above sea level to that of interplanetary space (5 atoms Hydrogen per cubic cm)? The equation to solve will be $0.00144e^{lambda 1000000}=d_{space}$
    $endgroup$
    – Bey
    Jan 18 at 4:02


















$begingroup$
Thanks! I knew it would be something silly. You are correct, the Earth is not made up of gas. Do you have any suggestions for how to go about calculating a correct value for lambda? I have been sitting here scratching my head over it. Does that invalidate my entire approach? I cannot think of a way to cause the exponential function to 'decay starting at 6,000,000. You mention a second data point but I do not see how that would help the situation. This is intended to be an extremely rough approximation, I can invent an approximate data point.
$endgroup$
– N Lowe
Jan 18 at 3:35




$begingroup$
Thanks! I knew it would be something silly. You are correct, the Earth is not made up of gas. Do you have any suggestions for how to go about calculating a correct value for lambda? I have been sitting here scratching my head over it. Does that invalidate my entire approach? I cannot think of a way to cause the exponential function to 'decay starting at 6,000,000. You mention a second data point but I do not see how that would help the situation. This is intended to be an extremely rough approximation, I can invent an approximate data point.
$endgroup$
– N Lowe
Jan 18 at 3:35












$begingroup$
@NLowe yes, it invalidates your approach. Why not set the density at 1000km above sea level to that of interplanetary space (5 atoms Hydrogen per cubic cm)? The equation to solve will be $0.00144e^{lambda 1000000}=d_{space}$
$endgroup$
– Bey
Jan 18 at 4:02






$begingroup$
@NLowe yes, it invalidates your approach. Why not set the density at 1000km above sea level to that of interplanetary space (5 atoms Hydrogen per cubic cm)? The equation to solve will be $0.00144e^{lambda 1000000}=d_{space}$
$endgroup$
– Bey
Jan 18 at 4:02




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077767%2fcalculating-the-mass-of-the-atmosphere-using-a-triple-integral%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Human spaceflight

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

File:DeusFollowingSea.jpg