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guitarstorm
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Homework Statement
Consider an isothermal atmosphere with temperature [itex]T_{0}[/itex]. Show that the mass of this atmosphere on a planet with the same radius at earth, a, and the same gravity, g, is greater than [itex]\frac{4\pi a^{2}p_{s}}{g}[/itex].
If the scale height of this isothermal atmosphere is 7 km, by how much does the atmospheric mass exceed the value given in the text on page 27?
Homework Equations
[itex]m_{a}=\frac{4\pi a^{2}p_{s}}{g}[/itex]
[itex]H=\frac{RT_{0}}{g}[/itex]
[itex]\rho (z)=\frac{p_{s}}{RT_{0}}e^{-z/H}[/itex]
The Attempt at a Solution
Basically, our professor wants us to disprove the atmospheric mass formula given in our textbook. He said in lecture that the actual mass is greater because the method in the textbook doesn't account for the mass between the columns due to the nature of spherical geometry.
I'm trying two different methods.
First, I thought maybe multiplying by the solid angle would account for this missed mass, and when multiplying [itex]\frac{4\pi a^{2}p_{s}}{g}[/itex] by [itex]\omega =\frac{4\pi a^{2}}{a^{2}}[/itex], I get [itex]\frac{16\pi^{2} a^{2}p_{s}}{g}[/itex].
Plugging in a = 6.37x10^6 m, [itex]p_{s}[/itex]=100000 Pa, and g=9.81 m/s^2, I get a mass of 6.53x10^19 kg, greater than the 5.20x10^18 kg obtained by the original equation... However I'm not sure if my logic is quite right here.
The other method would be to use a shell method instead... But my calculus skills are rusty and I'm not sure how to go about this.
I know that to get the volume you just integrate [itex]4\pi a^{2}[/itex] to get [itex]\frac{4}{3}\pi a^{3}[/itex].
Volume multiplied by density would give the mass... However, I believe this has to be integrated over a certain height to account for the depth of the atmosphere... This is where I'm unsure of how to proceed.
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