- #1

rlmurra2

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The only thing I can think of is to subtract something from the mass of the entire Earth or something...

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- Thread starter rlmurra2
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- #1

rlmurra2

- 19

- 0

The only thing I can think of is to subtract something from the mass of the entire Earth or something...

- #2

Astronuc

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rlmurra2 said:What is the mass of the Earth's atmosphere? The radius of the Earth is 6.4E6m.

The atmosphere is a thin band of gas surrounding a solid/liquid earth, but one can use thin shell method of calculating the thickness of that band.

So V = [itex]\int_{R_i}^{R_o} 4\pi\,\rho(r)\,r^2\,dr[/itex]

or V = [itex]4\pi\,R^2\,\int_0^H \rho(z)\,dz[/itex], where R would be the mean radius of the atmosphere referenced from the center of the earth.

Then one needs to integrate as a function of altitude, since density decreases with increase in altitude.

Height of Earth's atmosphere - http://www.rcn27.dial.pipex.com/cloudsrus/atmosphere.html

http://en.wikipedia.org/wiki/Earth's_atmosphere

That should give you enough information.

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- #3

Bystander

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- #4

NateTG

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Easy (this is probably what you want to do)- determine the difference in the acceleration due to gravity at the high and low ends of the atmosphere. This will (probably) allow you to make a very nice simplifying assumption so you can get a good approximation quickly and easily using the surface air pressure, the acceleration of gravity, and the surface area of the earth.

Medium - Integrate by shells assuming that the Earth is spherical, and the temperature of the atmosphere is constant. Remember that the density is proportional to the pressure.

Hard - Integrate but account for the fact that the Earth is a spinning elipsoid and for temperature with respect to lattititude and altititude.

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