Homework Statement
We have to calculate the mass concentration for a particle size distribution. I'm just doing the first size range here, since once I get that the rest is just repetition.
FYI, the particles are spherical.
Size range (diameter) = 0.005-0.5 μm
Average particle size...
Homework Statement
A small, perfectly black, spherical satellite is in orbit around the Earth. If the Earth radiates as a black body at an equivalent blackbody temperature T_{E} = 255 K, calculate the radiative equilibrium temperature of the satellite when it is in the Earth’s shadow. Start by...
Yeah, I used the scale height formula H=\frac{RT_{0}}{g} and rearranged to plug in for T_{0} because he didn't give a specific temperature value.
The only problem I see arising is that even with the previous integral I evaluated, I couldn't get an answer because when I tried to evaluate...
I set up the integral as \int_{a}^{\infty }4\pi a^{2}\rho\, da, but when I evaluated it, it came out to zero.
I spoke to my professor and he said my error is in the density function I plugged in...
I was plugging it in as \rho =\frac{p_{s}}{Hg}e^{-a/H}, while the textbook has it as \rho...
When I run through the integration, I simply get what I had... So apparently I didn't even need to integrate to get the volume, and just assume V=\frac{4\pi a^{3}}{3}.
But then I have the density distribution, which I believe I have to integrate from zero to infinity?
No, actually he said the mass is too small given that equation... He showed an image where you integrate in cylindrical columns from the surface to the top of the atmosphere, and due to the curvature of the Earth, they diverge as you get higher up so that there are spaces in between the...
Running through the integration in my previous post didn't work out, as it became zero when I integrated phi and evaluated it from 0 to 2*pi... Guess I'll take another stab at this question tomorrow. :frown:
I've attempted to set up the integral to calculate the volume of the entire atmosphere...
\int_{\theta =0}^{2\pi }\int_{\phi =0}^{2\pi}\int_{a}^{\infty }4\pi a^{2}\sin \phi\, da\, d\phi\, d\theta
Are my limits of integration for a correct? What's throwing me off is that we're...
Homework Statement
Consider an isothermal atmosphere with temperature T_{0}. 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 \frac{4\pi a^{2}p_{s}}{g}.
If the scale height of this isothermal atmosphere is 7 km, by...
I was incorrect in my previous post saying that the units didn't work out... Solving for {\rho _{v}} when e=e_{s} does produce an answer in \frac{kg}{m^{3}}... However I'm getting 1,103,248.397 \frac{kg}{m^{3}}, for the first case where T=288 K, which is way off from what I should be getting...
Homework Statement
Compute the maximum amount of water vapor per unit volume that air can hold at the surface, where Ts = 288 K, and at a height of 10 km where T = 220 K. Express your answers in kg m-3.
Homework Equations
e_{s}=Ae^{\beta T}
e=\rho _{v}R_{v}T
The Attempt at a...
I guess I could only use the 2nd equation then, because I have the molar masses...
Is there a way I could have determined the number of moles of each gas at that height in the atmosphere?
Well, I did a lot of work on this question today, including going to my professor's office hours, and I've gotten through most of it but still have a few questions.
I ended up having to use the equation p=p_{0}e^{-\frac{h}{h_{0}}} to relate the pressure and height. Assume p_{0}=100,000 Pa...
Homework Statement
At which minimum height atoms or molecules are able to leave Earth’s
atmosphere to space? To obtain an approximate estimate use the assumption that molecules can leave if the mean free path, λ, is larger than the scale height H. Write λ and H as a function of height and...
Thanks for the help!
The final equation I got was:
(3.67/(0.7679-h)) + 133357.14h=100591
Putting it into Mathematica, I got h=0.7525 m=752.5 mm=29.63 in., which makes sense...
However, I tried solving it algebraically as a quadratic equation and did not get that answer... I got...