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Fermi Estimate with atmospheric pressure

  1. Sep 23, 2009 #1
    1. There are many different definitions of the location of the top of the atmosphere – the near edge of
    outer space.
    • Until around 1600, it was believed that the Earth’s atmosphere extended throughout space
    (recall that the Earth was also considered to be the center of the Universe until around then).
    • In the BOREALIS ballooning program, we claim we fly student payloads to the edge of space,
    since we reach altitudes of around 110,000 feet (approximately 30 km).
    • The X-prize competition awarded Burt Rutan on October 4, 2004, $10,000,000 for the first
    spaceship to carry three people to an altitude of greater than 100 km twice within two weeks.

    Look up the radius of the Earth and atmospheric pressure at sea level, and estimate the height, H, of the atmosphere, assuming the following:
    • At sea level, the weight of an object of mass m is mg, where g = 9.80665m/s^2 is the acceleration due to gravity. Atmospheric pressure is the weight of the air above us per unit area.
    • The density of air at sea level is 1.3 kg/m^3, about 1000 times less than the density of water.
    • Take the atmosphere to be a spherical shell of constant density up to H after which the density rapidly falls to zero. (In reality, the density of the atmosphere decreases smoothly with height.)






    This is a Fermi Estimate, and it should be quite easy, as all the other homework problems were. I think I am just missing something.



    The diameter of the Earth is 6,378km
    The pressure at sea level is 101.325kPa

    I am not sure where to start, and I don't see why it matters that the density of air at sea level is about 1000 times less than the density of water.

    If you can help me get started, then thanks a lot!
     
  2. jcsd
  3. Sep 24, 2009 #2

    Redbelly98

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    Welcome to PF.

    The key here seems to be:

    • You know the density of air.
    • Can you find the total weight of the entire atmosphere, using the pressure information?
     
  4. Sep 24, 2009 #3
    weight = m*g = (1.3kg/m^3)(9.80665m/s) = 12.748645
    What unit is this weight (is it kg?), and is it for a cubic meter?
    I'm thinking that the weight is 12.748645kg/m^3, but I am not sure where the m/s comes in.
     
  5. Sep 24, 2009 #4

    Redbelly98

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    Weight is a force , so its units are Newtons or N. A Newton is equivalent to kg*m/s^2.

    Since weight is a force, try using the usual equation that relates pressure, force and area. The "force" here will be the weight of the air.

    p.s. I'm logging off for the night, good luck!
     
  6. Sep 25, 2009 #5
    OK, so I got the surface area of the Earth:

    (4)(pi)(r2) at 40,678,884km2.

    I multiplied Earth's surface area by the atmospheric pressure at sea level to get the weight of Earth's atmosphere.

    40,678,884km2 x 101.325kPa = 4,121,787,921.3

    Dividing the weight by the acceleration due to gravity at the Earth's surface should give the mass of Earth's atmosphere, 420,305,396.98

    Knowing the weight of a m3 I can tell the Height of the atmosphere.

    Am I on the right track?
     
  7. Sep 25, 2009 #6

    Redbelly98

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    You're getting there, but there are some things to "clean up" in your calculation. See my following comments.

    That is r^2. You need to calculate 4 pi r^2.
    Also, I recommend:
    • Using m, not km.
    • Using scientific notation for large numbers like this. For example, the above number is 4.068x107 km^2, or one can write 4.068e7 km^2, in scientific notation.

    In what units? We tend to let the units slide when we agree to use MKS (m, kg, s, which implies N for force or weight), but since this is not in Newtons it would be good to say what the units are ... OR do the calculation in m, kg, s so that the weight IS in Newtons.

    Units?

    Actually, we know the mass of a m3. That's what the density of air tells us.

    Pretty much, yes.
     
  8. Sep 28, 2009 #7
    I converted to metres.
    Here's what I have:
    Density of air at sea level: [tex]1.3kg/m^{3}[/tex]
    Surface area of the Earth: 5e14m
    Air Pressure: 101.325kPa = [tex]101,325N/m^{2}[/tex]
    Total weight (force) of atmosphere (pressure x area): 5.06625e19 N
    Convert newtons to kilograms: 5.06625e19 N / 9.8 = 5.17e18kg
    Total weight of Earth's atmosphere: 5.17e18 kg

    Can I just divide the total weight in kg by the density of a cubic metre of air at sea level to get the Height H of our theoretical atmosphere in metres?

    5.17e18 / 1.3 = 3.98e18 m

    This number seems way to large.
     
  9. Sep 28, 2009 #8
    You need to start keeping track of your units. Doing that would have let you on to the fact that that answer is nonsense since it is in m³. You found the volume of earth's
    atmosphere.

    Now you need to apply some basic geometry to see what the height of the atmosphere is.
    Remember your original data, that the atmosphere forms a spherical shell around the earth.
     
  10. Sep 28, 2009 #9
    Thanks, so I will back up a step.
    I will divide the total weight of the atmosphere by the surface area of Earth:
    5.17e18kg / [tex]5e14m^{2}[/tex] = [tex]10340kg/m^{2}[/tex]
    So if I'm right, now I have the weight of a column of atmosphere on a square metre.

    So my next step is to figure out how high the column is.
    I guess there is a few ways of doing this, but I am hellbent on using volume. So:

    The volume of this column is weight/density:

    [tex]10340kg/m^{2}[/tex]/[tex]1.3kg/m^{3}[/tex] = (about) [tex]7953m^{3}[/tex]

    If I did it right, a column will be a square metre around, and [tex]7953m[/tex] tall.
     
  11. Sep 28, 2009 #10
    Again, be sure to keep track of your units. You got an answer in meters, but wrote it as m³

    I'd also like to mention that your method is much smarter than mine.

    I calculated the volume of the atmosphere as the difference between two spheres. One of radius [tex]R_e+h[/tex] and one of radius [tex]R_e[/tex].
    What I ended up with was a 3rd degree polynomial, whose one real solution was around 7770m. I suspect that the difference between the two answers is merely a question of truncation and rounding.

    Well done. :)
     
  12. Sep 28, 2009 #11
    Ah, that's a clever way to do it.

    Thanks for all the help. I love how simple physics can calculate epic things.
     
  13. Sep 29, 2009 #12
    Actually, upon further thought, I think the difference in our answers is the result of an approximation you made, but which I avoided.

    Let's just start with what we can agree on:

    [tex]V=4\pi R_E^2 \frac{P}{\rho g}[/tex]

    You said that the volume equals the surface area times the height:

    [tex]V=4\pi R_E^2*h_1[/tex]

    [tex]h_1=\frac{V}{4\pi R_E^2}=\frac{P}{\rho g}[/tex]

    Plugging in our values we get:
    [tex]h_1\approx 7953 m[/tex]

    This is an interesting result, as it is independent of the radius of the earth.
    This would be true, if the earth were flat and had a total surface area of [tex]S=4\pi R_E^2[/tex]

    But the earth is round, and not flat:

    [tex]V=\tfrac{4\pi}{3}(R_E+h_2)^3-\tfrac{4\pi}{3}R_E^3[/tex]

    [tex]h_2=(\tfrac{V}{\tfrac{4\pi}{3}}+R_E^3)^{1/3}-R_E[/tex]

    [tex]h_2=R_E(\sqrt[3]{\frac{3P}{\rho g R_E}+1}-1)[/tex]

    Plugging in our values we get:

    [tex]h_2\approx 7935 m[/tex]

    [tex]\frac{h_1}{h_2}=1.0022[/tex]

    As you can see, there is a critical difference between the two approaches, though the difference is fairly small. They are not equivalent.

    The difference stems from spherical surface area compared to flat surface area. I don't know much about the subject, but the way I see it, a column of height [tex]r[/tex] resting on an area of [tex]1 m^2[/tex] will have a bigger volume if that area is part of a sphere as opposed to a flat surface. Think of how a column starting from the surface of a sphere will sort of 'fan out' as opposed to just going straight up from a straight surface.
     
    Last edited: Sep 29, 2009
  14. Sep 29, 2009 #13
    Yeah, I see what you mean.

    In your equation you used

    [tex]\frac{P}{\rho g}[/tex]

    Is p volume and pg surface area?
     
  15. Sep 29, 2009 #14
    [tex]P[/tex] is the atmospheric pressure (101325 Pa)
    [tex]\rho[/tex] is the density of earth's atmosphere (I took your data of 1.3 kg/m³)
    [tex]g[/tex] is gravity (9.8 m/s²)
     
  16. Sep 29, 2009 #15

    Redbelly98

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    [/URL]

    Interestingly, that is less than the height of Mt. Everest (8848 m above sea level).
     
    Last edited by a moderator: May 4, 2017
  17. Sep 29, 2009 #16
    Well, I tried it your way, and got a different number.
    Here's my step by step:

    The radius of the Earth is 6,370,000m^2
    Surface area of Earth is 4πr^2 = 5.1e14m^2
    Now I will multiply Earth's surface area by the atmospheric pressure at sea level to get the weight of Earth's atmosphere:
    5.1e14m^2 x 101,325N/m^2 = 5.167575e19N
    Next I will convert Newtons to kilograms:
    5.167575e19N / 9.8m/s = 5.27e18kg
    Total weight of the Earth’s atmosphere is 5.27e18kg.
    Next I want the atmosphere’s volume:
    5.27e18kg / 1.3kg/m^3 = 4.05e18m^3
    Now I will find the Earth’s volume by using the formula for a sphere:
    (4/3)πr^3 = 1.08e21
    By adding the Earth’s volume to the atmosphere’s volume, I will get the total volume of both:
    4.05e18m^3 + 1.08e21m^3 = 1.08e21m^3
    I will use the volume formula for a sphere to figure out the radius of Earth and the atmosphere:
    V = (4/3)πr^3
    1.08e21m^3 = (4/3)πr^3
    Next I will divide both sides by (4/3)π to isolate r3:
    1.08e21m^3 / [(4/3)π] = (4/3)πr^3 / [(4/3)π]
    1.08e21m^3 / 4.19 = r^3
    2.587e20 = r^3
    [cube root](2.587e20) = [cube root](r^3)
    r = 6371849m
    Finally I will subtract Earth’s radius from the radius of both to get h, the height of the atmosphere:
    6,371,849m – 6,370,000m = 1849m
     
  18. Sep 29, 2009 #17

    Redbelly98

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    Hope you realize that you did have a valid answer back in post #9, though you'd need to round that off given that the density is known to just 2 significant figures.

    Here is where this method runs into problems. Because of roundoff error, any information about the atmosphere's volume is completely lost at this point!

    The next part of your calculation -- using this volume to get the radius of a sphere -- just reproduces the Earth's radius, correct to 3 significant figures (since the volume here is given to 3 significant figures).
     
  19. Sep 30, 2009 #18
    So you have Riedel's class too? xD Fun stuff, fun stuff.
     
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