Atmospheric physics 1: Mass of the atmosphere
The atmosphere is a layer of gas that surrounds our planet and is several hundred kilometers thick. Its main components are nitrogen, oxygen, argon, and water vapor.
What is the approximate mass m of the entire Earth's atmosphere?
Details and Assumptions:
- The circumference of the Earth is about 4 0 , 0 0 0 km .
- Gravitational acceleration is approximately 1 0 m/s 2 .
- The mean air pressure on the Earth's surface is about 1 bar ( 1 b a r = 1 0 5 P a ) .
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3 solutions
Finally you post a manageable problem (at least for me)! I really liked it. :)
Note: This doesn't reflect the reality of the mass of the earth's atmosphere, since our assumptions are simple, and only provides a rough estimation.
It is quite close though http://www.wolframalpha.com/input/?i=mass+of+earth+atmosphere
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Wow, that was waaay closer than I expected to be!
the difference being that small clearly indicates that the assumptions are valid
Earth is flat.
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TOTALLY agree with you
obviously, we know
I introduce to you: The earth really is in the shape of a cube. The one who made Minecraft got the idea from somewhere.
Can't we do it with the formula..... HDg
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The mass density $D$ is not constant in the atmosphere, so that would be more complicated.
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No this is not the reason behind why we cannot use hdg, because a figure of mean density of the atmosphere would also yield the correct answer. The actual reason is that, in questions from brilliant, you have to use magnitudes from the question itself and not your previous knowledge or the internet. So, as h (or height of the atmosphere) was not provided, we could not use this formula.
The density isn't constant in the atmosphere and this formula is used to find pressure in liquid column not in atmosphere
Ye, I got the same solution: 5* 10^18 But why is 5* 10^21 marked as correct? Did I miss something?
P=f/A A=area where force is applied and in the solution he took the curved surface area. So the solution is wrong. In fact the question is wrong in practical way
No need of complicated calculations.
Area of Earth's Surface = 4 x 3.14 x (6.370.000 ^2) sqm = 5 x 10^14 sqm
Atmospheric pressure = a column of water of 10 m = 10 ton/sqm
Total weight of atmosphere = 5 x 10^15 tons
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Whats complicated about the solution? Putting number in before the final calculation never gives a simplest solution. Math is about logic creativity and symbols, not numbers and calculations
you are assuming a constant gravitational field where it actually varies by 20% from beginning to end. Great explanation otherwise.
oh my god. i made a mistake in calculation. we have to consider the change in gravitational field.the effect cannot be ignored.
How did you get pi mg/C^2 from the m g/4 pi r^2 ?
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Circumference is given by 2 π r , so 4 π R 2 m g = 4 π 2 R 2 π m g = ( 2 π R ) 2 π m g = C 2 π m g
But C^2 is not 4pir^2 is it its 4(pir)^2
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Note that he has added a factor of π in the numerator.
Doh missed the pi on the numerator!
Wait. How is area = 4 pi R^2 ?
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This is the surface area of a sphere with radius R. The formula was first discovered by Archimedes and engraved on his tombstone.
Consider powers of 10 for a shortcut.
C = 40, 000, 000 m implies radius has 7 zeroes.
Squaring radius for surface area gives 14 zeroes.
Multiply by pressure for force gives 19 zeroes.
Divide by g for answer - 18 zeroes. Answer is B.
Very good end around! You da bomb! My math is sooooooooooo weak! The concepts and visualizing, I have, but them lil' symbols and stuff stymie me. Need more HARD work to compensate for my shortfalls!
Thanks a lot for this answer.
I kinda think there is a useless condition ,but lack the R of the earth.
Circumference can let you find the radius of the earth.
Pressure is force per area, so that P = F / A . The force F = m g is the weight of the air in the gravitational field of the earth. The area A = 4 π R 2 is the earth's surface, which we assume as a sphere with radius R and circumference C = 2 π R . Therefore,
P ⇒ m = 4 π R 2 m g = C 2 π m g = π g C 2 P ≈ π ⋅ 1 0 m / s 2 ( 4 ⋅ 1 0 7 m ) 2 ⋅ ( 1 0 5 Pa ) = π 1 6 ⋅ 1 0 1 8 kg ≈ 5 ⋅ 1 0 1 8 kg