Consider the Earth’s atmosphere by assuming the constituent particles comprise an ideal gas, such that P “ nkBT , where n is the number density of particles (with units cm ́3), k “ 1.4ˆ10 ́16 erg K ́1 is the Boltzmann constant. We’ll use this ideal gas law in just a bit, but first
(a) Think of a small, cylindrical parcel of gas, with the axis running vertically in the Earth’s at- mosphere. The parcel sits a distance r from the Earth’s center, and the parcel’s size is defined by a height ∆r ! r and a circular cross-sectional area A (it’s okay to use r here, because it is an intrinsic property of the atmosphere). The parcel will feel pressure pushing up from gas below (Pup “ P prq) and down from above (Pdown “ P pr ` ∆rq).
(a) Think of a small, cylindrical parcel of gas, with the axis running vertically in the Earth’s at- mosphere. The parcel sits a distance r from the Earth’s center, and the parcel’s size is defined by a height ∆r ! r and a circular cross-sectional area A (it’s okay to use r here, because it is an intrinsic property of the atmosphere). The parcel will feel pressure pushing up from gas below (Pup “ P prq) and down from above (Pdown “ P pr ` ∆rq).
Make a drawing of this, and discuss the situation and the various physical parameters with your group.
(b) What other force will the parcel feel, assuming it has a density ρprq and the Earth has a mass MC?
(c) If the parcel is not moving, give a mathematical expression relating the various forces, remem- bering that force is a vector and pressure is a force per unit area.
(d) Give an expression for the gravitational acceleration, g, at at a distance r above the Earth’s center in terms of the physical variables of this situation.
(e) Show that
dP(r) “ ́gρprq (1) dr
This is the equation of hydrostatic equilibrium.
(f) Now go back to the ideal gas law described above. Derive an expression describing how the density of the Earth’s atmosphere varies with height, ρ(r)? (HINT: It may be useful to recall that dx{x “ d ln x.)
(g) Show that the height, H, over which the density falls off by an factor of 1{e is given by
H “ kT (2)
m ̄g
where m ̄ is the mean (average) mass of a gas particle. This is the “scale height.” First, check the units. Then do the math. Then make sure it makes physical sense, e.g. what do you think should happen when you increase m ̄ ? Finally, pat yourselves on the back for solving a first-order differential equation and finding a key physical result!
(h) What is the Earth’s scale height, HC? The mass of a proton is 1.7 ˆ 10 ́24 g, and the Earth’s atmosphere is mostly molecular nitrogen, N2, where atomic nitrogen has 7 protons, 7 neutrons.
(b) What other force will the parcel feel, assuming it has a density ρprq and the Earth has a mass MC?
(c) If the parcel is not moving, give a mathematical expression relating the various forces, remem- bering that force is a vector and pressure is a force per unit area.
(d) Give an expression for the gravitational acceleration, g, at at a distance r above the Earth’s center in terms of the physical variables of this situation.
(e) Show that
dP(r) “ ́gρprq (1) dr
This is the equation of hydrostatic equilibrium.
(f) Now go back to the ideal gas law described above. Derive an expression describing how the density of the Earth’s atmosphere varies with height, ρ(r)? (HINT: It may be useful to recall that dx{x “ d ln x.)
(g) Show that the height, H, over which the density falls off by an factor of 1{e is given by
H “ kT (2)
m ̄g
where m ̄ is the mean (average) mass of a gas particle. This is the “scale height.” First, check the units. Then do the math. Then make sure it makes physical sense, e.g. what do you think should happen when you increase m ̄ ? Finally, pat yourselves on the back for solving a first-order differential equation and finding a key physical result!
(h) What is the Earth’s scale height, HC? The mass of a proton is 1.7 ˆ 10 ́24 g, and the Earth’s atmosphere is mostly molecular nitrogen, N2, where atomic nitrogen has 7 protons, 7 neutrons.
Let's take it from the top:
(a) Think of a small, cylindrical parcel of gas, with the axis running vertically in the Earth’s at- mosphere. The parcel sits a distance r from the Earth’s center, and the parcel’s size is defined by a height ∆r ! r and a circular cross-sectional area A (it’s okay to use r here, because it is an intrinsic property of the atmosphere). The parcel will feel pressure pushing up from gas below (Pup “ P prq) and down from above (Pdown “ P pr ` ∆rq).
Make a drawing of this, and discuss the situation and the various physical parameters with your group.
PUp is the pressure below the slice. PDown is the pressure above the slice. Δr is the infinitesimal thickness of the slice. r is the height above the earth's surface.
(b) What other force will the parcel feel, assuming it has a density ρprq and the Earth has a mass MC?
The parcel will feel a force of gravity given by FG=ρ(r)Ag where ρ(r) is the density of the gas a a function of distance from the Earth's surface and A is the area of the slice.
(c) If the parcel is not moving, give a mathematical expression relating the various forces, remem- bering that force is a vector and pressure is a force per unit area.
If the parcel is not moving then the sum of forces upon it equals zero. So:
FPUp=FPDown+Fg
(d) Give an expression for the gravitational acceleration, g, at a distance r above the Earth’s center in terms of the physical variables of this situation.
Newton's gravitational formula states that:
Fg=GMobjectMEarthr2
We also know that Fg=Mobjectg so:
g=GMEarthr2
(e) Show that
dP(r)/r =-gp(r) dr
This is the equation of hydrostatic equilibrium.
dP(r)/r =-gp(r) dr
This is the equation of hydrostatic equilibrium.
The standard gauge pressure relation states that P=ρgh
Substituting our expression for g into this equation gives us:
P(r)=GMEarthr
Taking the first derivative of this with respect to r yields:
dP(r)dr=−GMEarthρ(r)r2
Substituting g back in gives:
dP(r)dr=−gρ(r)
(f) Now go back to the ideal gas law described above. Derive an expression describing how the density of the Earth’s atmosphere varies with height, ρ(r)?
Ideal gas law states that P=nkbT where n is the number of particles per unit volume, kb is the Boltzmann Constant which equals 1.4 times10−16erg/K, and T is the temperature in degrees Kelvin.
We can modify this by introducing the term ¯m which will be the average mass of a particle to get it back in terms of ρ(r).
P=ρ(r)¯mkbT
From above we know dP(r)dr=−gρ(r), so:
dρ(r)¯mkbTdr=−gρ(r)
dρ(r)ρ(r)=−g¯mkbTdr
Integrating both sides yields:
ln(ρ(r))=−g¯mrkbT
Integrating both sides yields:
ln(ρ(r))=−g¯mrkbT
Solving for ρ(r) we get:
ρ(r)=e−g¯mrkbT
(g) Show that the height, H, over which the density falls off by an factor of 1/e is given by
H=kT/mg
where m ̄ is the mean (average) mass of a gas particle. This is the “scale height.” First, check the units. Then do the math. Then make sure it makes physical sense, e.g. what do you think should happen when you increase m ̄ ? Finally, pat yourselves on the back for solving a first-order differential equation and finding a key physical result!
This means that a height difference of H will cause pressure to change by a factor of 1/e. Let's test this.
1e=e(−g¯mrkbT−−g¯m(r−H)kbT)
e−1=e−g¯mHkbT
ρ(r)=e−g¯mrkbT
(g) Show that the height, H, over which the density falls off by an factor of 1/e is given by
H=kT/mg
where m ̄ is the mean (average) mass of a gas particle. This is the “scale height.” First, check the units. Then do the math. Then make sure it makes physical sense, e.g. what do you think should happen when you increase m ̄ ? Finally, pat yourselves on the back for solving a first-order differential equation and finding a key physical result!
This means that a height difference of H will cause pressure to change by a factor of 1/e. Let's test this.
1e=e(−g¯mrkbT−−g¯m(r−H)kbT)
e−1=e−g¯mHkbT
1=g¯mHkbT
H=kbTg¯m
(h) What is the Earth’s scale height, HC? The mass of a proton is 1.7 ˆ 10 ́24 g, and the Earth’s atmosphere is mostly molecular nitrogen, N2, where atomic nitrogen has 7 protons, 7 neutrons.
It's time to plug and chug.
It's time to plug and chug.
H=kbTg¯m=(1.4×10−16ergs/K)×(273K)(980cm/s2)×(1.7×10−24g/atom)×(28atoms/particle)=8.2×105cm
I collaborated with G. Grell and S. Morrison on this problem.
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