Consider the Earth's atmosphere by assuming the constituent particles comprise an ideal gas, such that \(P=nk_BT\), where n is the number density of particles (with units \(cm^{-3}\)). \(k=1.4\times 10^{-16}\text{ erg K}^{-1}\) 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 atmosphere. The parcel sits a distance r from the Earth's center, and the parcel's size is defined by a height \(\Delta r\ll r\) 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 form gas below \((P_{up}=P(r))\) and down from above (\(P_{down}=P(r+\Delta r)\)).
Make a drawing of this, and discuss the situation and the various physical parameters.
This cylindrical parcel of gas represents a small volume of the Earth's atmosphere where \(\Delta r\) is much smaller than the distance to the center of the Earth, r. This parcel experiences an upward force from the pressure of the gas below it on the cross-sectional area, A, and a downward force from the pressure of the gas above it on the cross-sectional area. Additionally, since the particles of the gas have a mass, the parcel will experience a downward force of gravity from the Earth.
b) What other force will the parcel feel, assuming it has a density \(\rho (r)\) and the Earth has a mass \(M_\oplus\)?
As mentioned in part (a), the parcel has a mass, so in addition to forces caused by pressure, it will also experience a gravitational force from the Earth. The force on the parcel, \(m_{parcel}\), due to gravity is given by:
\(F_g = \dfrac{GM_\oplus m_{parcel}}{r^2}\)
Where G is the gravitational constant, \(M_\oplus\) is the mass of the Earth, and r is the distance between the parcel and the center of the Earth.
c) If the parcel is not moving, give a mathematical expression relating the various forces, remembering that force is a vector and pressure is force per unit area.
Since the parcel isn't moving, we know that it is in equilibrium, meaning that the upward forces exactly cancel out the downward forces so that there is a net force of zero. We know that the three forces acting on the parcel are a downward force from gravity, a downward force due to pressure, and an upward force due to pressure. To find the forces from pressure, we know that pressure is in units of force per area, so we can multiply the pressure (which is a function of r) by the cross-sectional area, A. So we get the two pressure forces to be:
Using these forces and the force of gravity, we can write the expression:
\(F_{P(r)}=P(r)A\)
And,
\(F_{P(r+\Delta r)}=P(r+\Delta r)A\)
Using these forces and the force of gravity, we can write the expression:
\(F_g+F_{P(r+\Delta r)}-F_{P(r)}=0\)
And from part (b), we solved for the force due to gravity, so we have:
\(\dfrac{GM_\oplus m_{parcel}}{r^2}+P(r+\Delta r)A-P(r)A=0\)
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.
The equation for the force of gravity can be simplified into an expression of the form:
And to find the gravitational acceleration, g, at a distance r from the center of the Earth, we can use our expression for \(F_g\) from part (b) to set it equal to the expression above:
The equation for the force of gravity can be simplified into an expression of the form:
\(F_g=m_{parcel}g\)
And to find the gravitational acceleration, g, at a distance r from the center of the Earth, we can use our expression for \(F_g\) from part (b) to set it equal to the expression above:
\(F_g =m_{parcel}g= \dfrac{GM_\oplus m_{parcel}}{r^2}\)
Solving for g, we get:
\(g =\dfrac{GM_\oplus }{r^2}\)
We have found the acceleration due to gravity as a function of r.
e) Show that
\(\dfrac{dP(r)}{dr}=-g\rho (r)\)
This is the equation of hydrostatic equilibrium.
Returning to the equation we created in part (c), and adding our expression for g, we have:
Returning to the equation we created in part (c), and adding our expression for g, we have:
\(m_{parcel}g+P(r+\Delta r)A-P(r)A=0\)
We can also find \(m_{parcel}\) since we know the density of gas particles in the parcel and the volume, V, of the parcel:
\(m_{parcel}=V\rho (r)\)
\( =A \Delta r \rho (r) \)
So now our equation can be further modified to:
\(A\Delta r\rho (r) g+P(r+\Delta r)A-P(r)A=0\)
We can rearrange this equation to the following expression:
\(\dfrac{P(r+\Delta r)-P(r)}{\Delta r}=-g\rho (r)\)
This is the definition of the derivative of \(P(r)\) with respect to r, so we have:
\(\dfrac{dP(r)}{dr}=-g\rho (r)\)
And we have now solved for the equation of hydrostatic equilibrium. We can see from this equation that rate at which the pressure changes with distance from the center of the Earth is directly proportional to the density of the gas and to the gravitational acceleration. So for denser atmospheres and larger gravitational accelerations (which occur at smaller r), we will have a greater rate of change in pressure.
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, \(\rho (r)\)?
The ideal gas law is given by:
\(P=nk_BT\)
Where \(P\) is pressure, \(k_B\) is the Boltzmann constant, T is temperature, and n is the number density of particles in the gas. Number density can also be written as:
\(n=\dfrac{\rho (r)}{\overline{m}}\)
Where \(\overline{m}\) is the mean mass of a gas particle. Using the equation of hydrostatic equilibrium, we get:
\(\dfrac{dP(r)}{dr}=\dfrac{d}{dr}\left(\dfrac{\rho (r)}{\overline{m}}k_BT\right)=-g\rho (r)\)
Which gives us:
\(\dfrac{d\rho (r)}{\rho (r)}=\dfrac{\overline{m}g}{k_BT}\)
Integrating over both sides, we get:
\(\displaystyle\int\dfrac{d\rho (r)}{\rho (r)}=\displaystyle\int\dfrac{\overline{m}g}{k_BT}\)
\(\ln{\rho (r)}=-\dfrac{\overline{m}g}{k_BT}r+C\)
And we find that:
\(\rho (r)=\rho _0e^{-\frac{\overline{m}g}{k_BT}r}\)
Where \(\rho_0\) represents the density at the center of the Earth.
g) Show that the height, \(H\), over which the density falls off by a factor of \(1/e\) is given by
\(H=\dfrac{k_BT}{\overline{m}g}\)
where \(\overline{m}\) 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 \(\overline{m}\)?
Checking units, we know that the units of \(k_B\) are erg/K, which is equal to:
\([k_B]=\dfrac{\text{g cm}^2}{\text{K s}^2}\)
And the unit for T is Kelvins, for g is \(\text{cm/s}^2\) and for \(\overline{m}\) is grams. So we combine these to get:
\([H]=\dfrac{\text{g cm}^2\text{ K s}^2}{\text{ g cm K s}^2}=\text{cm}\)
As desired.
Solving for scale height, we get:
\(\dfrac{1}{e}=\dfrac{\rho (H)}{\rho_0}\)
\(\dfrac{1}{e}=\dfrac{\rho _0 e^{-\frac{\overline{m}gH}{k_BT}}}{\rho_0}\)
And we find that:
\(H=\dfrac{k_BT}{\overline{m}g}\)
As the temperature increases and the mass of the particles decreases, the scale height increases. This makes sense since temperature is directly related to pressure and the mass of the particles is indirectly related to pressure. So for larger T and smaller \(\overline{m}\), we have more pressure and therefore a larger scale height.
h) What is the Earth's scale height, \(H_\oplus\)? The mass of a proton is \(1.7\times 10^{-24}\text{g}\), and the Earth's atmosphere is mostly molecular nitrogen, \(N_2\), where atomic nitrogen has 7 protons and 7 neutrons.
To find the Earth's scale height, we need to estimate the temperature in the Earth's atmosphere, the mass of a single particle in the atmosphere, and the acceleration due to gravity at the scale height. To find the mass of a particle, we know that the atmosphere is mostly molecular nitrogen, so we want to find the mass of molecular nitrogen. We know that atomic nitrogen has 7 protons and 7 neutrons, and that the mass of protons and neutrons are approximately equal. So the mass of atomic nitrogen, \(M_N\), would be:
We can approximate g using the value at Earth's surface since it is safe to assume that g doesn't change very rapidly with distance away from Earth in our atmosphere:
\(m_N=7m_p+7m_n\approx 14m_p\approx 14\times1.7\times 10^{-24}\text{g}\approx2.38\times10^{-23}\)g
But this is the mass of atomic nitrogen. To find the mass of molecular nitrogen, \(M_{N_2}\), we simply double the mass of atomic nitrogen:
\(\overline{m}=m_{N_2}\approx 28m_p\approx 28\times 1.7\times 10^{-24}\text{g}\approx 4.76\times 10^{-23}\)g
We can approximate g using the value at Earth's surface since it is safe to assume that g doesn't change very rapidly with distance away from Earth in our atmosphere:
\(g =\dfrac{GM_\oplus }{r^2}\approx 10^3\text{cm s}^{-2}\)
Finally, we can also assume that temperature doesn't change (relatively) very much from Earth's surface. The following diagram shows the temperature as a function of distance from Earth:
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| Image Credit: http://www.windows2universe.org/earth/images/profile.jpg |
As we can see, there is relatively little change in temperature, so we can use the temperature near Earth's surface, which is approximately -50 degrees Celsius, or 200 degrees Kelvin.
Plugging these values into our equation for scale height, we get:
\(H_\oplus=\dfrac{k_BT}{\overline{m}g}\approx \dfrac{(1.7\times 10^{-16}\text{ erg K}^{-1})(200\text{ K})}{(4.76\times 10^{-23}\text{g})(10^3\text{cm s}^{-2})}\approx 7.14\times 10^5\text{cm}\approx 7\text{ km}\)
So we have found a scale height of approximately 7 km.
I would like to thank Charles Law and Daniel Chen for their help in solving this problem.
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