Atmospheric pressure as a function of altitude

[Karl Karlsson]

Summary:: i) Set up a differential equation that describes how the pressure $p$ varies with the distance
r from the center of the planet. Hint: You can base your reasoning on static
equilibrium and Archimedes' principle.

ii)Calculate how the atmospheric pressure p and the density of the atmosphere $ρ$ depend on r.
Assume that pressure and density only depend on r.

Consider a planet with a thin atmosphere. The planet is assumed to have radius $r_0$ and mass M.
The gravitational field outside the planet is given by: $$\vec g = - \frac {GM} {r^2} \vec e_r$$ where G is Newton's gravitational constant. The atmosphere near the surface of the planet can be considered as a stationary linearly compressible fluid, which means that a relationship $ρ = ρ_0 + α(p - p_0)$ applies between the density of the fluid $ρ$ and the pressure $p$. Here $ρ_0$ and $p_0$ are the density of the atmosphere respective pressure at the planetary surface.

i) Set up a differential equation that describes how the pressure $p$ varies with the distance
r from the center of the planet. Hint: You can base your reasoning on static
equilibrium and Archimedes' principle.

ii)Calculate how the atmospheric pressure p and the density of the atmosphere $ρ$ depend on r.
Assume that pressure and density only depend on r.

I am not quite sure how to start on i) since equilibrium in the radial direction is given by $-m\frac {GM} {r^2} + ρVg = 0$ and $ρV = m$ the previous expression is just zero and that doesn't give me anything

Thanks in advance!

 

[Chestermiller]

Are you familiar with the hydrostatic equation? If so, please write it down.

 

[Karl Karlsson]

Are you familiar with the hydrostatic equation? If so, please write it down.

Hi Chestermiller!
I am not familiar with the hydrostatic equation. The course where I got this problem from is a mathematical course I am studying. I don't think you are supposed to be familiar with physics formulas except basic ones like archimedes principle

 

[Chestermiller]

Static equilibrium requires that $$\frac{dp}{dr}=-\rho g$$

 

[vela]

I am not quite sure how to start on i) since equilibrium in the radial direction is given by $-m\frac {GM} {r^2} + ρVg = 0$ and $ρV = m$ the previous expression is just zero and that doesn't give me anything.

What you need to do is express the buoyancy force in terms of pressure. After all, you're looking for a differential equation for the pressure. Instead of using $\rho g V$, consider a differential volume of cross-sectional area $A$ and thickness $dr$ and use $dF_B = -p(r+dr) A + p(r)A$.

 

[Chestermiller]

What has this problem got to do with buoyancy?

 

[Karl Karlsson]

Static equilibrium requires that $$\frac{dp}{dr}=-\rho g$$

Why does static equilibrium require $\frac{dp}{dr}=-\rho g$?
But if so then I guess I am supposed to $\vec e_r: - \frac {GM} {r^2} \cdot dm + \rho\cdot g\cdot dV=0 \implies \vec e_r: - \frac {GM} {r^2} \cdot\frac {dm} {dV} - \frac {dp} {dr} = - \frac {GM} {r^2} \cdot\rho - \frac {dp} {dr} = 0$. Is this correct?

 

[Chestermiller]

Why does static equilibrium require $\frac{dp}{dr}=-\rho g$?
But if so then I guess I am supposed to $\vec e_r: - \frac {GM} {r^2} \cdot dm + \rho\cdot g\cdot dV=0 \implies \vec e_r: - \frac {GM} {r^2} \cdot\frac {dm} {dV} - \frac {dp} {dr} = - \frac {GM} {r^2} \cdot\rho - \frac {dp} {dr} = 0$. Is this correct?

No. You would write: $$\frac{dp}{dr}=-[\rho_0+\alpha(p-p_0)]\frac{GM}{r^2}$$

 

[mpresic3]

Static equilibrium requires that $$\frac{dp}{dr}=-\rho g$$

You might now see if you can apply the ideal gas law from high school chemistry. I expect the problem assumes the atmosphere is isothermal so T is a constant as is k (Boltzmanns constant) or R (the ideal gas constant)

 

[Chestermiller]

You might now see if you can apply the ideal gas law from high school chemistry. I expect the problem assumes the atmosphere is isothermal so T is a constant as is k (Boltzmanns constant) or R (the ideal gas constant)

As a scientist with years of actual working experience in atmospheric science, including numerous publications in this area, my first inclination would always be to use the ideal gas law for the atmospheric equation of state (including dependence of temperature on altitude). However, in this problem, the OP was specifically told to use the relationship given in the problem statement. Since he was not even aware of the barotropic equation, I judged that it would be even more confusing for him and irresponsible of me to mention this additional point.

 

[Charles Link]

It might interest the OP where the formula in post 4 originates from. There is a force per unit volume on a fluid element that is $f_v=-\nabla P$. To have static equilibrium, this must balance the gravitational force per unit volume, which is $f_g=\rho g$ that points downward. With spherical symmetry, $f_v=-\frac{dP}{dr}$.
In addition, $g=GM/r^2$, and $\rho=\frac{P \cdot MW}{RT}$, where MW=molecular weight (approximately 30).
If the atmosphere is a thin layer, (e.g. if it extends 20 miles and the radius of the planet is 4000 miles), $g$ can be taken as constant, independent of $r$, so that $g=GM/R_e^2$.
(Reading post 10), instead of using $\rho=\alpha P$, it appears they want the OP to use $\rho=\rho_o +\alpha (P-P_o)$.

 

[Chestermiller]

It might interest the OP where the formula in post 4 originates from. There is a force per unit volume on a fluid element that is $f_v=-\nabla P$. To have static equilibrium, this must balance the gravitational force per unit volume, which is $f_g=\rho g$ that points downward. With spherical symmetry, $f_v=-\frac{dP}{dr}$.
In addition, $g=GM/r^2$, and $\rho=\frac{P \cdot MW}{RT}$, where MW=molecular weight (approximately 30).
If the atmosphere is a thin layer, $g$ can be taken as constant, independent of $r$, so that $g=GM/R_e^2$.
(Reading post 10), instead of using $\rho=\alpha P$, it appears they want the OP to use $\rho=\rho_o +\alpha (P-P_o)$.

I think by a "thin atmosphere," they meant an atmosphere of low density.

 

[mpresic3]

I suppose I overlooked the part where the OP was given the equations to use. Sorry. mpresic3

 

[vela]

It might interest the OP where the formula in post 4 originates from.

I think the point of section i) of the problem

i) Set up a differential equation that describes how the pressure p varies with the distance
r from the center of the planet. Hint: You can base your reasoning on static
equilibrium and Archimedes' principle.

was to derive this equation. There's a pretty simple line of reasoning typically covered in intro physics which leads to the equation.

 

[Karl Karlsson]

Hi!

Thanks for all good answers I solved it by sort of using what Chestermiller wrote. I considered a small box in the atmosphere with area A and height dr which gave me the static equilibrium equation in the radial direction: $\rho\cdot (dr\cdot A)g+p\cdot A-(p+dp)\cdot A=0$. The rest was straight forward

 

[Chestermiller]

Hi!

Thanks for all good answers I solved it by sort of using what Chestermiller wrote. I considered a small box in the atmosphere with area A and height dr which gave me the static equilibrium equation in the radial direction: $\rho\cdot (dr\cdot A)g+p\cdot A-(p+dp)\cdot A=0$. The rest was straight forward

This is the right idea, but it seems to me there is a sign error.