Pressure higher up in the atmosphere

[brotherbobby]

Homework Statement

Suppose the pressure in NY city is $P_{\text{atm}}$ and the air density is $\rho_0$. We are also given that the the density of air as well the acceleration due to gravity decrease as one moves vertically up. We wished to calculate the pressure $P$ at a height of $1\;\text{km}$ above the city using $P_0-P=\rho_0 gz$, where $z$ is the vertical distance. Answer the following :
1. Will the pressure we obtain be accurate, less than, or more than the actual if we ignored variation in density?
2. Will the pressure we obtain be accurate, less than, or more than the actual if we ignored variation in $g$?

Relevant Equations

1. Change of pressure with height $\mathbf{\dfrac{dP}{dz}=-\rho(z) g(z)}$, is true for a given point.

Attempt : 1. Pressure at height $z$ : $P(z) = P_{\text{atm}}-\rho_0 gz$, ignoring density variation. But actually, we have $\rho(z)<\rho_0\;\forall z>0$. Hence, we are subtracting a bigger value from $P_{\text{atm}}$ than we actually should, meaning that we would end up what a smaller value of $P(z)$ than what is actual. Thus $\boxed{P(z)_{\text{actual}}> P(z)_{\text{calculated}}\;\;\text{when density decline is ignored}}$.

2. Similar is the case if we ignore variation in $g(z)$ as we do in $P(z) = P_{\text{atm}}-\rho_0 gz$. Since $g(z)<g\;\forall z>0$, we would be subtracting a larger value from $P_{\text{atm}}$ than we should in our calculation, implying that our calculated value of $P(z)$ will be lower than the actual. Thus $\boxed{P(z)_{\text{actual}}> P(z)_{\text{calculated}}\;\;\text{when decline of gravity with height is ignored}}$.

Are my arguments correct? The text provides no answers.

 

[TSny]

Looks good to me.

 

[TSny]

A fun exercise is to calculate at what height the atmospheric pressure would fall all the way to zero under the assumption that $g$ and $\rho$ don’t vary with altitude. Compare to the height of Mt. Everest.

 

[Delta2]

Hmm, I am not so sure as @TSny, seems you solve that ODE by ignoring both variations in density and g. In my opinion for

1. you should solve the ODE as $$P(z)=P_{atm}-\rho_0\int g(z) dz$$ and see what conclusions you can get from this solution
2. you should solve the ODE as $$P(z)=P_{atm}-g_0\int \rho(z) dz$$ and see what conclusions you can get from this solution

 

[brotherbobby]

A fun exercise is to calculate at what height the atmospheric pressure would fall all the way to zero under the assumption that $g$ and $\rho$ don’t vary with altitude. Compare to the height of Mt. Everest.

Let me see.

Attempt : To calculate the value of the height $z$ in the atmosphere for which the pressure there $P(z) = 0$.
We have $P_{\text{atm}}-P(z) = \rho_0 gz\Rightarrow P(z) = 0 = P_{\text{atm}} - \rho_0 gz\Rightarrow z = \dfrac{P_{\text{atm}}}{\rho_0 g}=\dfrac{1.013\times 10^5}{1.3\times 9.8} \approx \boxed{7950\;\text{m}}$.

Mount Everest is almost 8900 m tall. Which means that the topmost almost 1 km of the mountain would be above the atmosphere, i.e. in vacuum.

Clearly this is not the case. It underscores the result that by neglecting the variation (decline) of air density with height, we calculate air pressures to fall off faster than what they otherwise do $\left[ \left(\dfrac{dP}{dz}\right)_{\text{calculated}}> \left(\dfrac{dP}{dz}\right)_{\text{actual}}\right]$

 

[PeroK]

I'm not sure it's expected to solve an ODE. We could for example improve the model by resetting the density at an intermediate height:
$$P_1 = P_0 - \rho_0gz_1$$$$P = P_1 - \rho_1g(z -z_1)= P_0 -\rho_0gz_1 -\rho_1g(z -z_1)$$And, as $\rho_1 < \rho_0$ we see that this estimate is higher.

Taking more intermediate points or in the limit taking an integral will only increase the estimate further.

 

[brotherbobby]

Hmm, I am not so sure as @TSny, seems you solve that ODE by ignoring both variations in density and g. In my opinion for

1. you should solve the ODE as $$P(z)=P_{atm}-\rho_0\int g(z) dz$$ and see what conclusions you can get from this solution
2. you should solve the ODE as $$P(z)=P_{atm}-g_0\int \rho(z) dz$$ and see what conclusions you can get from this solution

1. I know how $g(z)$ varies with height (for small heights) : $g(z) = g_0\left( 1-\dfrac{2z}{R} \right)$ where $z\ll R$, the radius of the earth.
2. But what does $\rho(z)$ look like?

 

[PeroK]

1. I know how $g(z)$ varies with height (for small heights) : $g(z) = g_0\left( 1-\dfrac{2z}{R} \right)$ where $z\ll R$, the radius of the earth.
2. But what does $\rho(z)$ look like?

I assumed that air density reduces with height. If air higher up temporarily got more dense it would fall under gravity.

 

[brotherbobby]

I assumed that air density reduces with height. If air higher up temporarily got more dense it would fall under gravity.

Yes, but what does the density function $\rho(z)$ look like?

 

[PeroK]

Yes, but what does the density function $\rho(z)$ look like?

It decreases with height. That's all that matters.

 

[brotherbobby]

It decreases with height. That's all that matters.

I cannot carry out the integral $$P(z)=P_{atm}-g_0\int \rho(z) dz$$ till I know how does it vary with the height $z$. I was just looking online and saw a function that included temperature along with height.

 

[PeroK]

I cannot carry out the integral

You don't need to do the integral. That's the point. You know it must be less than $\rho_0z$

 

[haruspex]

I cannot carry out the integral $$P(z)=P_{atm}-g_0\int \rho(z) dz$$ till I know how does it vary with the height $z$. I was just looking online and saw a function that included temperature along with height.

Yes, because the density depends on pressure and temperature, and the relationship is not always the same. You could assume adiabatic neutrality (the 'lapse rate'), but a) that rate depends on humidity and b) the temperature can decline with height more slowly than that (temperature 'inversion').

 

[Frabjous]

As an aside, people have calculated “standard” atmospheres.
https://en.wikipedia.org/wiki/International_Standard_Atmosphere
https://en.wikipedia.org/wiki/U.S._Standard_Atmosphere
https://ntrs.nasa.gov/api/citations/19770009539/downloads/19770009539.pdf?attachment=true