My first post, I am sure I will have many more! I am so excited to have found this community!

I have a game called Rise: The Vieneo Province and a very old issue that I was hoping to get help with.

According to our wiki for our fictitious planet, the surface pressure is 2631 mb or 2.596 times that of Earth’s. However, in the aerodynamic work-up for the aircraft/spacecraft the scale height that is used is 8.0 km or equal to that of Earth’s.

I found a formula in which you can solve for the scale-height which requires the known variables of rho (density at a given altitude of z) and surface pressure. I don’t know how to reverse engineer from the function of e (is it ln?) or how to determine the density at a given height above the ground.

  • 7
    $\begingroup$ If the composition of the atmosphere is the same as Earth's, the gravitational acceleration is the same, and the temperature is also similar then the scale height should be the same. H = RT / mg, H = scale height, R = specific gas constant (depends on atmospheric composition), T = absolute temperature, m = average molecular mass, g = gravitational acceleration. As you can see it does not depend on the pressure or density at sea level. Basically, your atmosphere is just like Earth's but 2.6 times more dense at any altitude. $\endgroup$
    – AlexP
    Commented Jun 10, 2017 at 21:01
  • $\begingroup$ @AlexP thanks for the fast response! So we have a Nitrogen (ipp: 2006), Helium (ipp: 578) Oxygen (ipp: 11) and Argon (ipp: 6) composition... but that is probably the topic of another post because questions have been raised about the viability of field crops on the surface with no CO2 and lack of animals causing a gap in the carbon cycle, etc. $\endgroup$ Commented Jun 11, 2017 at 0:22
  • 1
    $\begingroup$ Your world will have a somewhat higher scale height due to all that helium. $\endgroup$ Commented Jun 11, 2017 at 19:17
  • $\begingroup$ @AlexP gravitational acceleration has NOTHING to do with atmosphere. What you mean is atmospheric resistance $\endgroup$
    – Fl.pf.
    Commented Jun 14, 2017 at 11:18
  • $\begingroup$ so we came up with a new composition from worldbuilding.stackexchange.com/questions/85033/… which is N, SO2, O, and trace CH4, NH3, H2O, Ne, N2, CO... Nitrogen 98.4% 2589 mb Sulfur Dioxide 1.0% 26 mb and Oxygen 0.6% 16 mb ... running that through the formula from @KareemElashmawy gives us 6.923 km thanks for all the help everyone! $\endgroup$ Commented Jul 2, 2017 at 2:20

1 Answer 1


How to derive the Scale Height

The Nebraska Astronomy Applet Project of the University of Nebraska-Lincoln provides a complete derivation of Scale Height with respect to Planetary atmospheres which I will duplicate below. Please note that I'll hold of on explaining what each symbol means until Part 3: Calculating the Scale Height.

First we begin with the equation of hydrostatic equilibrium.

(1). $~~ dP = - \rho g dz$

Then we use the ideal gas law to simplify (1). We may use one of two forms:

(2). $~~ P = \cfrac{kT}{\mu m_\mu} \rho$

(3). $~~ \rho = \cfrac{ kT}{\mu m_\mu} P$

to eliminate $\rho$ or P from (1). Using (2) returns (4) and using (3) returns (5):

(4). $~~ dP = - \left( \cfrac{\mu m_\mu P}{kT} \right) g dz$

(5). $~~ d\rho = - \left( \cfrac{\mu m_\mu \rho}{kT} \right) g dz$

Shift P or /rho to the left hand side and you'll arrive at:

(6). $~~ \cfrac{dP}{P} = - \left( \cfrac{\mu m_\mu}{kT} \right) g dz$

(7). $~~ \cfrac{d\rho}{\rho} = - \left( \cfrac{\mu m_\mu }{kT} \right) g dz$

If you look closely, you'll notice that the formulas are identical, and therefore equivalent. It then follows that whichever we solve for, will have an identical form to the other. Taking after the University of Nebraska-Lincoln, we'll derive the scale height from pressure and continue with (6).

Integrate from the surface of the planet ($P_i = P_0$, $z_i = 0$) to some height ($P_f = P$, $z_f =z$).

$\int{\cfrac{dP}{P}} = - \cfrac{\mu m_\mu }{kT} g \int dz $

$\ln P |_{P_0}^P = \left(- \cfrac{\mu m_\mu }{kT} g \right) z |_{0}^z $

$\ln P - \ln P_0 = \left(- \cfrac{\mu m_\mu }{kT} g \right) (z-0) $

Use the logarithmic identity for subtraction to simplify the left hand side.

$\ln \cfrac{P}{P_0} = \left(- \cfrac{\mu m_\mu g}{kT} \right) z $

Raise the entire formula to an exponent to eliminate the logarithm, then rearrange as a function of pressure.

$e^{\ln \cfrac{P}{P_0}} = e^{\left(- \cfrac{\mu m_\mu g}{kT} \right) z} $

(8). $~~ P = P_0 e^{\left(- \cfrac{\mu m_\mu g}{kT} \right) z} $

Lump the coefficients on z into a single coefficient. We'll call it H.

(9). $~~ H = \left(\cfrac{kT}{\mu m_\mu g} \right)$

Then (1) simplifies to:

$P = P_0 e^{-\cfrac{z}{H}} $

Since H is a constant coefficient, if we take z to equal it (z = H), then we'll get: $ P = P_0 e^{- 1} = \cfrac{P_0}{e} $

Therefore we formally define $H$ as the height where the pressure drops to $\frac{1}{e}$ of the surface pressure, i.e. The Scale Factor.

Determine Air density at a given height.

Recall earlier how the Scale Factor may be derived either from pressure or density? Using density instead of pressure, i.e. formula (7) to derive the scale factor gives us:

(10). $~~ \rho = \rho_0 e^{\left(- \cfrac{\mu m_\mu g}{kT} \right) z} $

using the same logic as before:

$H = \left( \cfrac{kT}{\mu m_\mu g} \right)$

which is the scale height. Therefore (10) may be used to calculate the air density at a given height, or in your case, at the scale height.

Calculating your scale height.

$m_\mu$ is the atomic mass constant: $1.66 \times 10^−27 \frac{kg}{amu}$.

k is the Boltzmann Constant: $1.38 \times 10^−23 \frac{J}{K}$.

g is the surface gravity: $11.127 \frac{m}{s^2}$.

T is the temperature of the gas (average temperature of the atmosphere):

$\overline T = \frac{8.0°C + 2.9°C}{2} = 5.45 °C = 278.6 K $

$\mu$ is the average particle mass of the gas (our atmosphere). Similarly to The University of Nebraska-Lincoln, we compute this from the weighted composition of the atmosphere:

$\mu = 0.771 m_{N_2} + 0.222 m_He + 0.004 m_{O_2} + 0.002 m_{Ar}$

$\mu = (0.771)(28) + (0.222)(4) + (0.004)(32) + (0.002)(39)$

$\mu = (0.771)(28) + (0.222)(4) + (0.004)(32) + (0.002)(39) = 22.682 amu$

Plugging these values into formula (9) returns:

$H = 9176.83 m $ or $H = 9.177 km $

Therefore the air density at a given height, z, is

$\rho = \rho_0 e^{- \cfrac{z}{9176.83 m}} $

Sample densities:

Since the scale factor is the distance required to reduce the pressure or density by $\frac{1}{e}$ sample pressures and densities at integer multiples of the scale factor ($nH$) are simple to calculate.

$\rho = \rho_0 e^{- \cfrac{nH}{H}} =\rho_0 e^{-n} = \cfrac{\rho_0}{e^n} $

$$\begin{array}{|c|c|c|c|c|} \hline \text{Elevation} & \text{n} & \text{coefficient} & \rho & P\\ \hline 0 & \text{0} & \left(\cfrac{1}{1}\right) & 1.000 \rho_0 & 1.000 P_0\\ \hline H & \text{1} & \left(\cfrac{1}{e}\right) & 0.368 \rho_0 & 0.368 P_0\\ \hline 2H & \text{2} & \left(\cfrac{1}{e^2}\right) & 0.135 \rho_0 & 0.135 P_0\\ \hline 3H & \text{3} & \left(\cfrac{1}{e^3}\right) & 0.050 \rho_0 & 0.050 P_0\\ \hline 4H & \text{4} & \left(\cfrac{1}{e^4}\right) & 0.018 \rho_0 & 0.018 P_0\\ \hline \end{array}$$


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