Atmosphere Temperature and Pressure
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Rama
physics Atmosphere Temperature and Pressure |
The goal of this section is to compute the temperature, density and pressure of the atmosphere inside Rama. Computing their "exact" value is a complex task because the atmosphere interacts with the ground in the short and long wave ranges, thus modifying the lighting and temperature profiles at the ground level (computed in the previous section in the absence of atmosphere). In return, the lighting and temperature profiles at the ground level influence the atmospheric properties.
However, we saw in the previous section that the lighting and temperature at the ground level, without atmosphere, are almost constant. Assuming that the atmosphere does not change this, we can compute approximate temperature, density and pressure values by supposing that all physical quantities are axisymetric, i.e. independent of $\theta$ (we also suppose an infinitely long cylinder, as in the previous sections). This hypothesis greatly simplifies computations because, as shown previously, the lighting and temperature values at the ground level are then independent of the atmospheric properties. In other words the feedback loop between the atmosphere and the ground disappears, and we can simply compute the atmospheric properties directly from the ground properties, already known.
This is what we do in the following, by supposing that the atmosphere density and pressure at the ground level are the same as on Earth (recall that we supposed Rama to be designed to provide an environment as close as possible to the one on Earth).
2 Temperature
Intuitively, since we assumed that the temperature of the gound is the same everywhere, and since the atmosphere is completely enclosed by the ground, each air parcel should be "heated in the same way" from all directions, whatever its location within the atmosphere. In other words, the temperature of the atmosphere should be the same everywhere, and equal to the ground temperature. This section shows that this intuition is correct.
The radiance $L(\bp,\bw)$ traveling at $\bp$ in direction $\bw$ in the atmosphere is given by the radiative transfer equation as: \begin{equation} \begin{split} \bw\cdot\nabla L(\bp,\bw)=&-(k_a(\bp)+k_s(\bp))L(\bp,\bw)\\ &+k_s(\bp)\int_{4\pi}P(\bw,\bw')L(\bp,\bw')\diff\bw'+k_e(\bp) B(\bp) \end{split} \end{equation} This equation states that the variation of $L$ in direction $\bw$, the left hand side term, is equal to the sum of
Note that all the terms in this equation depend on the wavelength $\lambda$. Assuming that we can replace the radiance terms with their integral over all wavelengths, and the coefficients with their average over all frequencies, we get the same equation for these integrated or averaged quantities, where $B(\bp)$ can now be replaced with $\sigma T^4(\bp)/\pi$. Also, according to Kirchhoff's law of thermal radiation, we have $k_a(\bp)=k_e(\bp)$. We thus get: \begin{equation} \begin{split} \bw\cdot\nabla L(\bp,\bw)=&-(k_e(\bp)+k_s(\bp))L(\bp,\bw)\\ &+k_s(\bp)\int_{4\pi}P(\bw,\bw')L(\bp,\bw')\diff\bw'+k_e(\bp) \frac{\sigma T^4(\bp)}{\pi} \end{split}\label{eq:radiative1} \end{equation}
At radiative equilibrium, the radiance $L(\bp,\bw)$ also satisfies another equation, which states that the radiative energy can't accumulate in any given volume: all the radiative energy that enters the volume per unit of time should be equal to the energy that leaves the volume in the same time interval. Mathematically, this means that the divergence of the radiative vector flux density $\mathbf{L}(\bp)\eqdef\int_{4\pi}L(\bp,\bw)\bw\diff\bw$ must be 0 [1]. Substituting the above equation into $\nabla\cdot\mathbf{L}=0$ we get: \begin{align*} 0&=\nabla\cdot\mathbf{L}(\bp)=\int_{4\pi}\nabla L(\bp,\bw)\cdot\bw\diff\bw\\ &=-(k_e(\bp)+k_s(\bp))\int_{4\pi}L(\bp,\bw)\diff\bw+k_s(\bp)\int_{4\pi}\int_{4\pi}P(\bw,\bw')L(\bp,\bw')\diff\bw'\diff\bw+4k_e(\bp)\sigma T^4(\bp)\\ &=-k_e(\bp)\int_{4\pi}L(\bp,\bw)\diff\bw+4k_e(\bp)\sigma T^4(\bp) \end{align*} Or, in other words: \begin{equation} \frac{1}{4\pi}\int_{4\pi}L(\bp,\bw)\diff\bw=\frac{\sigma T^4(\bp)}{\pi}\label{eq:radiative2} \end{equation}
It is then easy to see that a uniform temperature $T_a$, independent of $\bp$, and a uniform radiance $L=\sigma T_a^4/\pi$, independent of $\bp$ and of $\bw$, are solutions of Eqs. \eqref{eq:radiative1} and \eqref{eq:radiative2}. Indeed the left hand side in Eq. \eqref{eq:radiative1} is then 0, the absorption term cancels with the emission term, and the out-scattering term cancels with the in-scattering term. Moreover, this remains true whatever the dependence of the emissivity and scattering coefficients with $\bp$, i.e. whatever the density profile of the atmosphere ($k_e$ and $k_s$ are proportional to the atmosphere density).
Finally, note that a uniform radiance $L=\sigma T_a^4/\pi$ everywhere in the atmosphere is also consistent with a uniform radiance at the surface $\sigma T_s^4/\pi$, provided the surface temperature $T_s$ and the atmosphere temperature $T_a$ are equal. Which is what we wanted to demonstrate.
The pressure $p(r)$ in $Pa$ and the density $\rho(r)$ in $kg.m^{-3}$ at radius $r$ are given by two equations:
Combining these two equations gives $p'(r)=r\,p(r)\frac{\omega^2}{R_sT_a}$, whose solution is \begin{equation} p(r)=p(0)\exp\left(\frac{\omega^2}{R_sT_a}\frac{r^2}{2}\right) \end{equation} or, using the pressure $p_s$ at the surface: \begin{equation} p(r)=p_s\exp\left(-\frac{\omega^2}{R_sT_a}\frac{r_s^2-r^2}{2}\right) \end{equation} which, substituted in the ideal gas law equation, gives: \begin{equation} \rho(r)=\frac{p_s}{R_sT_a}\exp\left(-\frac{\omega^2}{R_sT_a}\frac{r_s^2-r^2}{2}\right) \end{equation}
This pressure profile is shown in Fig. 1 for 3 different values of $T_a$, namely 5°C, 15°C and 25°C. Note that the pressure profile is almost insensitive to "small" temperature variations (in fact ten times larger than the actual temperature variations computed in the previous section): a 10°C difference results in less than 2% pressure difference on the Rama axis. Note also that the pressure decreases less quickly with altitude than on Earth: in Rama the pressure at 8000 m is about 60% of the ground level pressure, while on Earth, at the same altitude, it is less than 40% of the ground level value.
In the following we assume that the scattering and emissivity / absorptivity coefficients in the atmosphere are simply proportional to the atmosphere density. In reality these coefficients can also depend on the pressure. However, assuming that the temperature variations are small (as it is the case in the absence of atmosphere, cf the previous section), then the pressure is itself proportional to the density, and thus, at first order, the scattering and emissivity / absorptivity coefficients only depend on the density.
Following our hypothesis that Rama is designed to provide an environment as close as possible to the one on Earth, we assume that the scattering and emissivity / absorptivity coefficients in Rama, at the ground level, have the same value as on Earth:
In summary we assume the following values for the scattering coefficient $k_s(\bp)$ and emissivity / absorptivity coefficient $k_e(\bp)$ in Rama: \begin{align} k_s(r)&=1.50\,10^{-5} \exp\left(-\frac{\omega^2}{R_sT_a}\frac{r_s^2-r^2}{2}\right)\,m^{-1}\\ k_e(r)&=2.58\,10^{-4} \exp\left(-\frac{\omega^2}{R_sT_a}\frac{r_s^2-r^2}{2}\right)\,m^{-1} \end{align}
Under the hypothesis that the lighting and temperature variations remain small in Rama when the atmosphere is taken into account, then the temperature of the atmosphere is approximatively the same everywhere, and equal to the surface temperature. Moreover, the radiance inside the atmosphere is also uniform, and isotropic. The pressure profile can then be computed from this constant temperature profile, and the results show that the pressure and density profiles are almost insensitive to small temperature variations.
As a concluding remark, note that we can compute the total mass of the Rama atmosphere with \begin{equation} l_s\int_0^{r_s}\rho(r)2\pi r \diff r=\frac{2\pi l_s p_s}{\omega^2}\left(1-\exp\left(-\frac{\omega^2}{R_sT_a}\frac{r_s^2}{2}\right)\right) \end{equation} using $T_a=288\,K$, $l_s=40\,km$ and $p_s=101325\,Pa$, we get 7.86 GT, i.e. 7.86 billion of tons (recall that the mass of the shell would be 17090 GT if it was made of plain steel).