Surface Lighting and Temperature Profile
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Rama
physics Surface Lighting and Temperature Profile |
The Rama lighting and temperature values computed in the previous section are only averages. They would be exact only if Rama was lit by a single linear light source placed on its axis, i.e. only with a fully axisymmetric spacecraft. However, Rama is lit by 3 linear light sources placed on the ground, 120° appart from each other (in fact 3 in the north plain, and 3 in the south plain). Thus, there are in fact lighting and temperature variations inside Rama. The goal of this section is to compute these variations, to see if they are large or not.
For this we keep the infinite cylinder model used throughout this document, with 3 infinite linear light sources at the surface. Also, to simplify computations, we suppose there is no atmosphere. Indeed, as shown in the previous section, the atmosphere does not change the average lighting and temperature. It does change the lighting and temperature variations, but these effects are studied in the next sections.
Several designs can be imagined for the linear sources. Some examples are show in Fig. 1, using cross sections in a plane perpendicular to the Rama axis. In all these cases, we assume that the light source surface is purely diffuse, i.e. that each point of the light source emits light equally in all directions (the emitted radiance is the same in all directions).
The half-cylindrical and cylindrical cross sections have the drawback that points on the surface near the light source receive a very large amount of light, and would thus probably be very hot. The flat design does not have this drawback, because points on the ground near the light source see it under a very small solid angle. We therefore assume in the following that the light sources use this design.
In the "incandescent" scenario, the light sources emit a total of $S=14.9\,MW.m^{-1}$ (see the previous section). Assuming they are black bodies with a temperature of $T=5800\,K$ like the Sun, they emit $\sigma T^4=64.1\,MW$ per square meter. We deduce that the width of each of the 3 light sources is $(14.9/64.1)/3=0.077\,m$, which is extremely small compared to the radius of Rama. We therefore assume in the following that, in both the "incandescent" and "energy-saving" scenarios, the width of the light sources is much smaller than Rama's radius.
The goal of this section is to compute the irradiance received on the ground in the short wave range, based on the position inside Rama. Thanks to the infinite cylinder hypothesis, this position can be represented with a single angle $\theta$ in cylindrical coordinates. To simplify computations, we compute this irradiance for a single light source located at $\theta=0$. The irradiance for 3 light sources can then be obtained by simply adding 3 copies of this irradiance function, shifted by 0, 120° and 240°.
3.1 Lighting profile equation
Noting $E(\theta)$ the irradiance received on the ground, $L(\theta)$ the radiance reflected or emitted by the ground, and assuming the ground has a Lambertian reflectance (i.e. that $L(\theta)$ is independent of the outgoing direction), we have, according to the formula for the irradiance in an infinite cylinder: \begin{equation} E(\theta)=\frac{\pi}{4}\int_{2\pi}L(\theta')\vert\sin\frac{\theta'-\theta}{2}\vert\diff\theta'\label{eq:light1} \end{equation} Moreover, by definition, $L(\theta)$ is the sum of the reflected light and of the emitted light:
Since the light source width is very small, i.e. $\Delta\theta\ll 1$, we can approximate the second integral with $\vert\sin(-\frac{\theta}{2})\vert\Delta\theta$, yielding: \begin{equation} E(\theta)=\frac{\alpha}{4}\int_{2\pi}E(\theta')\vert\sin\frac{\theta'-\theta}{2}\vert\diff\theta'+\frac{S}{4r_s}\vert\sin\frac{\theta}{2}\vert\label{eq:light2} \end{equation}
This is the equation we must solve in order to get the variations of the irradiance $E$, and we do this in the next section. Before that, note that the average value of $E$, $\frac{1}{2\pi}\int_{2\pi}E(\theta)\diff\theta$, can be directly computed from this equation. Doing this gives $S/2\pi r_s(1-\alpha)$, i.e. precisely the value we obtained in the previous section.
3.2 Solution
The solution to Eq. \eqref{eq:light2} can be expressed by using the linear operator \begin{equation} \mathcal{E}[f](\theta)\eqdef \frac{1}{4}\int_{2\pi}f(\theta')\vert\sin\frac{\theta'-\theta}{2}\vert\diff\theta' \end{equation} and the direct irradiance \begin{equation} E_0(\theta)\eqdef\frac{S}{4r_s}\vert\sin\frac{\theta}{2}\vert \end{equation} as the infinite series \begin{equation} E(\theta)=E_0(\theta)+\alpha\mathcal{E}[E_0](\theta)+\alpha^2\mathcal{E}[\mathcal{E}[E_0]](\theta)+\alpha^3\mathcal{E}[\mathcal{E}[\mathcal{E}[E_0]]](\theta)+\ldots \end{equation} as can be seen by checking that $\alpha\mathcal{E}[E]+E_0=E$, i.e. Eq. \eqref{eq:light2}. Note that this series has a simple physical interpretation: the $n$-th term corresponds to the irradiance received on the ground after exactly $n$ bounces on the ground. The first few terms, plotted in Fig. 2, can be computed analytically, and we get (for $0<\theta<2\pi$): \begin{align*} E_1(\theta)&\eqdef\mathcal{E}[E_0](\theta)=\frac{S}{16r_s}\int_0^{2\pi}\sin\frac{\theta'}{2}\vert\sin\frac{\theta'-\theta}{2}\vert\diff\theta'\\ &=\frac{S}{16r_s}\left((\pi-\theta)\cos\frac{\theta}{2}+2\sin\frac{\theta}{2}\right)\\ E_2(\theta)&\eqdef\mathcal{E}[\mathcal{E}[E_0]](\theta)=\mathcal{E}[E_1](\theta)=\frac{S}{64r_s}\int_0^{2\pi}\left((\pi-\theta')\cos\frac{\theta'}{2}+2\sin\frac{\theta'}{2}\right)\vert\sin\frac{\theta'-\theta}{2}\vert\diff\theta'\\ &=\frac{S}{64r_s}\left(3(\pi-\theta)\cos\frac{\theta}{2}+(6+\frac{\pi^2}{2}-\frac{(\pi-\theta)^2}{2})\sin\frac{\theta}{2}\right)\\ E_3(\theta)&\eqdef\mathcal{E}[\mathcal{E}[\mathcal{E}[E_0]]](\theta)=\mathcal{E}[E_2](\theta)=\ldots\\ &=\frac{S}{256r_s}\left(\left(\frac{\pi^3}{3}+10(\pi-\theta)-\frac{\pi\theta^2}{2}+\frac{\theta^3}{6}\right)\cos\frac{\theta}{2}+(20+4\pi\theta-2\theta^2)\sin\frac{\theta}{2}\right) \end{align*}
As shown in Fig. 2, the terms of the series quickly converge towards a constant function. This allows us to approximate all the remaining terms with their average, i.e. $E_n(\theta)\approx S/2\pi r_s$ for $n>3$. The solution to Eq. \eqref{eq:light2} can now be rewritten as: \begin{equation} E(\theta)=\sum_{i=0}^\infty \alpha^i E_i(\theta)\approx E_0(\theta)+\alpha E_1(\theta)+\alpha^2 E_2(\theta)+\alpha^3 E_3(\theta)+\frac{\alpha^4}{1-\alpha}\frac{S}{2\pi r_s}\label{eq:light3}\end{equation} with, for Rama, $\alpha=0.15$. Finally, to get the actual lighting variations inside Rama it suffice to average three copies of $E(\theta)$, shifted by 0, 120° and 240°. Doing this gives the irradiance values shown in Fig. 3, from which we can see that the maximum illuminance is only $13\%$ larger than the minimal illuminance.
The goal of this section is to compute the irradiance received on the ground in the long wave range, based on the position inside Rama, in order to eventually compute the temperature variations.
4.1 Temperature profile equation
Due to the lighting variations, the irradiance received at $\theta$ in the long wave range, noted $\bar{E}^{\downarrow}(\theta)$, can not be supposed equal to the irradiance emitted at this same place, also in the long wave range, and noted $\bar{E}^{\uparrow}(\theta)$. If we look at the radiative fluxes in this case, and supposing we have a composite shell emitting $\Delta$ times more towards the inside than towards the outside in the long wave range, then we have (see Fig. 4):
The net flux must be 0, which gives $E(\theta)+\bar{E}^{\downarrow}(\theta)=\alpha E(\theta)+(1+\frac{1}{\Delta})\bar{E}^{\uparrow}(\theta)$. The received irradiance can be computed from the emitted one with the formula for the irradiance in an infinite cylinder, yielding $\bar{E}^{\downarrow}(\theta)=\frac{\pi}{4}\int_{2\pi}\bar{E}^{\uparrow}(\theta')/\pi\vert\sin\frac{\theta'-\theta}{2}\vert\diff\theta'$. Putting this together we get \begin{equation} \bar{E}^{\uparrow}(\theta)=\frac{\Delta}{\Delta+1}\mathcal{E}[\bar{E}^{\uparrow}](\theta)+(1-\alpha)\frac{\Delta}{\Delta+1}E(\theta)\label{eq:temp1} \end{equation}
This is the equation we must solve in order to get the variations of the irradiance in the long wave range, $\bar{E}^{\uparrow}$, and we do this in the next section. Before that, note that the average value of $\bar{E}^{\uparrow}$, $\frac{1}{2\pi}\int_{2\pi}\bar{E}^{\uparrow}(\theta)\diff\theta$, can be directly computed from this equation. Doing this gives $\Delta L/2\pi r_s$, i.e. precisely the value we obtained in the previous section.
Having the same form as Eq. \eqref{eq:light2}, the solution of Eq. \eqref{eq:temp1} can also be expressed as an infinite series: \begin{equation} \bar{E}^{\uparrow}(\theta)=\bar{E}^{\uparrow}_0(\theta)+\bar{\alpha}\mathcal{E}[\bar{E}^{\uparrow}_0](\theta)+\bar{\alpha}^2\mathcal{E}[\mathcal{E}[\bar{E}^{\uparrow}_0]](\theta)+\bar{\alpha}^3\mathcal{E}[\mathcal{E}[\mathcal{E}[\bar{E}^{\uparrow}_0]]](\theta)+\ldots \end{equation} with \begin{equation} \bar{E}^{\uparrow}_0(\theta)\eqdef(1-\alpha)\bar{\alpha} E(\theta)\quad\mathrm{and}\quad\bar{\alpha}\eqdef\frac{\Delta}{\Delta+1} \end{equation}
Substituting Eq. \eqref{eq:light3} in this series, we get: \begin{align*} \bar{E}^{\uparrow}(\theta)&=(1-\alpha)\bar{\alpha} \sum_{i=0}^\infty\bar{\alpha}^i\mathcal{E}^i\left[\sum_{j=0}^\infty \alpha^jE_j\right](\theta)=(1-\alpha)\bar{\alpha} \sum_{i=0}^\infty\bar{\alpha}^i\sum_{j=0}^\infty \alpha^j E_{i+j}(\theta)\\ &=(1-\alpha)\bar{\alpha} \sum_{k=0}^\infty\left(\sum_{i+j=k} \bar{\alpha}^i\alpha^j\right) E_k(\theta)\ \end{align*} or, in developped form: \begin{align*} \bar{E}^{\uparrow}(\theta)\approx(1-\alpha)\bar{\alpha} \Big[&E_0(\theta)\\ &+(\alpha+\bar{\alpha})E_1(\theta)\\ &+(\alpha^2+\alpha\bar{\alpha}+\bar{\alpha}^2)E_2(\theta)\\ &+(\alpha^3+\alpha^2\bar{\alpha}+\alpha\bar{\alpha}^2+\bar{\alpha}^3)E_3(\theta)\\ &+\left((1-\alpha-\bar{\alpha}+\alpha\bar{\alpha})^{-1}-\sum_{i+j\leq 3}\alpha^i\bar{\alpha}^j\right)\frac{S}{2\pi r_s}\Big] \end{align*} which is the solution for a single linear light source. Finally, to get the actual irradiance in the long wave range inside Rama, it suffice to average three copies of $\bar{E}^{\uparrow}(\theta)$, shifted by 0, 120° and 240°. Doing this gives the irradiance values shown in Fig. 5, for $\Delta=1.3$ and $\Delta=2.9$ (based on the two scenarios defined in the previous section).
From this we see that the maximum irradiance is only 5.4% (resp. 3%) larger than the minimum irradiance in the "incandescent" scenario $\Delta=1.3$ (resp. in the "energy-saving" scenario $\Delta=2.9$). Converted to temperatures with the relation $E=\sigma T^4$, and assuming an average temperature of $288\,K=15°C$, we get a temperature delta of at most 3.7°C in the incandescent scenario ($\Delta=1.3$) and less than 2.2°C in the energy saving scenario ($\Delta=2.9$) - see Fig. 6.
Even with 3 linear light sources, i.e. non axisymmetric lighting condititions, the lighting and temperature inside Rama are almost the same everywhere, if we ignore the atmosphere. Indeed the lighting variations are about 13%, and the temperature variations about 3.7°C or 2.2°C, depending on the scenario.