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IOAA 2014, Анализ на данни · предварителна тема — Задача 1. Atmospheric refraction around a planet

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8th IOAA · Preliminary data analysis · Atmospheric refraction · 50 т.

Внимание: Бележка към темата

This is the complete preliminary Problem 1 preserved in the archived solution booklet, not a question on the final photographed examination. The supplied text contains numerical, sign, molecular-label and unit inconsistencies, retained with source notes in the solution.

Условие

The light refraction in the atmosphere of a planet is an optical effect, which is due to the optical inhomogeneity of the atmosphere, thus the light propagation obey the Fermat principle. In this problem you have to find out if it is the possible that due to the atmospheric refraction an astronaut could see his back. The atmospheric absorption is negligible and the structure of the atmosphere is stable. Here you have some theoretical issues to help you to solve the problem:

The situation you have to investigate is sketched in the figure 1.

Fig. 1. Proposed light path and concentric atmospheric layers.

The non-homogeneity of the atmosphere can be simulated by considering the atmosphere consisted of concentric spherical thin layers of thickness $\Delta r$ . In one layer the refractive index is $n>1$ due to the presence of the gas molecules. The refractive index of vacuum is $n_{\rm vacuum}=1$ .

The plane wave-front of the light AB has to remain normal on the surface of the planet and at the same distance from it, despite that the propagation occurs through an inhomogeneous atmosphere consisted of concentric spherical thin, homogenous layers.

The consequence of the model is that near the planet surface the following relationship occurred

n-1=\alpha N,

whith $N$ the gas molecules concentration, and $\alpha$ a constant.

For Earth atmosphere near the surface:

n-1\approx11.35\cdot10^{-5}.

In the table are given the data for 10 planets (8 from Solar System and 2 exoplanets).

$R$ – planet radius in m; $m$ – the mass of on gas molecule of the dominant one near the surface of the planet in atomic mass unit ( $1\,\mathrm{a.m.u.}=1.66\cdot10^{-27}\,\mathrm{kg}$ ); $p$ – atmospheric pressure near the surface of the planet in bar ( $1\,\mathrm{bar}=10^5\,\mathrm{Pa}=10^5\,\mathrm{N\,m^{-2}}$ ); $T$ – atmosphere temperature near the surface of the planet in K; $g$ – gravitational acceleration in $\mathrm{m\,s^{-2}}$ ; You will assume that the molecules are spheres with an effective diameter d. The Boltzmann-constant is $k=1.38\cdot10^{-23}\,\mathrm{J/K}$ .

Planet	$R$ ( $10^5$ m)	The gas	m (amu)	$p$ ( $10^5$ Pa)	T (K)	$g$ ( $\mathrm{m/s^2}$ )	d (m)
Mercur	24	CO $_2$	44	$10^{-14}$	450	3.7	$3.75\cdot10^{-10}$
Venus	62	CO $_2$	44	100	700	8.5	$3.75\cdot10^{-10}$
Earth	64	N $_2$	28	1	300	9.8	$3\cdot10^{-10}$
Mars	34	CO $_2$	44	$10^{-2}$	200	3.7	$3.75\cdot10^{-10}$
Jupiter	715	H $_2$	2	1	340	23	$2\cdot10^{-10}$
Saturn	600	H $_2$	2	1000	400	10.4	$2\cdot10^{-10}$
Uranus	260	H $_2$	2	1000	50	8.8	$2\cdot10^{-10}$
Neptun	250	H $_2$	2	1000	70	11.1	$2\cdot10^{-10}$
Exoplanet X	72.5	H $_2$	28	15	500	20	$2\cdot10^{-10}$
Exoplanet Y	25.75	CH $_4$	16	20	600	15	$4\cdot10^{-10}$

By using the data, you have to analyze and justify the possibility of the occurrence of the described phenomenon for each planet, considering the propagation of the light in the near the planet surface layer of the atmosphere.

a)

Derive the necessary geometrical condition for producing the light refraction around a planet, if $\Delta n$ is the small variance of the refractive index of the gas medium correlated with the small gas layer thickness $\Delta r$ .

b)

Express the previous condition in terms of the thermodynamic parameters of the gas near the surface of the atmosphere i.e. the pressure, the temperature, the concentration of the gas molecules, the mass of a molecule, by calculating the value of coefficient $\alpha$ for the gas in the layer near the Earth surface. The refractive index $n$ of the gas, near the surface of each planet, obey this relation

\frac{\Delta n}{n}=\frac{\alpha\Delta N}{n}\approx\alpha N.

c)

Calculate the numerical values of those parameters which allow you to establish the possibility of the occurrence of the described optical phenomenon. Around which planets the phenomenon occurs. A table which centralize the data is an advantage for better understanding the solution.

Решение

The following is the complete supplied preliminary key. Source inconsistencies are retained and identified; this problem is absent from the final photographed data-analysis examination.

Marking scheme

Part	Criterion	Points
a	Any relations which drive to the correct answer	10
b	Total	20
b	Thermodynamic analysis	10
b	The value and signification of α	10
c	Total	20
c	Calculations for each planets	15
c	Correct identification of the planets	5

a) The optical condition for the light refraction in the planet atmosphere

As seen in the figure 1 the plane wave front AB has to remain perpendicular to the surface of the planet, and at the same distance from that. If we consider the thin layer near the planet which can be considered homogenous the time of light travel will be the same on the circle sectors AA′ și BB′ corresponding to the same angle $\varphi$ .

Fig. 1. Wave-front travel at radii R and R+Δr, with original velocity labels.

t=\frac{AA'}{v_R}=\frac{BB'}{v_{R+\Delta r}},

$v_R$ – speed of light in the thin spherical homogenous layer near the planet surface – radius R;
$v_{R+\Delta r}$ – speed of light in the thin spherical homogenous layer near the planet surface – radius $R+\Delta r$ ;

v_{R+\Delta r}>v_R;\qquad n_R=\frac c{v_R};\quad n_{R+\Delta r}=\frac c{v_{R+\Delta r}}<n_R,

$c$ speed of light in vacuum, $n_R$ refraction index of the atmosphere layer – radius R, $n_{R+\Delta r}$ refraction index of the atmosphere layer – radius $R+\Delta r$ .

Results:

AA'n_R=BB'n_{R+\Delta r};\quad n_{R+\Delta r}<n_R;

AA'=R\varphi;\quad BB'=(R+\Delta r)\varphi;

Rn_R=(R+\Delta r)n_{R+\Delta r};\quad n_{R+\Delta r}=n_R+\Delta n,

where $\Delta n<0$ is the variance of the refraction index of the atmospheric gas associated with the height $\Delta r$ ;

Rn_R=(R+\Delta r)(n_R+\Delta n);\quad\Delta r\,\Delta n\approx0;

-\frac{\Delta n}{n_R\Delta r}=\frac1R;

n_R=n;\quad-\frac{\Delta n}{n\Delta r}=\frac1R;\quad\frac{-\Delta n/n}{\Delta r}=\frac1R;\quad-\frac{\Delta n}{n\Delta r}=\frac1R.

The condition of the occurrence of the refraction around the planet:

-\frac{\Delta n/n}{\Delta r}=\frac1R.

Using the following notations:

theoretical curvature of the Earth surface $C_{\text{suprafata planeta}}=1/R$ ;
the curvature of the light ray in the atmosphere layer near the Earth surface $C_{\rm light}=-\Delta n/(n\Delta r)$ ;

Thus:

C_{\rm light}=C_{\text{surface of the planet}}.

b1) Relationship between the refractive index and the physical properties of the gas near the planet surface

According with the notations in figure 1 it can be obtain the equilibrium condition for the gas column near the Earth's surface.

Fig. 1. Atmospheric gas column and force equilibrium.

\vec F_1+\vec G+\vec F_2=0;\quad F_1-G-F_2=0;

p\Delta S-Mg-(p+\Delta p)\Delta S=0;

p\Delta S-Mg-p\Delta S-\Delta p\Delta S=0;

-\Delta p\Delta S=Mg;\quad-\Delta p\Delta S=Nmg\Delta r\Delta S;\quad-\Delta p=Nmg\Delta r,

where $m$ is the mass of a molecule, and $g$ is the gravitational acceleration at the base of the atmosphere:

p=NkT;\quad\Delta p=kT\Delta N;\quad-kT\Delta N=Nmg\Delta r;

-\frac{\Delta N}{\Delta r}=N\frac{mg}{kT}.\qquad\text{10 points}

According to the relationship $n-1=\alpha N$ , and the fact that $n_{\rm vid}=1$ the coefficient $\alpha$ represents the volume.

\langle N\rangle_{\rm SI}=\mathrm{m^{-3}};\qquad\langle\alpha\rangle_{\rm SI}=\mathrm{m^3}

and $n>1$ being proportional to the concentration of the molecules in the atmosphere.

n-1=\alpha N;\quad\Delta(n-1)=\Delta(\alpha N);\quad\Delta n=\alpha\Delta N;

\frac{\Delta n}{n}=\frac{\alpha\Delta N}{n};\quad n\approx1;\quad\frac{\Delta n}{n}=\frac{\alpha\Delta N}{n}\approx\alpha\Delta N.

Using the relation (1):

\frac{\Delta n}{n}=\alpha\Delta N;\quad\Delta N=\frac{\Delta n}{\alpha n};\quad-\frac{\Delta N}{\Delta r}=N\frac{mg}{kT};

-\frac{\Delta n/(\alpha n)}{\Delta r}=N\frac{mg}{kT};\qquad-\frac{\Delta n}{\alpha n\Delta r}=N\frac{mg}{kT}.

The relationship between the light curvature and the physical properties of the atmosphere will be:

C_{\rm light}=-\frac{\Delta n}{n\Delta r}=\alpha N\frac{mg}{kT}.\qquad (2)

This one can be experimental determined for each planet atmosphere and compared with the each planet surface curvature.

C_{\text{light,experimental}}=\alpha N\frac{mg}{kT};\qquad C_{\rm planet}=\frac1R.

The condition is

C_{\rm light}=C_{\rm planet}.\qquad (3)

b2) The value of α, corelated with the volume of the gas molecule

$\alpha$ represents the volume of a gas molecule.

For Earth:

n-1\approx11.35\cdot10^{-5}=\alpha N;\quad p=1\,\mathrm{bar}=10^5\,\mathrm{N\,m^{-2}};\quad T=300\,\mathrm K,

N=\frac p{kT}=\frac{10^5\,\mathrm{N\,m^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot300\,\mathrm K},

where $k=1.38\cdot10^{-23}\,\mathrm{J/K}$ (constanta lui Boltzman);

N\approx2.41\cdot10^{25}\,\mathrm{m^{-3}},

Is the nytrogen concentration in the Earth atmosphere;

11.35\cdot10^{-5}=\alpha\cdot2.41\cdot10^{25}\,\mathrm{m^{-3}};\quad\alpha\approx4.70\cdot10^{-30}\,\mathrm{m^3}.

The volume of an nytrogen molecule with the diameter $d_{\rm N_2}\approx3\cdot10^{-10}$ m is:

V_{1,\rm N_2}=\frac{4\pi}3\left(\frac{d_{\rm N_2}}2\right)^3=\frac{4\cdot3.14}3\frac{9\cdot10^{-30}}8\,\mathrm{m^3};

Thus

V_{1,\rm N_2}\approx4.71\cdot10^{-30}\,\mathrm{m^3}\approx\alpha.

c) Data analysis

20 points. Each original numerical substitution and printed result follows. The source occasionally interchanges the units of $\alpha$ and $N$ in their product; those units and the gas subscripts are retained below. Numerical inconsistencies are not silently recalculated.

Earth

For Earth atmosphere considered full of nitrogen where $p=1$ bar and temperature $T=300$ K, the key again writes

V_{1,\rm N_2}=\frac{4\pi}3\left(\frac{d_{\rm N_2}}2\right)^3=\frac{4\cdot3.14}3\frac{9\cdot10^{-30}}8\,\mathrm{m^3}.

N=\frac p{kT}=\frac{10^5\,\mathrm{N\,m^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot300\,\mathrm K};\qquad N\approx2.41\cdot10^{25}\,\mathrm{m^{-3}};

V_{1,\rm N_2}\approx4.71\cdot10^{-30}\,\mathrm{m^3}\approx\alpha;

\alpha N=4.71\cdot10^{-30}\,\mathrm{m^{-3}}\cdot2.41\cdot10^{25}\,\mathrm{m^3}=11.35\cdot10^{-5};

\alpha N\frac{mg}{kT}=11.35\cdot10^{-5}\frac{28\cdot1.66\cdot10^{-27}\,\mathrm{kg}\cdot9.8\,\mathrm{m\,s^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot300\,\mathrm K};

\alpha N\frac{mg}{kT}\approx0.12\cdot10^{-7}\,\mathrm{m^{-1}}.

Venus

For Venus atmosphere $T\approx700$ K, $p\approx100$ atm, gas CO $_2$ .

CO $_2$ molecule diameter $d_{\rm CO_2}=1.25d_{\rm N_2}$ .

CO $_2$ molecule volume:

d_{\rm CO_2}=3.75\cdot10^{-10}\,\mathrm m;

V_{1,\rm CO_2}=\frac{4\pi}3\left(\frac{d_{\rm CO_2}}2\right)^3=\frac{4\cdot3.14}3\frac{(3.75)^3\cdot10^{-30}}8\,\mathrm{m^3}.

N=\frac p{kT}=\frac{10^2\cdot10^5\,\mathrm{N\,m^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot700\,\mathrm K};\qquad N\approx10^{27}\,\mathrm{m^{-3}};

V_{1,\rm CO_2}=27.59\cdot10^{-30}\,\mathrm{m^3}=\alpha;

\alpha N=27.59\cdot10^{-30}\,\mathrm{m^{-3}}\cdot10^{27}\,\mathrm{m^3}=27.59\cdot10^{-3};

\alpha N\frac{mg}{kT}=27.59\cdot10^{-3}\frac{44\cdot1.66\cdot10^{-27}\,\mathrm{kg}\cdot8.5\,\mathrm{m\,s^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot700\,\mathrm K};

\alpha N\frac{mg}{kT}\approx17.7318\cdot10^{-7}\,\mathrm{m^{-1}}.

Mars

For Mars:

N=\frac p{kT}=\frac{10^{-2}\cdot10^5\,\mathrm{N\,m^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot200\,\mathrm K};\qquad N\approx3.62\cdot10^{23}\,\mathrm{m^{-3}};

V_{1,\rm CO_2}=27.59\cdot10^{-30}\,\mathrm{m^3}=\alpha;

\alpha N=27.59\cdot10^{-30}\,\mathrm{m^{-3}}\cdot3.62\cdot10^{23}\,\mathrm{m^3}=99.87\cdot10^{-7};

\alpha N\frac{mg}{kT}=99.87\cdot10^{-7}\frac{44\cdot1.66\cdot10^{-27}\,\mathrm{kg}\cdot3.7\,\mathrm{m\,s^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot200\,\mathrm K};

\alpha N\frac{mg}{kT}\approx9.77886\cdot10^{-10}\,\mathrm{m^{-1}}.

Jupiter

For Jupiter:

N=\frac p{kT}=\frac{10^5\,\mathrm{N\,m^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot340\,\mathrm K};\qquad N\approx2.13\cdot10^{25}\,\mathrm{m^{-3}};

V_{1,\rm H_2}=4.18\cdot10^{-30}\,\mathrm{m^3}=\alpha;

\alpha N=4.18\cdot10^{-30}\,\mathrm{m^{-3}}\cdot2.13\cdot10^{25}\,\mathrm{m^3}=8.9\cdot10^{-5};

\alpha N\frac{mg}{kT}=8.9\cdot10^{-5}\frac{2\cdot1.66\cdot10^{-27}\,\mathrm{kg}\cdot23\,\mathrm{m\,s^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot340\,\mathrm K};

\alpha N\frac{mg}{kT}\approx14.48\cdot10^{-10}\,\mathrm{m^{-1}}.

Uranus

For Uranus:

N=\frac p{kT}=\frac{10^3\cdot10^5\,\mathrm{N\,m^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot50\,\mathrm K};\qquad N\approx145\cdot10^{27}\,\mathrm{m^{-3}};

V_{1,\rm H_2}=4.18\cdot10^{-30}\,\mathrm{m^3}=\alpha;

\alpha N=4.18\cdot10^{-30}\,\mathrm{m^{-3}}\cdot145\cdot10^{27}\,\mathrm{m^3}=606.1\cdot10^{-3};

\alpha N\frac{mg}{kT}=606.1\cdot10^{-3}\frac{2\cdot1.66\cdot10^{-27}\,\mathrm{kg}\cdot8.8\,\mathrm{m\,s^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot50\,\mathrm K};

\alpha N\frac{mg}{kT}\approx2566.35\cdot10^{-8}\,\mathrm{m^{-1}}.

Saturn

For Saturn:

N=\frac p{kT}=\frac{10^3\cdot10^5\,\mathrm{N\,m^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot400\,\mathrm K};\qquad N\approx18\cdot10^{27}\,\mathrm{m^{-3}};

V_{1,\rm H_2}=4.18\cdot10^{-30}\,\mathrm{m^3}=\alpha;

\alpha N=4.18\cdot10^{-30}\,\mathrm{m^{-3}}\cdot18\cdot10^{27}\,\mathrm{m^3}=75.24\cdot10^{-3};

\alpha N\frac{mg}{kT}=75.24\cdot10^{-3}\frac{2\cdot1.66\cdot10^{-27}\,\mathrm{kg}\cdot10.4\,\mathrm{m\,s^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot400\,\mathrm K};

\alpha N\frac{mg}{kT}\approx470.63\cdot10^{-9}\,\mathrm{m^{-1}}.

Mercury

For Mercury:

N=\frac p{kT}=\frac{10^{-14}\cdot10^5\,\mathrm{N\,m^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot450\,\mathrm K};\qquad N\approx16\cdot10^{10}\,\mathrm{m^{-3}};

V_{1,\rm H_2}=27.59\cdot10^{-30}\,\mathrm{m^3}=\alpha;

\alpha N=27.59\cdot10^{-30}\,\mathrm{m^{-3}}\cdot16\cdot10^{10}\,\mathrm{m^3}=441.44\cdot10^{-20};

\alpha N\frac{mg}{kT}=441.44\cdot10^{-20}\frac{44\cdot1.66\cdot10^{-27}\,\mathrm{kg}\cdot3.7\,\mathrm{m\,s^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot450\,\mathrm K};

\alpha N\frac{mg}{kT}\approx1921\cdot10^{-25}\,\mathrm{m^{-1}}.

Neptun

For Neptun:

N=\frac p{kT}=\frac{10^3\cdot10^5\,\mathrm{N\,m^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot70\,\mathrm K};\qquad N\approx10^{29}\,\mathrm{m^{-3}};

V_{1,\rm H_2}=4.18\cdot10^{-30}\,\mathrm{m^3}=\alpha;

\alpha N=4.18\cdot10^{-30}\,\mathrm{m^{-3}}\cdot10^{29}\,\mathrm{m^3}=4.18\cdot10^{-1};

\alpha N\frac{mg}{kT}=4.18\cdot10^{-1}\frac{2\cdot1.66\cdot10^{-27}\,\mathrm{kg}\cdot11.1\,\mathrm{m\,s^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot70\,\mathrm K};

\alpha N\frac{mg}{kT}\approx15.95\cdot10^{-6}\,\mathrm{m^{-1}}.

Exoplanet X

For Exoplanet X:

N=\frac p{kT}=\frac{15\cdot10^5\,\mathrm{N\,m^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot500\,\mathrm K};\qquad N\approx2.17\cdot10^{26}\,\mathrm{m^{-3}};

V_{1,\rm N_2}=4.71\cdot10^{-30}\,\mathrm{m^3}=\alpha;

\alpha N=4.71\cdot10^{-30}\,\mathrm{m^{-3}}\cdot2.17\cdot10^{26}\,\mathrm{m^3}=10.22\cdot10^{-4};

\alpha N\frac{mg}{kT}=10.22\cdot10^{-4}\frac{28\cdot1.66\cdot10^{-27}\,\mathrm{kg}\cdot20\,\mathrm{m\,s^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot500\,\mathrm K};

\alpha N\frac{mg}{kT}\approx137.68\cdot10^{-9}\,\mathrm{m^{-1}}.

Exoplanet Y

V_{1,\rm CH_4}=\frac{4\pi}3\left(\frac{d_{\rm CH_4}}2\right)^3=\frac{4\cdot3.14}3\frac{4^3\cdot10^{-30}}8\,\mathrm{m^3}.

N=\frac p{kT}=\frac{20\cdot10^5\,\mathrm{N\,m^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot600\,\mathrm K};\qquad N\approx2.41\cdot10^{26}\,\mathrm{m^{-3}};

V_{1,\rm H_2}=33.5\cdot10^{-30}\,\mathrm{m^3}=\alpha;

\alpha N=33.5\cdot10^{-30}\,\mathrm{m^{-3}}\cdot2.41\cdot10^{26}\,\mathrm{m^3}=80.7\cdot10^{-4};

\alpha N\frac{mg}{kT}=80.7\cdot10^{-4}\frac{16\cdot1.66\cdot10^{-27}\,\mathrm{kg}\cdot15\,\mathrm{m\,s^{-2}}}{1.38\cdot10^{-23}\,\mathrm{J/K}\cdot600\,\mathrm K};

\alpha N\frac{mg}{kT}\approx3883\cdot10^{-10}\,\mathrm{m^{-1}}.

Centralized results

Planeta	p (bar)	m (uam)	T (K)	$g$ ( $\mathrm{m/s^2}$ )	$N$ ( $\mathrm{m^{-3}}$ )	$\alpha$ ( $\mathrm{m^3}$ )	$\alpha Nmg/kT$ ( $\mathrm{m^{-1}}$ )
Pământ	1	28	300	9.8	$2.41\cdot10^{25}$	$4.71\cdot10^{-30}$	$0.12\cdot10^{-7}$
Venus	100	44	700	8.5	$10^{27}$	$27.59\cdot10^{-30}$	$17.73\cdot10^{-7}$
Marte	$10^{-2}$	44	200	3.7	$3.62\cdot10^{23}$	$27.59\cdot10^{-30}$	$9.7\cdot10^{-10}$
Jupiter	1	2	340	23	$2.13\cdot10^{25}$	$4.18\cdot10^{-30}$	$14.48\cdot10^{-10}$
Uranus	1000	2	50	8.8	$145\cdot10^{27}$	$4.18\cdot10^{-30}$	$2566.35\cdot10^{-8}$
Saturn	1000	2	400	10.4	$18\cdot10^{27}$	$4.18\cdot10^{-30}$	$470.63\cdot10^{-9}$
Mercur	$10^{-14}$	44	450	3.7	$16\cdot10^{10}$	$27.59\cdot10^{-30}$	$1921\cdot10^{-25}$
Neptun	1000	2	70	11.1	$10^{29}$	$4.18\cdot10^{-30}$	$15.95\cdot10^{-6}$
Exo X	15	28	500	20	$2.17\cdot10^{26}$	$4.71\cdot10^{-30}$	$137.68\cdot10^{-9}$
Exo Y	20	16	600	15	$2.41\cdot10^{26}$	$33.5\cdot10^{-30}$	$388.3\cdot10^{-9}$

Planet	$\alpha Nmg/kT$ ( $\mathrm{m^{-1}}$ )	R (m)	$1/R$ ( $\mathrm{m^{-1}}$ )	Comparison
Earth	$0.12\cdot10^{-7}$	$64\cdot10^5$	$1.56\cdot10^{-7}$	$C_{\text{light,exp}}<C_{\text{surface planet}}$
Venus	$17.73\cdot10^{-7}$	$62\cdot10^5$	$1.61\cdot10^{-7}$	$C_{\text{light,exp}}>C_{\text{surface planet}}$
Marte	$9.7\cdot10^{-10}$	$34\cdot10^5$	$2.94\cdot10^{-6}$	$C_{\text{light,exp}}<C_{\text{surface planet}}$
Jupiter	$14.48\cdot10^{-10}$	$715\cdot10^5$	$1.3\cdot10^{-8}$	$C_{\text{light,exp}}<C_{\text{surface planet}}$
Uranus	$2566.35\cdot10^{-8}$	$26\cdot10^6$	$3.8\cdot10^{-8}$	$C_{\text{light,exp}}>C_{\text{surface planet}}$
Saturn	$470.63\cdot10^{-9}$	$60\cdot10^6$	$1.6\cdot10^{-9}$	$C_{\text{light,exp}}>C_{\text{surface planet}}$
Mercur	$1921\cdot10^{-25}$	$2.4\cdot10^6$	$416\cdot10^{-9}$	$C_{\text{light,exp}}>C_{\text{surface planet}}$
Neptun	$15.95\cdot10^{-6}$	$25\cdot10^6$	$4\cdot10^{-9}$	$C_{\text{light,exp}}>C_{\text{surface planet}}$
Exo X	$137.68\cdot10^{-9}$	$72.5\cdot10^5$	$137\cdot10^{-9}$	$C_{\text{light,exp}}\cong C_{\text{surface planet}}$
Exo Y	$388.3\cdot10^{-9}$	$25.75\cdot10^5$	$388\cdot10^{-9}$	$C_{\text{light,exp}}\cong C_{\text{surface planet}}$

e) Conclusion

For all the Solar System planets

C_{\text{light,exp}}\ne C_{\text{surface planet}}.

Thus the condition is not fullfiled.

For the exoplanets X and Y

C_{\text{raza lumina,exp}}\approx C_{\text{suprafata planeta}}.

Thus the astronaut can see its back.

Source notes: the original question prints $\Delta n/n\approx\alpha N$ ; the derivation later uses $\alpha\Delta N$ . The nitrogen-volume substitution prints 9 where cubing the given 3 would require a different factor. The source table calls Exoplanet X's gas H₂, with mass 28 and diameter $2\cdot10^{-10}$ m, while its key uses nitrogen. Mercury and Exoplanet Y calculations print an H₂ volume subscript despite their other molecular data. The curvature table contains inconsistent reciprocal values and prints a greater-than sign for Mercury despite its own numerical entries. These original values, labels and conclusion are retained, not endorsed as corrected physical calculations.

Оригинал в Архива: DA_Solution.pdf