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Условие A. Thermodynamic test B. Observer on an extrasolar star Data for B Решение

IOAA 2014, Анализ на данни · предварителна тема — Задача 3. Extra solar tests

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8th IOAA · Preliminary data analysis · Extra solar tests · 50 т.

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This is the archived preliminary whole Problem 3, with Parts A and B together. The final photographed exam separates thermodynamics into Problem 2 and replaces the seven-star Big Dipper data with a four-star Mizar table in Problem 3. The preliminary edition and its complete supplied key are preserved separately; source calculation discrepancies are retained.

Условие

A. Thermodynamic test

A shuttle was launched to investigate the CO $_2$ atmosphere of two planets P $_1$ and P $_2$ . When the shuttle was near the each planet, a radio probe was launched toward each of the planet, in vertical direction (the planet radius direction). When the radio-probe movement becomes uniform, it starts sending values of the pressure of the atmosphere. In fig. 3.1 was plotted the atmospheric pressure values (conventional units) of the planet P $_1$ as function of the time of descending. When the probe touched the surface of the planet P $_1$ it sends the value of the temperature $T_0=700\,\mathrm K$ and respectively the value of gravitational acceleration $g_0=10\,\mathrm{m\,s^{-2}}$ .

Fig. 3.1. Pressure versus descent time on P₁.

a Find out the altitude $h_0$ from where the radio-probe R $_1$ starts the uniform descending and thus starts the transmitting information.

b Find out the temperature of the planet P $_1$ at the altitude $h=39.6\,\mathrm{km}$ . You know: The universal gas constant $R=8.3\,\mathrm{J/molK}$ ; the molar mass of CO $_2$ , $\mu=44\,\mathrm{g/mol}$ .

c In fig. 3.2. was plotted the atmospheric pressure values (conventional units) as function of the time of descending in the planet P $_2$ atmosphere. When the probe touched the surface of the planet P $_2$ , it sends the value of the temperature $T_0=750\,\mathrm K$ and respectively the value of gravitational acceleration $g_0=8\,\mathrm{m\,s^{-2}}$ .

Draw the following dependency graphs for $p=f(h)$ and $T=f(h)$ in the CO $_2$ atmosphere of the planet P $_2$ .

The gravitational acceleration on each planet is constant during uniform descending of the radio-probes.

Fig. 3.2. Pressure versus descent time on P₂.

B. Observer on an extrasolar star

The Sirius star, located in the Canis Majoris constellation is the brightest star which can be observed on the sky with naked eye. What the observer's eye see as a single star is in matter of fact a binary star system.

The luminosity of Sirius is a consequence of two facts: its intrinsic luminosity and its closeness to Earth.

Let's assume that an observer is located on one of the planets of the Sirius system.

In the figure below there are represented the Sun, Sirius and the Big Dipper asterism from Ursa Major constellation.

Original illustration: Sun, Sirius and Big Dipper.

Determine:

a The magnitude of Sun as seen from the observer from the planet of Sirius $m_{\text{Sun,Planet}}$ .

b The magnitude of Sirius star system seen by the observer on one of the planets of Sirius, $m_{\text{SY,Planet}}$ .

c The luminosity luminozitatea of “Big Dipper”, $L_{\text{Big Dipper}}$ ;

d the distance between Big Dipper and Earth,

e The geocentric angular distance between Big Dipper and Sirius, $\Delta\theta$ ;

f The distance between Big Dipper and the observer from the planet of Sirius $d_{\text{Big Dipper,Planet}}$ ;

g The magnitude of Big Dipper as seen by the observer from the planet of Sirius $m_{\text{Big Dipper,Planet}}$

Data for B

The following data are given:

$d_{\text{Sirius,Earth}}=2.6\,\mathrm{pc}$ – the Sirius–Earth distance;
$m_{\text{Sirius,Earth}}=-1.46^{\mathrm m}$ – the apparent magnitude of Sirius measured from Earth;
$L_{\text{Sirius}}=25L_{\text{Sun}}$ – the luminosity of Sirius;
$d_{\text{Sun,Earth}}=1\,\mathrm{AU}$ – distance Sun–Earth;
$m_{\text{Sun,Earth}}=-26.78^{\mathrm m}$ – the apparent magnitude of Sun seen from Earth;
$d_{\text{Sirius,Planet}}=10\,\mathrm{AU}$ – the distance between Sirius and its planet where the observer is located;
$m_{\text{Big Dipper,Pamant}}\approx2^{\mathrm m}$ – the apparent magnitude of the Big Dipper asterism for an observer from Earth.

In the table bellow there are information for the stars from Big Dipper asterism regardless to an observer on Earth.

Star number	Name of the star	Apparent magnitude	Distance (light year)
1	Dubhe	1.8	124
2	Merak	2.4	79
3	Phecda	2.4	84
4	Megrez	3.3	58
5	Alioth	1.8	81
6	Mizar	2.1	78
7	Alkaid	1.9	101

The equatorial of Big Dipper ( $\sigma_1$ ) asterism and respectively of Sirius ( $\sigma_2$ ), located on the geocentric map are:

\alpha_{\text{Big Dipper}}=\alpha_1=13^{\mathrm h}23^{\mathrm{min}}55.5^{\mathrm s};\quad\delta_{\text{Big Dipper}}=\delta_1=54^\circ55'31'';

\alpha_{\text{Sirius}}=\alpha_2=6^{\mathrm h}45^{\mathrm{min}};\qquad\delta_{\text{Sirius}}=\delta_2=-16^\circ43'.

Original geocentric celestial sphere and coordinate labels.

Решение

Official preliminary Problem 3 marking schemes: Part A 25 points and Part B 25 points. Original numerical and notational inconsistencies remain visible in the transcription.

A. Thermodynamic test

Marking scheme: a) 5 points, b) 10 points, c) 10 points.

a The pressure and the density of the atmosphere increase with the decrease of the altitude. The pressure at the surface of the planet is $p_{\max}$ .

In figure 1 is represented a thin atmospheric layer with thickness $\Delta h$ near the surface of the planet P $_1$ where the atmosphere density can be considered $\rho_{\max}$ . Inside this layer the pressure and the temperature are variable, reaching the values p $_{\max}$ and T $_0$ near the surface.

Fig. 1. Thin atmospheric layer and pressure difference.

The following relations can be written :

p_{\max}=p+\Delta p;

\Delta p=\rho_{\max}g_0\Delta h,

$\Delta p$ - the variation of the pressure due to the altitude variation $\Delta h$ , $g_0$ the gravitational acceleration near the surface of planet P $_1$ ;

\Delta h=\frac{\Delta p}{\rho_{\max}g_0}.

The superior layer where the density is constant $\rho$ is represented too. At its base the pressure and temperature have the values $p$ and respectively $T$ .

For a gas volume inside the near surface layer of the atmosphere,near the surface of the planet for $\nu$ mol of de gas according with Mendeleev – Klapeyron law

p_{\max}V=\nu RT_0=\frac{m}{\mu}RT_0;

p_{\max}\mu=\frac{m}{V}RT_0=\rho_{\max}RT_0;

\rho_{\max}=\frac{p_{\max}\mu}{RT_0}.

Just before touching the surface of the planet P $_1$ the radio- probe moves uniformly, so the distance $\Delta h$ is spend with speed $\vec v$ , in time $\Delta t$ :

\Delta h=v\Delta t;

\Delta h=\frac{\Delta p}{\rho_{\max}g_0};\ \rho_{\max}=\frac{p_{\max}\mu}{RT_0};

\frac{\Delta p}{\rho_{\max}g_0}=v\Delta t;\ v=\frac{\Delta p}{\rho_{\max}g_0\Delta t};\ v=\frac{\Delta p}{\frac{p_{\max}\mu}{RT_0}g_0\Delta t};

v=\frac{RT_0\Delta p}{g_0\mu p_{\max}\Delta t},(1)

Where accordingly to fig .2 in the point F $(\tau;p_{\max})$ , the trigonometric tangent of $\alpha$ is:

\tan\alpha=\frac{\Delta p}{\Delta t}.

Fig. 2. Tangent at the final pressure on P₁.

The rapport :

\frac{\Delta p}{p_{\max}\Delta t},

Can be determined in the graph in the point F :

v=\frac{\Delta p}{p_{\max}\Delta t}\frac{RT_0}{g_0\mu};

$\dfrac{\Delta p}{p_{\max}\Delta t}$	$\dfrac{15\ \mathrm{u.c.}}{55\ \mathrm{u.c.}\,250\ \mathrm{s}}=\dfrac{3}{2750}\dfrac{1}{\mathrm{s}}$	$\dfrac{45\ \mathrm{u.c.}}{55\ \mathrm{u.c.}\,750\ \mathrm{s}}=\dfrac{3}{2750}\dfrac{1}{\mathrm{s}}$

v=\frac{3}{2750}\frac{1}{\mathrm{s}}\frac{8{,}3\frac{\mathrm{J}}{\mathrm{molK}}700\ \mathrm{K}}{10\frac{\mathrm{m}}{\mathrm{s}^2}44\frac{10^{-3}\ \mathrm{kg}}{\mathrm{mol}}}=14{,}4\frac{\mathrm{m}}{\mathrm{s}};

h_0=v\tau;\ \tau=3.750\ \mathrm{s};

h_0=54.000\ \mathrm{m}=54\ \mathrm{km}.

b Descending from the initial altitude $h_0=54\ \mathrm{km}$ , with speed $v=14{,}4\frac{\mathrm{m}}{\mathrm{s}}$ , the radio-probe , after 1000 s arrives at the altitude $h=39{,}6\ \mathrm{km}$

t=\frac{h_0-h}{v}=\frac{14400\ \mathrm{m}}{14{,}4\frac{\mathrm{m}}{\mathrm{s}}}=1000\ \mathrm{s}.

On Graph at altitude $h=39{,}6\ \mathrm{km}$ , the atmospheric pressure is $p_h=4\ \mathrm{u.c.}$ ,- see point C on graph.

Fig. 3. Tangent at C and surface point F.

In this point

\tan\beta=\left(\frac{\Delta p}{\Delta t}\right)_{\mathrm{C}};

\left(\frac{\Delta p}{p\Delta t}\right)_{\mathrm{C}}=\frac{1}{p_h}\left(\frac{\Delta p}{\Delta t}\right)_{\mathrm{C}}=\frac{1}{4\ \mathrm{u.c.}}\frac{3\ \mathrm{u.c.}}{500\ \mathrm{s}}=\frac{3}{2000}\frac{1}{\mathrm{s}};

\left(\frac{p\Delta t}{\Delta p}\right)_{\mathrm{C}}=p_h\left(\frac{\Delta t}{\Delta p}\right)_{\mathrm{C}}=4\ \mathrm{u.c.}\frac{500\ \mathrm{s}}{3\ \mathrm{u.c.}}=\frac{2000}{3}\mathrm{s};

v=\left(\frac{\Delta p}{p\Delta t}\right)_C\frac{RT_h}{g_0\mu};

v=\frac{1}{p_h}\left(\frac{\Delta p}{\Delta t}\right)_C\frac{RT_h}{g_0\mu};

T_h=\frac{vg_0\mu}{\frac{1}{p_h}\left(\frac{\Delta p}{\Delta t}\right)_C R};\ T_h=p_h\left(\frac{\Delta t}{\Delta p}\right)_{\mathrm{C}}\frac{vg_0\mu}{R};

T_h=\frac{2000}{3}\mathrm{s}\frac{14{,}4\frac{\mathrm{m}}{\mathrm{s}}10\frac{\mathrm{m}}{\mathrm{s}^2}44\frac{10^{-3}\ \mathrm{kg}}{\mathrm{mol}}}{8{,}3\frac{\mathrm{J}}{\mathrm{molK}}};

T_h=508{,}9\ \mathrm{K}.

c By using the same algorithm, corresponding to the points F, D, C, B şi A in graph on figure 4 :

Fig. 4. Linear pressure–time graph for P₂.

v=\frac{\Delta p}{p_{\max}\Delta t}\frac{RT_0}{g_0\mu};

v=\frac{10\ \mathrm{u.c}}{50\ \mathrm{u.c.}\,500\ \mathrm{s}}\frac{8{,}3\frac{\mathrm{J}}{\mathrm{molK}}750\ \mathrm{K}}{8\frac{\mathrm{m}}{\mathrm{s}^2}44\frac{10^{-3}\ \mathrm{kg}}{\mathrm{mol}}}\approx7\frac{\mathrm{m}}{\mathrm{s}};

h_0=v\tau=7\frac{\mathrm{m}}{\mathrm{s}}2000\ \mathrm{s}=14.000\ \mathrm{m}=14\ \mathrm{km};

h_{\mathrm{F}}=0;

\mathrm{F}(p=50\ \mathrm{u.c.};h=0\ \mathrm{km};T=750\ \mathrm{K});

T_{\mathrm{D}}=p_{\mathrm{D}}\left(\frac{\Delta t}{\Delta p}\right)_{\mathrm{D}}\frac{vg_0\mu}{R}=593{,}7\ \mathrm{K};

h_{\mathrm{D}}=h_0-7\frac{\mathrm{m}}{\mathrm{s}}1500\ \mathrm{s}=14\ \mathrm{km}-10.500\ \mathrm{m}=3{,}5\ \mathrm{km};

\mathrm{D}(p=40\ \mathrm{u.c.};h=3{,}5\ \mathrm{km};T=593{,}7\ \mathrm{K});

T_{\mathrm{C}}=p_{\mathrm{C}}\left(\frac{\Delta t}{\Delta p}\right)_{\mathrm{C}}\frac{vg_0\mu}{R}=445{,}3\ \mathrm{K};

h_{\mathrm{C}}=h_0-7\frac{\mathrm{m}}{\mathrm{s}}1000\ \mathrm{s}=14\ \mathrm{km}-7.000\ \mathrm{m}=7\ \mathrm{km};

\mathrm{C}(p=30\ \mathrm{u.c.};h=7\ \mathrm{km};T=445{,}3\ \mathrm{K});

T_{\mathrm{B}}=p_{\mathrm{B}}\left(\frac{\Delta t}{\Delta p}\right)_{\mathrm{B}}\frac{vg_0\mu}{R}=296{,}8\ \mathrm{K};

h_{\mathrm{B}}=h_0-7\frac{\mathrm{m}}{\mathrm{s}}500\ \mathrm{s}=14\ \mathrm{km}-3.500\ \mathrm{m}=10{,}5\ \mathrm{km};

\mathrm{B}(p=20\ \mathrm{u.c.};h=10{,}5\ \mathrm{km};T=296{,}8\ \mathrm{K});

T_{\mathrm{A}}=p_{\mathrm{A}}\left(\frac{\Delta t}{\Delta p}\right)_{\mathrm{A}}\frac{vg_0\mu}{R}=148{,}4\ \mathrm{K};

h_{\mathrm{A}}=14\ \mathrm{km};

\mathrm{A}(p=10\ \mathrm{u.c.};h=14\ \mathrm{km};T=148{,}4\ \mathrm{K}).

The data are in table below which allows to plot de graph $p=f(h)$ and $T=f(h)$ for planet P $_2$ figure 5 and figure 6.

$t$	0	500 s	1000 s	1500 s	2000 s
$p$	10 u.c.	20 u.c.	30 u.c.	40 u.c.	50 u.c.
$h$	14 km	10,5 km	7 km	3,5 km	0
$T$	148,4 K	296,8 K	445,3 K	593,3 K	750 K

Fig. 5. Pressure versus altitude, from the official key.

Fig. 6. Temperature versus altitude, from the official key.

B. Observer on an extrasolar star

Marking scheme: a) 3, b) 3, c) 4, d) 4, e) 4, f) 3, g) 4 points; total 25 points. All printed intermediate numbers and source discrepancies are retained.

a The magnitude of the SUN as seen from the observer on the Sirius planet:

\log\frac{E_{\text{SUN,EARTH}}}{E_{\text{SUN,PLANET}}}=-0.4(m_{\text{SUN,EARTH}}-m_{\text{SUN,PLANET}}),

Where: $E$ represents the apparent brightness of the body at the corresponding distance to the observer.

\log\frac{L_S/(4\pi d_{S,E}^2)}{L_S/(4\pi d_{S,P}^2)}=-0.4(m_{S,E}-m_{S,P});

d_{S,E}=1\,\mathrm{AU};\quad d_{S,P}=2.6\,\mathrm{pc}=2.6\cdot206265\,\mathrm{AU};

m_{S,E}=-26.78^{\mathrm m};\qquad m_{S,P}=1.87^{\mathrm m}.

Apparent magnitude of Sun seen from the observer on one of the Sirius planet.

b Magnitude of Sirius, seen from the observer from its planet:

\log\frac{E_{\text{Sirius,Planeta}}}{E_{\text{Sun,Planeta}}}=-0.4(m_{\text{Sirius,Planeta}}-m_{\text{Sun,Planeta}});

d_{\text{Sun,Planeta}}=2.6\,\mathrm{pc}=2.6\cdot206265\,\mathrm{AU};\qquad d_{\text{Sirius,Planeta}}=10\,\mathrm{AU};

L_{\text{Sirius}}=25L_{\text{SUN}};\qquad m_{\text{Sun,Planeta}}=1.87^{\mathrm m};

m_{\text{Sirius,Planeta}}=-25.255^{\mathrm m}.

c The luminosity of Big Dipper:

L_{\text{Big Dipper}}=L_1+L_2+L_3+L_4+L_5+L_6+L_7,

where $L_1,L_2,\ldots,L_7$ are the luminosities of the seven stars of the asterism:

\log\frac{E_{\text{Sun,Earth}}}{E_{1,\text{Earth}}}=-0.4(m_{\text{SUN,Earth}}-m_{1,\text{Earth}});

\log\frac{L_{\text{Sun}}/(4\pi d_{\text{Sun,Earth}}^2)}{L_1/(4\pi d_{1,\text{Earth}}^2)}=-0.4(m_{\text{Sun,Earth}}-m_{1,\text{Earth}});

d_{1,\text{Earth}}=124\cdot63240\,\mathrm{AU};\quad m_{\text{Sun,Earth}}=-26.78^{\mathrm m};\quad m_{1,\text{Earth}}=1.8^{\mathrm m};

L_1\approx225L_S.

The luminosity of star 1 (Dubhe) from the asterism Big Dipper.

Symilarly for the rest of stars in the asterism:

L_2\approx50L_S;\quad L_3\approx60L_S;\quad L_4\approx0.134L_S;

L_5\approx52L_S;\quad L_6\approx4L_S;\quad L_7\approx1.37L_S.

The luminosity of the asterism Big Dipper is:

L_{\text{Big Dipper}}=L_1+L_2+L_3+L_4+L_5+L_6+L_7;

L_{\text{Big Dipper}}=(225+50+60+0.134+52+4+1.37)L_{\text{Sun}};

L_{\text{Big Dipper}}=392.5L_{\text{Sun}}.

d The distance between Big Dipper and Earth, $d_{\text{Big Dipper,Earth}}$ :

\log\frac{E_{\text{Big Dipper,Earth}}}{E_{\text{Soare,Earth}}}=-0.4(m_{\text{Big Dipper,Earth}}-m_{\text{Soare,Earth}});

L_{\text{Big Dipper}}=392.5L_S;

d_{\text{Big Dipper,Earth}}=573\cdot10^3\sqrt{392.5}\,d_{\text{Soare,Earth}};

d_{\text{Big Dipper,Earth}}\approx11.3\cdot10^6\,\mathrm{AU}.

e Geocentric angular distance $\Delta\theta$ , between dintre Big Dipper and Sirius.

The equatorial coordinates of asterism Big Dipper ( $\sigma_1$ ) and Sirius ( $\sigma_2$ ) located on the sphere of geocentric celestial sphere are:

\alpha_{\text{Big Dipper}}=\alpha_1=13^{\mathrm h}23^{\mathrm{min}}55.5^{\mathrm s};\quad\delta_{\text{Big Dipper}}=\delta_1=54^\circ55'31'';

\alpha_{\text{Sirius}}=\alpha_2=6^{\mathrm h}45^{\mathrm{min}};\qquad\delta_{\text{Sirius}}=\delta_2=-16^\circ43'.

Original spherical-triangle construction from the key.

In the spherical dreptunghic triangle ABC:

\angle(\sigma_1O\sigma_2)=\Delta\theta.

Also

\angle(\gamma O\sigma_{01})=\alpha_{\text{Big Dipper}}=\alpha_1;\quad\angle(\sigma_{01}OB)=\delta_{\text{Big Dipper}}=\delta_1;

\angle(\gamma O\sigma_{02})=\alpha_{\text{Sirius}}=\alpha_2;\quad\angle(\sigma_{02}OC)=\delta_{\text{Sirius}}=\delta_2;

\angle(\sigma_{01}O\sigma_{02})=\angle(AOC)=\alpha_1-\alpha_2=\Delta\alpha;

\angle(\sigma_1OA)=\angle(\sigma_{01}O\sigma_1)+\angle(\sigma_{01}OA)=\delta_{\text{Big Dipper}}-\delta_{\text{Sirius}}=\Delta\delta;

\angle(\sigma_1O\sigma_2)=\Delta\theta.

Results

\cos a=\cos b\cos c+\sin b\sin c\cos A;\qquad A=90^\circ;

\cos a=\cos b\cos c;\qquad a=\Delta\theta;\quad b=\Delta\alpha;\quad c=\Delta\delta;

\cos\Delta\theta=\cos\Delta\alpha\cos\Delta\delta;

\Delta\alpha=\alpha_1-\alpha_2=13^{\mathrm h}23^{\mathrm{min}}55.5^{\mathrm s}-6^{\mathrm h}45^{\mathrm{min}}=6^{\mathrm h}38^{\mathrm{min}}55.5'\approx100^\circ;

\Delta\delta=\delta_1-\delta_2=54^\circ55'31''+16^\circ43'=71^\circ38'31''\approx72.4^\circ;

\cos\Delta\theta=\cos100^\circ\cos72.4^\circ=(-0.173648177)(0.30236989)=-0.05250598;

\Delta\theta\approx90.3^\circ\approx90^\circ.

f The distance between Big Dipper and the observer on sirius planet of Sirius, $d_{\text{Big Dipper,Planeta}}$ :

d_{\text{Big Dipper,Planeta}}\approx d_{\text{Big Dipper,Sirius}};

d_{\text{Sirius,Earth}}\approx d_{\text{Sirius,Soare}}\approx d_{\text{Soare,Planeta}};

d_{\text{Sirius,Pamant}}=2.6\,\mathrm{pc};

d_{\text{Soare,Planeta}}=2.6\,\mathrm{pc}=2.6\cdot206265\,\mathrm{UA}=d_{\text{Earth}};

d_{\text{Big Dipper,Earth}}\approx11.3\cdot10^6\,\mathrm{AU}.

Original approximate right-triangle distance construction.

d_{\text{Big Dipper,Sirius}}=\sqrt{d_{\text{Big Dipper,Pamant}}^2+d_{\text{Sirius,Earth}}^2}=d_{\text{Big Dipper,Planeta}}\approx11.312\cdot10^6\,\mathrm{UA}.

g The magnitude of Big Dipper for observer from the observer on planet of Sirius:

\log\frac{E_{\text{Big Dipper,Planeta}}}{E_{\text{Big Dipper,Earth}}}=-0.4(m_{\text{Big Dipper,Planeta}}-m_{\text{Big Dipper,Earth}});

m_{\text{Big Dipper,Earth}}\approx2^{\mathrm m};\quad d_{\text{Big Dipper,Earth}}\approx11.3\cdot10^6\,\mathrm{AU};

d_{\text{Big Dipper,Sirius}}=d_{\text{Big Dipper,Planeta}}\approx11.312\cdot10^6\,\mathrm{AU};

m_{\text{Big Dipper,Planeta}}=2^{\mathrm m}+5^{\mathrm m}\log\frac{11.312\cdot10^6\,\mathrm{AU}}{11.3\cdot10^6\,\mathrm{AU}};

m_{\text{Big Dipper,Planeta}}=2^{\mathrm m}+5^{\mathrm m}\log\frac{11.312}{11.3};\qquad m_{\text{Big Dipper,Planeta}}\approx2.0023^{\mathrm m}.

Source notes: the printed luminosities and their sum, angular formula, unit mark after 55.5 in the right-ascension subtraction, declination approximation and angular result are retained. They contain inconsistencies; these are the original key's calculations, not corrected answers.

Оригинал в Архива: DA_Solution.pdf