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Earth Moon Sun Jupiter Saturn Physical constants Условие Решение

IOAA 2017, Теория — Задача 7. Life on Other Planets

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Всички материали от IOAA 2017

IOAA 2017 — Theoretical Examination · 20 т.

Original Examination Instructions

The theoretical examination lasts for 5 hours and is worth a total of 300 marks.
Dedicated IOAA Summary Answer Sheets are provided for writing your answers. Enter the final answers into the appropriate boxes in the corresponding Summary Answer Sheet. On each Summary Answer Sheet, please fill in
- Student Code (Country Code and 1 digit)
There are Answer Sheets for carrying out detailed work/rough work. On each Answer Sheet, please fill in
- Student Code (Country Code and 1 digit)
- Question no.
- Page no. and total number of pages.
Start each problem on a separate Answer Sheet. Please write only on the printed side of the sheet. Do not use the reverse side. If you have written something on any sheet which you do not want to be evaluated, cross it out.
Use as many mathematical expressions as you think may help the evaluator to better understand your solutions. The evaluator may not understand your language. If it is necessary to explain something in words, please use short phrases (if possible in English).
You are not allowed to leave your work desk without permission. If you need any assistance (malfunctioning calculator, need to visit a restroom, need more Answer Sheets, etc.), please draw the attention of the proctor using the Help card.
The beginning and end of the examination will be indicated by a long sound signal. Additionally, there will be a buzzer sound, fifteen minutes before the end of the examination (before the final sound signal).
At the end of the examination you must stop writing immediately. Sort and put your sheets in separate stacks, a) Stack 1: Summary Answer Sheets, Answer Sheets of part 1
b) Stack 2: Summary Answer Sheets, Answer Sheets of part 2
c) Stack 3: Summary Answer Sheets, Answer Sheets of part 3
d) Stack 4: question papers and paper sheets you do not want to be graded.
Wait at your table until your envelope is collected. Once all envelopes are collected, your student guide will escort you out of the examination room.
A list of constants and a table of the mark distribution for this exam are given on the next two pages.

Original List of Constants

Earth

Quantity	Value
Mass ( $M_\oplus$ )	$5.98\times10^{24}$ kg
Radius ( $R_\oplus$ )	$6.38\times10^6$ m
Acceleration of gravity ( $g$ )	$9.8 \mathrm{m\cdot s^{-1}}$
Obliquity of Ecliptic	$23^\circ27'$
Length of Tropical Year	365.2422 mean solar days
Length of Sidereal Year	365.2564 mean solar days
Albedo	0.39

Moon

Quantity	Value
Mass (M with subscript ☾)	$7.35\times10^{22}$ kg
Radius (R with subscript ☾)	$1.74\times10^6$ m
Mean distance from Earth	$3.84\times10^8$ m
Orbital inclination with the Ecliptic	$5.14^\circ$
Albedo	0.14
Apparent magnitude (mean full moon)	−12.74

Sun

Quantity	Value
Mass ( $M_\odot$ )	$1.99\times10^{30}$ kg
Radius ( $R_\odot$ )	$6.96\times10^8$ m
Luminosity ( $L_\odot$ )	$3.83\times10^{26}$ W
Absolute Magnitude ( $\mathcal M_\odot$ )	4.80 mag
Angular diameter	0.5 degrees
Rotational velocity in the Galaxy	$220 \mathrm{km\,s^{-1}}$
Distance from Galactic centre	8.5 kpc

Jupiter

Quantity	Value
Mass	$1.89\times10^{27}$ kg
Orbital semi-major axis	5.20 au
Orbital period	11.86 year

Saturn

Quantity	Value
Mass	$5.68\times10^{26}$ kg
Orbital semi-major axis	9.58 au
Orbital period	29.45 year

Physical constants

Quantity	Value
1 au	$1.50\times10^{11}$ m
1 pc	206,265 au
Gravitational constant ( $G$ )	$6.67\times10^{-11} \mathrm{N\cdot m^2\cdot kg^{-2}}$
Planck constant ( $h$ )	$6.62\times10^{-34} \mathrm{J\cdot s}$
Boltzmann constant ( $k_B$ )	$1.38\times10^{-23} \mathrm{J\cdot K^{-1}}$
Stefan-Boltzmann constant ( $\sigma$ )	$5.67\times10^{-8} \mathrm{W\cdot m^{-2}\cdot K^{-4}}$
Hubble constant ( $H_0$ )	$67.8 \mathrm{km\,s^{-1}\,Mpc^{-1}}$
Speed of light in vacuum ( $c$ )	$299,792,458 \mathrm{m\cdot s^{-1}}$
Proton mass	$938.27 \mathrm{MeV\cdot c^{-2}}$
Deuterium mass	$1875.60 \mathrm{MeV\cdot c^{-2}}$
Neutron mass	$939.56 \mathrm{MeV\cdot c^{-2}}$
Helium-3 mass	$2808.30 \mathrm{MeV\cdot c^{-2}}$
Helium-4 mass	$3727.40 \mathrm{MeV\cdot c^{-2}}$

Original Mark Distribution

Problem number	Marks
T1	10
T2	10
T3	10
T4	10
T5	10
T6	15
T7	20
T8	20
T9	20
T10	25
T11	50
T12	40
T13	60
Total	300

Условие

One place to search for life is on planets orbiting main sequence stars. A good starting point is the planets that have an Earth-like temperature range and a small temperature fluctuation. Assume that for a main sequence star, the relation between the luminosity $L$ and the mass $M$ is given by

L\propto M^{3.5}.

You may assume that the total energy $E$ released over the lifetime of the star is proportional to the mass $M$ of the star. For the Sun, it will have a main sequence lifetime of about 10 billion years. The stellar spectral types are given in the table below. Assume that the spectral subclasses of stars (0-9) are assigned on a scale that is linear in $\log M$ .

Spectral Class	O5V	B0V	A0V	F0V	G0V	K0V	M0V
Mass ( $M_\odot$ )	60	17.5	2.9	1.6	1.05	0.79	0.51

a) If it takes at least $4\times10^9$ years for an intelligent life form to evolve, what is the spectral type (accurate to the subclass level) of the most massive star in the main sequence around which astronomers should look for intelligent life? [6 т.]

b) Assume that the target planet has the same emissivity $\varepsilon$ and albedo $a$ as the Earth. In order to have the same temperature as the Earth, express the distance $d$ , in au, of the planet to its parent main sequence star, of mass $M$ . [6 т.]

c) The existence of a planet around a star can be shown by the variation in the radial velocity of the star about the star-planet system centre of mass. If the smallest Doppler shift in the wavelength detectable by the observer is $(\Delta\lambda/\lambda)=10^{-10}$ , calculate the lowest mass of such a planet in b), in units of Earth masses, that can be detected by this method, around the main sequence star in a). [8 т.]

Решение

a) Solution

Since $E\propto M\to L\tau\propto M$ and $L\propto M^{3.5}$ ,

\tau=\tau_\odot\left(\frac{M_\odot}{M}\right)^{2.5}.

[2.0]

The maximum mass of the star should be

M=M_\odot\left(\frac{\tau_\odot}{\tau}\right)^{1/2.5}=M_\odot\left(\frac{10^{10}}{4\times10^9}\right)^{1/2.5},

[0.5]

M=1.44M_\odot.

[0.5]

Use of spectral class-log $M$ linear relation

\frac{10}{(\log1.6-\log1.05)}\times(\log1.6-\log1.44)=2.5

[2.0]

which corresponds to the star of type F2V to F3V. [1.0]

0.5 mark for the answer “between F0V and G0V”.

b) Solution

The received power by a planet of a radius $r$ and a distance $d$ from the star is

P_{input}=(1-a)\frac{\pi r^2L}{4\pi d^2}=(1-a)\frac{r^2L}{4d^2},

where $a$ is the planet’s albedo. [1.0]

The released power of blackbody radiation is

P_{output}=4\pi r^2\varepsilon\sigma T^4,

where $\varepsilon$ is emissivity. [1.0]

Since we assume that the planet emits heat as blackbody radiation,

(1-a)\frac{r^2L}{4d^2}=4\pi r^2\varepsilon\sigma T^4.

[1.0]

The temperatures of the planet and the Earth are similar,

T^4=\frac{(1-a)L}{16\pi\varepsilon\sigma d^2}=\frac{(1-a)L_\odot}{16\pi\varepsilon\sigma d_\odot^2}.

\frac L{d^2}=\frac{L_\odot}{d_\odot^2}.

[1.0]

\left(\frac d{d_\odot}\right)^2=\frac L{L_\odot}=\left(\frac M{M_\odot}\right)^{3.5}.

[1.0]

The distance of the planet to the star is $d=\left(\frac M{M_\odot}\right)^{3.5/2}$ au. [1.0]

c) Solution

Angular speed around the center of mass can be derived by Kepler’s law.

T^2=\frac{4\pi^2d^3}{G(M+m)}.

[1.0]

\omega=\sqrt{\frac{G(M+m)}{d^3}}.

[1.0]

According to the centre of mass of the planet and the star,

Md_s=md,

d_s=\frac mMd.

[1.0]

and

v=\omega d_s,

[1.0]

v=\omega d\frac mM=\frac mM\sqrt{\frac{G(M+m)}d}\approx m\sqrt{\frac G{Md}}.

From $\frac{\Delta\lambda}{\lambda}=\frac vc=\frac mc\sqrt{\frac G{Md}}$ : [1.0]

Use of previously derived expression The distance of the planet to the star is

d=\left(\frac M{M_\odot}\right)^{3.5/2}\ \mathrm{au}.

[1.0]

m=\frac{\Delta\lambda}{\lambda}c\sqrt{\frac{Md}G}=\frac{\Delta\lambda}{\lambda}c\sqrt{\frac{M\left(\frac M{M_\odot}\right)^{3.5/2}\times1\ \mathrm{au}}G}

=\frac{\Delta\lambda}{\lambda}c\sqrt{\frac{M(1.44)^{3.5/2}\times1\ \mathrm{au}}G}

=(299792458\ \mathrm{m/s})\times10^{-10}\sqrt{\frac{1.44\times1.99\times10^{30}\ \mathrm{kg}\ (1.44)^{3.5/2}\times1.50\times10^{11}\ \mathrm m}{6.67\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}}}.

[1.0]

m=3.31\times10^{24}\ \mathrm{kg}=0.554M_\oplus.

[1.0]

Бележки за изданието

The original question and key contain several differing expressions. The Galactic-arm question uses a flat curve beyond 5 kpc and its worked key uses 2 kpc. The HI survey key prints differing flux-threshold exponents and an unusual Hubble-constant unit. The Bethlehem key has differing multiplication and remainder values. The Galactic Outflow key heading says T2 while its full prompt matches T9. GOTO assigns a different pair of marks to parts a) and d) in the question and key. The Shipwreck key prints differing sidereal-time values in its main and alternative methods. Visible missing-font glyph squares and the Exomoon key’s grading dependency are preserved. The original constants table prints the acceleration unit as m·s^-1. These printed forms remain literal.

Оригинал в Архива: Theory.pdf · официални решения: Theory_Solution.pdf