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Earth Moon Sun Jupiter Saturn Physical constants Условие Решение

IOAA 2017, Теория — Задача 11. Mass of the Local Group

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Всички материали от IOAA 2017

IOAA 2017 — Theoretical Examination · 50 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

The dynamics of M31 (Andromeda) and the Milky Way (MW) can be used to estimate the total mass of the Local Group (LG). The basic idea is that galaxies currently in a binary system were at approximately the same point in space shortly after the Big Bang. To a reasonable approximation, the mass of the local group is dominated by the masses of the MW and M31. Via Doppler shifts of the spectral lines, it was found that M31 is moving towards the MW with a speed of $118\ \mathrm{km\,s^{-1}}$ . This may be surprising, given that most galaxies are moving away from each other with the general Hubble flow. The fact that M31 is moving towards the MW is presumably because their mutual gravitational attraction has eventually reversed their initial velocities. In principle, if the pair of galaxies is well-represented by isolated point masses, their total mass may be determined by measuring their separation, relative velocity and the time since the universe began. Kahn and Woltjer (1959) used this argument to estimate the mass in the LG.

In this problem we will follow this argument through our calculation as follows.

a) Consider an isolated system with negligible angular momentum of two gravitating point masses $m_1$ and $m_2$ (as observed by an inertial observer at the centre of mass).

Complete original two-point-mass geometry, m1/m2 disks, distance braces r1/r2 and Centre of mass arrow/label.

Write down the expression of the total mechanical energy ( $E$ ) of this system in mathematical form connecting $m_1$ , $m_2$ , $r_1$ , $r_2$ , $v_1$ , $v_2$ , and the universal gravitational constant $G$ , where $v_1$ and $v_2$ are the radial velocities of $m_1$ and $m_2$ , respectively. [5 т.]

b) Re-write the equation in a) in terms of $r$ , $v$ , $\mu$ , $M$ and $G$ where $r=r_1+r_2$ is the separation distance between $m_1$ and $m_2$ , $v$ is the changing rate of the separation distance, $\mu\equiv\frac{m_1m_2}{m_1+m_2}$ is the reduced mass of the system, and $M=m_1+m_2$ is the total mass of the system. [10 т.]

c) Show that the equation in b) yields

v^2=(2GM)\left(\frac1r-\frac1{r_0}\right),

where $r_0$ is a new constant. Find $r_0$ in terms of $\mu$ , $M$ , $G$ and $E$ . [5 т.]

The solution of the equation in b) is given below in parametric form, under the initial condition $r=0$ at $t=0$ :

r(\theta)=\frac{r_0}{2}(1-\cos\theta),

t(\theta)=\left(\frac{r_0^3}{8GM}\right)^{1/2}(\theta-\sin\theta),

where $\theta$ is in radians.

d) From the above parametric equations, show that an expression for $\frac{vt}{r}$ is

\frac{vt}{r}=\frac{(\sin\theta)(\theta-\sin\theta)}{(1-\cos\theta)^2}.

[10 т.]

e) Now we consider $m_1$ and $m_2$ as the MW and M31, respectively, such that the current values of $v$ and $r$ are $v=-118\ \mathrm{km\,s^{-1}}$ and $r=710\ \mathrm{kpc}$ , and $t$ may be taken to be the age of the Universe (13700 million years). Find $\theta$ using numerical iteration. [10 т.]

f) Use the value of $\theta$ from e) to calculate the maximum distance between M31 and the MW, $r_{\max}$ , and hence also obtain the value of $M$ in solar masses. [10 т.]

Решение

a) Solution

\frac12m_1v_1^2+\frac12m_2v_2^2-\frac{Gm_1m_2}{r_1+r_2}=E.

Note that the total energy is negative quantity. [5]

Detailed Marking Scheme: Students get 1.5 marks for each kinetic energy term. Students get 1.0 mark for correct sign in from of the potential energy term and another 1.0 mark for the correct expression of the potential energy.

b) Solution

From $m_1r_1=m_2r_2$ and $r_1+r_2=r$ , we have $r_1=\frac{m_2}{m_1+m_2}r$ and $r_2=\frac{m_1}{m_1+m_2}r$ .

Substituting them into the solution obtained in a), we get

\frac12v^2-\frac{G(m_1+m_2)}r=\frac E\mu,

v^2-\frac{2GM}r=\frac{2E}\mu.

Therefore, $E=\frac\mu2\left(v^2-\frac{2GM}r\right)$ . [10]

Detailed Marking Scheme:

Students get 2.0 marks for using the conservation of momentum or center of mass to get $m_1r_1=m_2r_2$ .
Students get 1.0 mark for each of the relationship $r_1=\frac{m_2}{m_1+m_2}r$ or $r_2=\frac{m_1}{m_1+m_2}r$ . (total 2.0 marks)
Students get 1.0 mark each for the relationships between $v$ and $v_1$ , and between $v$ and $v_2$ . (total 2.0 marks)
For correct substitution of suitable variables into the expression for $E$ , students get 2.0 marks.
Final expression for $E$ is worth 2.0 marks.

c) Solution

From b), we have $E=\frac\mu2\left(v^2-\frac{2GM}r\right)$ or $v^2-\frac{2GM}r=\frac{2E}\mu$ .

v^2=\frac{2GM}r+\frac{2E}\mu=(2GM)\left\{\frac1r+\frac1{\left(\frac{GM\mu}E\right)}\right\}=(2GM)\left\{\frac1r-\frac1{r_0}\right\},

where $r_0\equiv-\frac{GM\mu}E$ .

Thus $r_0=-\frac{GM\mu}E$ . [5]

Detailed Marking Scheme: Students get 4.0 marks for correct algebra and obtain the expression in the correct form. By correctly identifying that $r_0=-\frac{GM\mu}E$ , students get 1.0 mark.

d) Solution

Substituting $r(\theta)$ in the equation of part c):

v^2=2GM\left(\frac1r-\frac1{r_0}\right)=2GM\left(\frac1{\frac12r_0(1-\cos\theta)}-\frac1{r_0}\right)=\frac{2GM}{r_0}\left(\frac{1+\cos\theta}{1-\cos\theta}\right).

v^2=\frac{2GM}{r_0}\frac{(1-\cos^2\theta)}{(1-\cos\theta)^2}=\frac{2GM}{r_0}\frac{\sin^2\theta}{(1-\cos\theta)^2}\Rightarrow v(\theta)=\left(\frac{2GM}{r_0}\right)^{1/2}\frac{\sin\theta}{1-\cos\theta}.

Combining $v(\theta)$ , $t(\theta)$ , and $r(\theta)$ :

\frac{v(\theta)t(\theta)}{r(\theta)}=\frac{\left(\frac{2GM}{r_0}\right)^{1/2}\frac{\sin\theta}{1-\cos\theta}\cdot\left(\frac{r_0^3}{8GM}\right)^{1/2}(\theta-\sin\theta)}{\frac{r_0}2(1-\cos\theta)}=\frac2{r_0}\left(\frac{2GMr_0^3}{r_08GM}\right)^{1/2}\frac{(\sin\theta)(\theta-\sin\theta)}{(1-\cos\theta)^2},

\frac{vt}r=\frac{(\sin\theta)(\theta-\sin\theta)}{(1-\cos\theta)^2}.

Alternative solution: From $r(\theta)=\frac{r_0}2(1-\cos\theta)$ , we have

v=\left(\frac{r_0}2\sin\theta\right)\omega.

From $t(\theta)=\left(\frac{r_0^3}{8GM}\right)^{1/2}(\theta-\sin\theta)$ , we have

1=\left(\frac{r_0^3}{8GM}\right)^{1/2}(1-\cos\theta)\omega.

\therefore\frac{vt}r=\frac{\left(\frac{r_0}2\sin\theta\right)\left(\frac{r_0^3}{8GM}\right)^{1/2}(\theta-\sin\theta)}{\left(\frac{r_0^3}{8GM}\right)^{1/2}(1-\cos\theta)\frac{r_0}2(1-\cos\theta)}=\frac{(\sin\theta)(\theta-\sin\theta)}{(1-\cos\theta)^2}.

[10]

Detailed Marking Scheme: Student gets 3.0 marks for substituting $r(\theta)$ in $v^2$ and obtaining $\frac{2GM}{r_0}\left(\frac{1+\cos\theta}{1-\cos\theta}\right)$ . Student gets 3.0 marks for using trigonometry to simplify the expression for $v(\theta)$ . For the correct derivation of the final answer, student gets 4.0 marks.

e) Solution

v=-118\ \mathrm{km\,s^{-1}}=-118\times10^3\ \mathrm{m\,s^{-1}},

r=710\ \mathrm{kpc}=2.19\times10^{22}\ \mathrm m,

t=13700\times10^6\ \mathrm{years}=4.3233\times10^{17}\ \mathrm s.

\therefore\frac{vt}r=-\frac{118\times10^3\times4.3233\times10^{17}}{2.1908\times10^{22}}=-2.33=\frac{(\sin\theta)(\theta-\sin\theta)}{(1-\cos\theta)^2}.

[5]

The negative value of the left-hand side of this equation implies that $\theta$ is greater than $\pi$ .

$\theta$	$(\sin\theta)(\theta-\sin\theta)/(1-\cos\theta)^2$
$200^\circ=3.49$ radians	−0.348
$210^\circ=3.67$ radians	−0.598
$240^\circ=4.19$ radians	−1.95
$244^\circ=4.26$ radians	−2.24
$245^\circ=4.28$ radians	−2.32
$246^\circ=4.29$ radians	−2.40
$250^\circ=4.36$ radians	−2.77

This gives the value of $245^\circ=4.28$ radians. [5]

Detailed Marking Scheme: Students get 5.0 marks if numerical value and sign (2.5 marks for each) are both correct. Students get 5.0 marks if the obtained angle ( $\theta$ ) is within +/− 1 degree (or equivalent in radian), or 2.0 marks for answer within +/− 3 degrees (or equivalent in radian).

f) Solution

From $r(\theta)=\frac{r_0}2(1-\cos\theta)$ , we have $r_{\max}=r_0$ , and hence

r_{\max}=\frac{2r(\theta)}{1-\cos\theta}=\frac{2\times710\ \mathrm{kpc}}{1-\cos245^\circ}=998\ \mathrm{kpc}.

[5]

(This is the maximum separation between the Milky-Way Galaxy and Andromeda Galaxy; they started to move towards each other afterwards.)

The value of $M$ can be calculated from the relation $t(\theta)=\left(\frac{r_0^3}{8GM}\right)^{1/2}(\theta-\sin\theta)$ .

M=\left(\frac{r_{\max}^3}{8G}\right)\left\{\frac{\theta-\sin\theta}{t(\theta)}\right\}^2.

From the maximum separation of $r_{\max}=998\ \mathrm{kpc}=3.0795\times10^{22}\ \mathrm m$ , $\theta=245^\circ=4.276$ radians, $t=t(\theta)=13700\times10^6\ \mathrm{years}=4.3233\times10^{17}\ \mathrm s$ , and $G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$ , we obtain

\therefore M=7.86\times10^{42}\ \mathrm{kg}=\frac{7.86\times10^{42}}{1.99\times10^{30}}M_{\square}=3.95\times10^{12}M_{\square}.

[5]

The estimated number of stars, most of which are of less than a solar mass, for Andromeda and Our Galaxy are $4\times10^{11}$ and $2\times10^{11}$ respectively. This implies that most of mass of the system is DARK.

Detailed Marking Scheme: First part which is counting for 5.0 marks for determining $r_{\max}$ is detailed as follow:

Students get 2.0 marks for realizing that $r_{\max}=r_0$ .
Students get 1.0 mark by re-arranging $r_{\max}$ in term of $r(\theta)$ and $\theta$ .
Students get 1.0 mark if they substitute the correct values given in the question.
Students get 1.0 mark for correct numerical answer.

Second part that is worth 5.0 marks for determining $M$ is detailed as follow:

Students get 2.0 marks for successfully expressing M in term of known variables.
Students get 1.0 mark by substituting the correct numerical values given in the question into the expression from previous step.
Students get 2.0 marks for obtaining the correct answer in requested unit.

Note that

v=\frac d{dt}r=+(2GM)^{1/2}\left(\frac{r_0-r}{r_0r}\right)^{1/2}\quad\text{for }r_0>r,

\int(2GM)^{1/2}dt=+\int\left(\frac{r_0r}{r_0-r}\right)^{1/2}dr.

Putting $\frac r{r_0}\equiv\sin^2\phi$

t=\left\{\frac{r_0^3}{8GM}\right\}^{1/2}\{2\phi-\sin2\phi\}+D.

We choose the initial condition such that $m_1$ and $m_2$ are close together, that is that $r=0$ at $t=0$ . This implies that $\phi=0$ at $t=0$ , hence $D$ must be zero.

t=\left\{\frac{r_0^3}{8GM}\right\}^{1/2}(\theta-\sin\theta),\quad\theta\equiv2\phi.

And from $r=r_0\sin^2\phi=\frac12r_0(1-\cos2\phi)$ , we have:

r=\frac12r_0(1-\cos\theta).

Equation (5) and (6) form the parametric solution of equation (4).

r_{\max}=r(\theta=\pi)=r_0.

Native key solar-mass missing-glyph boxes remain $M_{\square}$ . The source’s supplementary derivation and equation-number references are retained.

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