Съдържание

IOAA 2024, Теория — Задача 8. T8. Binary Hardening

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA 2024

IOAA 2024 — Theory Examination, English (Official) · 25 т.

Внимание: Бележка към темата

The printed instructions refer to blank Detailed Worksheets and a summary Answer Sheet. Those forms are absent from the available 2024 archive packet. The complete question booklet, solution booklet and separate marking scheme are included. Printed source inconsistencies are identified in the relevant tasks and retained without silently changing the source.

T2: The solution first gives $R=2.3143\times10^{23}\ \mathrm m$ and later substitutes $3.857\times10^{23}\ \mathrm m$ , while retaining its boxed $3.9\times10^{44}\ \mathrm{kg}\simeq2\times10^{14}M_\odot$ . The main double-sum potential expression also omits the factor $1/2$ used for unique pairs in the appendix. These printed expressions are retained; the marking scheme allows valid alternative estimates.

T6: The question names Palomar 4; the solution's repeated question names Palomar 6. The solution gives $V_{sky}=12.9$ mag but substitutes $12.8$ in the following surface-brightness line. The native question and printed calculation are retained.

T7: The printed expansion on solution page 13 has $-\sin(a)\cos(\phi)$ where the preceding cosine-law equation would give $-\sin(a)\sin(\phi)$ . It is retained as printed; the subsequent sunrise calculation sets $a=0$ .

T10: Solution page 25 prints the unit of $|\vec M|^2-R_\oplus^2=1.288\times10^{17}$ as m rather than $\mathrm m^2$ . In (j), the final vector prints 77.59 m/s after previously using 77.76 m/s; the magnitude uses 77.76 m/s. These printed values are retained.

T11: The question gives Maceió's longitude as $35.7^\circ W$ , while solution (e) uses $36.4^\circ W$ . The solution's printed margin marks total 11 points in (g), versus 12 points in the question and separate marking scheme; its (j) heading and marks total 7 points, versus 6 in the question and separate marking scheme. All three sources are retained with their original point values.

Common solar-system table: The native solar-system table labels the semi-major-axis column AU, while explicitly giving the Moon's entry in km; its printed unit is retained.

Указания

The theoretical competition will be 5 hours in duration and is marked out of a total of 300 points.
There are Detailed Worksheets for carrying out detailed work / rough work. On each of the Detailed Worksheets, please fill in Student Code, Question No., Page no. and total number of pages.
Start each problem on a new page of the Detailed Worksheets. 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 marked, cross it out.
There is a summary Answer Sheet with your student ID code for your final answers.
Please remember that the graders may not understand your language. As far as possible, write your solutions only using mathematical expressions and numbers. If it is necessary to explain something in words, please use short phrases (if possible in English).
You are not allowed to leave your exam desk without permission. If you need any assistance (malfunctioning calculator, need to visit a restroom, need more Detailed Worksheets, etc.), please put up your hand to signal the invigilator.
The beginning and end of the competition will be indicated by a long sound signal. Additionally, there will be a short sound signal fifteen minutes before the end of the competition (before the final long sound signal).
Wait at your table until your envelope is collected. Once all envelopes are collected, your student guide will escort you out of the competition room.
A list of constants for this competition is given on the next page.

Fundamental Constants

Constant	Value
Speed of light in a vacuum	$c=2.998\times10^8\ \mathrm{m\,s^{-1}}$
Planck constant	$h=6.626\times10^{-34}\ \mathrm{J\,s}$
Boltzmann constant	$k_B=1.381\times10^{-23}\ \mathrm{J\,K^{-1}}$
Stefan-Boltzmann constant	$\sigma=5.670\times10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$
Elementary charge	$e=1.602\times10^{-19}\ \mathrm C$
Universal Gravitational constant	$G=6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Universal gas constant	$R=8.315\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Avogadro constant	$N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}$
Wien's displacement law	$\lambda_mT=2.898\times10^{-3}\ \mathrm{m\,K}$
Mass of the electron	$m_e=9.11\times10^{-31}\ \mathrm{kg}$
Mass of the proton	$m_p=1.67\times10^{-27}\ \mathrm{kg}$
Mass of the neutron	$m_n=1.67\times10^{-27}\ \mathrm{kg}$

Astronomical Data and Solar Magnitudes

Astronomical Data	Value
1 parsec	$1\ \mathrm{pc}=3.086\times10^{16}\ \mathrm m=206\,265\ \mathrm{AU}=3.262\ \mathrm{ly}$
1 Astronomical Unit (AU)	$1\ \mathrm{AU}=1.496\times10^{11}\ \mathrm m$
1 Jansky	$1\ \mathrm{Jy}=10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$
Hubble constant	$H_0=70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$
Solar luminosity	$L_\odot=3.826\times10^{26}\ \mathrm W$
Angular diameter of the Sun (from Earth)	$32'$
Effective temperature of the Sun	$T_{\mathrm{eff}}=5778\ \mathrm K$
Obliquity of the ecliptic	$23.5^\circ$
Inclination of the Moon's orbit (to ecliptic)	$5^\circ08'43''$
Magnitude of full Moon	$-12.74$
Coordinates of North Ecliptic Pole	$(18^{\mathrm h}00^{\mathrm m}00^{\mathrm s},+66^\circ33'39'')$
Coordinates of North Galactic Pole	$(12^{\mathrm h}51^{\mathrm m}26^{\mathrm s},+27^\circ07'42'')$
Sidereal day	$23^{\mathrm h}56^{\mathrm m}04^{\mathrm s}$
Tropical year	$365.2422$ solar days
Sidereal year	$365.2564$ solar days

Solar Magnitudes

Apparent Visual	Absolute Visual	Apparent Bolometric	Absolute Bolometric
$-26.75$	$+4.82$	$-26.83$	$+4.74$

Solar System, Spherical Trigonometry and Calculus

Solar System Data

Object	Radius (km)	Mass (kg)	Semi-major axis (AU)	Orbital eccentricity	Albedo
Sun	695500	$1.988\times10^{30}$	n/a	n/a	n/a
Mercury	2440	$3.301\times10^{23}$	0.387	0.206	0.088
Venus	6052	$4.867\times10^{24}$	0.723	0.007	0.76
Earth	6378	$5.972\times10^{24}$	1.000	0.016710	0.31
Moon	1737	$7.346\times10^{22}$	$3.844\times10^5\ \mathrm{km}$	0.054900	0.11
Mars	3390	$6.417\times10^{23}$	1.524	0.093	0.25
Jupiter	69911	$1.898\times10^{27}$	5.203	0.048	0.51
Saturn	58232	$5.683\times10^{26}$	9.537	0.054	0.34
Uranus	25362	$8.681\times10^{25}$	19.189	0.047	0.30
Neptune	24622	$1.024\times10^{26}$	30.070	0.009	0.29

Spherical Trigonometry

Cosine formula:

\cos a=\cos b\cos c+\sin b\sin c\cos A

Sine formula:

\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C}

Four parts formula:

\cot b\sin a=\cos a\cos C+\sin C\cot B

Native common spherical-trigonometry reference sphere and triangle A, B, C with sides a, b, c.

Calculus

Chain Rule:

\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

Product Rule:

\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}

\int x^r\,dx=\frac{x^{r+1}}{r+1}+C\qquad(r\ne-1)

\frac{d}{dx}\ln(x)=\frac1x,\qquad\frac{d}{dx}\tan(x)=\sec^2(x)

Условие

Consider a binary system of black holes, both of equal mass $M$ , separated by a distance $a$ , and revolving around their common centre of mass (CM) in circular orbits. This binary system moves against, and interacts with, a very large, uniform field of stars (each of mass $m \ll M$ ) with number density $n$ . Consider a star that approaches the system from infinity with speed $v$ and impact parameter $b$ , in the reference frame of the CM (as shown in the figure below). Its closest approach distance to the CM is $r_p \approx \tfrac{1}{2}a$ . For tasks (a) and (c), you should make use of the fact that $v^2 \ll \tfrac{GM}{a}$ .

Binary masses M in circular orbit with separation a and CM, and incident star m, velocity v and impact parameter b.

(a) Obtain an expression for $b$ , in terms of $M$ , $a$ , $v$ , and physical constants. In this task, assume that the star interacts with the binary as if its total mass was fixed at the CM. [5 т.]

After a complex interaction with the binary, the star is slingshot from the system. The exact calculation of its ejection speed is complex, but the result can be estimated by considering that the star only interacts with one of the components when near the system. As such, consider, in part (b), only the gravitational interaction between the star and one of the components in the binary.

(b) The star approaches the component with an initial speed negligible compared to the component's orbital speed, and both are moving directly towards each other. After interacting with the system, when the star is again far away from the black hole, we find that the direction of its velocity vector is reversed and the final speed is $v_f$ . Determine $v_f$ , in terms of $M$ , $a$ , and physical constants. Assume that linear momentum and mechanical energy are conserved during this interaction and that it takes place in a timescale much smaller than the binary's period. Recall that $m \ll M$ . [6 т.]

For the following task, assume that all stars approaching the system from infinity with an impact parameter $\tfrac{1}{2}b_0 \leq b \leq \tfrac{3}{2}b_0$ (where $b_0$ is the impact parameter of a star whose speed at infinity is $v_0$ ) attain a closest-approach distance $r_p \approx \tfrac{1}{2}a$ to the CM. Also, assume that all stars exit the system with the speed found in (b).

(c) Upon each encounter, part of the total energy of the binary is transferred to the kinetic energy of the star. Assume that the binary orbit remains circular. Knowing this, using your results from previous tasks, and taking into account only encounters with the stars within the specified range of impact parameters, show that the reciprocal of the binary's separation increases at a constant rate:

\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{a}\right) = H\frac{G\rho}{v_0}

Here, $\rho = nm$ is the mass density of the star field, and $G$ is the universal gravitational constant. Find the dimensionless constant $H$ , which refers to hardening. [14 т.]

Решение

(a)

Solution: Let $v_p$ be the closest approach velocity of the star. From conservation of mechanical energy:

-\frac{GM_Tm}{r_p} + \frac{1}{2}mv_p^2 = \frac{1}{2}mv^2

[2.0]

where $M_T = 2M$ is the total mass of the binary. From conservation of angular momentum, we obtain $v_p$ :

v_p = \frac{b}{r_p}v

[2.0]

setting $r_p \approx a/2$ and inserting $v_p$ into the first equation, we solve for $b$ :

b = a\sqrt{\frac{1}{4} + \frac{2GM}{av^2}}

\boxed{b \approx \frac{\sqrt{2GMa}}{v}}

[1.0]

Where we have used $v^2 \ll GM/a$ in the last step.

(b)

Solution: The speed of the black hole in circular orbit can be found by equating the gravitational force to the centripetal net force: [1.0]

\frac{GM^2}{a^2} = \frac{MV^2}{\frac{a}{2}}

\therefore\ V = \sqrt{\frac{GM}{2a}}

For the rest of the problem, we propose two different solutions:

Solution I:
Additionally, Let $v_f$ be the ejection speed of the star, and $V'$ the final speed of the black hole. To find $v_f$ , we use conservation of linear momentum and conservation of mechanical energy for the two bodies: [1.5]

MV = MV' + mv_f

[1.5]

\frac{1}{2}MV^2 = \frac{1}{2}mv_f^2 + \frac{1}{2}MV'^2

\frac{1}{2}MV^2 = \frac{1}{2}mv_f^2 + \frac{1}{2}M\left(V - \frac{m}{M}v_f\right)^2

0 = v_f^2 - 2Vv_f + \frac{m}{M}v_f^2

Since $m \ll M$ , we neglect the last term, yielding: [1.0]

v_f = 2V = \sqrt{\frac{2GM}{a}}

[1.0]

Because we are considering two moments where the distance between the star and the black hole is very large, we disregard potential energy terms.

Solution II:
In the reference frame of the component, the star approaches the component with a velocity $-\vec{V}$ . [1.5]
And, since mechanical energy and linear momentum are conserved, it can be shown that relative speed between the components is the same before and after the interaction– that is, restitution coefficient equal to unity). Furthermore, since $M \gg m$ , the velocity of the component can be considered unchanged (notice this is not valid in Solution I, as there we deal with second order terms arising from energy considerations) in the interaction. Therefore, after the interaction, the star is ejected back with a velocity $\vec{v}_f' = \vec{V}$ in the opposite direction. [1.5]
To return to the CM frame, we write $\vec{v}_f = \vec{v}_f' + \vec{V}$ , such that

v_f = 2V

[2.0]

As we have previously found.

For the following task, assume that all stars approaching the system from infinity with an impact parameter $\frac{1}{2}b_0 \leqslant b \leqslant \frac{3}{2}b_0$ (where $b_0$ is the impact parameter of a star whose speed at infinity is $v_0$ ) attain a closest-approach distance $r_p \approx \frac{1}{2}a$ to the CM. Also assume that all stars exit the system with the speed found in (b).

(c) (14 points) Upon each encounter, part of the total energy of the binary is lost as kinetic energy acquired by a star. Assume that the binary orbit remains circular. Knowing this, using your results from previous tasks and taking into account only encounters with the stars within the specified range of impact parameters, show that the reciprocal of the binary's separation increases at a constant rate:

\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{a}\right) = H\frac{G\rho}{v_0}

Here, $\rho = nm$ is the mass density of the star field, and $G$ the universal gravitational constant. Find the dimensionless constant $H$ , which refers to hardening.

Solution: First, we know the kinetic energy acquired by a star during an encounter: [1.0]

\Delta K_* = \frac{1}{2}mv_f^2 - \frac{1}{2}mv^2 \approx \frac{GMm}{a}

Where we used that $v^2$ is negligible with respect to $v_f^2$ , since $v_f^2 = \frac{2GM}{a}$ , and $v^2 \ll GM/a$ . This is also the energy lost by the binary during an encounter.
In order to find the rate of energy extraction, we must first obtain the rate of encounters, which can be estimated as follows:
imagine the binary travelling with velocity $v$ in the reference frame of a far away star of impact parameter $b$ . During a small interval $\mathrm{d}t$ , all the stars included inside the annular cylinder of length $L = v\,\mathrm{d}t$ bounded by radii of $b$ and $b + \mathrm{d}b$ will interact with the star. Call this number of stars $\mathrm{d}N$ . It is given by: [4.0]

\mathrm{d}N = n \cdot V_{cylinder}

= n \cdot A_{annulus} \cdot L

= n \cdot \pi \cdot \left[(b + \mathrm{d}b)^2 - b^2\right] \cdot v \cdot \mathrm{d}t

\approx 2\pi n\, b\, v\ \mathrm{d}b\ \mathrm{d}t

Defining $\mathrm{d}\phi = \frac{\mathrm{d}N}{\mathrm{d}t}$ to be the rate of encounters with stars of impact parameter between $b$ and $b + \mathrm{d}b$ and using $b$ from part (a), [1.0]

\mathrm{d}\phi = 2\pi n\, b\, v\ \mathrm{d}b

= 2\pi n \cdot \frac{\sqrt{2GMa}}{v} \cdot v \cdot \mathrm{d}b

\mathrm{d}\phi = 2\pi n\sqrt{2GMa} \cdot \mathrm{d}b

To find the total rate of encounters ( $\phi = \sum \mathrm{d}\phi$ ), we sum over all impact parameters in the interval $\frac{1}{2}b_0 \leqslant b \leqslant \frac{3}{2}b_0$ . Notice $\sum \mathrm{d}b = \frac{3}{2}b_0 - \frac{1}{2}b_0 = b_0$ , so that: [3.0]

\phi = 2\pi n\sqrt{2GMa} \cdot \sum \mathrm{d}b

= 2\pi n\sqrt{2GMa} \cdot \frac{\sqrt{2GMa}}{v_0}

= \frac{4\pi nGMa}{v_0}

Since each encounter takes away $\Delta K_*$ and there are $\phi$ encounters per unit time, the rate of change of energy for the binary is: [2.0]

\frac{\mathrm{d}E_{bin}}{\mathrm{d}t} = \left(-\Delta K_*\right) \cdot \phi = -\frac{4\pi nG^2M^2m}{v_0}

A change in the binary total energy can also be written as: [2.0]

\frac{\mathrm{d}E_{bin}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\left(-\frac{GM^2}{2a}\right) = -\frac{GM^2}{2}\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{a}\right)

Therefore. equating the two:

\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{a}\right) = \frac{8\pi Gnm}{v_0}

\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{a}\right) = 8\pi\frac{G\rho}{v_0}

Thus, $H = 8\pi$ [1.0]

Separate Marking Scheme — T8

(a)

For Solution I - Directly using energy-momentum conservation:

2 points: correctly writing down conservation of mechanical energy for the star.
2 points: correctly writing down conservation of angular momentum for the star.

For Solution II - Exploring the geometry of the hyperbolic orbit:

2 points: writing an expression that relates the eccentricity of the orbit to v and b (from the angular momentum).
1 point: correctly relating v to the semi-major axis of the hyperbolic orbit (using the total mechanical energy).
1 point: correctly relating the periastron distance $r_p$ to the eccentricity e.

For either solution:

1 point: Solving for b and correct final answer. No points deducted if the final answer is not in approximate form $b\approx\sqrt{2GMa}/v$ as in the solution.

(b)

For Solution I - Energy and momentum conservation in the CM frame

1.5 point: correctly writing down conservation of linear momentum for the star + component.
1.5 point: correctly writing down conservation of mechanical energy for the star + component.
1 point: using that $m\ll M$ to simplify the answer.

For Solution II - Simpler solution in the frame of a component:

1.5 point: noticing that the star approaches a component with velocity $-\vec V$ in this frame.
1.5 point: noticing that the star bounces back with a final speed $\vec V$ in this frame.
1 point: return to the frame of the CM by adding $\vec V$ to the velocity calculated.

For either solution:

1 point: calculating orbital velocity of the component.
1 point: correct final answer for $v_f$ .

(c)

1 point calculating energy $\Delta K_*$ acquired by a star during an encounter with the binary.
4 points: finding number of stars dN in the annular cylinder that interact with the star during time dt: $2\pi n b v\,db\,dt$ . If the student considers that the binary moves with the same speed for all stars in the frame of a far away star (say, $v_0$ ) and writes $dN=\pi(b_{\max}^2-b_{\min}^2)v_0\,dt$ , deduct -2 points.
1 point: using answer from (a) to write $d\phi$ in terms of db and given data in order to eliminate v dependence.
3 points: summing/integrating $d\phi$ over the given range of impact parameters to obtain the total rate of encounters $\phi$ .
2 points correctly writing $dE_{\mathrm{bin}}/dt$ (or just $dE_{\mathrm{bin}}$ ) using $\Delta K_*$ and $\phi$ .
2 points: correctly writing $dE_{\mathrm{bin}}/dt$ (or just $dE_{\mathrm{bin}}$ ) using the total energy of the binary.
1 point: equating the two to find $H=8\pi$ .

General Rules for Marking

These are general guidelines for the grading of all problems. All of the rules listed below refer to deductions on specific items in the marking scheme.

Case	Marking
Using absurd physical or mathematical concepts (despite correct answers)	No points given
Giving correct answer but the method is unclear	Deduct 50% of the marks for final answer
Minor mistakes in the calculations, e.g., wrong signs, numerical constants	Deduct 20% of the marks for that part, unless there is a physical incoherence
Too many or too few significant figures ( $n\pm1$ where $n$ is the expected number of significant figures)	Deduct 50% of the marks for final answer
Correct answer but missing or wrong units	No points given for final answer
For numerical values, the answer is not within a 5% precision range	No points given for final answer, unless differently stated in the problem or marking scheme
Error resulting from an earlier part error, for which the student already lost marks, if the answer is physically reasonable	No double penalization, full points given
Error resulting from an earlier part error, if the answer is physically unreasonable	No points given

First example: If due to an error in an earlier item, the student calculates the speed of a satellite as $4.0\times10^4\ \mathrm{m/s}$ instead of $4.5\times10^4\ \mathrm{m/s}$ , they will only lose marks for the earlier item. However, if, for the same reason, a student calculates the velocity of the satellite as $4.5\times10^{12}\ \mathrm{m/s}$ , they should realize this is wrong (higher than the speed of light) and thus should lose some marks for this item as well.

Second example: Consider a question that requires a student to determine the temperature of a planet that orbits a star with luminosity $L$ at a distance $r$ . To answer correctly, the student must calculate the star's radiant flux, the power incident on the planet, the power radiated by the planet, and then, using the condition of thermal equilibrium, the planet's temperature. If the student only calculates the radiant flux and arrives at the correct final answer without clearly explaining the other steps, they will receive marks only for the flux calculation, but not for the other steps, neither for the final answer.

Оригинал в Архива: theory_examination_304b6eb66a8940d79fccc404cf1ba9ab_preview.pdf · официални решения: theory solutions.pdf · допълнителни данни: theory marking scheme.pdf