Съдържание

SAO 2016, Теория и практика — Задача 4. Question 4

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от SAO 2016

4th Singapore Astronomy Olympiad · 19 март 2016 г. · 12 т.

Внимание: Бележка към темата

Suggested answers preserve native scoring and scientific differences: Q5 question 25 marks/key 24 marks; Q7 question 5 marks/key 6 marks. Original units, intermediate formulas, assumptions and printed descriptions are retained. The complete source-supported discrepancy list is bound in the source-reader evidence.

Native cover, rules, constants and formulae

4th Singapore Astronomy Olympiad — 19 March 2016

Organized by Astronomy.SG. Sponsored by DSO National Laboratories.

I hereby consent to provide my details to the organizing and sponsoring organisations, Astronomy.SG and DSO National Laboratories:

Full Name: __________; School: __________; NRIC/FIN: __________; Email address: __________; Signature: __________.

For official use only (the cover prints two identical tables):

Question	Score	Question	Score
1	/14	1	/14
2	/8	2	/8
3	/20	3	/20
4	/12	4	/12
5	/25	5	/25
6	/16	6	/16
7	/5	7	/5

Rules

Students may use pen or pencil to answer the paper.
Any working or answers on the question papers will not be considered.
Only the use of scientific calculators is allowed. Graphing/programmable calculators are not allowed.
Students are to bring their own stationery.
Students have 3 hours to complete the paper. If a student is late, no time extension is granted.
Students should clearly state their assumptions in their working, if any.
Students may leave any time upon submission of their solutions and question paper.
Cheating or allowing others to cheat is grounds for immediate disqualification.
No notes are allowed.
Question papers must be returned together with their scripts. The cover page and Figure 2 are to be attached to the answer script.

Constants / Values

Constant / value	Printed value
Gravitational constant, $G$	$6.67\times10^{-11}\,\mathrm{m^3\,kg^{-1}\,s^{-2}}$
Speed of light in vacuum, $c$	$3.0\times10^8\,\mathrm{m\,s^{-1}}$
Hubble constant, $H_0$	$67.8\,\mathrm{km\,s^{-1}\,Mpc^{-1}}$
Radius of Earth, $R_e$	$6.371\times10^6\,\mathrm m$
Mass of Earth, $M_e$	$5.972\times10^{24}\,\mathrm{kg}$
Solar mass	$2.0\times10^{30}\,\mathrm{kg}$
Nominal solar luminosity	$3.828\times10^{26}\,\mathrm W$
Thomson-cross section for an electron, $\sigma_T$	$6.65\times10^{-29}\,\mathrm{m^2}$
Mass of a proton, $m_p$	$1.67\times10^{-27}\,\mathrm{kg}$
Current density parameter for matter (baryonic and dark), $\Omega_{m,0}$	$0.31$
Current density parameter for radiation	$\Omega_{r,0}\simeq9.1\times10^{-5}$
[Errata] Apparent magnitude of the Sun	$-26.74$

Formulae

Hawking Temperature of a black hole, $T_H=\hbar c^3/(8\pi GMk_B)$ .

Spatial curvature density, $\Omega_{k,0}=\Omega_{total,0}-\Omega_{r,0}-\Omega_{m,0}-\Omega_{\Lambda,0}$ .

Условие

Black holes are objects in space with a gravitational field so large that not even light can escape from its grasp. It has been thought that black holes do not emit radiation until in 1974 when Professor Stephen Hawking gave theoretical proof that black holes do emit blackbody radiation according to the Stefan-Boltzmann law due to quantum effects near the event horizon, and therefore will evaporate its mass away. This radiation was called Hawking Radiation, named after the professor.

Black holes have an equivalent temperature that can be used in the Stefan-Boltzmann law; this is called the Hawking Temperature. In this question, we will attempt some very crude black hole thermodynamics and consider the abnormal behaviour of black holes when it comes to thermodynamics. Consider only non-rotating and non-charged black holes.

a The Schwarzschild radius of a black hole is the radius at which objects travelling at the speed of light cannot escape. It also doubles as the “radius” of a black hole. Given this, relate the surface area of a black hole, A, to its mass, M. [2m] [2 т.]

b Given a black hole of mass M, calculate its luminosity. Let the emissivity of a black hole be 1. [2m] [2 т.]

c Consider a star of absolute magnitude $m_0$ . For some reason it decides to split into N identical stars of the same temperature as the initial star, conserving mass (i.e. each new star has a mass of initial mass/N). What is the final combined absolute magnitude of all N stars? [4m] [4 т.]

d Consider now, that a black hole of absolute magnitude $m_0$ also decides to split into N identical black holes, conserving mass only (i.e. each new black hole has a mass of initial mass/N). What is the final combined absolute magnitude of all N black holes? [4m] [4 т.]

Решение

Physics of stars and planets (12m)

a)

First, derive the Schwarzchild radius by setting $v=c$ and $R=R_s$ into the equation for escape velocity.

\because v=\sqrt{\frac{2GM}{R}},\qquad\therefore R_s=\frac{2GM}{c^2}.

[B1]

Therefore, the surface area of a black hole (assumed spherical) is,

A=4\pi R_s^2=\frac{\mathbf{16\pi G^2M^2}}{\mathbf{c^4}}.

[B1]

Common Mistakes:

Failing to show clear derivation of the Schwarzchild radius.
Calculating area of the black hole instead of surface area, which is physically meaningless
Incorrect expression for the surface area A of a sphere

b)

Since the emissivity $\varepsilon=1$ , i.e. the black hole is a perfect blackbody radiator,

L=A\varepsilon\sigma T^4=4\pi R_s^2\sigma T_H^4=\left(\frac{16\pi G^2M^2}{c^4}\right)\sigma T_H^4

[M1]

=\left(\frac{16\pi G^2M^2}{c^4}\right)\left(\frac{\hbar c^3}{8\pi GMk_B}\right)^4=\frac{\mathbf{\sigma\hbar^4c^8}}{\mathbf{256\pi^3G^2M^2k_B^4}}.

[B1]

Common Mistakes:

Some participants did not understand emissivity – the ratio of the electromagnetic radiation from a body to the radiation from an ideal blackbody at a given temperature
Not simplifying the expression to obtain the final answer
Incorrect expression for luminosity $L$

c)

Assume the luminosities of the stars stacks linearly, i.e. none of the stars obstruct our line of sight to other stars, and that the surface temperature of the stars remain unchanged after splitting.

Let the luminosity of the parent star be $L_0$ , and its radius and surface temperature be $R$ and $T$ respectively.

L_0=4\pi R^2\sigma T^4.

[A1]

Let the radius of each daughter star be $r$ , thus by conservation of mass, assuming density remains unchanged,

\frac43\pi R^3=N\left(\frac43\pi r^3\right)\quad\Rightarrow r^2=N^{-2/3}R^2.

[B1]

The combined luminosity of the $N$ daughter stars is

L=N(4\pi r^2\sigma T^4)=N(4\pi N^{-2/3}R^2\sigma T^4)=4\pi N^{1/3}R^2\sigma T^4=N^{1/3}L_0.

[M1]

Hence, the final apparent magnitude is

m=m_0-2.5\log\left(\frac L{L_0}\right)=\mathbf{m_0-\frac{2.5}3\log(N)}.

[B1]

Common Mistakes:

Incorrect expression for comparing apparent magnitudes (Pogson’s law)
Incorrect coefficient of the $lg$ term in Pogson’s law; it is2.5, not2.512 or $100^{1/5}$
Forgetting to account for the $N$ daughter stars of smaller radius in combined luminosity
Missing or incomplete assumption(s)

d)

Use the same variable notation as part (c) above.

Define $\xi=\frac{\hbar c^3}{8\pi Gk_B}$ such that $T_H=\xi/M$ .

The surface area of each daughter black hole is

A=\frac{16\pi G^2(M/N)^2}{c^4}=\frac{A_0}{N^2}.

[M1]

The luminosity of the parent black hole is

L_0=A_0\sigma T_{H,0}^4=A_0\sigma\left(\frac\xi M\right)^4.

[B1]

Thus the combined luminosity of the $N$ daughter black holes is

L=N(A\sigma T_H^4)=N\left(\frac{A_0}{N^2}\sigma\left(\frac\xi{M/N}\right)^4\right)=N\left(N^2A_0\sigma\left(\frac\xi M\right)^4\right)=N^3\left(A_0\sigma\left(\frac\xi M\right)^4\right)=N^3L_0.

[M1]

Hence, the final apparent magnitude is

m=m_0-2.5\log\left(\frac L{L_0}\right)=m_0-2.5\log(N^3)=\mathbf{m_0-7.5\log(N)}.

[B1]

Common Mistakes:

Refer to part(c) comments.

Оригинал в Архива: SAO4_q.pdf · официални решения: SAO4_sol.pdf