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USAPhO 2018, Полуфинал — Задача 6. Question B3

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Всички материали от USAPhO 2018

USA Physics Olympiad Exam — Part A / Part B · 25 т.

Внимание: Бележка към темата

The supervisor and student security dates differ in the original packet. The printed diode key retains its positive negative-voltage exponential and positive discharge derivative despite surrounding text/graphs. The radiation-pressure key retains inconsistent intensity/energy-density terminology and its classical electromagnetic definitions. Exact original sources and detailed defect evidence accompany the transcription.

Important Instructions for the Exam Supervisor — DO NOT DISTRIBUTE THIS PAGE

This examination consists of two parts: Part A has three questions and is allowed 90 minutes; Part B also has three questions and is allowed 90 minutes.
The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam.
The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining.
Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B.
Allow 90 minutes to complete Part B. Do not let students go back to Part A.
Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-5), Part B (pages 7-9), and several answer sheets for one of the questions in Part A (pages 11-13). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!
The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 21, 2018.
Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.

USA Physics Olympiad Exam — Instructions

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN - Work Part A first. You have 90 minutes to complete all three problems. Each question is worth 25 points. Do not look at Part B during this time. - After you have completed Part A you may take a break. - Then work Part B. You have 90 minutes to complete three problems. Each question is worth 25 points. Do not look at Part A during this time. - Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. - Start each question on a new sheet of paper. Put your AAPT ID number, your proctor’s AAPT ID number, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example,

Student AAPT ID # Proctor AAPT ID # A1 - 1/3

A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas. - Questions with the same point value are not necessarily of the same difficulty. - In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 13, 2018.

Possibly Useful Information. You may use this sheet for both parts of the exam.

$g$	$9.8\ \mathrm{N/kg}$
$k=1/(4\pi\varepsilon_0)$	$8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$
$c$	$3.00\times10^8\ \mathrm{m/s}$
$N_A$	$6.02\times10^{23}\ (\mathrm{mol})^{-1}$
$\sigma$	$5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
$1\ \mathrm{eV}$	$1.602\times10^{-19}\ \mathrm J$
$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
$\sin\theta$	$\approx\theta-\frac16\theta^3$ for $\vert \theta\vert \ll1$
$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k_m=\mu_0/(4\pi)$	$10^{-7}\ \mathrm{T\cdot m/A}$
$k_B$	$1.38\times10^{-23}\ \mathrm{J/K}$
$R=N_Ak_B$	$8.31\ \mathrm{J/(mol\cdot K)}$
$e$	$1.602\times10^{-19}\ \mathrm C$
$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$(1+x)^n$	$\approx1+nx$ for $\vert x\vert \ll1$
$\cos\theta$	$\approx1-\frac12\theta^2$ for $\vert \theta\vert \ll1$

Important Instructions for the Exam Supervisor — DO NOT DISTRIBUTE THIS PAGE

Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-10), Part B (pages 12-18), and several answer sheets for one of the questions in Part A (pages 20-22). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!

Условие

Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles.

a.

The force of radiation on a spherical particle of radius $r$ is given by

F=PQ\pi r^2

where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$ . Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max}$ ; unless told otherwise, leave your answers in terms of symbolic variables.

i. Given that the total power radiated from the sun is given by $L_\odot$ , find an expression for the radiation pressure a distance $R$ from the sun.

ii. Assuming that the particle has a density $\rho$ , derive an expression for the ratio $F_{\mathrm{radiation}}/F_{\mathrm{gravity}}$ in terms of $L_\odot$ , mass of sun $M_\odot$ , $\rho$ , particle radius $r$ , and quality factor $Q$ .

iii. The quality factor is given by one of the following

If $r\ll\lambda$ , $Q\sim(r/\lambda)^2$
If $r\sim\lambda$ , $Q\sim1$ .
If $r\gg\lambda$ , $Q=1$

Considering the three possible particle sizes, which is most likely to be blown away by the solar radiation pressure?

b.

The Poynting-Robertson effect acts as another mechanism for cleaning out the solar system.

i. Assume that a particle is in a circular orbit around the sun. Find the speed of the particle $v$ in terms of $M_\odot$ , distance from sun $R$ , and any other fundamental constants.

ii. Because the particle is moving, the radiation force is not directed directly away from the sun. Find the torque $\tau$ on the particle because of radiation pressure. You may assume that $v\ll c$ .

iii. Since $\tau=dL/dt$ , the angular momentum $L$ of the particle changes with time. As such, develop a differential equation to find $dR/dt$ , the rate of change of the radial location of the particle. You may assume the orbit is always quasi circular.

iv. Develop an expression for the time required to remove particles of size $r\approx1\ \mathrm{cm}$ and density $\rho\approx1000\ \mathrm{kg/m^3}$ originally in circular orbits at a distance $R=R_{\mathrm{earth}}$ , and use the numbers below to simplify your expression.

Some useful constants include

$M_\odot$	$1.989\times10^{30}\ \mathrm{kg}$
$L_\odot$	$3.828\times10^{26}\ \mathrm W$
$R_{\mathrm{earth}}$	$1.5\times10^{11}\ \mathrm m$
$\lambda_{\max}$	$500\ \mathrm{nm}$

Решение

a.

i. We will assume that light from the sun is completely absorbed, and then re-radiated as blackbody isotropically. In that case,

P=\frac Ic=\frac{L_\odot}{4\pi R^2c}.

The relationship between the pressure $P$ and the energy density $I$ can be derived from the equation $p=E/c$ for photons, or simply postulated by dimensional analysis, since there is no other relevant speed.

Alternatively, this relation can be derived using classical electromagnetism, though this was not required. For a plane wave, the momentum density (i.e. pressure) of the electromagnetic field is $P=\epsilon_0|\mathbf E\times\mathbf B|$ , while the energy density is $I=|\mathbf S|/c$ where $\mathbf S=\mathbf E\times\mathbf B/\mu_0$ . Then $P=I/c$ since $c^2=1/\epsilon_0\mu_0$ .

ii. We have

F_{\mathrm{gravity}}=\frac{GM_\odot}{R^2}\frac43\pi\rho r^3,\qquad F_{\mathrm{radiation}}=\frac{L_\odot}{4\pi R^2c}Q\pi r^2=\frac{L_\odot}{4R^2c}Qr^2

which gives

\frac{F_{\mathrm{radiation}}}{F_{\mathrm{gravity}}}=\frac{3L_\odot}{16\pi GcM_\odot\rho}\frac Qr.

iii. In order to be blown away, the ratio should be greater than one. Since it is independent of distance from the sun, if it is blown away, it will be blown away at any distance. For $r\gg\lambda$ , the ratio is proportional to $1/r$ , so smaller particles are more likely to be blown away. For $r\ll\lambda$ , the ratio is proportional to $r$ , so larger particles are more likely to be blown away. Thus particles of size near $\lambda$ are most likely to be blown away, and even then, only if the density is small enough.

b.

i. Using the circular motion equation

\frac{GM_\odot}{R^2}=\frac{v^2}R

we have

v=\sqrt{\frac{GM_\odot}R}.

ii. Work in the reference frame of the particle. In this frame, the radiation hits the particle at an angle $\theta=v/c$ from the radial direction. The particle then re-emits the radiation isotropically, contributing no additional radiation pressure. (We ignore relativistic effects because they occur at second order in $v/c$ , while the effect we care about is first order.)

The tangential component of the force is

F=\frac vc\frac{L_\odot}{4\pi R^2c}Q\pi r^2=\frac vc\frac{L_\odot}{4R^2c}Qr^2

\tau=-\frac vc\frac{L_\odot}{4Rc}Qr^2

where the negative sign is because this tends to decrease the angular momentum.

The problem can also be solved in the reference frame of the Sun. In this frame, the radiation hits the particle radially, but the particle does not re-emit the radiation isotropically since it is moving; instead the radiation is Doppler shifted. This eventually leads to the same result, after a much more complicated calculation.

iii. The angular momentum is

L=mvR=\frac43\pi\rho r^3\sqrt{\frac{GM_\odot}R}R=\frac43\pi\rho r^3\sqrt{GM_\odot R}.

Differentiating both sides with respect to time,

-\frac vc\frac{L_\odot}{4Rc}Qr^2=\frac43\pi\rho r^3\sqrt{\frac{GM_\odot}R}\frac12\frac{dR}{dt}

which simplifies to

-\frac1{c^2}\frac{L_\odot}RQ=\frac83\pi\rho r\frac{dR}{dt}.

iv. Integrating both sides,

T\frac{L_\odot Q}{c^2}=\frac43\pi\rho rR^2.

Since $r\gg\lambda$ , we have $Q=1$ and

T=\frac43\frac{\pi c^2}{L_\odot}\rho rR^2\approx2\times10^{14}\ \mathrm s\approx7\times10^6\ \mathrm y.

Оригинал в Архива: USAPhO-2018.pdf · официални решения: USAPhO-2018-Solutions.pdf