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USAAAO 2021, II кръг — Long 2

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Всички материали от USAAAO 2021

2021 National Astronomy Competition · 35 т.

Внимание: Бележка към темата

All ten whole original Short/Medium/Long problems, 30 lettered parts, 9 nested roman requests and 7 numbered satellite quantities, all original worked keys and marking notes are retained. Printed scientific discrepancies are preserved in attributed local source notes. Medium 3 original calls both masses m; its key introduces a distinct M for the station and assumes $m\lt M$ . A separate undated common archive constants/reference table is fully included with its own source identity; its use as the exact 2021 provided attachment is not established, and its constants do not replace native exam values. The native cover specifies blank paper and a signed declaration. March 20th is the end of the examination period, not an established single sitting date.

1 Instructions (Please Read Carefully)

The top 5 eligible scorers on the NAC will be invited to represent USA at the next IOAA. In order to qualify for the national team, you must be a high school student with US citizenship or permanent residency.

This exam consists of 3 parts: Short Questions, Medium Questions and Long Questions.

The maximum number of points is 180.

The test must be completed within 2.5 hours (150 minutes).

Please solve each problem on a blank piece of paper and mark the number of the problem at the top of the page. The contestant’s full name in capital letters should appear at the top of each solution page. If the contestant uses scratch papers, those should be labeled with the contestant’s name as well and marked as “scratch paper” at the top of the page. Scratch paper will not be graded. Partial credit will be available given that correct and legible work was displayed in the solution.

This is a written exam. Contestants can only use a scientific or graphing calculator for this exam. A table of physical constants will be provided. Discussing the problems with other people is strictly prohibited in any way until the end of the examination period on March 20th. Receiving any external help during the exam is strictly prohibited. This means that the only allowed items are: a calculator, the provided table of constants, a pencil (or pen), an eraser, blank sheets of papers, and the exam. No books or notes are allowed during the exam. Exam is proctored and recorded. You are expected to have your video on at all times.

After reading the instructions, please make sure to sign, affirming that:

All work on this exam is mine.
I did not receive any external aids besides the materials provided.
I am not allowed to discuss the test with others throughout the period of this examination.
Failure to follow these rules will lead to disqualification from the exam.

Signed: ____________________

Common archive reference — Table of Physical Constants

This separate common USAAAO archive reference is undated in the catalogue. Its use as the exact 2021 exam attachment is not established.

Quantity	Symbol	Value
Electron rest mass	$m_e$	$9.109\times10^{-31}\ \mathrm{kg}$
Proton rest mass	$M_p$	$1.6726\times10^{-27}\ \mathrm{kg}$
Electronic charge	$e$	$1.6022\times10^{-19}\ \mathrm C$
Speed of light in free space	$c$	$2.9979\times10^8\ \mathrm{m\ s^{-1}}$
Permeability of free space	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\ m^{-1}}$
Permittivity of free space	$\epsilon_0$	$8.854\times10^{-12}\ \mathrm{F\ m^{-1}}$
Planck’s constant	$h$	$6.626\times10^{-34}\ \mathrm{J\ s}$
Reduced Planck’s constant	$\hbar=h/2\pi$	$1.0546\times10^{-34}\ \mathrm{J\ s}$
	$\hbar c$	$197.33\ \mathrm{MeV\ fm}$
Boltzmann’s constant	$k_B$	$1.3807\times10^{-23}\ \mathrm{J\ K^{-1}}$
Gas constant	$\mathcal R=k_B/m_H$	$8.250\times10^3\ \mathrm{J\ kg^{-1}\ K^{-1}}$
Molar gas constant	$R$	$8.315\ \mathrm{J\ mol^{-1}\ K^{-1}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Standard molar volume		$22.414\times10^{-3}\ \mathrm{m^3\ mol^{-1}}$
Unified atomic mass unit ( $^{12}\mathrm C$ scale)	$u$	$931.5\ \mathrm{MeV}/c^2=1.660538\times10^{-27}\ \mathrm{kg}$
Mass of hydrogen atom	$m_H$	$1.0078u=1.6735\times10^{-27}\ \mathrm{kg}$
Bohr magneton	$\mu_B$	$9.274\times10^{-24}\ \mathrm{A\ m^2}$ or $\mathrm{J\ T^{-1}}$
Nuclear magneton	$\mu_N$	$5.051\times10^{-27}\ \mathrm{A\ m^2}$ or $\mathrm{J\ T^{-1}}$
Proton magnetic moment	$\mu_p$	$2.7928\mu_N$
Neutron magnetic moment	$\mu_n$	$-1.9130\mu_N$
Bohr radius	$a_0$	$5.292\times10^{-11}\ \mathrm m$
Fine structure constant	$\alpha=e^2/(4\pi\epsilon_0\hbar c)$	$(137.04)^{-1}$
Compton wavelength of electron	$\lambda_C=h/(m_ec)$	$2.4263\times10^{-12}\ \mathrm m$
Rydberg’s constant	$R_{\infty}$	$1.0974\times10^7\ \mathrm{m^{-1}}$
	$R_{\infty}hc$	$13.606\ \mathrm{eV}$
Stefan-Boltzmann constant	$\sigma$	$5.671\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}$
Radiation density constant	$a=4\sigma/c$	$7.561\times10^{-16}\ \mathrm{J\ m^{-3}\ K^{-4}}$
Gravitational constant	$G$	$6.673\times10^{-11}\ \mathrm{N\ m^2\ kg^{-2}}$

Rest masses of some leptons and hadrons in $\mathrm{MeV}/c^2$ :

$e^{\pm}$ 0.5110, $\mu^{\pm}$ 105.66, $\tau^{\pm}$ 1777, $\pi^0$ 134.98, $\pi^{\pm}$ 139.57, $\mathrm K^{\pm}$ 493.7, $\mathrm K^0$ 497.7, $\eta$ 547, $\mathrm D^0$ 1865, $\mathrm D^{\pm}$ 1869, p 938.3, n 939.6, $\Lambda^0$ 1115.7, $\Sigma^+$ 1189, $\Sigma^0$ 1193, $\Sigma^-$ 1197, $\Xi^0$ 1315, $\Xi^-$ 1321, $\Omega^-$ 1672, $\mathrm Z^0$ $91.187\times10^3$ , $\mathrm W^{\pm}$ $80.41\times10^3$ .

Quark	Charge	$I_3$	$S$	$C$	$B$	$T$	Mass ( $\mathrm{GeV}/c^2$ )
u	$+\frac23$	$\frac12$	0	0	0	0	$\sim0.003$
d	$-\frac13$	$-\frac12$	0	0	0	0	$\sim0.006$
c	$+\frac23$	0	0	+1	0	0	$\sim1.25$
s	$-\frac13$	0	-1	0	0	0	$\sim0.11$
t	$+\frac23$	0	0	0	0	+1	174.3
b	$-\frac13$	0	0	0	-1	0	4.2

Common archive reference — Astrophysical Data and identities

Astrophysical Data

Quantity	Symbol	Value
1 astronomical unit	AU	$1.496\times10^{11}\ \mathrm m$
1 parsec	pc	$3.086\times10^{16}\ \mathrm m$
Luminosity of Sun	$L_{\odot}$	$3.85\times10^{26}\ \mathrm W$
Mass of Sun	$M_{\odot}$	$1.989\times10^{30}\ \mathrm{kg}$
Radius of Sun	$R_{\odot}$	$6.96\times10^8\ \mathrm m$
Mass of Earth	$M_E$	$5.9742\times10^{24}\ \mathrm{kg}$
Radius of Earth	$R_E$	$6.3781\times10^6\ \mathrm m$

Other data and conversion factors

Quantity	Symbol	Value
1 ångstrom	Å	$10^{-10}\ \mathrm m$
1 fermi	fm	$10^{-15}\ \mathrm m$
1 barn	b	$10^{-28}\ \mathrm{m^2}$
1 pascal	Pa	$1\ \mathrm{Nm^{-2}}$
1 standard atmosphere		$1.0132\times10^5\ \mathrm{Pa}$
Standard acceleration due to gravity	$g$	$9.807\ \mathrm{m\ s^{-2}}$
1 electron volt	eV	$1.6022\times10^{-19}\ \mathrm J$
	$\mathrm{eV}/hc$	$8.065\times10^5\ \mathrm{m^{-1}}$
	$\mathrm{eV}/k_B$	$1.1604\times10^4\ \mathrm K$
Wavelength of 1 eV photon		$1.2399\times10^{-6}\ \mathrm m$

Trigonometrical identities

\sin(\theta+\phi)=\sin(\theta)\cos(\phi)+\cos(\theta)\sin(\phi)

\cos(\theta+\phi)=\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)

\sin\alpha+\sin\beta=2\sin\frac12(\alpha+\beta)\cos\frac12(\alpha-\beta)

\cos\alpha+\cos\beta=2\cos\frac12(\alpha+\beta)\cos\frac12(\alpha-\beta)

\cos\alpha-\cos\beta=2\sin\frac12(\alpha+\beta)\sin\frac12(\beta-\alpha)

In a triangle ABC, $a/\sin A=b/\sin B=c/\sin C$

and $a^2=b^2+c^2-2bc\cos A$

Prefixes

Prefix	Name	Factor
T	tera	$10^{12}$
G	giga	$10^9$
M	mega	$10^6$
k	kilo	$10^3$
c	centi	$10^{-2}$
m	milli	$10^{-3}$
$\mu$	micro	$10^{-6}$
n	nano	$10^{-9}$
p	pico	$10^{-12}$
f	femto	$10^{-15}$

Условие

In 2020, during the day of the winter solstice for the Northern hemisphere, Jupiter and Saturn were at their minimum angular separation (approximately 6.11′) during the Great Conjunction.

(a) (6 points) Consider a system with three planets in circular, concentric, and coplanar orbits around a star. Suppose that the three planets and the star are initially aligned. Will they necessarily align again after this moment? Prove your answer with quantitative arguments. Assume that the sidereal periods of all planets are rational numbers in terms of some unit period.

(b) (4 points) Suppose that there were N planets instead of three in the system from item A. N is an integer greater than 3. If the orbits were still circular, concentric, and coplanar, and the planets and star were all initially aligned, would they necessarily align again afterwards? Assume that the sidereal periods of all planets are rational numbers in terms of some unit period.

(c) (8 points) In the system from (a), if the three planets were not initially aligned with respect to the star, would they necessarily be perfectly aligned at some point? Again, use quantitative arguments to prove your answer.

(d) (4 points) Suppose that you are an astronomer who wants to use a telescope to observe the conjunction. Since you are a very skilled astronomer, you are going to build your own telescope. The only basic requirement you want to meet is that your telescope must be able to resolve the planets at the minimum separation during the conjunction. Calculate the value of all parameters of your telescope that are relevant for this goal. Do not try to calculate the values of any parameters that are not related to this requirement.

(e) (8 points) Calculate the total apparent magnitude of the planets together in the conjunction. Assume that the observers see Jupiter and Saturn as a single point in the sky, but Saturn is not covered (totally or partially) by Jupiter. For this item, neglect the atmospheric extinction, consider that the planets reflect isotropically, and consider that the albedos of both Jupiter and Saturn are equal to one. Also, in order to make the calculations simpler, assume that both Jupiter and Saturn were almost in opposition with respect to the Earth (even though this was not the case for this conjunction).

(f) (5 points) Calculate the difference in the magnitude of the conjunction at the zenith and at a zenith distance of $15^\circ$ . Assume that the zenith optical depth of Earth’s atmosphere for visible light is 0.50.

Mean orbital radius of Jupiter: 5.2 AU
Mean orbital radius of Saturn: 9.5 AU
Radius of Jupiter: $7.1492\times10^7$ meters
Radius of Saturn: $5.8232\times10^7$ meters
Apparent magnitude of the Sun: $-26.74$
Central wavelength of visible light: 550 nm

Решение

(a) It is possible to write the following expression for the synodic period between the first two planets ( $S_{1,2}$ , in which planet 1 is the closest to the star, and $T_n$ represents the sidereal period of the $n^{th}$ planet):

\frac1{S_{1,2}}=\frac1{T_1}-\frac1{T_2}

S_{1,2}=\frac{T_1T_2}{T_2-T_1}

Likewise, the synodic period between planets 1 and 3 ( $S_{1,3}$ ) is the following:

S_{1,3}=\frac{T_1T_3}{T_3-T_1}.

It is important to notice that both synodic periods are rational numbers. All sidereal periods are rational numbers, and by definition, the subtraction or division of two rational numbers must result in a rational number, so both $S_{1,2}$ and $S_{1,3}$ are rational.

If we assume that the planets are aligned at $t=0$ , planets 1 and 2 must be aligned at all instants $t=m\cdot S_{1,2}$ , in which $m\in\mathbb N$ . Likewise, planets 1 and 3 must be aligned at all instants $t=n\cdot S_{1,3}$ , in which $n\in\mathbb N$ . Therefore, the following condition must be met for a triple alignment:

m\cdot S_{1,2}=n\cdot S_{1,3}

\frac mn=\frac{S_{1,3}}{S_{1,2}}

\frac mn=r.

Since r is the ratio between two rational numbers, it must also be a rational number. By definition, every rational number might be expressed as the ratio between two integers. In this case, since r must be positive, the two integers must have the same sign. Therefore, this equation has solutions in which $m,n\in\mathbb N$ , so the planets will align periodically after $t=0$ .

(b) It is possible to expand the alignment condition from item A to more planets:

m_1\cdot S_{1,2}=m_2\cdot S_{2,3}=\ldots=m_n\cdot S_{n,n+1}.

In this equation, all $m_n$ terms are natural numbers greater than zero.

Again, it is possible to use the definition of a rational number to solve this problem. As demonstrated on item A, if the sidereal periods are rational numbers, the synodic periods must be rational as well. Since a rational number might be expressed as a ratio between two integers, the product $m_n\cdot S_{n,n+1}$ is an integer for certain values of $m_n$ . Therefore, there are values of $m_n$ for which all $m_n\cdot S_{n,n+1}$ terms in the alignment equality are natural numbers. By definition, any set of natural numbers has a least common multiple (LCM), and an infinite number of common multiples. Therefore, it is possible to multiply all $m_n\cdot S_{n,n+1}$ terms by an integer factor to obtain the LCM. If all terms are equal to the LCM, the alignment equality is true, which proves that all planets are aligned at an instant later than $t=0$ .

(c) In order to obtain an expression for the instants $t_{1,2}$ in which the first two planets are aligned, it is possible to write the following formula, in which $m\in\mathbb N$ and $\theta_{i,n}$ corresponds to the initial angular position of the $n^{th}$ planet:

\theta_{i,1}+\frac{2\pi}{T_1}t_{1,2}+2\pi m=\theta_{i,2}+\frac{2\pi}{T_2}t_{1,2}

t_{1,2}\left(\frac1{T_1}-\frac1{T_2}\right)=\left(\frac{\theta_{i,2}-\theta_{i,1}}{2\pi}-m\right)

t_{1,2}=\frac1{S_{1,2}}\left(\frac{\theta_{i,2}-\theta_{i,1}}{2\pi}-m\right)

Likewise, the formula for the alignment between planets 1 and 3 will be the following:

t_{1,3}=\frac1{S_{1,3}}\left(\frac{\theta_{i,3}-\theta_{i,1}}{2\pi}-n\right).

In this formula, n is a natural number.

Therefore, it is possible to write the following equality for the alignment between three planets:

\frac1{S_{1,2}}\left(\frac{\theta_{i,2}-\theta_{i,1}}{2\pi}-m\right)=\frac1{S_{1,3}}\left(\frac{\theta_{i,3}-\theta_{i,1}}{2\pi}-n\right)

In order to simplify the formula, it is possible to group a few variables together:

\alpha=\frac{S_{1,3}}{S_{1,2}}

\beta=\frac{\theta_{i,2}-\theta_{i,1}}{2\pi}

\gamma=\frac{\theta_{i,3}-\theta_{i,1}}{2\pi}

Since $\alpha$ , $\beta$ , and $\gamma$ are the result of consecutive subtractions and divisions of rational numbers, they are also rational numbers. However, they are not necessarily rational.

It is possible to rewrite the formula using $\alpha$ , $\beta$ , and $\gamma$ :

\alpha(\beta-m)=(\gamma-n)

\alpha m-n=\alpha\beta-\gamma

It is possible to define a new variable $\delta$ to simplify the expression even more:

\alpha m-n=\delta

From this expression, it is clear that there are values of $\alpha$ and $\delta$ for which there are no solutions in which both m and n are natural numbers.

The easiest way to demonstrate this is with a simple counterexample. Suppose that $\alpha=2$ and $\delta=3.5$ .

2m-n=3.5

If both 2 n and m are integers, the result of the subtraction should necessarily be an integer. However, since 3.5 is not an integer, there are no solutions for which both m and n are integers. Therefore, if the three planets are not initially aligned, they might never have a triple conjunction.

It is important to highlight that $\alpha$ is a function only of the planets’ periods, but $\delta$ is a function of both the periods and the initial angular positions. Therefore, the method of choosing arbitrary values for a counterexample is valid.

Note: For the first three items of this question, students were allowed to assume that the periods of the planets are rational numbers.

However, the period of a body in a circular motion is not necessarily a rational number. It must be a real number, but it might also be an irrational number. For instance, consider a circular movement with an angular velocity of 1 rad/s. In this case, the period would be equal to $2\pi$ , which is an irrational number.

The reason why this question assumes that the periods are rational numbers is that it is impossible to measure periods with an infinite number of significant figures. On the aforementioned example, although $2\pi$ is an irrational number,6.28319 is rational. Since we can only measure periods with a limited number of significant figures, even our most precise measurement for the period of any planet will still be a rational number.

It is also important to notice that in a real life situation, it is essential to consider that planets are not point particles and take into account the radius of each planet. Consider a hypothetical system with three planets with periods of 0.5 year,1 year, and 2.000000000000000000000001 years. It is clear that if the planets are initially aligned, they will be pretty much aligned again in two years (considering that the third period is equal to approximately 2 years). However, because the third period is not exactly 2 years, the time interval until the next perfect alignment is very long. In this case, it makes much more sense to consider that the period between the alignments is equal to two years, not to a very large number.

(d) In this the, the angular resolution must be less of equal to 6.11′. Besides the wavelength, the diameter is the only parameter that affects the angular resolution. It is possible to use the following formula to calculate the diameter:

\theta=1.22\frac\lambda D

D=1.22\frac\lambda\theta

Since $\theta\leq6.11'$ and visible light is centered at 550 nm:

D\geq1.22\frac{5.50\times10^{-7}}{6.11\pi/(60\times180)}

D\geq3.78\times10^{-4}\ \mathrm m

$\therefore$ The diameter of the telescope must be greater or equal to $3.78\times10^{-4}\ \mathrm m$ . In other words, basically any telescope you could possibly build will meet this requirement.

(e) Solar flux that arrives at Jupiter and Saturn:

F_J=\frac{L_\odot}{4\pi r_J^2}

F_S=\frac{L_\odot}{4\pi r_S^2}

Flux from Jupiter and Saturn that arrives at the Earth:

F_{\oplus1}=\frac{F_J\pi R_J^2}{4\pi(r_J-r_\oplus)^2}+\frac{F_S\pi R_S^2}{4\pi(r_S-r_\oplus)^2}

F_{\oplus1}=\frac{L_\odot R_J^2}{16\pi r_J^2(r_J-r_\oplus)^2}+\frac{L_\odot R_S^2}{16\pi r_S^2(r_S-r_\oplus)^2}

F_{\oplus1}=\frac{L_\odot}{16\pi}\left(\frac{R_J^2}{r_J^2(r_J-r_\oplus)^2}+\frac{R_S^2}{r_S^2(r_S-r_\oplus)^2}\right)

Solar flux that arrives at the Earth:

F_{\oplus2}=\frac{L_\odot}{4\pi r_\oplus^2}

Ratio between the fluxes:

\frac{F_{\oplus1}}{F_{\oplus2}}=\frac{r_\oplus^2}4\left(\frac{R_J^2}{r_J^2(r_J-r_\oplus)^2}+\frac{R_S^2}{r_S^2(r_S-r_\oplus)^2}\right)

\frac{F_{\oplus1}}{F_{\oplus2}}=\frac{(1.496\times10^{11})^2}4\left(\frac{(7.1492\times10^7)^2}{5.2^2\times4.2^2\times(1.496\times10^{11})^4}+\frac{(5.8232\times10^7)^2}{9.5^2\times8.5^2\times(1.496\times10^{11})^4}\right)

\frac{F_{\oplus1}}{F_{\oplus2}}=1.2551\times10^{-10}

Using Pogson’s Law:

\begin{aligned} m_{\mathrm{Conjunction}}-m_\odot&=-2.5\times\log\left(\frac{F_{\oplus1}}{F_{\oplus2}}\right)\\ m_{\mathrm{Conjunction}}&=-2.5\times\log(1.2551\times10^{-10})-26.74\\ m_{\mathrm{Conjunction}}&=-1.99 \end{aligned}

$\therefore$ The apparent magnitude of the great conjunction is equal to $-1.99$ .

Note: While students were supposed to assume that an observer would see Jupiter and Saturn as a single point in the sky, the minimum separation of 6.11′ corresponds to about 1/5 of the Moon’s angular diameter, so the human eye can easily resolve this angular separation.

(f) The optical depth ( $\tau$ ) is defined by the following expression, in which $\kappa$ is the opacity coefficient, $\rho$ is the density of the medium, and s is the distance travelled in the medium:

\tau=\kappa\rho s

Using the flat atmosphere approximation, which works well for small zenith distances, it is possible to calculate the optical depth of the Earth’s atmosphere at a zenith distance of $15^\circ$ ( $\tau_{15}$ ). The only factor that varies in this case is the distance travelled, which is equal to $d_z\times\sec(15^\circ)$ , in which $d_z$ is the distance travelled for a zenith distance of $0^\circ$ .

Therefore, is is possible to use the zenith optical depth $\tau_z$ to calculate $\tau_{15}$ :

\begin{aligned} \tau_{15}&=\tau_z\times\sec(15^\circ)\\ \tau_{15}&=\tau_z\times\sec(15^\circ)\\ \tau_{15}&=0.5\times\sec(15^\circ)\\ \tau_{15}&=0.5176 \end{aligned}

Considering that the ratio between the flux outside the atmosphere and the flux after atmospheric extinction is by definition the exponential function of the optical depth, it is possible to calculate the ratio between the fluxes for the zenith and for a zenith distance of $15^\circ$ :

\begin{aligned} \frac{F_{15}}{F_z}&=\frac{F_0\cdot e^{-\tau_{15}}}{F_0\cdot e^{-\tau_z}}\\ &=e^{\tau_z-\tau_{15}}\\ &=e^{0.5-0.5176}\\ &=0.9825. \end{aligned}

Now, it is possible to use Pogson’s Law to calculate the difference in magnitude:

\begin{aligned} \Delta m&=-2.5\cdot\log\left(\frac{F_{15}}{F_z}\right)\\ &=-2.5\cdot\log(0.9825)\\ &=1.92\times10^{-2}. \end{aligned}

$\therefore$ The difference in magnitude between the Great Conjunction seen at a zenith distance of $15^\circ$ and the Great Conjunction seen at the zenith corresponds to $1.92\times10^{-2}$ .

Source note: The key (c) prints 1/S in the solved alignment times, asserts alpha/beta/gamma are rational and immediately states they need not be rational, and refers to 2 n and m after the displayed 2 m−n counterexample. The entire original proof and two-page source Note are preserved without mathematical normalization. Part (f) repeats the tau 15= tauz sec 15 line twice, as printed.

Теми и трудност

Оценена трудност: 4/5.

Небесна механика · Уреди и наблюдения · Светлина и астрономически измервания · Орбитално движение · Телескопи · Звездни величини · Междузвездно поглъщане

Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка

Оригинал в Архива: usaaao_second_exam_2021_final.pdf · официални решения: usaaao_second_exam_2021_solutions.pdf · допълнителни данни: physical-constants.pdf