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USAAAO 2024, II кръг — Medium 3. Spiff’s adventures

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Всички материали от USAAAO 2024

2024 National Astronomy Competition · 30 т.

Внимание: Бележка към темата

All eight whole original problems, their native sections, lettered parts, three roman subparts of Long 1(f), full official worked keys and original marking/acceptance notes are retained. Printed scientific discrepancies are preserved in local source notes. The official key adds a Friedmann-variable clarification and a solar apparent magnitude assumption absent from the question booklet, and changes Long1(f)(iii) from a factor to a percentage; these source variants are explicitly attributed. A separate undated common archive constants/reference table is included with its own source identity; its use as the exact 2024 provided attachment is not established, and its constants are not substituted into the printed key. The native cover specifies blank paper rather than a designated answer sheet. April 13th 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 225 points.

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 April 13th. 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.

We acknowledge the following people for their contributions to this year’s exam:

Abhay Bestrapalli, Orion Foo, Hagan Hensley, David Lee, Sandesh Kalantre, Andrew Liu, Joe McCarty, Leo Yao

Common archive reference — Table of Physical Constants

This separate common USAAAO archive reference is undated in the catalogue. Its use as the exact 2024 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	$10^{-15}$

Source note: This undated common table prints $G=6.673\times10^{-11}\ \mathrm{N\ m^2\ kg^{-2}}$ , $M_E=5.9742\times10^{24}\ \mathrm{kg}$ and $R_E=6.3781\times10^6\ \mathrm m$ . The 2024 eclipse key instead uses $G=6.674\times10^{-11}$ , Earth mass $5.976\times10^{24}\ \mathrm{kg}$ and Earth radius $6.371\times10^6\ \mathrm m$ . Both original sets of values are preserved separately; this reference is not an established exact dated 2024 attachment.

Условие

You may attempt subsections without attempting the previous parts by making appropriate assumptions.

Spaceman Spiff is exploring the distant reaches of the Milky Way. He ends up crashing on a very cold planet. He calls the corresponding star Burner and the planet Ash. Though there aren’t signs of life nearby, Spiff wants to determine if Ash is habitable.

(a) (5 points) The first thing Spiff observes is that Burner (seen through safe equipment!) is not uniformly bright over its disc in the sky. Indeed, the edges seem to be less bright than the center. Spiff opens up an astrophysics book (his computer is not working) and find the following equation:

I(\tau,\mu)=\frac3{4\pi}\left(\tau+\mu+\frac23\right)F(0)

Here, $I(\tau,\mu)$ gives the intensity of light at optical depth $\tau$ and $\mu=\cos\theta$ where the angle $\theta$ is the angle between the direction of the ray and the radial direction. $F(0)$ is the net flux at the surface. Help Spiff find the ratio of intensity of the center of Burner to the edge (as it appears to Spiff). What is this phenomenon called?

(b) (15 points) Spiff walks for a few days but doesn’t find any life nearby. Over the many nights and days, Spiff makes several observations about Burner and Ash:

The peak wavelength of Burner appears to be 500 nm.
Before the ship crashed, Spiff had found the planet has a radius twice that of Earth and that it has a circular orbit around Burner.
Precisely calculation yields that each solar (Burnar?) day is 192 hours long and that Burner moved 1 degree against the background stars over a day.
The mass and radius of Burner are similar to that of the Sun.
Comparing the rotation of Burner (found by tracking a Burnerspot) to the sky, Spiff finds out that Ash has an axial tilt of 60 degrees! Spiff hypothesises that, because of a thin atmosphere and other effects, the heat received from the star never redistributes to the part of Ash that is not exposed to the star, and is only emitted by the exposed part. Also somewhat peculiarly, the unexposed part always remains unexposed (Ash is always titled towards Burner). Assume that Ash is at thermal equilibrium.

What is the equilibrium temperature of Ash? Is it theoretically possible to have life on this planet?

(c) (4 points) Spiff, being the genius he is, manages to fix the spaceship on his own, weaves through a dangerous meteor field and escape Burner’s system. When he is around 10 parsecs away, he stops and looks back at Burner. He realises that he is still on Ash’s plane of revolution when it starts transiting across Burner’s disc. How long will the transit of Ash across Burner take in seconds?

(d) (6 points) Now Spiff has moved away further to 100 parsecs and has finally reached home. He looks back at Burner one last time. What is the apparent magnitude of the star (Assume that the transit of Ash is not happening) for Spiff? Can he see it with his naked eye (Spiff is indeed human)? Assume that extinction is significant, and is equal to 2 mag/kpc.

Отговори

Решение

Key-only statement addition (solution PDF page 11): In addition to constants from the TOC, you can assume the apparent magnitude of the Sun is -26.74. This assumption is absent from the question booklet. The repeated key statement also prints “Precise calculation”, where the question prints “Precisely calculation”. Both original question and key retain “Ash is always titled towards Burner” and “weaves through a dangerous meteor field and escape Burner’s system”.

(a) Let’s first consider a point on the edge of Burner. The optical depth is by definition 0 (as it is on the surface) while $\mu=\cos(90^\circ)$ . Thus, the intensity/brightness is given by the radiative transfer equation above: $I(0,\cos(90^\circ))=\frac{F(0)}{2\pi}$ . (1 point)

Next, for a point at the center of Burner, we have $\mu=\cos(0)$ with $\tau=0$ . It follows that $I(0,\cos(0))=\frac{5F(0)}{4\pi}$ . (1 point)

This gives a ratio of

\boxed{\frac52}

(2 points)

Notably, the ratio between the brightness at the center of Burner and the surface is $>1$ ; this is a phenomenon called limb darkening. (1 point)

(b) The temperature of the star, from Wien’s law is: $T_B=\frac b{\lambda_{\max}}=5795.54\mathrm K$ (1 point)

Since the radius of the star is the same as the Sun’s radius, we can find the net luminosity of Burner with Stephen Boltzmann as: $L_B=4\pi R_B^2\sigma T_B^4=3.891\times10^{26}\ \mathrm W$ . (1 point)

Burner moves 1 degree in the sky every day and it needs to move 360 degrees to come back to the same place in the sky. There are thus 360 days in a year. Each such day is 192 hours. So each year is $\frac{192\times360}{24\times365.25}=7.885$ Earth years. (3 point)

Since the mass of Burner is the same as Sun, we can use Kepler’s second law in earth years and A.U. as $a^3\propto T^2$ . Thus $a=(7.885)^{2/3}=3.961A.U.$ .

Thus, we find that radius $a=d$ of the orbit is: $5.92557\times10^{11}$ meters. (1 point)

Using the fact that Ash’s radius $R_A$ is known (in terms of Earth’s radius), we can write that the net radiant power absorbed by the planet is: $\frac{\pi R_A^2}{4\pi d^2}L_B=4.508\times10^{16}\ \mathrm W$ . (2 points)

Now, Ash will also emit some radiation in order to be in thermal equilibrium. However, it does not emit in all directions: only in the area exposed to the star. The solid angle not exposed to the star is the cone of angle 60 degrees. The solid angle is given by $2\pi(1-\cos\theta)=\pi$ . This $\frac\pi{4\pi}=\frac14$ of the entire sphere that does not emit. (3 points)

Thus if the temperature of Ash is $T$ , then the net flux emitted by Ash by Stephen Boltzman’s law can be written as: $\frac34\times4\pi R_A^2\sigma T^4=8.696\times10^7T^4$ . (1 point)

Equating the two expressions we can get: $4.508\times10^{16}=8.696\times10^7T^4$

So, $\boxed{T=150.892\ \mathrm K.}$ (2 points)

Water freezes at $273\mathrm K$ , much higher than the temperature above. Thus, this planet is probably too cold for life. Note that the unexposed part is probably even colder. (1 point)

(c) Note: Some of the calculations in the previous part may be done here if the previous one is skipped. Appropriate points should be given in the previous part.

Ash has to cover the diameter of Burner as well as the diameter of itself. The total distance that Ash must cover is $2(R_B+R_A)$ . (1 point)

The angle that the star subtends is given by $2*\arcsin\left(\frac{R_A+R_B}{d}\right)=0.137$ degrees. (1 point)

Thus the time it takes to cover this is $\frac{0.137}{360}\times7.885$ Earth years or $\boxed{9.46\times10^4\ \mathrm{seconds}}$ . (2 points)

(d) Now that Spiff is 100 pc away from the system, we note that his line of sight must travel through $\frac2{1000}*100=0.2$ magnitudes of ISM extinction. (1 point)

To find the apparent magnitude of Burner, we use the Sun-Earth system as a reference:

m_1+26.74=-2.5\log\frac{F_{Burner\to Spiff}}{F_{\odot\to Earth}}+0.2

(1 point)

\Longrightarrow\quad m_1=-26.74-2.5\log\frac{\frac{L_B}{4\pi(100\mathrm{pc})^2}}{\frac{L_\odot}{4\pi(1\mathrm{AU})^2}}+0.2

(1 point)

=-26.74+36.55+0.2=\boxed{10.01}

(2 points)

The limit magnitude of the human eye is 6, so he cannot see it. (1 point)

Source note: The key calls the a^3 proportional to T^2 relation “Kepler’s second law”; this printed name is retained. Its spelling “Stephen Boltzmann” and “Stephen Boltzman’s” also remains literal. The solar apparent magnitude -26.74 is a key-only question supplement, not silently added to the original statement.

Теми и трудност

Оценена трудност: 4/5.

Светлина и астрономически измервания · Екзопланети и среди · Абсолютно черно тяло · Звездни величини · Планетни среди · Междузвездно поглъщане

Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка

Оригинал в Архива: second_exam_2024.pdf · официални решения: second_exam_2024-solutions.pdf · допълнителни данни: physical-constants.pdf