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2026 National Astronomy Competition 1 Instructions (Please Read Carefully)Table of Physical Constants Astrophysical Data Other data and conversion factors Trigonometrical identities Prefixes Условие Решение

USAAAO 2026, II кръг — Задача 2. Celestial Sphere: Night at the Observatory

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Всички материали от USAAAO 2026

2026 National Astronomy Competition · 15 т.

Внимание: Бележка към темата

Questions 3 and 9 refer to designated answer sheets, and question 6 to provided grid paper; these sheets were not obtained in the two native 2026-II PDFs or their two-row same-folder archive catalogue. The separately attributed physical-constants.pdf is an undated common USAAAO archive reference; its identity as the table supplied for this exam is unproven.

Instructions (Please Read Carefully)

2026 National Astronomy Competition

1 Instructions (Please Read Carefully)

The top 15 eligible scorers on the NAC will be invited to the training camp. A third round will be conducted during the camp to select the national team for the IOAA 2026. In order to qualify for camp, you must be a high school student with US citizenship or permanent residency.

The test must be completed within 2.5 hours (150 minutes) and the maximum number of points is 300 points. The point breakdown per problem is as follows.

Problem	P1	P2	P3	P4	P5	P6	P7	P8	P9	Total
Points	10	15	15	20	20	40	40	55	85	300

Please solve each problem on a blank piece of paper and mark the number of the problem at the top of the page. Some problems may require a designated answer sheet; these will be provided. 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 calculator (non-programmable and non-graphing) 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 20, 2026. 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. Students must be proctored during the entire duration of the exam.

This exam sheet and your answer sheets (including scratch papers) should be returned to your proctors once the exam ended. Your proctors should upload your answer sheets to the Google Form provided to them.

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

Srihari (Hari) Balaji, Feodor Yevtushenko, Ferdinand, Hagan Hensley, Joe McCarty, Lucas Pinheiro, Vincent Bian, Elizabeth Lee

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

All work on this exam has been completed by me.
I took this exam under the supervision of a proctor.
I did not receive any external help beyond the materials provided.
I will not discuss the contents of this exam with anyone until April 20, 2026.
Failure to follow these rules will result in disqualification from the exam.

Student Signature: ____________________ Date: ____________________

Undated common archive reference · Table of Physical Constants

Table of Physical Constants

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 ( $\vphantom{C}^{12}\mathrm C$ 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

Undated common archive reference · Astrophysical Data and Conversion Factors

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	$1\,\mathrm{N\,m^{-2}}$
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

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

Условие

An observer is located at latitude $\phi=+42.0^\circ$ (Northern Hemisphere). Assume the Earth is a perfect sphere and ignore atmospheric refraction.

Star A has equatorial coordinates

\alpha=6^{\mathrm h}40^{\mathrm m},\qquad\delta=+62.0^\circ.

On a particular night, the local sidereal time (LST) is

\mathrm{LST}=5^{\mathrm h}10^{\mathrm m}

at 10:00 PM local clock time.

You may assume that one sidereal day equals $23^{\mathrm h}56^{\mathrm m}$ of clock time.

(a) (2 points) Compute the hour angle $H$ of star A at 10:00 PM. State whether the star is east or west of the meridian at that moment.

(b) (5 points) Compute the altitude $h$ of star A at 10:00 PM. Give your answer in degrees.

(c) (3 points) How much clock time will pass after 10:00 PM until star A reaches upper culmination? Express your answer in hours, minutes, and seconds.

Suppose star B has declination $\delta=+30.0^\circ$ .

(d) (5 points) For star B, compute the total clock-time duration during which the star is above the horizon during one sidereal day. Express your final answer in hours and minutes of clock time.

Решение

Solution:

Given $\phi=+42.0^\circ$ , $\delta=+62.0^\circ$ , and $\alpha=6^{\mathrm h}40^{\mathrm m}$ . At 10:00 PM, $\mathrm{LST}=5^{\mathrm h}10^{\mathrm m}$ . Recall $H=\mathrm{LST}-\alpha$ and $H>0$ means west of the meridian.

(a) Hour angle and sky side. (2 points)

Convert to hours:

\alpha=6^{\mathrm h}40^{\mathrm m}=6.666\overline6\,\mathrm h,\qquad\mathrm{LST}=5^{\mathrm h}10^{\mathrm m}=5.166\overline6\,\mathrm h.

Thus

H=5^{\mathrm h}10^{\mathrm m}-6^{\mathrm h}40^{\mathrm m}=-1^{\mathrm h}30^{\mathrm m}.

Since $H<0$ , the star is east of the meridian.

Equivalently, in degrees ( $15^\circ$ per hour):

H=-1.5\,\mathrm h\times15^\circ/\mathrm h=-22.5^\circ.

Grading:

2 pts: Correct $H$ and correct east/west interpretation.
1 pt: Correct subtraction but wrong east/west interpretation.
1 pt: Arithmetic error in conversion but correct method.
0.5 pt: Correct formula $H=\mathrm{LST}-\alpha$ but major arithmetic mistake.
0 pts: Incorrect formula or no meaningful progress.

Accept answers in range: $-22.5^\circ\pm0.3^\circ$ .

(b) Altitude at 10:00 PM. (5 points)

Using spherical law of cosines,

\sin h=\sin\phi\sin\delta+\cos\phi\cos\delta\cos H.\qquad (3)

Substitute $\phi=42^\circ$ , $\delta=62^\circ$ , $H=-22.5^\circ$ :

\sin h=\sin42^\circ\sin62^\circ+\cos42^\circ\cos62^\circ\cos(22.5^\circ).

Numerically,

\sin h\approx0.9131

h\approx\arcsin(0.9131)\approx65.9^\circ.

\boxed{h\approx65.9^\circ}

Accept answers between $65^\circ$ and $67^\circ$ .

Grading:

5 pts: Correct substitution, trig evaluation, and final altitude.
4 pts: Correct setup with minor arithmetic error.
3 pts: Correct formula but computational errors.
2 pts: Some progress towards Eq. 3 but incorrect substitution and arithmetic errors.
1 pt: Partial substitution or wrong angle sign.
0 pts: No correct use of relation.

Note: If a student forgets $\cos(-H)=\cos H$ , do not deduct heavily unless the conceptual reasoning is affected.

Upper culmination occurs when $H=0$ , i.e. when $\mathrm{LST}=\alpha$ .

Currently $H=-1^{\mathrm h}30^{\mathrm m}$ , so the LST must increase by

\Delta H=1^{\mathrm h}30^{\mathrm m}=1.5\text{ sidereal hours}.

A sidereal day is 24 sidereal hours $=23^{\mathrm h}56^{\mathrm m}$ of clock time.

Thus

1\text{ sidereal hour}=\frac{23^{\mathrm h}56^{\mathrm m}}{24}\approx59^{\mathrm m}50^{\mathrm s}.

Therefore

\Delta t=1.5\times59^{\mathrm m}50^{\mathrm s}\approx1^{\mathrm h}29^{\mathrm m}45^{\mathrm s}.

\boxed{1^{\mathrm h}29^{\mathrm m}45^{\mathrm s}}

Grading:

3 pts: Correct sidereal interval and correct elapsed time.
2 pts: Correct sidereal interval and recognizes conversion needed but performs it incorrectly.
1 pt: Correct sidereal interval but assumes 1 sidereal hour = 1 clock hour, leading to $1^{\mathrm h}30^{\mathrm m}$ .

(d) Time above the horizon. (5 points)

Use the horizon condition

\cos H_0=-\tan\phi\tan\delta,

which comes from Eq. 3 with altitude $h=0$ .

Substitute $\phi=42^\circ$ , $\delta=30^\circ$ :

\cos H_0=-\tan42^\circ\tan30^\circ.

Using

\tan42^\circ\approx0.900,\qquad\tan30^\circ\approx0.577,

\cos H_0\approx-0.519.

Thus

H_0\approx\cos^{-1}(-0.519)\approx121.3^\circ.

Convert to hours:

H_0=\frac{121.3^\circ}{15^\circ/\text{hour}}\approx8.09\text{ hours}.

The star is above the horizon for

2H_0\approx16.18\text{ sidereal hours}.

Convert to clock time using

1\text{ sidereal hour}\approx59^{\mathrm m}50^{\mathrm s}.

16.18\times59^{\mathrm m}50^{\mathrm s}\approx16^{\mathrm h}08^{\mathrm m}.

\boxed{\text{Star is above the horizon for about }16^{\mathrm h}08^{\mathrm m}.}

Grading:

5 pts: Correct setup of the horizon equation, correct computation of $H_0$ , and correct duration above the horizon.
4 pts: Correct method but minor arithmetic error in $H_0$ or time conversion.
3 pts: Correct horizon equation but significant computational errors.
2 pts: Partial substitution into the horizon condition.
1 pt: Incorrect derivation of horizon condition from Eq. 3.
0 pts: No correct use of the horizon relation.

Оригинал в Архива: usaaao2026.pdf · официални решения: usaaao2026_sol.pdf · допълнителни данни: physical-constants.pdf