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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 7. Interstellar Flight

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Всички материали от USAAAO 2026

2026 National Astronomy Competition · 40 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

Kirara is traveling to some stars in the constellation Crux, starting from Earth, and then visiting Gacrux ( $\gamma$ Crucis), Ginan ( $\varepsilon$ Crucis), Acrux ( $\alpha$ Crucis), and Mimosa ( $\beta$ Crucis) in that order. To pass the time, she does some astronomy on each leg of the trip.

(a) (5 points) Suppose Kirara starts at rest in the reference frame of the barycenter of the solar system, and that Gacrux, in the same reference frame, has a radial velocity of $20.6\,\mathrm{km/s}$ away from Kirara. She has a magic warp drive that can teleport her rocket straight to Gacrux (and preserves her velocity).

However, if she directly teleports, she’ll have a large velocity relative to Gacrux once she arrives! She plans to use a normal rocket to first accelerate to a point where she has no velocity relative to Gacrux. If her unfueled rocket weights $1.20\times10^5\,\mathrm{kg}$ , how much fuel would she need? The exhaust velocity of her rocket is $10.0\,\mathrm{km/s}$ .

The Tsiolkovsky rocket equation says that $\Delta v=v_e\ln\frac{m_0}{m_f}$ , where $\Delta v$ is the change in velocity, $v_e$ is the exhaust velocity, $m_0$ is the initial mass of the rocket (including fuel), and $m_f$ is the final mass of the rocket.

Ignore the proper motion of Gacrux.

(b) (10 points) When Kirara gets to Ginan, she takes a break on a (fictional) exoplanet orbiting the star. Ginan has a mass of $1.5M_\odot$ , and suppose the exoplanet has a mass of $2M_\oplus$ , a radius of $1.2R_\oplus$ , and a circular orbit with radius 10 AU. She rigs a catapult to give her rocket a one-time boost of $\Delta v$ in any direction. What is the minimum $\Delta v$ required for her to escape the combined gravitational field of both Ginan and the exoplanet?

(c) (5 points) Kirara arrives at Acrux, which is actually a system of 6 stars. She gets distracted taking measurements of them, and doesn’t realize that she’s drifting too close to one of the stars! Specifically, she is close to $\alpha$ Crucis Aa, which has a mass of $15.17M_\odot$ .

She currently is at rest, at a distance $R$ from $\alpha$ Crucis Aa. The free-fall time for an object at distance $R$ to fall into a star with mass $15.17M_\odot$ is 24.0 hours, meaning that if Kirara doesn’t act, her ship will be pulled into the star in 24 hours!

She repositions to a distance of $2R$ from $\alpha$ Crucis Aa, again with no velocity relative to the star. What is the free-fall time for an object at a distance $2R$ to fall into a star of mass $15.17M_\odot$ ?

Assume $\alpha$ Crucis Aa is at rest, and ignore the gravitational effects of the other stars in the system.

(d) (20 points) When Kirara is halfway from Acrux to Mimosa (in linear Euclidean distance), she looks at Mimosa through a telescope. What is the apparent magnitude she observes for Mimosa? The following information will be helpful:

Designation	Name	RA	Dec	Apparent magnitude	Distance to Earth
$\alpha$ Crucis	Acrux	$12^{\mathrm h}\ 26^{\mathrm m}\ 36^{\mathrm s}$	$-63^\circ\ 06'$	0.76	321 ly
$\beta$ Crucis	Mimosa	$12^{\mathrm h}\ 47^{\mathrm m}\ 43^{\mathrm s}$	$-59^\circ\ 41'$	1.25	352 ly

Assume that Acrux and Mimosa are stationary.

Решение

Solution:

(a) (5 points) Plugging in the given values directly into the rocket equation, we get

20.6\,\mathrm{km/s}=(10.0\,\mathrm{km/s})\ln\frac{m_0}{1.20\times10^5\,\mathrm{kg}}.

Then, the total mass of the fueled rocket is

\left(1.20\times10^5\,\mathrm{kg}\right)\exp\left(\frac{20.6\,\mathrm{km/s}}{10.0\,\mathrm{km/s}}\right)=9.42\times10^5\,\mathrm{kg}.

Subtracting the mass of the unfueled rocket, we get the mass of the fuel is $9.42\times10^5\,\mathrm{kg}-1.20\times10^5\,\mathrm{kg}=\boxed{8.22\times10^5\,\mathrm{kg}}$ .

(b) (10 points) We solve this problem in two stages: Kirara needs to first escape the gravity well of the planet, after which she will have some leftover velocity, with which she needs to escape the gravity well of Ginan.

First, we work in the reference frame of Ginan, and compute the escape velocity from Ginan, from a distance of 10 AU (ignoring the planet). This is

v_{esc}=\sqrt{\frac{2G(1.5M_\odot)}{10\,\mathrm{AU}}}=\sqrt{\frac{2\cdot(6.67\times10^{-11})\cdot(1.5\cdot1.989\times10^{30})}{10\cdot(1.496\times10^{11})}}=16.3\,\mathrm{km/s}.

Since the planet already has an orbital velocity of

\sqrt{\frac{G(1.5M_\odot)}{10\,\mathrm{AU}}}=\frac1{\sqrt2}v_{esc}=11.5\,\mathrm{km/s},

if Kirara launches her rocket in the same direction as the planet’s orbit, she only needs a velocity of $16.3-11.5=4.8\,\mathrm{km/s}$ relative to the planet once she escapes its gravity well.

Now, we work in the reference frame of the planet. If her rocket has mass $m$ and she launches with a velocity $v$ , then her kinetic energy will be $\frac12mv^2$ , and gravitational potential energy is

-\frac{Gm(2M_\oplus)}{1.2R_\oplus}=-\frac{6.67\times10^{-11}\cdot(2\cdot5.976\times10^{24})}{1.2\cdot(6.371\times10^6)}m=-(1.043\times10^8)m.

If she escapes the gravitational well of the planet, then we know her final velocity must be 4.8 km/s, meaning her final total energy (in the planet’s frame) is $\frac12m(4.8\,\mathrm{km/s})^2$ . Solving for $v$ , we get

\frac12mv^2-(1.043\times10^8)m=\frac12m(4.8\,\mathrm{km/s})^2.

Canceling out the $m$ and plugging in numbers, we are left with $v=\boxed{15.2\,\mathrm{km/s}}$ .

(c) (5 points) The path of an object starting at rest falling into a star can be thought of as half an “elliptical” orbit with eccentricity 1. By Kepler’s 3rd law, the orbital period is proportional to $r^{1.5}$ . Thus, doubling the starting distance multiplies the amount of time it takes to fall by $2^{1.5}=2.83$ , so from her new position, she has $24\cdot2.83=\boxed{68\text{ hours}}$ until she falls into the star.

(d) (20 points) We need to determine the distance from Acrux to Mimosa (since Kirara’s distance to Mimosa will be half of this distance). To do this, we will first find the angular separation between Acrux and Mimosa using the spherical law of cosines, and then use the triangular law of cosines to find the linear distance between the two stars.

On the celestial sphere, let $N$ be the north celestial pole, $A$ be the position of Acrux, and $M$ be the position of Mimosa. Then,

NA=90^\circ-(-63^\circ06')=153.1^\circ,

NM=90^\circ-(-59^\circ41')=149.7^\circ,

and

\angle ANM=(12^{\mathrm h}\ 47^{\mathrm m}\ 43^{\mathrm s})-(12^{\mathrm h}\ 26^{\mathrm m}\ 36^{\mathrm s})=21^{\mathrm m}\ 7^{\mathrm s}=5.3^\circ.

By the spherical law of cosines,

\cos(AM)=\cos(NA)\cos(NM)+\sin(NA)\sin(NM)\cos(\angle ANM),

AM=\arccos(\cos(153.1)\cos(149.7)+\sin(153.1)\sin(149.7)\cos(5.3))=4.4^\circ.

By the triangular law of cosines, using the distances from Earth to Acrux and Mimosa, we have that the distance from Acrux to Mimosa is

\sqrt{321^2+352^2-2\cdot321\cdot352\cos(4.4^\circ)}=41.6\,\mathrm{ly}.

Kirara’s distance to Mimosa is half of that, which is 20.8 ly.

Now, we calculate the apparent magnitude she observes for Mimosa. The shift in apparent magnitude from being closer to Mimosa is

5\log\frac{\text{distance from Earth to Mimosa}}{\text{Kirara’s distance to Mimosa}}=5\log_{10}\frac{352}{20.8}=6.14.

Thus, her observed apparent magnitude for Mimosa is $1.25-6.14=\boxed{-4.89}$ .

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