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USAAAO 2021, II кръг — Medium 3

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Всички материали от USAAAO 2021

2021 National Astronomy Competition · 30 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

A space station of mass $m$ is orbiting a planet of mass $M_0$ on a circular orbit of radius r. At a certain moment, a satellite of mass m is launched from the space station with a relative velocity $\vec w$ oriented towards the center of the planet. Assume that $w<\sqrt{\frac{GM_0}{r}}$ .

(a) Justify the shape of the satellite’s orbit after launching and, for the satellite-planet system, determine the following quantities:

(1) Satellite’s velocity relative to the planet, immediately after launch, v

(2) Total angular momentum of the satellite-planet system, $L_{P,Sat}$

(3) Satellite’s orbit semi-major and semi-minor axes, $a_{Sat}$ and $b_{Sat}$

(4) Satellite’s orbit eccentricity, $\epsilon_{Sat}$

(5) Apogee and perigee distances, $r_{max,Sat}$ and $r_{min,Sat}$

(6) Satellite’s minimum velocity, $v_{min,Sat}$ and maximum velocity $v_{max,Sat}$ on it’s orbit

(7) Total energy of the satellite-planet system, $E_{Sat,P}$ .

(b) Determine the shape of the space station’s orbit relative to the planet, after the satellite was launched.

Решение

(a) The absolute velocity of the satellite (relatively to the planet) at the moment of launching is:

\vec v=\vec w+\vec u

,where u is the orbital station’s velocity relatively to the planet.

v=\sqrt{w^2+u^2}

u=\omega_0r

,where $\omega_0$ is the angular velocity of the space station orbiting around the planet, before the satellite was launched.

T_0=\frac{2\pi r}{u}=\frac{2\pi}{\omega_0}=\frac ur

,where $T_0$ is the orbital period of the space station, before the satellite was launched.

\frac{mu^2}{r}=\frac{GmM_0}{r^2}

u=\sqrt{\frac{GM_0}{r}}\qquad\omega_0=\sqrt{\frac{GM_0}{r^3}}

Thus, we get:

v=\sqrt{w^2+\frac{GM_0}{r}}>w

So, after the satellite was launched at point Q, it will move on an elliptical orbit, having the planet as a one of the focal points. This happens because the total energy of the satellite - planet system, calculated for the moment when the satellite is injected on its orbit is:

E_{Q,S-P}=\frac{mv^2}{2}-\frac{GmM_0}{r}=\frac m2(w^2+u^2)-\frac{GmM_0}{r}

E_{Q,S-P}=\frac m2\left(w^2+\frac{GM_0}{r}\right)-\frac{GmM_0}{r}

E_{Q,S-P}=\frac m2\left(w^2-\frac{GM_0}{r}\right)

w<\sqrt{\frac{GM_0}{r}};\quad w^2<\frac{GM_0}{r};\quad E_{Q,S-P}<0

Having this, we can determine the angular momentum of the space station relatively to the fixed planet, corresponding to point Q (point of injection):

\overrightarrow{L_{Q,Sat}}=\vec r\times mv=\vec r\times m(\vec w+\vec u)

But the angle between $\vec r$ and $\vec w$ is $180^\circ$ . Thus, $\vec r\times\vec w=0$ . So:

\overrightarrow{L_{Q,Sat}}=\vec r\times m\vec u

Moreover, the angle between $\vec r$ and $\vec u$ is $90^\circ$ . Thus, we get:

L_{Q,Sat}=mru\qquad u=\omega_0r

L_{Q,Sat}=mr^2\omega_0=mr^2\sqrt{\frac{GM_0}{r^3}}=m\sqrt{GrM_0}

Evolving on an elliptical orbit around the planet, when the satellite will reach the minimum distance from the planet, its angular momentum will be:

\overrightarrow{L_{P,Sat}}=\overrightarrow{r_{min,Sat}}\times m\overrightarrow{v_{max,Sat}}

Because the angle between $\overrightarrow{r_{min,Sat}}$ and $\overrightarrow{v_{max,Sat}}$ is $90^\circ$ , we get:

L_{P,Sat}=mr_{min,Sat}v_{max,Sat}

Because the total angular momentum is conserved:

L_{P,Sat}=L_{Q,Sat}=L_{A,Sat}

mr_{min,Sat}v_{max,Sat}=m\sqrt{GrM_0}=mr_{max,Sat}v_{min,Sat}

Thus, we get:

r_{min,Sat}v_{max,Sat}=\sqrt{GrM_0}=r_{max,Sat}v_{min,Sat}

v_{max,Sat}=\frac{\sqrt{GrM_0}}{r_{min,Sat}};\quad v_{max,Sat}^2=\frac{GrM_0}{r_{min,Sat}^2}

v_{min,Sat}=\frac{\sqrt{GrM_0}}{r_{max,Sat}};\quad v_{min,Sat}^2=\frac{GrM_0}{r_{max,Sat}^2}

Also, the total mechanical energy of the satellite - planet system is conserved, thus:

E_{Q,S-P}=E_{P,S-P}

\frac12m\left(w^2+\frac{GM_0}{r}\right)-\frac{GmM_0}{r}=\frac{mv_{max,Sat}^2}{2}-\frac{GmM_0}{r_{min,Sat}}

\left(w^2+\frac{GM_0}{r}\right)-\frac{2GM_0}{r}=v_{max,Sat}^2-\frac{2GM_0}{r_{min,Sat}}

w^2-\frac{GM_0}{r}=v_{max,Sat}^2-\frac{2GM_0}{r_{min,Sat}}

But:

v_{max,Sat}^2=\frac{GrM_0}{r_{min,Sat}^2}

Thus:

w^2-\frac{GM_0}{r}=\frac{GrM_0}{r_{min,Sat}^2}-\frac{2GM_0}{r_{min,Sat}}

\left(w^2-\frac{GM_0}{r}\right)r_{min,Sat}^2+2GM_0r_{min,Sat}-GrM_0=0

By solving this, we get:

r_{min,Sat}=\frac{r(-GM_0\pm w\sqrt{GM_0r})}{w^2r-GM_0}

r_{min,Sat}=\frac{r(GM_0\mp w\sqrt{GM_0r})}{GM_0-w^2r}

Because we are searching for the minimum value, we will assign $r_{min,Sat}$ the negative solution. Thus, the positive solution will be the value for $r_{max,Sat}$ . Thus:

r_{min,Sat}=\frac{r(GM_0-w\sqrt{GM_0r})}{GM_0-w^2r}

v_{max,Sat}=\frac r{r_{min,Sat}}\sqrt{\frac{GM_0}{r}}

Thus, we get;

v_{max,Sat}=\frac{GM_0-w^2r}{GM_0-w\sqrt{GrM_0}}\sqrt{\frac{GM_0}{r}}

Consequently, for $r_{max,Sat}$ :

r_{max,Sat}=\frac{r(GM_0+w\sqrt{GM_0r})}{GM_0-w^2r}

v_{min,Sat}=\frac r{r_{max,Sat}}\sqrt{\frac{GM_0}{r}}

Thus:

v_{min,Sat}=\frac{GM_0-w^2r}{GM_0+w\sqrt{GrM_0}}\sqrt{\frac{GM_0}{r}}

From the properties of the ellipse:

r_{min,Sat}+r_{max,Sat}=2a_{Sat}

,where a is the semi-major axis of the ellipse. Thus:

\frac{r(GM_0+w\sqrt{GM_0r})}{GM_0-w^2r}+\frac{r(GM_0-w\sqrt{GM_0r})}{GM_0-w^2r}=2a_{Sat}

a_{Sat}=\frac{GrM_0}{GM_0-rw^2}>r

Using the conservation laws for the total energy and total angular momentum, we prove that:

L_{Sat}=mb_{Sat}\sqrt{\frac{GM_0}{a_{Sat}}}=L_{P,Sat}=mv_{max,Sat}r_{min,Sat}

b_{Sat}\sqrt{\frac{GM_0}{a_{Sat}}}=v_{max,Sat}r_{min,Sat}

b_{Sat}=v_{max,Sat}r_{min,Sat}\sqrt{\frac{a_{Sat}}{GM_0}}=\sqrt{GrM_0}\cdot\sqrt{\frac{a_{Sat}}{GM_0}}

Thus, the semi-minor axis is:

b_{Sat}=r\cdot\sqrt{\frac{GM_0}{GM_0-rw^2}}

Now, we can determine the eccentricity of the satellite’s orbit:

e_{Sat}=\sqrt{1-\frac{b_{Sat}^2}{a_{Sat}^2}}=\sqrt{1-\frac{r^2\cdot\frac{GM_0}{GM_0-rw^2}}{r^2\cdot\left(\frac{GM_0}{GM_0-rw^2}\right)^2}}=\sqrt{1-\frac{GM_0-rw^2}{GM_0}}

Thus, we get:

e_{Sat}=w\sqrt{\frac r{GM_0}}

The total energy of the satellite-Earth system is:

E_{Sat-E}=-\frac{GmM_0}{2a_{Sat}}

,where $a_{Sat}=r\frac{GM_0}{GM_0-rw^2}$ .

E_{Sat-E}=-\frac{GmM_0}{2r\frac{GM_0}{GM_0-rw^2}}=-GmM_0\frac{GM_0-rw^2}{2rGM_0}

E_{Sat-E}=-\frac{m(GM_0-rw^2)}{2r}

(b) After the satellite was launched with a radial relative velocity of $\vec w$ towards the planet, the space station also gained a radial velocity $\vec W$ oriented in the opposite direction of $\vec w$ because the total momentum is conserved:

\vec V=\vec W+\vec u

V=\sqrt{W^2+u^2}

Conservation of momentum:

M\vec W+m\vec w=0;\quad W=\frac mMw;\quad m<M;\quad W<w;

V=\sqrt{\frac{m^2}{M^2}w^2+u^2};\quad V<v;

In order to determine the shape of the orbit we need to calculate the total energy of the station-planet system:

E_{Sta,P}=\frac{MV^2}{2}-\frac{GMM_0}{r}=\frac M2\left(\frac{m^2}{M^2}w^2+u^2\right)-\frac{GMM_0}{r}

E_{Sta,P}=\frac M2\left(\frac{m^2}{M^2}w^2+\frac{GM_0}{r}\right)-\frac{GMM_0}{r}

E_{Sta,P}=\frac12M\frac{m^2}{M^2}w^2-\frac12\frac{GMM_0}{r}

E_{Sta,P}=\frac12M\frac{m^2}{M^2}\frac{GM_0}{r}-\frac12\frac{GMM_0}{r}

E_{Sta,P}=\frac12\frac{GMM_0}{r}\left(\frac{m^2}{M^2}-1\right)<0

Thus, after the satellite is launched, the space station will move on an elliptical orbit around the planet.

Source note: The original and repeated key question call both station and satellite masses m. The worked key uses m for the satellite and M for the station and assumes $m\lt M$ , a distinction absent from the original statement. The printed period chain ends with u/r; the energy paragraph changes the generic planet to Earth. Part (b) writes MW+mw=0 with radial W and changes an energy bound into an equality. All original notation, inequalities, labels and derivation steps remain literal.

Теми и трудност

Оценена трудност: 4/5.

Небесна механика · Орбитална енергия и напускане · Орбитално движение

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