Съдържание

OWAO 2025, Теория — Задача 1. Celestial Pause

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от OWAO 2025

2025 ◇ Theoretical Round ◇ Problems

Внимание: Бележка към темата

All eight whole original tasks and 29 lettered parts, complete official worked solutions, alternatives and marking schemes are retained. Constants are native question page 5 and repeated key page 31, fully transcribed once as a shared reference. No paper/time/task/part total is printed, so no totals are inferred from marking criteria. Source mathematical/subscript/endpoint/equality discrepancies are preserved with attributed local notes. The separately printed cooler-star Criterion (x) remains inside original task 8; no extra problem is created.

Константи

Universal

Gravitational constant	$G=6.67\cdot10^{-11}\ \mathrm{m^3/(kg\cdot s^2)}$
Speed of light	$c=3.00\cdot10^8\ \mathrm{m/s}$
Planck constant	$h=6.63\cdot10^{-34}\ \mathrm{J\cdot s}$
Boltzmann constant	$k_B=1.38\cdot10^{-23}\ \mathrm{J/K}$
Gas constant	$\mathfrak R=8.314\ \mathrm{J/(mol\cdot K)}$
Proton mass	$m_p=1.673\cdot10^{-27}\ \mathrm{kg}$

Astronomical

Astronomical unit	$1\ \mathrm{au}=149.6\cdot10^6\ \mathrm{km}$
Parsec	$1\ \mathrm{pc}=206265\ \mathrm{au}$
Hubble constant	$H_0=70\ \mathrm{(km/s)/Mpc}$

Earth

Radius	$R_\oplus=6371\ \mathrm{km}$
Mass	$\mathfrak M_\oplus=5.97\cdot10^{24}\ \mathrm{kg}$
Obliquity of ecliptic	$\varepsilon=23.4^\circ$
Surface gravity	$g=9.81\ \mathrm{m/s^2}$
Orbital period	$T_\oplus=365.26\ \mathrm{days}$
Orbital eccentricity	$e_\oplus=0.0167$

Hydrogen spectrum

Lyman $\mathrm L\alpha$	$\lambda_{\mathrm L\alpha}=1215.7\ \text{Å}$
Balmer $\mathrm H\alpha$	$\lambda_{\mathrm H\alpha}=6562.8\ \text{Å}$

Jupiter

Radius	$R_J=6.99\cdot10^4\ \mathrm{km}$
Mass	$\mathfrak M_J=1.90\cdot10^{27}\ \mathrm{kg}$
Orbital radius	$a_J=5.20\ \mathrm{au}$
Orbital period	$T_J=11.86\ \mathrm{yr}$

Sun

Radius	$R_\odot=6.96\cdot10^5\ \mathrm{km}$
Mass	$\mathfrak M_\odot=1.99\cdot10^{30}\ \mathrm{kg}$
Absolute magnitude	$M_\odot=4.74^{\mathrm m}\ \text{(bol.)}$
Effective temperature	$T_\odot=5.8\cdot10^3\ \mathrm K$
Luminosity	$L_\odot=3.828\cdot10^{26}\ \mathrm W$
Color index	$(B-V)_\odot=+0.65^{\mathrm m}$

Emission constants

Stefan-Boltzmann	$\sigma=5.67\cdot10^{-8}\ \mathrm{W/(m^2\cdot K^4)}$
Wien’s displacement	$b=2898\ \mathrm{\mu m\cdot K}$

UBV… system — Mean wavelengths

U band	$\lambda_U=364\ \mathrm{nm}$
B band	$\lambda_B=442\ \mathrm{nm}$
V band	$\lambda_V=540\ \mathrm{nm}$
K band	$\lambda_K=2190\ \mathrm{nm}$

Условие

From opposition to the nearest stationary point, the apparent angular diameter of an asteroid decreases by 10%. A stationary point is the point at which the asteroid’s apparent motion reverses. The observer is located on the surface of the Earth. Assume the asteroid’s orbit is circular and lies in the ecliptic plane.

a) Determine the radius of the asteroid’s orbit.

b) How much time passes between these two moments?

Отговори

Решение

a) A stationary point is a configuration in which the relative velocity of the observer and the asteroid is directed along the line of sight, meaning there is no relative tangential component. The stationary points are symmetric with respect to opposition.

Let $\lambda$ denote the difference in the heliocentric longitudes of the Earth and the asteroid. The distance between them is

AE=\sqrt{ES^2+AS^2-2ES\cdot AS\cos\lambda}=\sqrt{a_E^2+a^2-2a_Ea\cos\lambda}.

Express the tangential components of velocity:

v_{\perp E}=v_E\sin(\angle SEA-90^\circ)=-v_E\cos\angle SEA;

v_{\perp A}=v_A\sin(90^\circ-\angle SAE)=v_A\cos\angle SAE.

Complete native stationary-point orbital geometry with Sun S, Earth E, asteroid A, both native circular arcs, dashed triangle, lambda/aE/a and all red/blue velocity components and arrows. — Figure 1: On determining the position of the stationary point

Apply the law of cosines to $\triangle SEA$ :

SA^2=SE^2+AE^2-2SE\cdot AE\cos\angle SEA\quad\Longrightarrow\quad\cos\angle SEA=\frac{SE^2+AE^2-SA^2}{2SE\cdot AE},

SE^2=SA^2+AE^2-2SA\cdot AE\cos\angle SAE\quad\Longrightarrow\quad\cos\angle SAE=\frac{SA^2+AE^2-SE^2}{2SA\cdot AE};

\cos\angle SEA=\frac{a_E-a\cos\lambda}{\sqrt{a_E^2+a^2-2a_Ea\cos\lambda}},\qquad\cos\angle SAE=\frac{a-a_E\cos\lambda}{\sqrt{a_E^2+a^2-2a_Ea\cos\lambda}}.

Assuming circular orbits, we equate the tangential components of the orbital velocities:

-\sqrt{\frac{G\mathfrak M_\odot}{a_E}}\cdot\frac{a_E-a\cos\lambda}{\sqrt{a_E^2+a^2-2a_Ea\cos\lambda}}=\sqrt{\frac{G\mathfrak M_\odot}{a}}\cdot\frac{a-a_E\cos\lambda}{\sqrt{a_E^2+a^2-2a_Ea\cos\lambda}},

\frac{a\cos\lambda-a_E}{\sqrt{a_E}}=\frac{a-a_E\cos\lambda}{\sqrt a}\quad\Longrightarrow\quad\cos\lambda=\frac{a\sqrt{a_E}+a_E\sqrt a}{a\sqrt a+a_E\sqrt{a_E}}.

The distance to the asteroid at opposition is $r_o=a-a_E$ . The distance at the stationary point is derived using the geometrical relation involving $\cos\lambda$ . The ratio of the asteroid’s angular sizes at these two points leads to the following equation:

(a-a_E)^2=0.9^2\left(a_E^2+a^2-2a_Ea\cdot\frac{a\sqrt{a_E}+a_E\sqrt a}{a\sqrt a+a_E\sqrt{a_E}}\right).

Let us measure distances in au, so $a_E=1$ . The equation simplifies to:

a^2-2a+1=0.81\left(1+a^2-2a\cdot\frac{a+\sqrt a}{a\sqrt a+1}\right)\quad\Longrightarrow\quad a=5.6\ \mathrm{au}.

b) Next, determine the time interval between opposition and the stationary point. The difference in heliocentric longitudes at the stationary point is

\lambda=\arccos\frac{a+\sqrt a}{a\sqrt a+1}=56^\circ.

The change in the difference of heliocentric longitudes is related to the synodic period $S$ . Then, the time interval is proportional to the difference in heliocentric longitudes:

S=\frac{TT_\oplus}{T-T_\oplus}=\frac{a^{1.5}\cdot1}{a^{1.5}-1}=1.08^{\mathrm y},

\Delta T=S\cdot\frac{\lambda}{360^\circ}=0.17^{\mathrm y}=61^{\mathrm d}.

Marking Scheme:

Question (a) Orbital radius
1. Geometric and kinematic relations for the stationary point — 5 pt.
2. Valid technique to solve the equation for the orbital radius — 5 pt.
3. Correct numerical result — 3 pt.
Question (b) Time interval
1. Synodic period, difference in heliocentric longitudes at the stationary point — 5 pt.
2. Correct numerical result — 2 pt.

Теми и трудност

Оценена трудност: 3/5.

Небесна механика · Орбитално движение · Синодични периоди

Предпоставки: Гимназиална подготовка, Разширена олимпиадна подготовка

Оригинал в Архива: OWAO_2025_Theoretical_Problems.pdf · официални решения: OWAO_2025_Theoretical_Solutions.pdf