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IOAA 2015, Теория, дълги задачи, български — Задача 1

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Всички материали от IOAA 2015

IOAA 2015 — ДЪЛГИ ЗАДАЧИ

Внимание: Бележка към темата

Официалните решения и критерии са запазени на английски. В българската задача 3 няма въпрос за разстоянието: части d/e съответстват на e/f от английския финал. В задача 1(c) българският текст определя височината на точката от орбитата, а английският финал — на горната допирателна към спътника; решението следва английската формулировка. По-ранните варианти са показани отделно и не заместват финалните условия.

Instruction — Theoretical Round

Write your answer on the answer-sheet only. You can use more than one sheet for one problem. Please ask the organizer for the additional answer-sheets.
Different problems should be done on different sheets.
You can use your own calculator without program and graphic facility.
You can ask the organizer for rough paper.

Astronomical and Physical Constants — 20150729

Constant	Value
Astronomical unit (AU)	$1.4960\times10^{11}\ \mathrm m$
Light year (ly)	$9.4605\times10^{15}\ \mathrm m=63\,240\ \mathrm{au}$
Parsec (pc)	$3.0860\times10^{16}\ \mathrm m=206\,265\ \mathrm{au}$
Jansky (Jy)	$10^{-26}\ \mathrm{W\,m^{-2}\,Hz^{-1}}$
1 Sidereal year	365.2564 solar days
1 Tropical year	365.2422 solar days
1 Calendar year	365.2425 solar days
1 Sidereal day	$23^{\mathrm h}\ 56^{\mathrm m}\ 04^{\mathrm s}.091$
1 Solar day	$24^{\mathrm h}\ 03^{\mathrm m}\ 56^{\mathrm s}.555$ units of sidereal time
1 Å	$10^{-10}\ \mathrm m$
1 erg	$10^{-7}\ \mathrm J$
1 dyne	$10^{-5}\ \mathrm N$
Mass of Earth	$5.9736\times10^{24}\ \mathrm{kg}$
Mean radius of Earth	$6.371\times10^6\ \mathrm m$
Equatorial radius of Earth	$6.378\times10^6\ \mathrm m$
Mean velocity of Earth on its orbit	$29.783\ \mathrm{km\,s^{-1}}$
Mass of Moon	$7.3490\times10^{22}\ \mathrm{kg}$
Radius of Moon	$1.737\times10^6\ \mathrm m$
Mass of Jupiter	$1.89813\times10^{27}\ \mathrm{kg}$
Mean Earth – Moon distance	$3.844\times10^8\ \mathrm m$
Mass of Sun	$1.98892\times10^{30}\ \mathrm{kg}$
Radius of Sun	$6.96\times10^8\ \mathrm m$
Effective temperature of the Sun	5780 K
Luminosity of the Sun	$3.96\times10^{26}\ \mathrm{J\,s^{-1}}$
Solar constant	$1366\ \mathrm{W\,m^{-2}}$
Brightness of the Sun in V-band	−26.8 mag.
Absolute brightness of the Sun in V-band	4.75 mag.
Absolute bolometric brightness of Sun	4.72 mag.
Angular diameter of the Sun	30′
Speed of light in vacuum (c)	$2.9979\times10^8\ \mathrm{m\,s^{-1}}$
Gravitational constant (G)	$6.6738\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$
Boltzmann constant (k)	$1.381\times10^{-23}\ \mathrm{m\,kg\,s^{-2}\,K^{-1}}$
Universal gas constant (R)	$8.31\ \mathrm{J\,K^{-1}\,mol^{-1}}$
Stefan–Boltzmann constant (σ)	$5.6704\times10^{-8}\ \mathrm{kg\,s^{-3}\,K^{-4}}$
Planck constant (h)	$6.6261\times10^{-34}\ \mathrm{J\,s}$
Wien's constant (b)	$2.8978\times10^{-3}\ \mathrm{m\,K}$
Hubble constant (H₀)	$70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$

Astronomical and Physical Constants — continuation

Constant	Value
electron charge (e)	$1.602\times10^{-19}\ \mathrm C$
Mass of hydrogen atom	$1.67\times10^{-27}\ \mathrm{kg}$
Current inclination of the ecliptic (ε)	23º 26.3′
Coordinates of the northern ecliptic pole for epoch 2000.0 (αE, δE)	$18^{\mathrm h}\ 00^{\mathrm m}\ 00^{\mathrm s},+66^{\circ}\ 33.6^{\prime}$
Coordinates of the northern galactic pole for epoch 2000.0 (αG, δG)	$12^{\mathrm h}\ 51^{\mathrm m},+27^{\circ}\ 08^{\prime}$

You can try to solve an equation $x=f(x)$ using iteration: $x_{n+1}=f(x_n)$ .

Basic equations of spherical trigonometry

\sin a\sin B=\sin b\sin A

\sin a\cos B=\cos b\sin c-\sin b\cos c\cos A,

\cos a=\cos b\cos c+\sin b\sin c\cos A.

Rayleigh-Jeans formula (long wavelength approximation of Planck’s Law):

B_\nu(T)=\frac{2kT\nu^2}{c^2}

Условие

Спътник се движи в орбита около планета, като равнината на тази орбита е перпендикулярна на повърхността на планетата в точката в която се намира наблюдателят. След извършване на необходимото скалиране, приемаме, че орбитата на спътника се описва от следното уравнение:

9\left(\frac{x}{2}+\frac{\sqrt{3}y}{2}-4\right)^2+25\left(-\frac{\sqrt{3}x}{2}+\frac{y}{2}\right)^2=225

Разгледайте орбитата в правоъгълна координата система, в която остта Ox лежи в хоризонталната равнина, а Oy е насочена към зенита. Нека с $r$ да означим радиуса на спътника. Да предположим, че периодът на околоосно въртене на планетата е много по-дълъг от орбиталния период на спътника. Атмосферната рефракция се пренебрегва.

Наклонена елипса, представляваща орбитата на спътника в равнина x-y; означени са осите x (хоризонтална равнина) и y (зенит), точка на спътника с радиус r и ъгъл θ между хоризонталата и линията към спътника. — Фигура 1

a. Пресметнете голямата и малката полуос на елиптичната орбита

b. Пресметнете зенитното отстояние на перигея на орбитата, за наблюдателя.

c. Намерете $\tan\dfrac{\theta}{2}$ , където $\theta$ е височината над хоризонта на точката от орбитата в която спътникът изглежда най-голям за наблюдателя.

Решение

1. Polar version — (a)

Rewrite the orbit equation using polar coordinate

x=R(\alpha)\cos\alpha,\qquad y=R(\alpha)\sin\alpha

Thus, if we let

x^*=R(\alpha)\cos\left(\alpha-\frac\pi3\right)=\frac x2+\frac{y\sqrt3}2

y^*=R(\alpha)\sin\left(\alpha-\frac\pi3\right)=-\frac{x\sqrt3}2+\frac y2

10 points

the equation of the orbit can be written as

9\left(R(\alpha)\cos\left(\alpha-\frac\pi3\right)-4\right)^2+25\left(R(\alpha)\sin\left(\alpha-\frac\pi3\right)\right)^2=225

9(x^*-4)^2+25(y^*)^2=225

10 points

Determine semi major and semi minor axis of the orbit

Since $9(x^*-4)^2+25(y^*)^2=225$ is equivalent to

\frac{(x^*-4)^2}{25}+\frac{(y^*)^2}{9}=1

then, the semi major axis is 5, and (5 points) the semi minor axis is 3 (5 points).

1. Polar version — (b) (20 т.)

Determine the zenith angle at perigee

At the perigee, $\alpha-\frac\pi3=\pi$ . Thus, the zenith angle at the perigee is

\alpha-\frac\pi2=\pi+\frac\pi3-\frac\pi2=\frac{5\pi}6.

Ellipse and the zenith-angle construction at perigee.

1. Polar version — (c)

Characterize the coordinates of the moon when it looks largest to the observer.

Let $P(x_0,y_0)$ be the point where it looks largest to the observer. Since the distance between the moon and the observer decreases as it moves from apogee towards perigee, then the moon looks largest to the observer at the planet when it closest to the planet while the whole of the moon still can be seen in full. Hence, $y_0=r$ .

y_0=r=R(\pi-\alpha_0\alpha_0)\sin(\pi-\alpha_0)=R_0\sin(\pi-\alpha_0)

x_0=R_0\cos(\pi-\alpha_0)

10 points

Determine the coordinates of the point.

Let the coordinate be $(x_0,r)$ .

9\left(\frac{x_0}2+\frac{\sqrt3r}2-4\right)^2+25\left(-\frac{\sqrt3x_0}2+\frac r2\right)^2=225

Rewrite it as a quadratic equation in $x_0$ .

9(x_0+r\sqrt3-8)^2+25(-x_0\sqrt3+r)^2=900

84(x_0)^2+(-32\sqrt3r-144)x_0+(52r^2-144\sqrt3r+576=900)

Solving for $x$ , we get

x_0=\frac{4r\sqrt3+18}{21}\pm\frac{15\sqrt{-r^2+4r\sqrt3+9}}{21}

20 points

Then choose the smaller one

x_0=\frac{4r\sqrt3+18}{21}-\frac{15\sqrt{-r^2+4r\sqrt3+9}}{21}

10 points

Find $\tan\frac\theta2$ . Hence

\tan\frac{\theta_0}2=\tan\alpha_0=\frac r{x_0}=\frac{21r}{4r\sqrt3+18-15\sqrt{-r^2+4r\sqrt3+9}}

10 points

Polar construction with moon tangent to the horizon, radius r, R0 and alpha0=theta/2.

2. Cartesian version — (a)

Notice the standard version of the orbits

Simplify the equation by introducing new variables

x^*=\frac x2+\frac{\sqrt3y}2=x\cos\frac\pi3+y\sin\frac\pi3

y^*=-\frac{\sqrt3x}2+\frac y2=x\sin\left(-\frac\pi3\right)+y\cos\frac\pi3

10 points

Write the orbit equation in terms new variables

9(x^*-4)^2+25(y^*)=225

10 points

Determine semi major and semi minor axis of the orbit

Since $9(x^*-4)^2+25(y^*)^2=225$ is equivalent to

\frac{(x^*-4)^2}{25}+\frac{(y^*)^2}{9}=

The semi major axis is 5 (5 points)

The semi major axis is 3 (5 points)

Original rotated ellipse and dashed standard ellipse in transformed coordinates.

2. Cartesian version — (b) (20 т.)

Determine the zenith angle at perigee

Since the original ellipse may be obtained from standard ellipse

\frac{(x^*-4)^2}{25}+\frac{(y^*)^2}{9}=1

by $\pi/3$ counterclockwise rotation respect to the origin, the the zenith angle at perigee is equal to

\pi+\frac\pi3-\frac\pi2=\frac{5\pi}6.

2. Cartesian version — (c)

Characterize the coordinates of the moon when it looks largest to the observer.

10 points

Find $x_0$ .

9\left(\frac{x_0}2+\frac{\sqrt3r}2-4\right)^2+25\left(-\frac{\sqrt3x_0}2+\frac r2\right)^2=225

Rewrite it as a quadratic equation in $x$ .

9(x_0+r\sqrt3-8)^2+25(-x_0\sqrt3+r)^2=900

\left(9x_0^2+(18\sqrt3r-144)x_0+27r^2-144\sqrt3r+576\right)+\left(75x_0^2-50\sqrt3x_0r+25r^2\right)=900

84(x_0)^2+(-32\sqrt3r-144)x_0+(52r^2-144\sqrt3r+576)=900

Solving it for $x$ we get

x_0=\frac{4r\sqrt3+18}{21}\pm\frac{15\sqrt{-r^2+4r\sqrt3+9}}{21}

20 points

Then choose the smaller one

x_0=\frac{4r\sqrt3+18}{21}-\frac{15\sqrt{-r^2+4r\sqrt3+9}}{21}

10 points

Find $\tan\frac\theta2$ . Hence

\tan\frac\theta2=\tan\alpha_0=\frac r{x_0}=\frac{21r}{4r\sqrt3+18-15\sqrt{-r^2+4r\sqrt3+9}}

10 points

3. Cartesian version — (a)

Notice the standard version of the orbits

Simplify the equation by introducing new variables

x^*=\frac x2+\frac{\sqrt3y}2=x\cos\frac\pi3+y\sin\frac\pi3

y^*=-\frac{\sqrt3x}2+\frac y2=x\sin\left(-\frac\pi3\right)+y\cos\frac\pi3

10 points

Write the orbit equation in terms new variables

9(x^*-4)^2+25(y^*)=225

10 points

Determine semi major and semi minor axis of the orbit

Since $9(x^*-4)^2+25(y^*)^2=225$ is equivalent to

\frac{(x^*-4)^2}{25}+\frac{(y^*)^2}{9}=

The semi major axis is 5 (5 points)

The semi major axis is 3 (5 points)

3. Cartesian version — (b) (20 т.)

Notice the standard version of the orbits

Since the original ellipse may be obtained from standard ellipse

\frac{(x^*-4)^2}{25}+\frac{(y^*)^2}{9}=1

by counterclockwise $\pi/3$ rotation respect to the origin, the the zenith angle at perigee is $\pi+\pi/3-\pi/2=5\pi/6$ .

3. Cartesian version — (c)

Identify and characterized the coordinates when the planet looks largest.

Since the distance between the moon and the observer decreases as it moves from apogee towards perigee, then the moon looks largest to the observer at the planet when it closest to the planet while the whole of the moon still can be seen in full. Hence, it is at necessary that $y=r$ . Let the coordinates be $(x_0,r)$ .

10 points

Obtain expression for calculating distance from a point on an ellipse to the foci to set up an equation

Consider standard ellipse

\frac{(x^*-4)^2}{5^2}+\frac{(y^*)^2}{3^2}=1

Choose any point $(x^*,y^*)$ on the ellipse. Let $d_1$ and $d_2$ be the distances from any point $(x^*,y^*)$ on the ellipse to the foci $(0,0)$ and $d(8,0)$ ; $d_1^2=(x^*)^2+(y^*)^2$ and $d_2^2=(x^*-8)^2+(y^*)^2$ .

Thus,

d_1^2=(x^*)^2+(y^*)^2=(x^*)^2+\left(9-\frac{9(x^*-4)^2}{5^2}\right)=\frac{16(x^*)^2+72x^*+81}{25}

Therefore

d_1^2=\frac{16(x^*)^2+72x^*+81}{25}=x^2+r^2

where $x^*=\frac x2+\frac{r\sqrt3}2$ . Therefore, we can obtain the value of $x$ by solving the above equation for $x$ .

10 points

Solve equation $\frac{16(x^*)^2+72x^*+81}{25}=x^2+r^2$ to obtain value of $x$ .

Substituting $x^*=\frac x2+\frac{r\sqrt3}2$ to the equation, we have

25d_1^2=25x^2+25r^2=16\left(\frac x2+\frac{r\sqrt3}2\right)^2+72\left(\frac x2+\frac{r\sqrt3}2\right)+81=

25x^2+25r^2=4x^2+(8r\sqrt3+36)x+(12r^2+36\sqrt3r+81)

21x^2-(8r\sqrt3+36)x+(13r^2-36r\sqrt3-81)=0

Then use quadratic formula to obtain $x$ in term of $r$ . Choose smaller value of $x$ to get larger value of $\tan\frac\theta2=\frac rx$ . The smaller one is

x=\frac{4r\sqrt3+18}{21}-\frac{15\sqrt{-r^2+4r\sqrt3+9}}{21}

20 points

Find the expression of $\tan\frac\theta2$ . Hence

\tan\frac\theta2=\frac rx=\frac{21r}{4r\sqrt3+18-15\sqrt{-r^2+4r\sqrt3+9}}

10 points

Cartesian moon-at-horizon geometry with x,r,d0 and alpha0=theta/2.

Earlier image edition — marking scheme

Step	Points
1 Notice the standard version of the orbits	10
2 Established relation between coordinates before and after transformation	20
3 Obtain expression for calculating distance from a point on an ellipse to the foci	30
4 Identify and characterized the coordinates when the planet looks largest.	20
5 Solve equation to obtain value of $x$ .	15
6 Find the expression of $\tan\frac\theta2$ .	5

The ellipse may be obtained from a standard ellipse

\frac{(x^*-4)^2}{25}+\frac{(y^*)^2}{9}=1

by rotating the standard ellipse with respect to the origin, counterclockwise, (by $\pi/3$ radians). Then $c=\sqrt{a^2-b^2}=\sqrt{16}=4$ .

For any point $(x,y)$ on the ellipse, let $(x^*,y^*)$ be its coordinates before the transformation. From the equation of the ellipse, we can easily see that

x^*=\frac x2+\frac{\sqrt3y}2,\qquad y^*=-\frac{\sqrt3x}2+\frac y2

\begin{bmatrix}x^*\\y^*\end{bmatrix}=\begin{bmatrix}\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}\frac x2+\frac{\sqrt3y}2\\-\frac{\sqrt3x}2+\frac y2\end{bmatrix}

Earlier image edition — original page 1

Complete original earlier-edition key page 1 of 10, preserving draft questions, diagrams and marking weights.

Earlier image edition — original page 2

Complete original earlier-edition key page 2 of 10, preserving draft questions, diagrams and marking weights.

Earlier image edition — original page 3

Complete original earlier-edition key page 3 of 10, preserving draft questions, diagrams and marking weights.

Оригинал в Архива: Long_problem_Final_BG.pdf · официални решения: Long Problem Solution-Marking Scheme [Revised] - 310715 - 13.58.pdf · допълнителни данни: long_sol_1.png · допълнителни данни: long_sol_2.png · допълнителни данни: long_sol_3.png · допълнителни данни: long_sol_4.png · допълнителни данни: long_sol_5.png · допълнителни данни: long_sol_6.png · допълнителни данни: long_sol_7.png · допълнителни данни: long_sol_8.png · допълнителни данни: long_sol_9.png · допълнителни данни: long_sol_10.png · допълнителни данни: 9th IOAA Constants Table 20150729.pdf · допълнителни данни: constants.pdf · допълнителни данни: instructions.docx