Съдържание

IAO 2004, Всички кръгове — Задача 6. α6. Eclipse

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IAO 2004

IX International Astronomy Olympiad · 1 октомври 2004 г.

Внимание: Бележка към темата

Old photocopy scan. Two short passages of the official solutions are lost where the photocopied pages join (problem 4 between pages 4 and 5, the Sunrise solution between pages 13 and 14). The sky maps of the observational round are not in the archive file.

Условие

For the reconstruction of historical events, dates and places of the past solar eclipses are often calculated. However, the rotation of the Earth continually slows down and solar eclipses take place not there, where they should be in agreement with calculations on a model with constant rotation of the Earth with the present angular rate. In what century did such an eclipse take place in Paris ( $\lambda = 0^{\circ}$ , $\varphi = 45^{\circ}$ ) instead of Crimea ( $\lambda = 34^{\circ}$ , $\varphi = 45^{\circ}$ )? The length of a day increases by 0.0016 seconds per 100 years due to that rotation of the Earth slows down.

Отговори

Решение

The mechanical analogy for the process is a race of two trains. One of them moves with a constant velocity (moving without deceleration, hypothetical Earth), and the second one – with the constant acceleration "a" (as the real Earth). If the trains came to one point with identical speeds, and the acceleration of the second train is equal "a", in the past (time t ago) the distance between them was $S = at^2/2$ . The analog of S in our case is $\Delta\lambda = 360^{\circ} - 34^{\circ} = 326^{\circ}$ or $34/360 = 0.906$ revolutions of the planet. (In the opposite case $\Delta\lambda = 34^{\circ}$ or $34/360 = 0.094$ revolutions of the planet.) If S will be used in connection with revolutions of the planet (i.e. in days), and time is given in years, $a = 0.0016\ \mathrm{sec}/(\mathrm{day}\cdot100\ \mathrm{years})$ should be in units day/yrs $^2$ . Evidently for this purpose it is required to divide it by the number of seconds in a day and to multiply with the number of days in one year:

a = 0.0016\ \mathrm{sec}/(1\ \mathrm{day}\cdot100\ \mathrm{yrs}) = (1.6\cdot10^{-3}\cdot365.24)/(100\cdot86400)\ \mathrm{day/yrs^2} = 6.76\cdot10^{-8}\ \mathrm{day/yrs^2}.

Now we easily find the time, for which the difference 0.906 revolutions will be accumulated:

t = (2S/a)^{1/2} = (0.189\ \mathrm{day}/6.76\cdot10^{-8}\ \mathrm{day/yrs^2})^{1/2} \approx 5180\ \mathrm{years},

the 32-th century B.C.

(For the case $\Delta\lambda = 34^{\circ}$ $t = (2S/a)^{1/2} = (0.189\ \mathrm{day}/6.76\cdot10^{-8}\ \mathrm{day/yrs^2})^{1/2} \approx 1670$ years, the fourth century.)

Answer: in the 32-th century B.C. (in the fourth century).

Another way of middle part of the solution. If S will be used in degrees (i.e. in units of 1/360 revolution of the planet = 1/360 day), and time is given in years, $a = 0.0016\ \mathrm{sec}/(\mathrm{day}\cdot100\ \mathrm{years})$ should be in units degree/yrs $^2$ . Evidently for this purpose it is required to divide it by the number of seconds for one degree of the Earth rotation and to multiply with the number of days in one year:

a = 0.0016\ \mathrm{sec}/(1\ \mathrm{day}\cdot100\ \mathrm{yrs}) = (1.6\cdot10^{-3}\cdot365.24)/(100\cdot240)\ {}^{\circ}/\mathrm{yrs^2} = 2.43\cdot10^{-5}\ {}^{\circ}/\mathrm{yrs^2}.

Now we easily find the time, for which the difference 326° will be accumulated:

t = (2S/a)^{1/2} = (652^{\circ}/2.43\cdot10^{-5}\ {}^{\circ}/\mathrm{yrs^2})^{1/2} \approx 5180\ \mathrm{years}.

(For the case $\Delta\lambda = 34^{\circ}$ $t = (2S/a)^{1/2} = (68^{\circ}/2.43\cdot10^{-5}\ {}^{\circ}/\mathrm{yrs^2})^{1/2} \approx 1670$ years.)

Group α, theoretical round

Data from the “Table of planetary data” may be used for the solving of every problem.
And the White Bear and Penguin are waiting for you on the practical round.

Group β, theoretical round

Data from the “Table of planetary data” may be used for the solving of every problem.
And the White Bear and Penguin are waiting for you in the practical round.

Parameters of orbits. Physical characteristics of some planets, Moon and Sun

Элементы орбит. Физические характеристики некоторых планет, Луны и Солнца

Body, planet	Average distance to central body, in astr. units	Average distance to central body, in mln. km	Sidereal (or analogous) period, in trop. years	Sidereal (or analogous) period, in days	Eccentricity e	Equat. diameter, km	Mass, $10^{24}$ kg	Av. density, g/cm $^3$	Grav. acceler. at surf., m/s $^2$	Max. magn. from Earth **)	Albedo
Солнце Sun	$1{,}6\cdot10^{9}$	$2{,}5\cdot10^{11}$	$2{,}2\cdot10^{8}$	$8\cdot10^{10}$		1392000	1989000	1,409		$-26{,}8^{m}$
Меркурий Mercury	0,387	57,9	0,241	87,97	0,206	4 879	0,3302	5,43	3,70	$-2{,}2^{m}$	0,06
Венера Venus	0,723	108,2	0,615	224,70	0,007	12 104	4,8690	5,24	8,87	$-4{,}7^{m}$	0,78
Земля Earth	1,000	149,6	1,000	365,26	0,017	12 756	5,9742	5,515	9,81		0,36
Луна Moon	0,00257	0,38440	0,0748	27,3217	0,055	3 475	0,0735	3,34	1,62	$-12{,}7^{m}$	0,07

**) Для внешних планет и Луны – в среднем противостоянии.
**) For outer planets and Moon – in mean opposition.

Оригинал в Архива: iao2004.pdf