Съдържание

IAO 2007, Теория, Група β, английски — Задача 5. Visit.

Оригинал на условието (изображение)Съобщи за грешка

Всички материали от IAO 2007

Theoretical round. Problems to solve · 29 септември 2007 г.

Бележка

Maybe not all problems have correct questions. Some questions (maybe the main question of the problem, maybe one of the subquestions) may have no real sense. In this case you have to write in your answer (in English or Russian): «situation is impossible – ситуация невозможна». Of course, this answer has to be explained numerically or logically.

Source availability

The numerical table of planetary data is not present in the32-file2007 archive inventory. The file bgtheo6.jpg contains only a translation of column headings, without numerical planetary rows. No table from another year has been substituted. The separately archived airplane/Moon photograph is included in full with question4.

Note for jury and team leaders

The proposed sketches are not full; the team leaders have to give more detailed explanations for students. But the correct solutions in the students’ papers (enough for8pts) may be shorter.

Source photograph damage

The second and fourth official solution photographs lose some initial letters or words at the left edge. Ellipses inside the affected solution text mark those source losses. Legible formulas, results and grading remarks are retained; missing letters have not been silently supplied. The solution for question4 contains no explanatory drawing, although the question requests one.

Printed source annotations

The galaxy key prints the same final logarithm1+HL₂/c in its classical and relativistic final lines, despite different numerical totals22.46 and22.82. The sail key repeats Δp and uses the printed multiplication/division notation. The source results and formulas are reproduced without scientific correction; punctuation and multiplication spacing are normalized only for mathematical rendering.

Условие

Extraterrestrial animals (animaloids) decided to colonize the Earth. The navigation devices of the spacecrafts of vituloids (lat. vitulus = seal = тюлень) needs that sometimes after the landing the Polar star ( $\alpha=2^{\mathrm h}32^{\mathrm m}$ , $\delta=89^\circ16'$ at the epoch of starting the colonization) must be exactly in Zenith and the devices of crocodiloids needs that sometimes the Polar Star must be exactly on the Horizon. The devices are placed on the top of the spacecrafts and these can land on the Earth only in vertical positions. Also every device has to be invisible directly for the device of any other spacecraft. Estimate, how many spacecrafts of every type of animaloids can land on the Earth. Consider the Earth to be spherical, the spacecrafts’ bottoms to be on its surface (may stay either on ground or on the water) and the height of every spacecraft is $h=10$ m.

Решение

Внимание: Непълно решение

The official solution photograph clips some initial letters/words at the left edge. All legible content is retained and ellipses mark the genuine source loss.

At first we have to find the minimum distance D between two spacecrafts. From the equation

R_E^2+L^2=(R_E+h)^2

where $R_E$ is the Earth’s radius, we may find the distance L to the visible horizon from the top of the spacecraft,

L=(2R_Eh)^{1/2}\approx11.3\ \mathrm{km}.

The minimum distance between two spacecrafts equals to double of this distance, i.e.

D=2L=(8R_Eh)^{1/2}\approx22.6\ \mathrm{km}.

The navigation devices of the spacecrafts of vituloids should stay at the circle with latitude exactly equal to the declination of the Polar Star, i.e. $\varphi=89^\circ16'$ . The length of this circle is

S=2\pi\sin(90^\circ-\varphi)R_E=512\ \mathrm{km}.

So an estimate of the possible number of vituloids spacecrafts is the largest integer number not exceeding

S/D=2\pi\sin(90^\circ-\varphi)R_E/(8R_Eh)^{1/2}\approx22.7.

Thus,

N_V=[S/D]=[2\pi\sin(90^\circ-\varphi)R_E/(8R_Eh)^{1/2}]=22.

$[S/D]$ means the largest integer number not exceeding $S/D$ .

(For a more accurate solution we have to measure the distance not along the circle but directly. The correcting coefficient would be $\sin\alpha/\alpha$ , $\alpha$ is half the angle between meridians of two nearest spacecrafts. The coefficient is too near to 1, $\alpha\approx8.2^\circ$ , $k=0.997$ , so the integer value $[S/D]$ is still equal to 22.)

And the navigation devices of the spacecrafts of crocodiloids should stay not at the circle but in whole belt near equator. The length of this belt is about the equator length, i.e. $A=2\pi R_E=40076$ km, and the width of the belt is twice the value $\sin(90^\circ-\varphi)R_E$ ,

B=2\sin(90^\circ-\varphi)R_E=163.3\ \mathrm{km}.

Been placed in triangular net, there may stay $[B/(D(3)^{1/2}/2)]+1$ lines with $[A/D]$ spacecrafts in every line; or $[B/D]+1$ lines with $[A/(D(3)^{1/2}/2)]$ spacecrafts in every line.

[B/D]=[2\sin(90^\circ-\varphi)R_E/(8R_Eh)^{1/2}]=[\sin(90^\circ-\varphi)(R_E/2h)^{1/2}]=7,

[B/(D(3)^{1/2}/2)]=8,

[A/D]=[2\pi R_E/(8R_Eh)^{1/2}]=[\pi(R_E/2h)^{1/2}]=1774,

[A/(D(3)^{1/2}/2)]=2048.

Calculations give:

([B/(D(3)^{1/2}/2)]+1)\times[A/D]=9\times1774=15\,966,

([B/D]+1)\times[A/(D(3)^{1/2}/2)]=8\times2048=16\,384.

The second number is larger so it should be taken as the answer.

Note: It is not written in the text at what horizon should be the Polar Star: real or mathematical. But the answer does not depend on it. It may only move the belt but not change (as first approximation) its square and so not change the number of spacecrafts.

…e for jury: The important difference between these two situations should be understood by students and accordingly evaluated: …ecrafts of vituloids placed in line and spacecrafts of crocodiloids placed in area.

Answers. 22 spacecrafts of vituloids and 16 384 spacecrafts of crocodiloids.

Data

Data from the table of planetary data may be used for solving every problem.

Оригинал в Архива: betatheo.jpg · официални решения: theosol1.jpg · допълнителни данни: theosol2.jpg · допълнителни данни: theosol3.jpg · допълнителни данни: theosol4.jpg · допълнителни данни: photo.JPG