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IOAA 2014, Теория, предварителен вариант — Задача 2. Sun gravitational catastrophe!

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Всички материали от IOAA 2014

8th International Olympiad on Astronomy and Astrophysics, Suceava - Gura Humorului - August 2014 — THEORETICAL TEST, 15 short problems

Indications

The problems were elaborated concerning two aspects: a) To cover merely all the subjects from the syllabus;
b) The average time for solving the items is about 15 minutes per a short problem;
In your folder you will find out the following: a) Answer sheets
b) Draft sheets
c) The envelope with the subjects in English and the translated version of them in your mother tongue;
The solutions of the problems will be written down only on the answer sheets you receive on your desk. PLEASE WRITE ONLY ON THE PRINTED SIDE OF THE PAPER SHEET. DON’T USE THE REVERSE SIDE. The evaluator will not take into account what is written on the reverse of the answer sheet.
The draft sheets is for your own use to try calculation, write some numbers etc. BEWARE: These papers are not taken into account in evaluation, at the end of the test they will be collected separately . Everything you consider as part of the solutions have to be written on the answer sheets.
Each problem have to be started on a new distinct answer sheet.
On each answer sheet please fill in the designated boxes as follows: a) In PROBLEM NO. box write down only the number of the problem: i.e. 1 – 15 for short problems, 16 – 19 for long problems. Each sheet containing the solutions of a certain problem, should have in the box the number of the problem;
b) In Student ID – fill in your ID you will find on your envelope, consisted of 3 letters and 2 digits.
c) In page no. box you will fill in the number of page, starting from 1. We advise you to fill this boxes after you finish the test
We don’t understand your language, but the mathematic language is universal, so use as more relationships as you think that your solution will be better understand by the evaluator. If you want to explain in words we kindly ask you to use short English propositions.
Use the pen you find out on the desk. It is advisable to use a pencil for the sketches.
At the end of the test: a) Don’t forget to put in order your papers;
b) Put the answer sheets in the folder 1. Please verify that all the pages contain your ID, correct numbering of the problems and all pages are in the right order and numbered. This is an advantage of ease of understanding your solutions.
c) Verify with the assistant the correct number of answer sheets used fill in this number on the cover of the folder and sign.
d) Put the draft papers in the designated folder, Put the test papers back in the envelope.
e) Go to swim

GOOD LUCK !

Условие

In a gravitational catastrophe, the mass of the Sun mass decrease instantly to half of its actual value. If you consider that the actual Earth orbit is elliptical, its orbital period is $T_0 = 1\ \mathit{year}$ and the eccentricity of the Earth orbit is $e_0 = 0{,}0167$ .

Find the period of the Earth`s orbital motion, after the gravitational catastrophe, if it occurs on: a) 3rd of July b) 3rd of January.

Отговори

Решение

Problem 2. Marking scheme Sun gravitational catastrophe!

Correct analyze of the initial conditions when the catastrophe occurs ( A) — 5 points
Correct calculations (B) — 5 points
- Correct use of laws of conservation — 2 points
- Finding out that in the first case the orbit will be elliptic, relations (1) and (2) — 2 points
- Correct conduct of calculations — 1 point

Detailed solution

(A) The orbit of Earth is elliptical, so the shape of the orbit after the solar catastrophe will depend on the moment when the decrees of the mass of the Sun will occur.

Initial analysis of the problem

a) In $3^{\mathrm{rd}}$ July the Earth is at the aphelion. The speed of the Earth is smaller than the speed of Earth on a circular orbit with radius $r_{0,\max} = a_0\left(1 + e_0\right)$ ..

b) In $3^{\mathrm{rd}}$ January the Earth is at perihelion. The speed of the Earth is bigger than the speed of Earth on a circular orbit with radius $r_{0,\max} = a_0\left(1 - e_0\right)$ ..

Conclusion (A) the period should be calculated only for situation a). The expected trajectory in this case is an elliptic one. The possibility that Earth hit the Sun is available too.

Initial small elliptical Earth orbit around the Sun S₀; M₀ with velocities v₀,per and v₀,aph at perihelion and aphelion, labelled “Initial Earth orbit”; a much larger dashed ellipse labelled “Final Earth orbi[t]” around S; M with perihelion at the old aphelion, r₀,max = r₀,aph = r_per and r_min marked, and velocity v_per at the new perihelion. — Figure 2.1

(B) Calculations:

In $3^{\mathrm{rd}}$ July the distance from Sun is maximum: fig. 2.1,

r_{0,\max} = a_0\left(1 + e_0\right).

Before the catastrophe:

$v_{0,aph}$ - the speed of Earth on aphelion,

$a_0$ - big Earth’s elliptical orbit semi axis

$v_0$ - the speed of Earth if its orbit is circular with radius $r_0 = a_0$

According to Keppler’s second law and the law of energy conservation (see figure 2.1) the following relations can be written :

v_{0,per}\, r_{0,per} = v_{0,aph}\, r_{0,aph};

\frac{v_{0,per}^2}{2} - K\frac{M_0}{r_{0,per}} = \frac{v_{0,aph}^2}{2} - K\frac{M_0}{r_{0,aph}}

r_{0,\min} = r_{0,per} = a_0\left(1 - e_0\right)

r_{0,\max} = r_{0,aph} = a_0\left(1 + e_0\right)

KM_0 = v_0^2 r_0 = v_0^2 a_0

v_0 = \sqrt{K\frac{M_0}{r_0}} = \sqrt{K\frac{M_0}{a_0}}

v_{0,per} = v_0\sqrt{\frac{1 + e_0}{1 - e_0}} > v_0

v_{0,per} > v_0 \qquad \textbf{(1)}

v_{0,aph} = v_0\sqrt{\frac{1 - e_0}{1 + e_0}}

v_{0,aph} < v_0 \qquad \textbf{(2)}

Conclusion – According to the relations (1) and (2) the new orbit of the Earth could be an elliptic one.

For the new elliptical Earth orbit:

r_{per} = r_{0,aph};

r_{\min} = r_{per} = a\left(1 - e\right);

a_0\left(1 + e_0\right) = a\left(1 - e\right);\ a = a_0\frac{1 + e_0}{1 - e};

v_{per} = v_{0,aph};

v_{per} = v\sqrt{\frac{1 + e}{1 - e}},

Where v este is the Earth’s speed on a circular orbit with the radius $r = a$ , when the mass of the Sun becomes

M = M_0/2;

v\sqrt{\frac{1 + e}{1 - e}} = v_0\sqrt{\frac{1 - e_0}{1 + e_0}};

v = \sqrt{K\frac{M}{r}} = \sqrt{K\frac{M_0}{2a}} = \sqrt{K\frac{M_0}{a_0}}\sqrt{\frac{a_0}{2a}} = v_0\sqrt{\frac{a_0}{2a}};

e = 1 - 2e_0;\ a = a_0\frac{1 + e_0}{2e_0}.

Conclusion

T_0 = \frac{2\pi r_0}{v_0} = \frac{2\pi a_0}{v_0}

T = \frac{2\pi r}{v} = \frac{2\pi a}{v};

\frac{T}{T_0} = \frac{a}{a_0}\frac{v_0}{v} = \frac{1 + e_0}{2e_0}\sqrt{\frac{2a}{a_0}} = \frac{1 + e_0}{2e_0}\sqrt{2}\sqrt{\frac{1 + e_0}{2e_0}};

T = T_0\sqrt{2}\left(\frac{1 + e_0}{2e_0}\right)^{3/2} \approx 230\ \mathrm{years}

b) In $3^{\mathrm{rd}}$ of January the Earth is at perihelion. In that moment the Erath speed is larger than the speed necessary for an Earth’s circular orbit. Thus the trajectory of the Earth after the catastrophe will be an open trajectory, i.e. an hyperbolic or parabolic orbit.

Conclusion it is not necessary to calculate the period of revolution or could be issued as infinite

Оригинал в Архива: th_short_ver1.pdf · официални решения: th_short_ver1_sol.pdf