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Условие Отговори Решение

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

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Всички материали от IOAA 2014

8th International Olympiad on Astronomy and Astrophysics — Theoretical Test, 12 short problems (final edition)

Указания

In your folder you will find the following items:

a. Answer sheets

b. Rough work sheets

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b. In „Student ID” – fill in your ID that you will find on your envelope, consisting of 3 letters and 2 digits.

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Marking scheme

ALL THE CORRECT SOLUTIONS WHICH ARE DIFFERENT FROM THE AUTHOR’S SOLUTION WILL BE EVALUATED AND MARKED ACORDINGLY TO THE MARKING SCHEME.

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 $T_0=1\ \mathrm{year}$ 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 (aphelion)

b) 3rd of January.

Отговори

Решение

Problem 2. Marking scheme Sun gravitational catastrophe!

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

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.

For following explanation

a) In 3rd 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 $r_{0,\max}=a_0(1+e_0)$ .. 1p

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

According to Keppler’s second law and the law of energy conservation 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(1-e_0)

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

KM_0=v_0^2r_0=v_0^2a_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\mathbf{(1)}

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

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

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

For the new elliptical Earth orbit:

r_{per}=r_{0,aph};\quad r_{\min}=r_{per}=a(1-e);

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

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

e=1-2e_0;\quad 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},\quad 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}\sqrt2\sqrt{\frac{1+e_0}{2e_0}};

T=T_0\sqrt2\left(\frac{1+e_0}{2e_0}\right)^{3/2}\approx238\ \mathrm{years}

b) In 3rd 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 1p

Оригинал в Архива: Short Theoretical.pdf · официални решения: short_final_sol.pdf · допълнителни данни: ioaa1.jpg · допълнителни данни: ioaa2.jpg · допълнителни данни: ioaa3.jpg · допълнителни данни: ioaa4.jpg · допълнителни данни: ioaa5.jpg · допълнителни данни: ioaa6.jpg · допълнителни данни: ioaa7.jpg