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USAAAO 2022, II кръг — Задача 7

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Всички материали от USAAAO 2022

2022 National Astronomy Competition · 30 т.

Instructions (Please Read Carefully)

The top 5 eligible scorers on the NAC will be invited to represent USA at the next IOAA. In order to qualify for the national team, you must be a high school student with US citizenship or permanent residency.

This exam consists of 3 parts: Short Questions, Medium Questions and Long Questions.

The maximum number of points is 200.

The test must be completed within 2.5 hours (150 minutes).

Please solve each problem on a blank piece of paper and mark the number of the problem at the top of the page. The contestant’s full name in capital letters should appear at the top of each solution page. If the contestant uses scratch papers, those should be labeled with the contestant’s name as well and marked as “scratch paper” at the top of the page. Scratch paper will not be graded. Partial credit will be available given that correct and legible work was displayed in the solution.

This is a written exam. Contestants can only use a scientific or graphing calculator for this exam. A table of physical constants will be provided. Discussing the problems with other people is strictly prohibited in any way until the end of the examination period on March 26th. Receiving any external help during the exam is strictly prohibited. This means that the only allowed items are: a calculator, the provided table of constants, a pencil (or pen), an eraser, blank sheets of papers, and the exam. No books or notes are allowed during the exam. Exam is proctored and recorded. You are expected to have your video on at all times.

Условие

A meteorite that is radially approaching the Earth collides with a space station that revolves around the Earth in a circular orbit with radius $R$ . For all parts of the question, express your results in terms of the mass $M$ of the Earth, the gravitational constant $G$ , the mass $m_1$ of the meteorite, and the mass $m_2$ of the space station.

(a) Assume that, after the impact, the meteorite and the space station form a conglomerate that moves in a closed orbit which approaches the center of the Earth at a minimum distance $R/2$ . State what the shape of the orbit of the conglomerate is and determine:

(i) the speed of the meteorite just before the collision,

(ii) the minimum and maximum speeds of the conglomerate,

(iii) the maximum distance of the conglomerate from the center of the Earth.

Determine the condition that $m_1$ and $m_2$ must obey so that the aforementioned scenario is possible.

(b) Determine the minimum speed that the meteorite should have just before the collision so that the conglomerate moves in an open orbit after the impact. For this minimum value of the speed of the meteorite, state what the shape of the orbit of the conglomerate would be and determine

(i) the maximum speed of the conglomerate,

(ii) its minimum distance from the center of the Earth,

(iii) the angle traversed by the orbital radius from the moment of the collision until the moment when the conglomerate approaches the center of the Earth to minimum distance.

Отговори

Решение

(a) The orbit of the conglomerate after the collision will be an ellipse.
Conservation of momentum implies that the radial and tangential components ( $v_r$ and $v_t$ , respectively) of the speed of the conglomerate are:

v_r = \frac{m_1}{m_1 + m_2} v

and

v_t = \frac{m_2}{m_1 + m_2} \sqrt{\frac{GM}{R}}.

Here, $v$ is the speed of the meteorite prior to the collision.
Conservation of angular momentum of the conglomerate during its orbital motion implies

\frac{m_2}{m_1 + m_2} \sqrt{\frac{GM}{R}} R = v_{max} \frac{R}{2}.

Therefore, we deduce that

\boxed{v_{max} = \frac{2m_2}{m_1 + m_2} \sqrt{\frac{GM}{R}}}.

Finally, conservation of mechanical energy per unit mass of the conglomerate yields

\frac{1}{2} v_{max}^2 - \frac{GM}{R/2} = \frac{1}{2}(v_r^2 + v_t^2) - \frac{GM}{R}.

Combining all previous equations, we find

\left(\frac{2m_2}{m_1 + m_2}\right)^2 \frac{GM}{R} - \frac{4GM}{R} = \left(\frac{m_1}{m_1 + m_2}\right)^2 v^2 + \left(\frac{m_2}{m_1 + m_2}\right)^2 \frac{GM}{R} - \frac{2GM}{R},

which immediately yields

\boxed{v = \frac{\sqrt{3m_2^2 - 2(m_1 + m_2)^2}}{m_1} \sqrt{\frac{GM}{R}}},

provided that

\sqrt{3} m_2 > \sqrt{2}(m_1 + m_2) \Leftrightarrow \boxed{m_1 < \left(\sqrt{\frac{3}{2}} - 1\right) m_2}.

Now, one can find the semi-major axis of the orbit of the conglomerate:

\frac{1}{a} = \frac{2}{r_{min}} - \frac{v_{max}^2}{GM} = \frac{4}{R}\left(1 - \frac{m_2^2}{(m_1 + m_2)^2}\right) \Leftrightarrow a = \frac{R}{4} \frac{(m_1 + m_2)^2}{(m_1 + m_2)^2 - m_2^2}.

Moreover, if we denote the eccentricity of the orbit by $e$ , we find

a(1 - e) = \frac{R}{2} \Leftrightarrow ae = a - \frac{R}{2}.

It follows that

r_{max} = a(1 + e) = 2a - \frac{R}{2} \Leftrightarrow \boxed{r_{max} = \frac{R}{2} \frac{m_2^2}{(m_1 + m_2)^2 - m_2^2}}.

Conservation of angular momentum immediately yields

v_{min} = v_{max} \frac{r_{min}}{r_{max}} \Leftrightarrow \boxed{v_{min} = \frac{2m_1(m_1 + 2m_2)}{m_2(m_1 + m_2)} \sqrt{\frac{GM}{R}}}.

(b) The orbit of the conglomerate will have the shape of a parabola. This scenario will occur when

v_r^2 + v_t^2 = \frac{2GM}{R} \Leftrightarrow \boxed{v = \frac{\sqrt{2(m_1 + m_2)^2 - m_2^2}}{m_1} \sqrt{\frac{GM}{R}}}.

Conservation of mechanical energy gives

v_{max}^2 = \frac{2GM}{r_{min}},

while conservation of angular momentum implies

v_{max} r_{min} = \frac{m_2}{m_1 + m_2} \sqrt{\frac{GM}{R}} R.

Therefore,

\boxed{v_{max} = \frac{2(m_1 + m_2)}{m_2} \sqrt{\frac{GM}{R}}}

and

r_{min} = \frac{m_2^2}{2(m_1 + m_2)^2} R.

From the equation of a parabola in polar coordinates,

r = \frac{2r_{min}}{1 + \cos\theta},

one can find

\frac{1 + \cos\theta}{2} = \cos^2\frac{\theta}{2} = \frac{r_{min}}{r} \Leftrightarrow \theta = 2\arccos\sqrt{\frac{r_{min}}{R}} = 2\arccos\frac{m_2}{\sqrt{2}(m_1 + m_2)}.

Оригинал в Архива: usaaao_second_exam_2022_problems.pdf · официални решения: usaaao_second_exam_2022_solutions.pdf