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IOAA Jr. 2022, Теория, втора дълга задача — Задача 2

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOAA Jr. 2022

Problem 2. The rocket that destroys the threatening asteroid

Условие

With the help of a rocket, R, launched from a spaceship, N, an asteroid, A, must be destroyed, approaching the Earth threateningly, coming right in the direction of the center of the Earth, as shown in the drawing in Figure 1. There are positions of the ship spacecraft, N and the asteroid A, respectively, at the time of the launch of the rocket R, to meet the asteroid A, when the speed of the spacecraft, N, relative to Earth, is constant,, oriented along the spacecraft line - asteroid, and the speed of asteroid A, relative to Earth, is considered constant, $\vec{v}_N$ , oriented along the spacecraft line - asteroid, and the speed of asteroid A, relative to Earth, is considered constant, $\vec{v}_A$ , its direction forming an angle α with the direction of the spacecraft, N.

During the flight of the rocket, from its launch to the impact with the asteroid, it is considered that the Earth, relative to the Sun, is at rest, and the movements of the rocket, R, the spacecraft, N, and the asteroid, A, are rectilinear and uniform movements.

Diagram showing Earth E at top, rocket R launched from spacecraft N with velocity vector v_RN at angle β above the spacecraft's velocity direction v_N, dashed line of length L connecting N to asteroid A, and asteroid A's velocity vector v_A forming angle α with line L, pointing toward Earth. — Fig. 1

Knowing that the direction of the rocket launching R was chosen in such a way that the duration of the rocket's movement, R, from launch to impact with asteroid A, is minimal, determine the angle $\beta$ , at which the rocket R was launched, relative to the direction of movement spacecraft, N;

Отговори

Решение

Reaching the target, asteroid A, by the rocket R, is done in a minimum time, if the direction of absolute displacement of the rocket R, ie the direction of movement NC, is perpendicular to the direction of absolute movement of asteroid A, ie the direction of movement AC, as indicated the drawing in Figure 2, which shows:

D = v_A \cdot t_{\min} = L\cdot\cos\alpha;

d = v_R \cdot t_{\min} = L\cdot\sin\alpha,

Diagram with Earth E, point C reached by both rocket and asteroid, right angle at C between segment d (rocket path NC, length v_R·t) and segment D (asteroid path AC, length v_A·t); angle β between v_RN and v_N at N, angle α at A between v_A and line L; angle (β−90°) and (90°−α) marked with dashed auxiliary perpendicular lines at N. — Fig. 2

where $v_R$ is the absolute speed of the R rocket, from the moment of its launch;

t_{\min} = \dfrac{L\cdot\cos\alpha}{v_A};\quad t_{\min} = \dfrac{L\cdot\sin\alpha}{v_R};

\dfrac{L\cdot\cos\alpha}{v_A} = \dfrac{L\cdot\sin\alpha}{v_R};\quad \dfrac{\cos\alpha}{v_A} = \dfrac{\sin\alpha}{v_R};

v_R = v_A\cdot\tan\alpha,

where $v_R$ is the absolute speed of the rocket, after launch;

\vec{v}_{RN} = \vec{v}_R - \vec{v}_A,

where $\vec{v}_{RN}$ is the relative speed of the rocket R, relative to the spacecraft N;

v_{RN}^2 = v_R^2 + v_A^2 - 2\cdot v_R\cdot v_N\cdot\cos(90^0-\alpha);

v_{RN}^2 = v_R^2 + v_A^2 - 2\cdot v_R\cdot v_N\cdot\sin\alpha;

v_R\cdot\cos\alpha = v_{RN}\cdot\cos(\beta-90^0),

where $\beta$ is the angle at which the rocket R is launched from the spacecraft, N, relative to the flight direction of the spacecraft, N;

v_R\cdot\cos\alpha = v_{RN}\cdot\sin\beta;

\sin\beta = \dfrac{v_R}{v_{RN}}\cdot\cos\alpha;

v_R = v_A\cdot\tan\alpha;

\sin\beta = \dfrac{v_A\cdot\tan\alpha}{v_{RN}}\cdot\cos\alpha = \dfrac{v_A}{v_{RN}}\cdot\dfrac{\sin\alpha}{\cos\alpha}\cdot\cos\alpha;

\sin\beta = \dfrac{v_A}{v_{RN}}\cdot\sin\alpha;

v_{RN}^2 = v_R^2 + v_A^2 - 2\cdot v_R\cdot v_N\cdot\sin\alpha;

v_R = v_A\cdot\tan\alpha;

v_{RN}^2 = v_A^2\cdot\dfrac{\sin^2\alpha}{\cos^2\alpha} + v_A^2 - 2\cdot v_A\cdot\dfrac{\sin\alpha}{\cos\alpha}\cdot v_N\cdot\sin\alpha;

v_{RN}^2 = v_A^2\cdot\left(\dfrac{\sin^2\alpha}{\cos^2\alpha}+1\right) - 2\cdot v_A\cdot\dfrac{\sin^2\alpha}{\cos\alpha}\cdot v_N;

v_{RN}^2 = v_A^2\cdot\dfrac{1}{\cos^2\alpha} - 2\cdot v_A\cdot v_N\cdot\dfrac{\sin^2\alpha}{\cos\alpha};

v_{RN} = v_A\cdot\sqrt{\dfrac{1}{\cos^2\alpha} - 2\cdot\dfrac{v_N}{v_A}\cdot\dfrac{\sin^2\alpha}{\cos\alpha}};

v_{RN} = v_A\cdot\sqrt{\dfrac{1}{\cos^2\alpha} - 2\cdot\dfrac{v_N}{v_A}\cdot\dfrac{\cos\alpha\cdot\sin^2\alpha}{\cos^2\alpha}};

v_{RN} = \dfrac{v_A}{\cos\alpha}\cdot\sqrt{1-2\cdot\dfrac{v_N}{v_A}\cdot\sin^2\alpha\cdot\cos\alpha},

representing the relative speed of the rocket R, relative to the spacecraft N;

\sin\beta = \dfrac{v_R}{v_{RN}}\cdot\cos\alpha;\quad v_R = v_A\cdot\tan\alpha;\quad v_{RN} = \dfrac{v_A}{\cos\alpha}\cdot\sqrt{1-2\cdot\dfrac{v_N}{v_A}\cdot\sin^2\alpha\cdot\cos\alpha};

\sin\beta = \dfrac{v_A\cdot\dfrac{\sin\alpha}{\cos\alpha}}{\dfrac{v_A}{\cos\alpha}\cdot\sqrt{1-2\cdot\dfrac{v_N}{v_A}\cdot\sin^2\alpha\cdot\cos\alpha}}\cdot\cos\alpha;

\sin\beta = \dfrac{\sin\alpha\cdot\cos\alpha}{\sqrt{1-2\cdot\dfrac{v_N}{v_A}\cdot\sin^2\alpha\cdot\cos\alpha}},

$\beta$ being the angle at which the rocket R is launched from the spacecraft, N, relative to the direction of flight of the spacecraft, N, to destroy the asteroid.

Оригинал в Архива: Second_long_problem.pdf