A worked-problem walkthrough covering orbital mechanics and gravitation: comparing orbital speeds at perihelion and aphelion using angular momentum conservation, finding an asteroid's orbital period from Kepler's Third Law, computing gravitational force between two masses with Newton's law, and deriving the Sun's mass from Earth's orbital data. Each problem presents the question, relevant formula, and step-by-step numerical solution. Useful for physics students reviewing orbital mechanics and gravitation calculations for exams or homework.
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Scene 1On-Screen Title: Problem 1: Comet Speed RatioQuestion: A comet moves in an elliptical orbit such that its distance from the Sun at the furthest point is $5.0$ times its distance at the closest point. Find the ratio of the comet's speed at the furthest point to its speed at the closest point. Answer:By Kepler's Second Law (conservation of angular momentum): $r_1 v_1 = r_2 v_2$ Ratio: $\frac{v_{furthest}}{v_{closest}} = \frac{r_{closest}}{r_{furthest}} = \frac{1}{5.0} = \mathbf{0.20}$ Scene 2On-Screen Title: Problem 2: Asteroid Orbital PeriodQuestion: An asteroid orbits the Sun with a semi-major axis of $4.0\text{ AU}$. Find its orbital period $T$ in years. Answer:By Kepler's Third Law ($T^2 = a^3$): $T = \sqrt{a^3} = \sqrt{4.0^3} = \sqrt{64} = \mathbf{8.0\text{ years}}$ Scene 3On-Screen Title: Problem 3: Gravitational Force Between Two MassesQuestion: Two bodies, each of mass $60\text{ kg}$, are placed at a distance of $3.0\text{ m}$. Find the gravitational force $F$ between these two bodies ($G = 6.7 \times 10^{-11}\text{ N}\cdot\text{m}^2\cdot\text{kg}^{-2}$). Answer:Formula: $F = \frac{G m_1 m_2}{r^2}$ Calculation: $F = \frac{(6.7 \times 10^{-11}) \times 60 \times 60}{(3.0)^2} = \mathbf{2.68 \times 10^{-8}\text{ N}}$ Scene 4On-Screen Title: Problem 4: Calculating the Mass of the SunQuestion: Calculate the mass of the Sun, given that Earth's orbital period around the Sun is $3.156 \times 10^7\text{ s}$ and the semi-major axis of Earth's orbit is $1.496 \times 10^{11}\text{ m}$ ($G = 6.7 \times 10^{-11}\text{ N}\cdot\text{m}^2\cdot\text{kg}^{-2}$). Answer:From Kepler's Third Law: $M = \frac{4\pi^2 a^3}{G T^2}$ Calculation: $M = \frac{4 \times (3.14)^2 \times (1.496 \times 10^{11})^3}{(6.7 \times 10^{-11}) \times (3.156 \times 10^7)^2} \approx \mathbf{1.99 \times 10^{30}\text{ kg}}$ IMPORTANT LANGUAGE RULE: All on-screen text must be in Arabic EXCEPT mathematical equations, formulas, laws, symbols, variables, and numerical expressions. Keep every equation, formula, and law exactly as provided, without translating or changing it. Translate only the explanatory text, headings, labels, and descriptions into Arabic. Arabic text must use proper RTL direction and correct Arabic typography. Do not use English text anywhere else on the screen. Audience: student