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Homework 2 Author: Student, Amazing A. September 26, 2016 Chapter 3 Problem 26 Earth is closer to the Sun in January than in July. Therefore, in accord with Kepler’s second law: (a) Earth travels faster in its orbit around the Sun in July than in January. (b) Earth travels faster in its orbit around the Sun in January than in July. (c) It is summer in January and winter in July. Kepler’s Second Law says that the planet sweeps through equal areas of the orbit at equal times. As the Earth is closer to the Sun in January, it must move faster then in order to satisfy this requirement. Therefore (b) is correct. Problem 48 Eris Orbit: The recently discovered Eris, which is slightly larger than Pluto, orbits the Sun every 557 years. What is its average distance (semi-major axis) from the Sun? How does its average distance compare to that of Pluto? Using p2 = a3 with the semi-major axis a in units of AU will give the period p in years. If the orbit if 557 years long, then (557)2 = a3 = 310249. Therefore, a = 67.7 AU. Pluto has a semi-major axis of 40 AU, nearly 60% smaller. Extra Credit: Problem 50 Halley Orbit. Halley’s Comet orbits the Sun every 76.0 years and has an orbital eccentricity of 0.97. a) Find its average distance from the Sun (semi-major axis). b) Find its perihelion and aphelion distances. To find the semi-major axis a, we can use the formula p2 = a3 (with units of years and AU). If the period of the comet is 76.0 years, a = p2/3 , which gives a = 17.9 AU. The perihelion and aphelion distances are calculated by d = a(1 − e) and d = a(1 + e), respectively. This gives dperi = 17.9(1 − 0.97) = 0.54 AU and daphe = 17.9(1 + 0.97) = 35.3 AU. 1 Chapter 4 Problem 26 Suppose you visited another planet. (a) Your mass and weight would be the same as they are on earth. (b) Your mass would be the same as on Earth, but your weight would be different. (c) Your weight would be the same as on Earth, but your mass would be different. Weight is the gravitational force pulling your mass to the planet. Your mass is invariant. Therefore, (b) is the correct choice. However, if the planet you visited was identical to the Earth, your weight could be the same as well. Problem 32 If Earth were twice as far from the Sun, the force of gravity attracting Earth to the Sun would be (a) twice as strong. (b) half as strong. (c) one quarter as strong. The force of gravity can be calculated using We begin with the equation Fg = G M 1 M2 d2 (1) If the masses remain constant, but the distance d is doubled Fg ∝ 1 1 = 2 d2 2 (2) , and so doubling the distance between the Earth and the Sun will give a Fg of (c) one quarter as strong. Problem 51 Find the orbital period for the planet in each case. Hint: The calculations for this problem are so simple that you will not need a calculator. (a) A planet with twice the Earth’s mass orbiting at a distance of 1 AU from a star with the same mass as the Sun. (b) A planet with the same mass as the Earth orbiting at a distance of 1 AU from a star with four times the Sun’s mass. We begin with the equation P2 = 4π 2 a3 G(M1 + M2 ) (3) Where a is the semi-major axis of the orbit and M1 , M2 refer to the two orbiting bodies (Sun and Earth in this case). The problem states that we need not use a calculator, and that a = 1 AU. We can simplify this by plugging in the numbers for the Earth-Sun orbit, which has an orbital period of 1 year. P2 = 4π 2 (1 AU )3 = 1 year G(MSun + MEarth ) 2 (4) The mass of the Earth M ≈ 6 × 1024 kg is essentially negligible compared to that of the Sun (M ≈ 2 × 1030 kg). We can rewrite the denominator into Mtotal = MSun + MEarth ≈ MSun to make things easier. The rest of the terms in the equation are constant. So, in units of years, we can write s 1 P ∝ (5) (Mtotal ) For problem (a), the book asks for a doubling of Earth’s mass. We know that even by doubling the mass of the Earth, the planet’s mass will be less than 5 orders of magnitude smaller, and therefore Mt otal will remain constant. As our period only depends on the combined mass of the system, which remains unchanged, the period will not be significantly altered. In (b), we are asked to compute the period of a system with a star that is 4 times more massive than our own Sun. So Mtotal = 4 × MSun + MEarth ≈ 4 × MSun , and s 1 = 0.5 years (6) P ∝ (4Mtotal ) By quadrupling the mass of the star the planet is orbiting, the period decreases by half. Chapter 5 Problem 28 Blue light has higher frequency than red light. Thus, blue light has (a) higher energy and shorter wavelength. (b) Higher energy and longer wavelength than red light. (c) Lower energy and shorter wavelength than red light. The correct answer is (a). Blue light has more energy (E) than red light. Frequency (ν) and wavelength (λ) are inversely proportional to one another, so if blue light has higher frequency, it will have a lower wavelength and larger energy. hc E= = hν (7) λ 3