Variable gravity At Earths surface, the acceleration due

Chapter 6, Problem 68

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Variable gravity At Earths surface, the acceleration due to gravity is approximately g = 9.8 m>s2 (with local variations). However, the acceleration decreases with distance from the surface according to Newtons law of gravitation. At a distance of y meters from Earths surface, the acceleration is given by a1y2 = - g 11 + y>R22, where R = 6.4 * 106 m is the radius of Earth. a. Suppose a projectile is launched upward with an initial velocity of v0 m>s. Let v1t2 be its velocity and y1t2 its height (in meters) above the surface t seconds after the launch. Neglecting forces such as air resistance, explain why dv dt = a1y2 and dy dt = v1t2. b. Use the Chain Rule to show that dv dt = 1 2 d dy 1v22. c. Show that the equation of motion for the projectile is 1 2 d dy 1v22 = a1y2, where a1y2 is given previously. d. Integrate both sides of the equation in part (c) with respect to y using the fact that when y = 0, v = v0. Show that 1 2 1v2 - v20 2 = g R a 1 1 + y>R - 1b. e. When the projectile reaches its maximum height, v = 0. Use this fact to determine that the maximum height is ymax = Rv0 2 2gR - v20. f. Graph ymax as a function of v0. What is the maximum height when v0 = 500 m>s, 1500 m>s, and 5 km>s? g. Show that the value of v0 needed to put the projectile into orbit (called the escape velocity) is 12gR. Additional Exercises