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Dynamics_Part6

Dynamics_Part6 - Problem Set 1 Problem 6 Problem Earth's...

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Unformatted text preview: Problem Set 1: Problem 6. Problem: Earth's gravitational acceleration on an object is a = -gR2 /r2 , where r is distance from Earth's center, R is Earth's radius and g is standard gravitational acceleration at the planet's surface. (a) Compute the velocity, v, of an object launched from the North Pole with initial velocity vo as a function of vo , g, r and R. (b) For what value of vo will the object escape Earth's gravitational pull, i.e., for what value of vo does v 0 as r ? (c) Determine the value of vo computed in Part (b). Earth's radius is R = 3960 mi. Express your answer in ft/sec. . .. . ......... .......... .. . . . R . .. ........... .......... . . . . . . ........................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... ....................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... . .. ... .................. . ...................... vo a . . v . .. . ................................................... ..................................................... . .. . . .. .. r ........................................................................................................ . . Solution: Because the problem does not involve time, the easiest approach is to begin with the differential equation relating acceleration and velocity, viz., a=v (a) For the given acceleration, we have - Then, integrating, we have v vo r dv dr = a dr = v dv gR2 dr = v dv r2 = v dv = -gR2 v=v dr r2 r=r r=R v dv = -gR2 R dr r2 = 1 2 v 2 = v=vo gR2 r or, 1 2 2 v - vo = gR2 2 1 1 - r R R r Thus, the particle's velocity as a function of its distance from Earth's center is given by 2 v 2 = vo - 2gR 1 - (b) The object will rise indefinitely if its velocity approaches zero only as r . From the solution of Part (a), we have 2 lim v 2 = vo - 2gR r ...
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