121616949-math.222 - obeys an inverse square law F = k/r 2...

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208 Chapter 9 Applications of Integration and a ( t ) = v ( t ) = x ( t ). We can use v = x ( t ) as a substitution to convert the integral from “ dx ” to “ dv ” in the usual way, with a bit of cleverness along the way: dv = x ( t ) dt = a ( t ) dt = a ( t ) dt dx dx dx dt dv = a ( t ) dx v dv = a ( t ) dx. Substituting in the integral: W = - Z x 1 x 0 ma ( t ) dx = - Z v 1 v 0 mv dv = - mv 2 2 fl fl fl fl v 1 v 0 = - mv 2 1 2 + mv 2 0 2 . You may recall seeing the expression mv 2 / 2 in a physics course—it is called the kinetic energy of the object. We have shown here that the work done in moving the object from one place to another is the same as the change in kinetic energy. We know that the work required to move an object from the surface of the earth to infinity is W = Z r 0 k r 2 dr = k r 0 . At the surface of the earth the acceleration due to gravity is approximately 9.8 meters per second squared, so the force on an object of mass m is F = 9 . 8 m . The radius of the earth is approximately 6378.1 kilometers or 6378100 meters. Since the force due to gravity
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Unformatted text preview: obeys an inverse square law, F = k/r 2 and 9 . 8 m = k/ 6378100 2 , k = 398665564178000 m and W = 62505380 m . Now suppose that the initial velocity of the object, v , is just enough to get it to infinity, that is, just enough so that the object never slows to a stop, but so that its speed decreases to zero, i.e., so that v 1 = 0. Then 62505380 m = W =-mv 2 1 2 + mv 2 2 = mv 2 2 so v = √ 125010760 ≈ 11181 meters per second , or about 40251 kilometers per hour. This speed is called the escape velocity . Notice that the mass of the object, m , cancelled out at the last step; the escape velocity is the same for all objects. Of course, it takes considerably more energy to get a large object up to...
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