121616949-math.210 - W = Z r 1 r k r 2 dr =-k r fl fl fl...

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196 Chapter 9 Applications of Integration In reality few situations are so simple. The force might not be constant over the range of motion, as in the next example. EXAMPLE 9.13 How much work is done in lifting a 10 pound weight from the surface of the earth to an orbit 100 miles above the surface? Over 100 miles the force due to gravity does change significantly, so we need to take this into account. The force exerted on a 10 pound weight at a distance r from the center of the earth is F = k/r 2 and by definition it is 10 when r is the radius of the earth (we assume the earth is a sphere). How can we approximate the work done? We divide the path from the surface to orbit into n small subpaths. On each subpath the force due to gravity is roughly constant, with value k/r 2 i at distance r i . The work to raise the object from r i to r i +1 is thus approximately k/r 2 i Δ r and the total work is approximately n - 1 X i =0 k r 2 i Δ r, or in the limit W = Z r 1 r 0 k r 2 dr, where r 0 is the radius of the earth and r 1 is r 0 plus 100 miles. The work is
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Unformatted text preview: W = Z r 1 r k r 2 dr =-k r fl fl fl fl r 1 r =-k r 1 + k r . Using r = 20925525 feet we have r 1 = 21453525. The force on the 10 pound weight at the surface of the earth is 10 pounds, so 10 = k/ 20925525 2 , giving k = 4378775965256250. Then-k r 1 + k r = 491052320000 95349 ≈ 5150052 foot-pounds . Note that if we assume the force due to gravity is 10 pounds over the whole distance we would calculate the work as 10( r 1-r ) = 10 · 100 · 5280 = 5280000, somewhat higher since we don’t account for the weakening of the gravitational force. EXAMPLE 9.14 How much work is done in lifting a 10 kilogram object from the surface of the earth to a distance D from the center of the earth? This is the same problem as before in different units, and we are not specifying a value for D . As before W = Z D r k r 2 dr =-k r fl fl fl fl D r =-k D + k r ....
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