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# I-sol - Chapter I Problems Blinn College Physics 2425 Terry...

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Chapter I - Problems Blinn College - Physics 2425 - Terry Honan Problem I.1 A 200 kg mass sits at x = 0 and a 500 kg mass sits at x = 0.4 m. (a) What is the net gravitational force on a 50 kg mass at the midpoint of the two, x = 0.2 m. (b) Where on the x axis would the net force on a third mass be zero. Solution to I.1 m 1 = 200 kg and m 2 = 500 kg (a) The force on m = 50 kg is the sum of the forces of m 1 and m 2 on m . F 1 = G m 1 m r 1 2 = 6.67 μ 10 - 11 200 μ 50 0.2 2 = 1.6675 μ 10 - 5 N F 2 = G m 2 m r 1 2 = 6.67 μ 10 - 11 500 μ 50 0.2 2 = 4.1688 μ 10 - 5 N F net = F 1 + F 2 . Since the forces are in opposite directions we should subtract their magnitudes. F net = F 2 - F 1 = 2.50 μ 10 - 5 N (b) Take the distances from m 1 and m 2 to be x and x - d , where d = 0.4 m . F 2 = F 1 ï G m 1 m x 2 = G m 2 m H d - x L 2 ï d - x x = m 2 ê m 1 ï d = x + x m 2 ê m 1 ï x = d 1 + m 2 ë m 1 = 0.4 1 + 5 ê 2 = 0.155 m Problem I.2 Consider a 80000 kg uniform solid sphere with a 1.2 m. What is the gravitational field a distance of r from the center for the values: (a) r = 0, (b) r = 0.6 m, (c) r = 1.2 m, (d) r = 2.4 m Solution to I.2 The gravitational field g is defined as the force per test mass, g = F ë m 0 . Here we are just looking for the magnitude of the field. The shell theorem implies the force and field at r go as F = G M inside m 0 r 2 ï g = F m 0 = G M inside r 2 where M inside is the total mass inside a sphere of radius r . Here take M = 80000 kg and R = 1.2 m. When outside the sphere M inside is just M .

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r ¥ R ï M inside = M When inside the sphere M inside depends on the fraction of the total volume inside r : call this V inside . r < R ï M inside = M V inside V total = M H 4 ê 3 L p r 3 H 4 ê 3 L p R 3 = M J r R N 3 (a) r = 0 ï M inside = 0 ï g = 0 (b) r = 0.6 m ï M inside = 80000 K 0.6 1.2 O 3 = 10000 ï g = G M inside r 2 = 1.85 μ 10 - 6 m s 2 (c) r = 1.2 m ï M inside = 80000 ï g = G M inside r 2 = 3.70 μ 10 - 6 m s 2 (d) r = 2.4 m ï M inside = 80000 ï g = G M inside r 2 = 0.926 μ 10 - 6 m s 2 For the above calculations the value of G = 6.67 μ 10 - 11 N ÿ
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I-sol - Chapter I Problems Blinn College Physics 2425 Terry...

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