06_InstSolManual_PDF_Part9 - Circular Motion and...

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6.21. Set Up: When we take the ratio of the forces and at and the masses of the two objects divides out. Solve: (a) The orbit is sketched in Figure 6.21. Figure 6.21 (b) so Reflect: The force is greater when Sedna is closer to the sun. 6.22. Set Up: Use coordinates where is to the right. Each gravitational force is attractive, so is toward the mass exerting it. Treat the masses as uniform spheres, so the gravitational force is the same as for point masses with the same center-to-center distances. The free-body diagrams for (a) and (b) are given in Figures 6.22a and b. Figure 6.22 Solve: (a) The net force on A is to the right. (b) The net force on A is to the left. 6.23. Set Up: The two masses are sketched in Figure 6.23a. Let the 3 M mass be to the right of M . For the net force to be zero, the forces and due to M and 3 M must be equal in magnitude and opposite in direction. The two forces will be opposite in direction only at points between the two masses. Consider a point P that is a distance d from M , so from 3 M . Let the third object, at
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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