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Three deminsions. Three point particles are fixed in place in a xyz coordinate system. Particle A, at the origin, has mass m_A. Particle B, at xyz coordinates (2.00d,1.00d. 200), has mass 2.00m_A, and particle C, at coordinates (-1.00d,2.00d,-3.00d), has mass 3.00m_A. A fourth particle D,with mass 4.00 m_A,is to be placed near the other particles. In terms of distance d, at what (a)x, (b)y, and (c)z coordinate should D be placed so that the net gravitational force on A from B, C, and D is zero ? |
| Answer» SOLUTION :`(a) -1.88d, (b) -3.90d, (C ) 0.489d`. In this way, we are ABLE to deduce that (X,y,Z) = (1.88d,3.90d, 0.489d). | |