A particle of mass 3?m is located 1.00 m from a particle of mass ?m?. (a) Where should you put a third mass ?M so that the net gravitational force on ?M due to the two masses is exactly zero? (b) Is the equilibrium of ?M at this point stable or unstable (i) for points along the line connecting ?m and 3 ?m?, and (ii) for points along the line passing through ?M? and perpendicular to the line connecting ?m? and 3 ?m??
Solution 9E Step 1 a) Suppose, the distance between the mass m and M as “x” metres. Then, the gravitational force of attraction between these two masses would be, F = GMm 1 x2 The distance between the masses M and 3m is, (1-x) metres Therefore, the gravitational force acting between these two forces would be, 3GMm F =2 2 (1x) At equilibrium point, both these forces will be equal in magnitude and opposite in direction and we can find out that point. F1= F2 GMm = 3GMm x2 (1x) That is, 12 = 3 2 x (1x) 2 2 That is,(1 x) = 3x 1+x - 2x = 3x 2 That is, 2x + 2x - 1 = 0 Solve for x and we will get the point. 2± 4+8 x = 4 2±2 3 x = 4 x = 1± 3 2 We can take only positive values. x = 1+1.732 = 0.732/2 = 0.366 m 2 The equilibrium point for M is 0.366 m from the mass “m”.