The point masses ?m and 2 ?m lie along the x -axis, with m at the origin and 2 ?m at x = L. A third point mass ?M is moved along the x -axis. (a) At what point is the net gravitational force on ?M due to the other two masses equal to zero? (b) Sketch the x -component of the net force on ?M due to m and 2 ?m?, taking quantities to the right as positive. Include the regions x < 0, 0 < x < L, and x > L. Be especially careful to show the behavior of the graph on either side of x = 0 and x = L.

Solution 10E Introduction We have to first calculate the distance at which the force is equal. Then we have to draw the graph of the force on the all in all three region as asked in the question. Step 1 Let us consider that at x the net gravitational force is zero. Hence the gravitational force acting on the M at x is given by F = GMm 1 x2 Negative sign because the force will act towards the negative x axis. And the gravitational force acting on M due to 2m is F = GM(2m) 2 (Lx) So the total force is GMm + GM(2m) = 0 x2 (Lx) 2 2 (L x) = 2x x = 2.41L and 0.41L Now ignoring the negative sign, as the force is only attractive, the equilibrium position will be in between the two masses, we get x = 0.41L The net force on the mass M will be equal to zero at x = 0.41L