Question 6.11: Stability of equilibrium in a 3-D conservative field A parti...
Stability of equilibrium in a 3-D conservative field
A particle P of mass m can move under the gravitational attraction of two particles, of equal mass M, fixed at the points (0, 0, ±a). Show that the origin O is a position of equilibrium, but that it is not stable. [This illustrates the general result that no free-space static gravitational field can provide a position of stable equilibrium.]
When P is at O, the fixed particles exert equal and opposite forces so that the total force on P is zero. The origin is therefore an equilibrium position for P.
Just as in rectilinear motion, O will be a position of stable equilibrium if the potential energy function V(x, y,z) has a minimum at O. This means that the value of V at O must be less than its values at all nearby points. But at points on the z-axis between z = −a and z = a
V(0,0, z)=-\frac{m M G}{a-z}-\frac{m M G}{a+z}=-\frac{2 a m M G}{a^{2}-z^{2}},
which has a maximum at z = 0. Hence the equilibrium at O is unstable to disturbances in the z-direction.