Gravitation and the Principle of Superposition — Question 2

Question 2

A zero-field point can restore sideways displacements but amplify axial ones

Equal fixed masses $M$ lie at $x=\pm a$. Consider a test particle close to their midpoint. For a small displacement, keep only terms linear in position.

1. Find $g_x$ for a displacement $(x,0)$ with $|x|\ll a$.

2. Find $g_y$ for a displacement $(0,y)$ and classify the equilibrium.

Question 2 - Solution

A zero-field point can restore sideways displacements but amplify axial ones

(a) Derivation and calculation

For an axial displacement, $$g_x=GM\left[\frac1{(a-x)^2}-\frac1{(a+x)^2}\right].$$ Using $(1\pm u)^{-2}=1\mp2u+O(u^2)$, $$g_x\simeq\frac{GM}{a^2}\left[(1+2x/a)-(1-2x/a)\right] =\boxed{\frac{4GM}{a^3}x}.$$ The acceleration reinforces the displacement.

(b) Result and interpretation

For a transverse displacement, horizontal components cancel: $$g_y=-\frac{2GMy}{(a^2+y^2)^{3/2}} \simeq\boxed{-\frac{2GM}{a^3}y}.$$ Thus $$\ddot x-\frac{4GM}{a^3}x=0,\qquad \ddot y+\frac{2GM}{a^3}y=0.$$ Transverse motion is restoring, but axial growth makes the unconstrained equilibrium unstable.

Physical check

One unstable direction is sufficient for instability; transverse restoration does not stabilize the midpoint.