Gravitation Potential Energy — Question 7

Question 7

A midpoint with zero field is still gravitationally bound

Two equal fixed point masses $M$ lie at $x=\pm a$. A test particle starts at their midpoint and is launched along the positive $y$ axis. The source masses remain fixed; the particle follows the symmetry axis.

1. Find its potential energy along that axis and the minimum launch speed to reach infinity.

2. Explain how the midpoint can have zero field but a nonzero escape speed.

Question 7 - Solution

A midpoint with zero field is still gravitationally bound

(a) Derivation and calculation

On the symmetry axis both source distances are $s=\sqrt{a^2+y^2}$: $$\boxed{U(y)=-\frac{2GMm}{\sqrt{a^2+y^2}}},\qquad U(0)=-\frac{2GMm}a.$$ Marginal escape has $E=U(\infty)=0$, giving $$\frac12mv_{\mathrm{esc}}^2-\frac{2GMm}a=0, \qquad\boxed{v_{\mathrm{esc}}=\sqrt{4GM/a}}.$$

(b) Result and interpretation

Differentiation gives $$F_y=-\frac{dU}{dy}=-\frac{2GMmy}{(a^2+y^2)^{3/2}},\qquad F_y(0)=0.$$ For $y>0$, $dU/dy>0$: the potential increases monotonically to zero. The required increase is $$\Delta U=0-U(0)=\frac{2GMm}a>0.$$ Vector fields cancel at the midpoint while scalar potentials add.

Physical check

The zero-energy launch can climb the entire axis without an intervening potential barrier.