Gravitation Potential Energy — Question 4

Question 4

A force-free cavity can still have a nonzero escape energy

A thin uniform spherical shell has mass $M$ and radius $R$. Its field vanishes inside, and its exterior potential with zero at infinity is $\Phi=-GM/r$. A small particle begins at rest anywhere inside. Assume passage through an ideal negligible opening.

1. Find the interior potential and work needed to carry the particle to infinity quasistatically.

2. Explain why zero interior force does not imply zero required work to escape.

Question 4 - Solution

A force-free cavity can still have a nonzero escape energy

(a) Derivation and calculation

Inside the shell, $$\vec g=-\boldsymbol\nabla\Phi=\vec0\quad\Longrightarrow\quad\Phi=\text{constant}.$$ Continuity at the surface fixes that constant: $$\boxed{\Phi_{\mathrm{inside}}=\Phi(R)=-GM/R}.$$ Hence quasistatic removal requires $$\boxed{W_{\mathrm{ext}}=m[\Phi(\infty)-\Phi_{\mathrm{inside}}]=GMm/R}.$$

(b) Result and interpretation

Split the path into interior and exterior portions: $$W_{\mathrm{inside}}=0,\qquad W_{\mathrm{outside}}=\int_R^\infty\frac{GMm}{r^2}\,dr=\frac{GMm}{R}.$$ The field constrains a potential gradient; it does not require the potential value to vanish. All interior starting positions have the same escape work.

Physical check

The potential is negative with zero at infinity, so escape work is positive even from a force-free cavity.