Gravitation Potential Energy — Question 6

Question 6

Assembly energy counts each gravitational pair exactly once

Three point masses $m$, $2m$ and $3m$ are placed at the vertices of an equilateral triangle of side $a$. They are assembled quasistatically from infinite separation and then held fixed.

1. Find total gravitational potential energy.

2. Find the external work needed to remove only the $3m$ mass to infinity while the other two stay fixed.

Question 6 - Solution

Assembly energy counts each gravitational pair exactly once

(a) Derivation and calculation

Sum unordered pairs exactly once: $$U=-G\sum_{i<j}\frac{m_im_j}{r_{ij}} =-\frac G a[(m)(2m)+(m)(3m)+(2m)(3m)].$$ $$\boxed{U=-\frac{11Gm^2}{a}}.$$ Equivalently, $U=\tfrac12\sum_i m_i\Phi_{\mathrm{others}}(\vec r_i)$; the half prevents double counting.

(b) Result and interpretation

After removing only $3m$, the remaining energy is $$U_f=-\frac{G(m)(2m)}a=-\frac{2Gm^2}a.$$ Quasistatic external work is $$\boxed{W_{\mathrm{ext}}=U_f-U_i=(-2+11)\frac{Gm^2}a=\frac{9Gm^2}a}.$$ Only the pair energies $-3Gm^2/a$ and $-6Gm^2/a$ disappear.

Physical check

The remaining pair stays bound, so removing one mass does not raise the system energy all the way to zero.