D.1.7 (HL)—Two-body potential energy

Use the two-body expression

For two point masses, or spherical bodies represented at their centres, use centre-to-centre separation $r$. The zero reference is infinite separation, so a finite bound pair has negative potential energy.

Worked example — 1.0 kg at Earth’s surface

Using $M=6.0\times10^{24}\,\mathrm{kg}$, $m=1.0\,\mathrm{kg}$ and $r=6.4\times10^6\,\mathrm{m}$, $E_p=-(6.67\times10^{-11})(6.0\times10^{24})(1.0)/(6.4\times10^6)=-6.3\times10^7\,\mathrm{J}$. The negative result means energy must be supplied to separate the pair to infinity.

Interpret the negative sign

At every finite separation, $E_p<0$ because the masses form a bound configuration relative to infinity. Increasing $r$ makes $E_p$ less negative; decreasing $r$ makes it more negative. The change in potential energy is what matters when comparing two positions.

Connect to a circular orbit

For a satellite in a circular orbit, the gravitational potential energy is still $-GMm/r$. If the orbital relation gives $K=GMm/(2r)$, then the total mechanical energy is $E_T=K+E_p=-GMm/(2r)$. This orbit result is a consequence of the circular-orbit model, not a replacement for the general two-body potential-energy equation.

Common trap

Do not omit the minus sign or use $r^2$ in the potential-energy expression. $r^2$ belongs to force and field-strength laws; potential energy varies as $1/r$.