Binding Energy of an Orbiting Satellite

A satellite moving in a stable circular orbit possesses BOTH kinetic energy (from its orbital motion) and potential energy (from its position in the Earth's gravitational field), and the sum of the two -- its total mechanical energy -- reveals exactly how tightly it is bound to the Earth.

For a satellite of mass m orbiting at radius $r=R+h$ with critical (orbital) speed $v_c$, from section 5.8.1, $v_c^2=GM/r$, so the kinetic energy is $KE=\dfrac{1}{2}mv_c^2=\dfrac{1}{2}\dfrac{GMm}{r}.$ The gravitational potential at distance r from the Earth's centre is $-GM/r$ (section 5.7.3), so the potential energy of the satellite is $PE=-\dfrac{GMm}{r}.$

Adding these gives the satellite's total mechanical energy: $TE=KE+PE=\dfrac{1}{2}\dfrac{GMm}{r}-\dfrac{GMm}{r}=-\dfrac{1}{2}\dfrac{GMm}{r}.$ Notice this total energy comes out NEGATIVE -- exactly half the magnitude of the (negative) potential energy alone, and equal in magnitude but opposite in sign to the kinetic energy. This negative sign is not a mathematical curiosity; it is the mathematical signature of the satellite being gravitationally BOUND to the Earth, unable to escape to infinity on its own, in exactly the way that a satellite with zero or positive total energy (section 5.7.4's escape-velocity condition) is not bound and can reach infinity. …