Is the escape velocity at Geosynchronous Earth Orbit 0km/hr?

Question

Follow-up question to How long must escape velocity be maintained?

Is the escape velocity at GSO 0?

Answer 1

No. Any circular orbital velocity is about 70% ($1/\sqrt{2}$) of the required escape velocity.

To find circular orbital velocity, equate the centripetal force to the force of gravity: $$ \frac{m v^2}{r} = G \frac{ M m}{r^2} \rightarrow \boxed{ v_\textrm{circ} = \sqrt{\frac{GM}{r}}}$$

To find escape velocity, equate the magnitude of the potential energy to that of the kinetic energy (i.e. to find zero total energy): $$ \frac{1}{2}m v^2 = G \frac{ M m}{r} \rightarrow \boxed{ v_\textrm{esc} = \sqrt{2\frac{GM}{r}}}$$

Edit: where $G$ is Newton's Constant, $m$ is the mass of the object, $M$ is the mass of the gravitating body, and $r$ is the separation between the centers of mass.

Answer 2

No, it isn't. If it were, then Geosynchronous Earth Orbit wouldn't be an orbit.