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

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

Is the escape velocity at GSO 0?

• What do you mean by GEO? – Bernhard Jun 8 '12 at 22:16
• Geosynchronous Earth Orbit is what I had in mind (+: But we can do without the mention of Earth there – Everyone Jun 8 '12 at 22:30

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.