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

Is the escape velocity at GSO 0?

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

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    $\begingroup$ Where G = acceleration due to gravity, M is the mass of the body in orbit, and r is the altitude at which velocity is to be determined? $\endgroup$ – Everyone Jun 9 '12 at 6:59
  • $\begingroup$ Not quite---I added a clarification. Generally big-'G' is Newton's gravitational constant, while little-'g' is the acceleration due to gravity. And 'r' is the distance between the center of earth (the center of mass of the earth) and the altitude at which the velocity is desired. $\endgroup$ – DilithiumMatrix Jun 9 '12 at 8:34
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No, it isn't. If it were, then Geosynchronous Earth Orbit wouldn't be an orbit.

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