Follow-up question to How long must escape velocity be maintained?
Is the escape velocity at GSO 0?
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1What do you mean by GEO? – Bernhard Jun 08 '12 at 22:16
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Geosynchronous Earth Orbit is what I had in mind (+: But we can do without the mention of Earth there – Everyone Jun 08 '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.
DilithiumMatrix
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1Where 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? – Everyone Jun 09 '12 at 06:59
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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. – DilithiumMatrix Jun 09 '12 at 08:34
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No, it isn't. If it were, then Geosynchronous Earth Orbit wouldn't be an orbit.
Keith Thompson
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