What is the vector equation for escape velocity?

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SUMMARY

The vector equation for escape velocity can be expressed as v = (sqrt(2GM)/r)−hat{r}, where G is the gravitational constant, M is the mass of the celestial body, and r is the distance from the center of the body. Escape velocity is technically a scalar quantity, often referred to as "escape speed." The direction of escape velocity is flexible, provided the trajectory does not intersect the surface of the celestial body, which depends on the distance from the center and the radius of the body.

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  • Familiarity with vector mathematics and notation
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rich86
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hi, i have been looking for a while but stillnot managed to find a vector equation for the Escape Velocity. i know the escape velocity is:

v=sqrt(2GM/r)

i would like a vector form to extend a program i am making for a three body gravitational problem

many thanks

richard
 
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Would it not just be \bold{v}=\frac{\sqrt{2GM}}{r}\hat{r}?
 
The term "escape velocity" is a bit of a misnomer: it really should be called "escape speed" because it truly is a scalar quantity.
 
thanks, that helps a lot...

..but makes my project harder in finding the right direction for the escape velocity!
 
rich86 said:
thanks, that helps a lot...

..but makes my project harder in finding the right direction for the escape velocity!

The right direction is any direction, as long as the resulting trajectory doesn't intersect the surface of the body you are trying to escape from. Whether any given trajectory intersects the surface depends a couple of factors, The distance you are from center of the body and the radius of the body. If you are sitting right above the surface of the body you have a smaller range of possible directions that do not result in collision with the surface than you would if you started some greater distance away from the body.
 

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