Is the velocity of escape of a particle at rest on the surface of the earth the same as that of a particle just orbiting the earth? Explain.

Explanation

No, the velocity of escape arises from the energy required to take it from the present condition to infinity. This is equal to the difference between energy on the surface of the earth and infinity. For particle at rest
`1/2mv_(e)^(2)=(GM)/R` (only potential energy)
`:.v_(e)=sqrt((2GM)/R)`
For a particle in circular motion.
`E` (energy) `=-(GM)/R+1/2mv^(2)`
But `(mv_(e)^(2))/R=-(GMm)/(R^(2))`
(from dynamics of circular motion)
`:. E=-1/2(GMm)/R`
`:. 1/2mv^(2)=1/2(GMm)/R`
or `v_(c)=sqrt((GM)/R`