A particle is fired vertically upward with a speed `v_(0)`. Find
(a) Speed at distance `2R_(e)` from centre of earth.
(b) Calculated maximum height attained by particle.

Explanation

Applying energy conservation between `(1)` and `(2)`
`K_(1)+U_(1)=K_(2)+U_(2)`
`1/2mv_(0)^(2)-(GM_(e)m)/(R_(e))=1/2mv^(2)-(GM_(e)m)/((R_(e)+h))`
`R_(e)+h=2R_(e)`
`1/2mv^(2)=1/2mv_(0)^(2)-(GM_(e)m)/(R_(e))+(GM_(e)m)/(2R_(e))`
`1/2mv_(0)^(2)-(GM_(e)m)/(2R_(e))`
`v^(2)=v_(0)^(2)-(GM_(e))/(R_(e))`
`v=(v_(0)-(GM_(e))/(R_(e)))^(1//2)=(v_(0)^(2)-gR_(e)^(2))^(1//2)`
(b) `h=H_(max)` if `v=0`
`1/2mv_(0)^(2)=GM_(e)m[1/(R_(e))-1/(R_(e)+h)]`
`(v_(0)^(2))/(2GM_(e))=h/(R_(e)(R_(e)+h))`
`h/((R_(e)+h))=(v_(0)^(2)R_(e))/(2GM_(e))=(v_(0)^(2)R_(e))/(2gR_(e)^(2))=(v_(0)^(2))/(2gR_(e))`
`(R_(e)+h)/h=(2gR_(e))/(v_(0)^(2))`
`(R_(e))/h+1=(2gR_(e))/(v_(0)^(2))`
`(R_(e))/h=(2gR_(e))/(v_(0)^(2))-1`
`h=(R_(e))/(((2gR_(e))/(v_(0)^(2))-1))=(V_(0)^(2)R_(e))/((2gR_(e)-v_(0)^(2)))`