An object is projected vertically upward from the surface of the earth of mass `M` with a velocity such that the maximum height reached is eight times the radius `R` of the earth. Calculate:
(i) the initial speed of projection
(ii) the speed at half the maximum height.

Answer

(i) `(4)/(3) sqrt((Gm)/(R ))`, (ii) `(2)/(3) sqrt((2Gm)/(5R))`

Explanation

(i) From energy ewnergy conservation
`U_(i)+K_(i)=U_(f)+K_(f)`
`-(GMm)/(R )+(1)/(2)mv^(2)=-(GMm)/((R+8R))+0`
`(v^(2))/(2) = (GM)/(R )[1-(1)/(9)] rArr (v^(2))/(2) = (8)/(9)(GM)/(R ) rArr v = (4)/(3) sqrt((GM)/(R ))` …(i)
(ii) From energy conservation
`U_(i)+K_(i)=U_(f)+K_(f)`
`-(GM)/(R )+(1)/(2)mv^(2)=-(GMm)/((R+4R))+(1)/(2)mv^(2)`
`-(GM)/(R )+(1)/(2)((4)/(3)sqrt((GM)/(R )))^(2) =-(GM)/(5R)+(v'^(2))/(2)`
`(v'^(2))/(2) = (GM)/(R )[(-45+40+90)/(45)] rArr (v'^(2))/(2)=(GM)/(R )xx(4)/(45)`
`v' = (2)/(3) sqrt((2GM)/(5R))`