A body is lauched from the earth's surface a an angle `alpha=30^(@)` to the horizontal at a speed `v_(0)=sqrt((1.5 GM)/R)`. Neglecting air resistance and earth's rotation, find (a) the height to which the body will rise. (ii) the radius of curvature of trajectory at its top point.

Answer

(a) `h =((sqrt(7))/(2)+1)R`, (b) `1.13 R`

Explanation

(a) Applying conservation of angular momentum between point P and Q.

`m(v_(0)cos 30^(@))R=mvr`
`v=(sqrt(3))/(2)(v_(0)R)/(r )` …..(i)
Applying conservation of mechanical energy.
`-(GMm)/(R )+(1)/(2)mv_(0)^(2)=-(GMm)/(r )+(1)/(2)mv^(2)` ....(ii)
from (i) and (ii)
`-(GMm)/(R )+(1)/(2)mv_(0)^(2)=-(GMm)/(r )+(1)/(2)m(3v_(0)^(2)R^(2))/(4r^(2))` ....(iii)
given that `v_(0) = sqrt((3)/(2)(GM)/(R ))` .....(iv)
From (3) and (4)
`4r^(2)-16Rr+9R^(2)=0`
`r=2R+(sqrt(7))/(2)R rArr R+h=2R+(sqrt(7))/(2)R`
`h = ((sqrt(7))/(2)+1)R`
(b) Radius of curvature `= (v^(2))/(g) = ((sqrt(3)v_(0)R)/(2r))^(2)//(GM)/(R^(2))`
Radius of curvature `= (3)/(4)(v_(0)^(2)R^(2))/(GM)`
`=(3)/(4)xx(sqrt((3)/(2)(GM)/(R )))^(2)xx(R^(2))/(GM)=(9)/(8)R=1.125R`