An object is projected horizontally with speed 1 2 GM R , from a point at height 3 R [where R is radius and M is mass of earth, then object will:-


(1) Fall back on surface of earth by following parabolic path
(2) Fall back on surface of earth by following hyperbolic path
(3) Start rotating around earth in a circular orbit
(4) Escape from gravitational field of earth

Centrifugal force for the particle to rotate in that speed is $F_c=\frac{m{v}^2}{r}=\frac{m{\left(12GMR\right)}^2}{R}=144G^2{M}^{2}mR$
Now gravitational force on the particle will be $F_g=\frac{GMm}{{\left(3R+R\right)}^2}=\frac{GMm}{16R^2}$
Now $\frac{F_c}{F_g}=144\times 16\times GM{R}^3$. So the centrifugal force is much more compared to the gravitational force, it will escape from the earth gravitational field. You can imagine that also in the velocity is much greater than the escape velocity at the surface of the earth. So it will escape. So answer (4) is correct.