Question

What happens to the body when it is thrown upwards @ 11.5km/s?

Answer 1

The escape velocity on the earth surface is about 11.2 km/s.
When a body is projected from the surface of a planet with a velocity equal to or higher than the escape velocity, it escapes from the gravitational influence of that planet.
As in this case, the velocity of the body(11.5 km/s) is higher than the escape velocity(11.2 km/s) on the earth, the body escapes from the earth's gravitational influence, when projected upwards.

Answer 2

Let v be the velocity of projection of the body from Earth.

$Total\ mechanical\ energy\ of\ the\ body\ on\ Surface\ of\ Earth\\\\ E= \frac{1}{2}mv^2-\frac{GM_em}{R_e},\ \ m=mass\ of\ the\ body,\ \ M_e=mass\ of\ Earth\\\\.\ \ \ \ \ \ R_e=Radius\ of\ Earth,\ \ G=Unversal\Gravitational\ Constant$

Since the total mechanical energy is conserved by the gravitational force,  the energy at a distance d from the center of Earth is given by:

$E=\frac{1}{2}mv_d^2-\frac{GM_em}{d}=\frac{1}{2}mv^2-\frac{GM_em}{R_e}\\\\For\ d=>\infty,\ and\ v_{\infty}=nearly\ 0,\ minimum\ velocity\ v_e\ needed:\\\\ \frac{1}{2}mv_e^2=\frac{GM_em}{R}\\\\v_e=\sqrt{\frac{2GM_e}{R_e}}$

It is called the escape velocity of Earth, as a body projected with this velocity away into the space, just manages to travel to infinite distance.  Substituting the values numerically and calculating, ,we find  $v_e=11.2\ km/sec.$

If a body is projected with  a velocity 11.5 km/sec., then the body escapes from the gravitational field of Earth and it will have some kinetic energy still. After that it will move at uniform velocity:

$v_{\infty}=\sqrt{11.5^2-11.2^2}=2.61\ km/sec.$

Perhaps the body will get into gravitational field of another planet in space.