A space ship of mass $$2 \times 10^4$$ kg is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the space ship in the orbit to overcome the gravitational pull will be (if $$g = 10$$ m s$$^{-2}$$ and radius of earth = 6400 km):

A spaceship in a circular orbit close to Earth's surface needs additional velocity to escape. We need to find this additional velocity.

First, calculate the orbital velocity for a circular orbit close to Earth's surface.

For a circular orbit at radius $$R$$ (Earth's radius, since the orbit is close to the surface), the gravitational force provides the centripetal force:

$$ \frac{mv_o^2}{R} = \frac{mg}{1} \implies v_o = \sqrt{gR} $$

Substituting $$g = 10$$ m/s$$^2$$ and $$R = 6400$$ km = $$6.4 \times 10^6$$ m:

$$ v_o = \sqrt{10 \times 6.4 \times 10^6} = \sqrt{64 \times 10^6} = 8000\;\text{m/s} = 8\;\text{km/s} $$

Next, calculate the escape velocity from this orbit.

The escape velocity from the surface of Earth is:

$$ v_e = \sqrt{2gR} = \sqrt{2} \times \sqrt{gR} = \sqrt{2} \times v_o $$

$$ v_e = 8\sqrt{2}\;\text{km/s} $$

Now, find the additional velocity needed.

The spaceship is already moving at orbital velocity $$v_o$$. To escape, it needs to reach escape velocity $$v_e$$. The additional velocity required is:

$$ \Delta v = v_e - v_o = 8\sqrt{2} - 8 = 8(\sqrt{2} - 1)\;\text{km/s} $$

Note: The mass of the spaceship ($$2 \times 10^4$$ kg) is not needed for this calculation, as the velocities are independent of mass.

The correct answer is Option 2: $$8(\sqrt{2} - 1)$$ km s$$^{-1}$$.