The escape velocity of a body on a planet $$A$$ is 12 km s$$^{-1}$$. The escape velocity of the body on another planet $$B$$, whose density is four times and radius is half of the planet $$A$$, is

Escape velocity on planet A is 12 km s$$^{-1}$$. Planet B has density 4 times and radius half that of planet A.

First, we recall that the escape velocity formula is $$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2G \cdot \frac{4}{3}\pi R^3 \rho}{R}} = R\sqrt{\frac{8\pi G\rho}{3}}$$, which implies that $$v_e \propto R\sqrt{\rho}$$.

Next, we find the ratio of escape velocities as $$\frac{v_B}{v_A} = \frac{R_B}{R_A} \times \sqrt{\frac{\rho_B}{\rho_A}} = \frac{1}{2} \times \sqrt{4} = \frac{1}{2} \times 2 = 1$$.

Finally, substituting into this relationship gives $$v_B = v_A = 12 \text{ km s}^{-1}$$.

The correct answer is Option A: 12 km s$$^{-1}$$.