The escape velocity of a body on the surface of the earth is $$11.2\,km/s.$$   If the earth’s mass increases to twice its present value and the radius of the earth becomes half, the escape velocity would become

Solution :
Escape velocity on the earth's surface is given by $${v_{{\text{es}}}} = \sqrt {\frac{{2G{M_e}}}{{{R_e}}}} $$
where, $$G$$ is gravitational constant, $${M_e}$$ and $${R_e}$$ are the mass and radius of the earth respectively. By taking the ratios of two different cases
$$\eqalign{ & \therefore \frac{{{{v'}_{{\text{es}}}}}}{{{v_{{\text{es}}}}}} = \sqrt {\frac{{{{M'}_e}}}{{{M_e}}} \times \frac{{{R_e}}}{{{{R'}_e}}}} \cr & {\text{but}}\,\,{{M'}_e} = 2{M_e} \cr & {\text{and}}\,\,{{R'}_e} = \frac{{{R_e}}}{2} \cr & {v_{{\text{es}}}} = 11.2\,km/s \cr & \therefore \frac{{{{v'}_{{\text{es}}}}}}{{{V_{{\text{es}}}}}} = \sqrt {\frac{{2{M_e}}}{{{M_e}}} \times \frac{{{R_e}}}{{\frac{{{R_e}}}{2}}}} = \sqrt 4 = 2 \cr & \therefore {{v'}_{{\text{es}}}} = 2{v_{{\text{es}}}} = 2 \times 11.2 = 22.4\,km/s \cr} $$
NOTE
The escape velocity on moon's surface is only $$2.5\,km/s.$$   This is the basic fundamental on which, absence of atmosphere on moon can be explained.