Escape speed from the Earth is close to 11.2 km s-1. On another planet whose radius is half of the Earth's radius and whose mass density is four times that of the Earth, the escape speed in km s-1 will be close to :

The escape speed (v) of a planet is given by the formula: v = √(2GM/R), where G is the gravitational constant, M is the mass, and R is the radius.

Since mass (M) can be expressed in terms of density (ρ) and volume as M = ρ × (4/3)πR³, the escape speed formula becomes: v ∝ R√ρ.

According to the question:

• Radius of the planet (Rp) = 1/2 Re (Earth's radius)
• Density of the planet (ρp) = 4 ρe (Earth's density)

Substituting these values into the proportionality:

vp / ve = (Rp / Re) × √(ρp / ρe)
vp / ve = (1/2) × √4 = (1/2) × 2 = 1

Therefore, the escape speed of the planet is equal to that of the Earth, which is 11.2 km s-1. Thus, Option A is correct.