The escape velocity from Earth's surface is approximately 11.2 km/s. If a planet has twice Earth's radius and same density, its escape velocity will be: MCQ with Answer and Explanation

Escape velocity v_e = √(2GM/R). Mass M = density ρ × volume = ρ×(4/3)πR³. Thus v_e = √(2Gρ×4/3πR³ / R) = √(8GρπR²/3) ∝ R√ρ. Given same density and R_planet = 2R_earth, v_e ∝ R, so v_e_planet = 2 × 11.2 = 22.4 km/s. This derivation shows escape velocity scales linearly with radius for constant density. Memory tip: v_e ∝ √(M/R) and M ∝ R³ for constant ρ, so v_e ∝ R. Such proportional reasoning questions test conceptual grasp of gravitation formulas in competitive exams without heavy calculation.