The escape velocity of an object from the surface of the earth is 11.6 km/s. If the mass or the earth is increased four times, the resultant escape velocity would be

The escape velocity ($v_e$) of an object from a celestial body is determined by the formula $v_e = \sqrt{2GM/R}$, where $G$ is the gravitational constant, $M$ is the mass of the body, and $R$ is its radius [t2, t8, t10]. This formula demonstrates that escape velocity is directly proportional to the square root of the mass of the body ($\sqrt{M}$) when the radius remains constant [t2, t4]. In this scenario, the initial escape velocity is given as 11.6 km/s. If the mass of the Earth is increased four times ($4M$), the new escape velocity ($v'_e$) becomes $\sqrt{2G(4M)/R}$. Mathematically, this simplifies to $\sqrt{4} \times \sqrt{2GM/R}$, which is $2 \times v_e$ [t2]. Therefore, the resultant escape velocity is $11.6 \times 2$ km/s, effectively doubling the original value due to the square root relationship with mass.