Physics - Gravitation Question with Solution | TestHub

For a projectile to leave the gravitational field of the earth, its kinetic energy would be at least equal to its potential energy. That is $$\frac{1}{2}m{v}_e^2=G\frac{Mm}{R}\Rightarrow v_e=\sqrt{\frac{2GM}{R}}$$ where $G$ is the universal gravitational constant, $M$ is the mass of the earth and $R$ is the radius of the earth. Hence the escape velocity of a projectile is independent of its own mass.