Here we find the minimum initial speed required to give a body on the Earth's surface to escape from the gravitational attraction of the Earth. Since the only conservative force acting is gravity, the total mechanical energy is conserved. Note that we'll ignore the affects of air resistance. The conservation of mechanical energy tells us that the total initial mechanical energy is equal to the total final mechanical energy. If \(v\) is the initial speed required to escape the gravitational field of the Earth called escape speed, initial kinetic energy is
\[K_1 = \frac{mv^2}{2}\]
The body is initially on the Earth's surface so the gravitational potential energy after putting height \(h = 0\) is,
\[U=-\frac{Gm{{M}_{E}}}{({{R}_{E}}+h)}=-\frac{Gm{{M}_{E}}}{{{R}_{E}}}\]
The final potential energy \(U_2\) at an infinite distance from the centre of the Earth is zero. The final kinetic energy \(K_2\) is also zero at the infinite distance from the centre of the Earth. Note that the potential energy is maximum at the infinite distance from the centre of the Earth and gradually decreases (becomes more negative) when the body falls towards the centre of the Earth. But the kinetic energy is minimum which is zero at the infinite distance. The key to find the escape speed is that the final potential energy is maximum (zero) and the final kinetic energy is minimum (zero) at the infinite distance from the centre of the Earth. Therefore, \(U_2 = 0\) and \(K_2 = 0\). The law of conservation of mechanical energy gives
\[\begin{align*} {{K}_{1}}+{{U}_{1}}&={{K}_{2}}+{{U}_{2}} \\ \text{or,}\quad \frac{1}{2}m{{v}^{2}}-\frac{Gm{{M}_{E}}}{{{R}_{E}}+0}&=0+0 \\ \text{or,}\quad v&=\sqrt{\frac{2G{{M}_{E}}}{{{R}_{E}}}} \tag{7} \label{7} \end{align*}\]
The above equation for escape speed is the minimum speed required to give a body on the Earth's surface to reach the maximum distance or it means the infinite distance from the centre of the Earth. The escape speed does not depend on the mass of the body to be launched.
