Using the law of conservation of energy, obtain the expression for the escape velocity.

Explanation

Here, we shall not consider the effects of air. Suppose a body of mass m is thrown vertically upward from the surface of the earth. Let the initial velocity of the body be the escape velocity (Vesc). When the body is on the earth's surface, its total energy `E_(1)` kinetic energy + potential energy `=(1)/(2)mv^(2)" esc"=(-(GmM)/(R)),` where G= universal R gravitational constant, M = mass of the earth and R = radius of the earth.
Thus, `E_(1)(1)/(2)m"(zero)"+(-(MmM)/(oo))=0+0=0.`
According to the law of conservation of energy, `E_(1)=E_(2)`.
`therefore (1)/(2)mv^(2)"esc"-(GmM)/(R)=0`
`therefore(1)/(2)v^(2)"esc"=(GM)/(sqrt2)`
`therefore v_("esc")=sqrt((2GM)/(R))`
This is the required expression.