Derive an expression for escape speed.

Explanation

Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed `v_(1)`, the initial total energy of the object is
`E_(i)=(1)/(2)Mv_(i)^(2)-(GMM_(E))/(R_E)" "...(1)`
Where, `M_(E)` is the mass of the Earth and `R_(E)` - the radius of the Earth. The term `(GMM_E)/(R_E)` is the potential energy of the mass M.
When the object reaches a height far away from Earth and as approaching infinity, the gravitational potential energy becomes zero `[U(infty)=0]` and the kinetic energy becomes zero as well. Hence the final total energy of the object becomes zero. It holds good for minimum energy and for minimum speed to escape. Othewise Kinetic energy can be non-zero.
`E_(f)=0`
According to the law of energy conservation,
`E_(i)=E_(f)" "...(2)`
Substituting (1) in (2) we get,
`(1)/(2)Mv_(i)^(2)-(GMM_E)/(R_E)=0`
`(1)/(2)Mv_(i)^(2)=(GMM_E)/(R_E)" "...(3)`
The minimum speed required by an object to escape Earth's gravitational field is escape speed. Hence `v_(1)` is replaced `v_(e)` , i.e.,
`(1)/(2)Mv_(e)^(2)=(GMM_E)/(R_E)`
`v_(e)^(2)=(GMM_E)/(R_E)(2)/(M)`
`v_(e)^(2)=(2GM_E)/(R_E)`
Using, `g=(GM_E)/(R_(e)^(2)),`
`v_(e)^(2)=2gR_(E)`
`v_(e)=sqrt(2gR_E)`