Derive an expression for escape speed.

Explanation

Consider an object of mass M on the surface of Earth. When it is thrown up with an initial speed `v_(i)`, the initial energy of the object is
`E_(i) = 1/2 mv_(1)^(2) - (GMM_E)/(R_E)" "....(1)`
Where `M_(E)` is the mass of the Earth and `R_E` is 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 hence treated as approaching infinity, the gravitational potential energy become zero `[U(oo) = 0]` and the kinetic energy become zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise kinetic energy can be non-zero.
`E_(f) = 0`
Accroding 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 escape speed, the minimum speed is defined as required by an object to escape from Earths gravitational field,
`1/2 mv__(e)^(2) = (GMM_E)/(R_E) , v_(e)^(2) = (GMM_E)/(R_E) cdot 2/M`
`v_(e)^(2) = (2GM_E)/(R_E)" ".......(4)`
Using `g = (GM_E)/(R_E)" "........(5)`
`v_(e)^(2) = 2gR_(E)`
`v_(e) = sqrt(2g R_E)" "......(6)`
From equation (6) the escape speed depends on two factors: acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object.
`g = 9.8 ms^(-2) and R_(e) = 6400 km`,
The escape speed of the Earth is `v_(e) = 11.2 kms^(-1)`.