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, the initial total energy of the object is
`E_(i)= (1)/(2)Mv_(i)^(2)- (GMM_(E))/(R_(E))`
where, `M_(E)` the is the mass of the Earth and R the radius of the Earth. The term 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 becomes zero `[U(oo) =0]` and the kinetic energy becomes 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 nonzero.
`E_(f)=0`
According to the law of energy conservation, `E_(i)= E_(f)`
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))`
Consider the escape speed, the minimum speed required by an object to gravitational escape Earth.s field, hence replace `v_(i)` with `v_(e)` i.e.,
`(1)/(2)Mv_(e)^(2)= (GMM_(E))/(R_(E))`
`v_(e)^(2)= (GMM_(E))/(R_(E))`
`v_(e)^(2)= (2GM_(E))/(R_(E))`
`v_(e)^(2)= 2gR_(E)`
`v_(e)= sqrt(2gR_(E))`
From equation (4) 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. By substituting the values of g (`9.8 ms^(-2)` ) and `R_(e) = 6400 km`, the escape speed of the Earth is `v_(e) = 11.2 kms^(-1)` The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth.s gravity.