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_(i)` the initial total energy of the object is
`E_(i) = 1/2 Mv_(i)^(2)- (GMM_(E))/(R_(E))....(i) `
where, `M_(E)` is the mass 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 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 the object becomes can be non zero
`E_(f) = 0 `
According to the law of energy conservation
`E_(i) = E_(f)` ...(2)
Substituting (1) and (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 escape E and gravitational 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)) . 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))` ...(4)
From equation (4) 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 .