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_t` the initial total energy of the object is
` E_i = 1/2 mv_(I )^(2) = (GM M_E )/( R_E )`
where `M_E` is the mass of the Earth and ` R_E ` the radius of the earth . the term `- (GM M_E )/( R_E )` is the potential energy of the mass M .
when the object reaches a height far away from Earth and hence and hence treated as approching 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 non - zero
`E_(f ) =0`
According to the law energy conservation ,
` E_i = E_f`
subsituting (1) in (2) we get ,
` 1/2 Mv_(i)^(2) - (GM M_E)/(R_E ) =0`
`1/2 MV_(i)^(2) = (GM M_E)/(R_E)`
consider the escape speed , the minimum speed required by an object to escape Earth .s gravitational field , hence replace `V_i` with ` V_e ` i.e.,
`1/2 Mv_(i) ^(2 ) = (GM M_E)/(R_E) `
`v_(e )^(2) = (GM M_E)/(R_(E )).(2)/(M )`
` V_(e )^(2) = (2G M M_E)/(R_E )`
Using `g= (GM_E )/(R_(e )^(2)) , v_(e )^(2) = 2 g R_E`
` V_(E ) = sqrt( 2g R_(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 substiting 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 object is thrown Irrespective of whether the object is thrown vertically up radially ourwards or tangentially it requires the same initial speed to escape Earth .s gravity .