Using the law of conservation of energy, obtain the expression for the escape velocity.

Explanation

Expression for escape velocity :
(i) Consider an object of mass m moving with initial velocity equal to escape velocity `v_(esc)` on the surface of the earth
The kinetic energy of the object is given as,
`K.E. = 1/2 mv_(esc)^2`
The potential energy of the object is given as ,
Potential energy = `-(GMm)/(R)`
`therefore " Total energy " = E_1= K.E+P.E`
`=1/2mv_(esc)^2 - (GMm)/(R) " " ...(1)`
(ii) The object escapes the gravitational force of the earth and comes to rest at infinite distance from the earth.
The kinetic energy of the object is given as K.E =0
The potential energy of the object is given as ,
Potential energy `=-(GMm)/(oo) = 0`
`therefore " Total energy " = E_2 = K.E + P.E = 0 " " ...(2)`
(iii) From the principle of conservation of energy,
`E_1 = E_2`
`1/2 mv_(esc)^2 - (GMm)/(R) = 0`
`Therefore v_(esc)^2 = (2GM)/(R)`
`therefore v_(esc) = sqrt((2GM)/(R)) " "...(3)`
(iv) Also , we know, accelertion due to gravity is given as
`g= (GM)/(R^2)`
`therefore GM=gR^2 " "...(4)`
(v) Substituting eqn (4) in (3) , we get ,
`v_(esc) = sqrt((2gR^2)/(R)) = sqrt(2gR) " " ...(5)`
Equations (3) and (5) represent the equations ofr escape velocity from the surface of the earth .