Explain escape speed.

Explanation

`implies` Suppose a body is at distance `r (r gt R_E)` from the centre of earth. This body is stationary. E = K.E. + Potential energy
`E = - (GM_(E)m)/r`
`E=-(GM_(E) m)/((R_(E) +h))" " ....(1)`
`implies` If by supply this amount of energy, its total energy becomes zero. It will go to infinite distance from the earth. This energy is known as escape energy.
`implies` To speed to be given to the body to give the kinetic energy equal to the escape energy `:. 1/2 mv_(e)^(2) = (GM_(E)m)/((R_(E)+h) )" "...(2)`
`implies` Here, `v_(e)` is called escape speed.
`implies` "The minimum amount of energy with which a body should be projected from the surface of the earth, so that it escape from the earth.s gravitational field is called escape speed (velocity).
`implies` From equation (2),
`v_(e)^2 =(2GM_E)/((R_(E)+h))`
`:. v_(e) = sqrt((2GM_E)/((R_(E)+h)))" "...(3)`
`implies` If body is lying on the surface of earth h = 0, hence escape speed,
`:. v_(e) = sqrt((2GM_E)/(R_(E)))" "...(4)`
` :. v_(e)=sqrt((2GM_(E))/(R_E^2).R_E)`
`:. v_(e)= sqrt(2g R_E)" "...(5)`
where g acceleration due to gravity of earth ,
`implies` In equation (4), taking mass of earth
`M_(E) = 4/3 pi R_(E)^3 rho`
( `rho` = density of earth)
`v_(e) = sqrt((2G)/R_(E)(4/3piR_(E)^2))`
`:.v_(e) = sqrt((8pi)/3(GR_(E)^(2))rho)" ".....(6)`
`implies` Equation 3, 4, 5, are the necessary equations for escape velocity.
`implies` It is cleared from all equation that escape velocity does not depend on the mass of the body. Heavier or lighter, both bodies as same escape of velocity.
`implies` In equation (4), `G = 6.67 x 10^(-11) (Nm^2)/(kg^2)`
Putting the value of `M_E = 6 x 10^(24) kg, R_E = 6400` km escape velocity will be `v_e = 11.2 km//s.`