what is escape velocity?obtain the expression for the escape velocity on earth ?find escape velocity on the surface of the earth.
The minimum velocity with which a projectile has to be projected to escape the earth’s gravitational field is called escape velocity.
Let a body of mass ‘m’ be projected with velocity ‘v’.
At the ground, PE = GMm/R
And the KE = ½ mv2
To overcome earth’s gravitational field,
=> ½ mv2 > GMm/R
=> v > (2GM/R)1/2
g = GM/R2
v > [2(GM/R2)R]1/2
=> v > [2gR]1/2
Thus, to escape the earth’s gravitational field, the minimum velocity must be,
ve = [2gR]1/2
For the last part please refer the answer posted by Ashley.
@ Ashley: Good effort keep it up!!
Consider an object of mass 'm' placed on the surface of the earth. According to Newton's law of gravitation, the gravitational pull of the earth on that object is given by
Where F is the force of attraction between the earth and the object,
Me is the mass of the earth and Re is the radius of the earth.
Then, work required to overcome the gravitational pull = F x Re
If v is the velocity of the projectile (object), then its kinetic energy = 1/2 mv2.
For the projectile to escape from earth's gravitational pull the kinetic energy of the projectile has to be greater than the work required to overcome the gravitational pull.
Let us now get an expression for vesc in terms of acceleration due to gravity.
(Multiply both numerator and denominator by Re)
Substituting for ge and Re in equation (2) we get,