Derive an expression for escape velocity and orbital velocity

Gravitational force of attraction provides the centripetal force for the earth to revolve around sun.

Hence we have

where G is gravitational constant, M is mass of Sun, m is mass of earth, R is orbital radius of earth and v is orbital velocity.

from eqn.(1), we get

**Escape velocity :-**

Suppose the vertically projected object reach infinity. Let its speed at infinity is v_{f} .

The energy of an object is sum of potential and kinetic energy.

Let W_{∞} denotes the gravitational potential energy of the object at infinity. The total energy of the projectile at infinity is

E(∞) = W_{∞}+(1/2)m×v_{f}^{2} …(1)

If the object is thrown initially with a speed vi from earth surface, its initial total energy is

E( R ) = (1/2)m×v_{i}^{2} - [ G×M×m / R ] + W_{∞} … (2)

By energy conversion, (1) and (2) are same.

(1/2)m×v_{i}^{2} - [ G×M×m / R ] = (1/2)m×v_{f}^{2} …(3)

eqns.(1) and (2) are equated to get eqn.(3) by cancelling out W_{∞} on both sides

RHS of eqn.(3) is always greater than or equal to zero to make vi as escape velocity hence we get

hence we get escape velocity as