Earth Escape Velocity

When an object is protected upwards from the surface of a planet, it may return to the surface of the planet or it may revolve around the planet in a circular orbit or an elliptical orbit or it may escape from the gravitational influence of the planet following a parabolic or rectangular  hyperbolic path.

Introduction to the Earth Escape Velocity

The velocity required by an object to revolve round a planet in a circular orbit is orbital velocity and the velocity required for an object to escape from the gravitational influence of the planet is escape velocity.

The minimum vertical velocity required for an object to escape from the gravitational influence of a planet is known as escape velocity. It is denoted by Ve . Having problem with Longitudinal Waves Definition keep reading my upcoming posts, i will try to help you.

Escape Velocity of from Earth : Derivation

Consider an object of mass 'm' at rest on the surface of a planet of mass m and radius R.

The gravitational potential on the surface of a planet = - `(GM)/(R)`

If we imagine that the object of mass m is brought from infinity to the surface of the lanet, the work done is stored as gravitational potential energy of the planet-object system.

`:.` The gravitational potential energy of the system = gravitational potential x mass of the object

=  `(-GMm)/(R)` .

Negative sign indicates that the object is attracted by the planet and the object is bound to the planet. The object can be made to escape from the gravitational field of the planet by imparting certain minimum speed to the object. This minimum speed to be imparted to the bound object at rest is escape velocity ve . The kinetic energy imparted to the object must be equal and opposite to the potential energy of the system so that the total energy is equal to zero. Please express your views of this topic Friction Formula by commenting on blog.

`(1)/(2)` mve2   =   - [  `(-GMm)/(R)` ].

`(1)/(2)` mve2  = `(GMm)/(R)`

`:.`  Ve=  `sqrt((2GM)/(R))`

In terms of g, the expression for Ve  is

Ve=  `sqrt((2GM)/(R))`         =  `sqrt((2gR^2)/(R))`

Ve = `sqrt(2gR)`

Earth Escape Velocity : Characteristics

The escape velocity does not depend on the mass of the body. Hence from a minute atom to mighty rocket, every object will have the same escape velocity irrespective of its mass, for a particular planet. It is also independent of angle of projection. But it depends on the mass of the planet, radius of the planet.

Molecules present in the atmosphere always move with a certain mean velocity. It depends upon the nature and temperature of surroundings. The escape velocity of an object on planet earth is as follows :

Radius of earth = 6.4 x 106 m ,             Acceleration due to gravity(g) = 9.8 m/s2 .,

Escape velocity Ve  =  `sqrt(2gR)`

= `sqrt(2 xx 9.8 xx 6.4 xx 10^6 m/s)`

= 11.2  x  103  m/s  =  11.2 km/s

If a body is projected with a velocity greater than or equal to 11.2 km/s, it will never return to the earth.